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Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 273052 14 pageshttpdxdoiorg1011552013273052
Research ArticleAnalytical Solutions for Steady Heat Transfer inLongitudinal Fins with Temperature-Dependent Properties
Partner L Ndlovu and Raseelo J Moitsheki
Center for Differential Equations Continuum Mechanics and Applications School of Computational and Applied MathematicsUniversity of the Witwatersrand Private Bag 3 Johannesburg 2050 South Africa
Correspondence should be addressed to Raseelo J Moitsheki raseelomoitshekiwitsacza
Received 30 January 2013 Accepted 24 April 2013
Academic Editor Ireneusz Zbicinski
Copyright copy 2013 P L Ndlovu and R J Moitsheki This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
Explicit analytical expressions for the temperature profile fin efficiency and heat flux in a longitudinal fin are derived Here thermalconductivity and heat transfer coefficient depend on the temperature The differential transform method (DTM) is employed toconstruct the analytical (series) solutions Thermal conductivity is considered to be given by the power law in one case and by thelinear function of temperature in the other whereas heat transfer coefficient is only given by the power lawThe analytical solutionsconstructed by the DTM agree very well with the exact solutions even when both the thermal conductivity and the heat transfercoefficient are given by the power law The analytical solutions are obtained for the problems which cannot be solved exactly Theeffects of some physical parameters such as the thermogeometric fin parameter and thermal conductivity gradient on temperaturedistribution are illustrated and explained
1 Introduction
Fins are surfaces that extend from a hot object (body) toincrease the rate of heat transfer to the surrounding fluidIn particular fins are used extensively in various industrialapplications such as the cooling of computer processors airconditioning and oil carrying pipe lines A well-documentedreview of heat transfer in extended surfaces is presented byKraus et al [1] The problems on heat transfer particularlyin fins continue to be of scientific interest These problemsare modeled by highly nonlinear differential equations whichare difficult to solve exactly However Moitsheki et al [2ndash4] have attempted to construct exact solutions for the steadystate problems arising in heat flow through fins A numberof techniques for example Lie symmetry analysis [2] Hersquosvariational iteration method [5] Adomain decompositionmethods [6] homotopy perturbationmethods [7] homotopyanalysis methods [8] methods of successive approximations[9] and other approximation methods [10] have been usedto determine solutions of the nonlinear differential equationsdescribing heat transfer in fins
Recently the solutions of the nonlinear ordinary differ-ential equations (ODEs) arising in extended surface heattransfer have been constructed using the DTM [11ndash19] TheDTM is an analytical method based on the Taylor seriesexpansion and was first introduced by Zhou [20] in 1986TheDTM approximates the exact solution by a polynomial andprevious studies have shown that it is an efficient means ofsolving nonlinear problems or systems with varying parame-ters [21] Furthermore DTM is a computational inexpensivetool for obtaining analytical solution and it generalizes theTaylor method to problems involving procedures such asfractional derivative (see eg [22ndash24]) Also this methodconverges rapidly (see eg [25])
Models arising in heat transfer through fins may containtemperature-dependent properties such as thermal conduc-tivity and heat transfer coefficient The dependency of ther-mal conductivity and heat transfer coefficient on temperaturerenders such problems highly nonlinear and difficult to solveparticularly exactly Thermal conductivity may be modeledfor many engineering applications by the power law and bylinear dependency on temperature On the other hand heat
2 Mathematical Problems in Engineering
transfer coefficient can be expressed as a power law for whichvalues of the exponent represent different phenomena (seeeg [26])
In this paper the DTM is employed to determine theanalytical solutions to the nonlinear boundary value problemdescribing heat transfer in longitudinal fins of rectangularexponential and convex parabolic profiles Both thermalconductivity and heat transfer coefficient are temperaturedependentWe adopt the terminology exact solutions to referto solutions given in terms of fundamental expressions suchas logarithmic trigonometric and exponential Howeveranalytical solutions will be series solutions and in partic-ular those constructed using the DTM The mathematicalmodelling of the problem under consideration is describedin Section 2 A brief discussion on the fundamentals of theDTM is be provided in Section 3 The comparison of theexact and analytical solutions constructed byDTM is given inSection 4 In Section 5 we provide analytical solutions for theheat transfer in longitudinal fins of various profilesHere heattransfer coefficient is given by the power law and we considertwo cases of the thermal conductivity namely the powerlaw and the linear function of temperature Furthermorewe describe the fin efficiency and the heat flux in Section 6Some exciting results are discussed in Section 7 Lastly theconcluding remarks are provided in Section 8
2 Mathematical Models
We consider a longitudinal one dimensional fin of cross-sectional area 119860
119888 The perimeter of the fin is denoted by
119875 and its length by 119871 The fin is attached to a fixed primesurface of temperature 119879
119887and extends to an ambient fluid
of temperature 119879119886 The fin thickness at the prime surface is
given by 120575119887and its profile is given by 119865(119883) Based on the one
dimensional heat conduction the energy balance equation isthen given by (see eg [1])
119860119888
119889
119889119883(120575119887
2119865 (119883)119870 (119879)
119889119879
119889119883) = 119875119867 (119879) (119879 minus 119879
119886)
0 le 119883 le 119871
(1)
where 119870 and 119867 are nonuniform temperature-dependentthermal conductivity and heat transfer coefficients respec-tively119879 is the temperature distribution119865(119883) is the fin profileand119883 is the space variable The length of the fin is measuredfrom the tip to the prime surface as shown in Figure 1Assuming that the fin tip is adiabatic (insulated) and the basetemperature is kept constant then the boundary conditionsare given by
119879 (119871) = 119879119887
119889119879
119889119883
10038161003816100381610038161003816100381610038161003816119883=0
= 0 (2)
Introducing the following dimensionless variables (see eg[1])
119909 =119883
119871 120579 =
119879 minus 119879119886
119879119887minus 119879119886
ℎ =119867
ℎ119887
119896 =119870
119896119886
1198722
=119875ℎ1198871198712
119860119888119896119886
119891 (119909) =120575119887
2119865 (119883)
(3)
119883 = 0 119883 = 119871
Fin tip
Fin profile
Prime surface
120575
119871
119882
Figure 1 Schematic representation of a longitudinal fin of anunspecified profile
with 119896119886being defined as the thermal conductivity at the
ambient temperature and ℎ119887as the heat transfer at the prime
surface (fin base) reduces (1) to
119889
119889119909[119891 (119909) 119896 (120579)
119889120579
119889119909] minus119872
2
120579ℎ (120579) = 0 0 le 119909 le 1 (4)
Here 120579 is the dimensionless temperature 119909 is the dimension-less space variable 119891(119909) is the dimensionless fin profile 119896 isthe dimensionless thermal conductivity ℎ is the dimension-less heat transfer coefficient and119872 is the thermogeometricfin parameter The dimensionless boundary conditions thenbecome
120579 (1) = 1 at the prime surface (5)
119889120579
119889119909
10038161003816100381610038161003816100381610038161003816119909=0
= 0 at the fin tip (6)
We assume that the heat transfer coefficient is given by thepower law
119867(119879) = ℎ119887(119879 minus 119879119886
119879119887minus 119879119886
)
119899
(7)
where the exponent 119899 is a real constant In fact the values of 119899may vary as between minus66 and 5 However in most practicalapplications it lies between minus3 and 3 [27] In dimensionlessvariables we have ℎ(120579) = 120579119899 We consider the two distinctcases of the thermal conductivity as follows
(a) the power law
119870 (119879) = 119896119886(119879 minus 119879119886
119879119887minus 119879119886
)
119898
(8)
with119898 being a constant and(b) the linear function
119870 (119879) = 119896119886[1 + 120574 (119879 minus 119879
119886)] (9)
Mathematical Problems in Engineering 3
The dimensionless thermal conductivity given by the powerlaw and the linear function of temperature is 119896(120579) = 120579119898 and119896(120579) = 1 + 120573120579 respectively Here the thermal conductivitygradient is120573 = 120574(119879
119887minus119879119886) Furthermore we consider a various
fin profiles including the longitudinal rectangular 119891(119909) =1 the longitudinal convex parabolic 119891(119909) = radic119909 and theexponential profile 119891(119909) = e119886119909 with 119886 being the constant (seealso [16])
3 Fundamentals of the DifferentialTransform Method
In this section the basic idea underlying the DTM is brieflyintroduced Let 119910(119905) be an analytic function in a domain DThe Taylor series expansion function of 119910(119905) with the centerlocated at 119905 = 119905
119895is given by [20]
119910 (119905) =
infin
sum
120581=0
(119905 minus 119905119895)120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=119905119895
forall119905 isin D (10)
The particular case of (10) when 119905119895= 0 is referred to as
the Maclaurin series expansion of 119910(119905) and is expressed as
119910 (119905) =
infin
sum
120581=0
119905120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=0
forall119905 isin D (11)
The differential transform of 119910(119905) is defined as follows
119884 (119905) =
infin
sum
120581=0
H120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=0
(12)
where 119910(119905) is the original analytic function and 119884(119905) is thetransformed function The differential spectrum of 119884(119905) isconfined within the interval 119905 isin [0H] where H is aconstant From (11) and (12) the differential inverse transformof 119884(119905) is defined as follows
119910 (119905) =
infin
sum
120581=0
(119905
H)
120581
119884 (120581) (13)
and if 119910(119905) is expressed by a finite series then
119910 (119905) =
119903
sum
120581=0
(119905
H)
120581
119884 (120581) (14)
Some of the useful mathematical operations performedby the differential transform method are listed in Table 1
The delta function 120575(120581 minus 119904) is given by
120575 (120581 minus 119904) = 1 if 120581 = 1199040 if 120581 = 119904
(15)
4 Comparison of Exact andAnalytical Solutions
In this section we consider a model describing temperaturedistribution in a longitudinal rectangular fin with both ther-mal conductivity and heat transfer coefficient being functions
Table 1 Fundamental operations of the differential transformmethod
Original function Transformed function119910(119905) = 119909(119905) plusmn 119911(119905) 119884(119905) = 119883(119905) plusmn 119885(119905)
119910(119905) = 120572119909(119905) 119884(119905) = 120572119883(119905)
119910(119905) =119889119910(119905)
119889119905119884(119905) = (120581 + 1)119884(120581 + 1)
119910(119905) =1198892
119910(119905)
1198891199052119884(119905) = (120581 + 1)(120581 + 2)119884(120581 + 2)
119910(119905) =119889119904
119910(119905)
119889119905119904119884(119905) = (120581 + 1)(120581 + 2) sdot sdot sdot (120581 + 119904)119884(120581 + 119904)
119910(119905) = 119909(119905)119911(119905) 119884(119905) =
120581
sum
119894=0
119883(119894)119885(120581 minus 119894)
119910(119905) = 1 119884(119905) = 120575(120581)
119910(119905) = 119905 119884(119905) = 120575(120581 minus 1)
119910(119905) = 119905119904
119884(119905) = 120575(120581 minus 119904)
119910(119905) = exp(120582119905) 119884(119905) =120582120581
120581
119910(119905) = (1 + 119905)119904
119884(119905) =119904(119904 minus 1) sdot sdot sdot (119904 minus 120581 + 1)
120581
119910(119905) = sin(120596119905 + 120572) 119884(119905) =120596120581
120581sin(120587120581
2+ 120572)
119910(119905) = cos(120596119905 + 120572) 119884(119905) =120596120581
120581cos(120587120581
2+ 120572)
of temperature given by the power law (see eg [2])The exactsolution of (4) when both the power laws are given by thesame exponent is given by [2]
120579 (119909) = [
cosh (119872radic119899 + 1119909)
cosh (119872radic119899 + 1)]
1(119899+1)
(16)
We use this exact solution as a benchmark or validation ofthe DTM The effectiveness of the DTM is determined bycomparing the exact and the analytical solutionsWe comparethe results for the cases 119899 = 1 and 119899 = 2 with fixed values of119872
41 Case 119899=1 Applying the DTM to (4) with the power lawthermal conductivity 119891(119909) = 1 (rectangular profile) andgivenH = 1 one obtains the following recurrence relation
120581
sum
119894=0
[Θ (119894) (120581 minus 119894 + 1) (120581 minus 119894 + 2)Θ (120581 minus 119894 + 2)
+ (119894 + 1)Θ (119894 + 1) (120581 minus 119894 + 1)Θ (120581 minus 119894 + 1)
minus1198722
Θ (119894)Θ (120581 minus 119894)] = 0
(17)
Exerting the transformation to the boundary condition (6) ata point 119909 = 0
Θ (1) = 0 (18)
The other boundary conditions are considered as follows
Θ (0) = 119888 (19)
4 Mathematical Problems in Engineering
where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows
Θ (2) =1198722
119888
2
(20)
Θ (3) = 0 (21)
Θ (4) = minus1198724
119888
24
(22)
Θ (5) = 0 (23)
Θ (6) =191198726
119888
720
(24)
Θ (7) = 0 (25)
Θ (8) = minus559119872
8
119888
40320
(26)
Θ (9) = 0 (27)
Θ (10) =29161119872
10
119888
3628800
(28)
Θ (11) = 0 (29)
Θ (12) = minus2368081119872
12
119888
479001600
(30)
Θ (13) = 0 (31)
Θ (14) =276580459119872
14
119888
87178291200
(32)
Θ (15) = 0
(33)
These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
241199094
+191198726
119888
7201199096
minus559119872
8
119888
403201199098
+29161119872
10
119888
362880011990910
minus2368081119872
12
119888
47900160011990912
+276580459119872
14
119888
8717829120011990914
+ sdot sdot sdot
(34)
To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have
120579 (1) = 119888 +1198722
119888
2minus1198724
119888
24+191198726
119888
720minus559119872
8
119888
40320
+29161119872
10
119888
3628800minus2368081119872
12
119888
479001600
+276580459119872
14
119888
87178291200+ sdot sdot sdot = 1
(35)
Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07
119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000
Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05
119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000
We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)
42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
81199094
+231198726
119888
2401199096
minus1069119872
8
119888
134401199098
+9643119872
10
119888
13440011990910
minus1211729119872
12
119888
1774080011990912
+217994167119872
14
119888
322882560011990914
+ sdot sdot sdot
(36)
The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)
Mathematical Problems in Engineering 5
08
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(a)
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(b)
Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05
5 Analytical Solutions
It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion
Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product
119891 (119909) sdot 1198911015840
(119909) = 120572 (37)
holds Here 120572 is a real constant
Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes
120572119889120585
119889119909
119889
119889120585[119896 (120579)
119889120579
119889120585] minus119872
2
120579119899+1
= 0 (38)
The most general solution of condition (37) is
119891 (119909) = (2120572119909 + 120574)12
(39)
where 120574 is an integration constant
Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into
119889
119889120585[119896 (120579)
119889120579
119889120585] minus 4119872
2
120585120579119899+1
= 0 (40)
only if 120590 = 12This example implies that the DTM may only be appli-
cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)
51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2
6 Mathematical Problems in Engineering
and applying the DTM one obtains the following recurrencerevelation
120581
sum
119894=0
120581minus119894
sum
119897=0
120581minus119894minus119897
sum
119901=0
[120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)
+ 2120590119901
119901Θ (119897) (119894 + 1)Θ (119894 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
+ 2120590120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
minus1198722
Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0
(41)
One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows
Θ (2) =1198722
1198882
2
Θ (3) = minus1205901198722
1198882
3
Θ (4) =
1198722
1198882
(31205902
minus 21198722
119888)
24
Θ (5) = minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
Θ (6) =
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720
(42)
The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution
120579 (119909) = 119888 +1198722
1198882
21199092
minus1205901198722
1198882
31199093
+
1198722
1198882
(31205902
minus 21198722
119888)
241199094
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
301199095
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
7201199096
+ sdot sdot sdot
(43)
To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is
120579 (1) = 119888 +1198722
1198882
2minus1205901198722
1198882
3+
1198722
1198882
(31205902
minus 21198722
119888)
24
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720+ sdot sdot sdot = 1
(44)
Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution
120579 (119909) = 119888 +1198722
21199092
minus1205901198722
31199093
+
1198722
(1205902
119888 minus 21198722
)
81198881199094
minus
1205901198722
(21205902
119888 minus 211198722
)
601198881199095
+
1198722
(1801198724
minus 1861205902
1198722
119888 + 51205904
1198882
)
7201199096
+ sdot sdot sdot
(45)
Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively
52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +1198722
21199092
minus1198724
41198881199094
+1198726
411988821199096
minus331198728
12211988831199098
+127119872
10
336119888411990910
+ sdot sdot sdot
(46)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +1198882
1198722
21199092
minus1198883
1198724
121199094
+291198884
1198726
3601199096
minus3071198885
1198728
50401199098
+23483119888
6
11987210
45360011990910
+ sdot sdot sdot
(47)
The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
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2 Mathematical Problems in Engineering
transfer coefficient can be expressed as a power law for whichvalues of the exponent represent different phenomena (seeeg [26])
In this paper the DTM is employed to determine theanalytical solutions to the nonlinear boundary value problemdescribing heat transfer in longitudinal fins of rectangularexponential and convex parabolic profiles Both thermalconductivity and heat transfer coefficient are temperaturedependentWe adopt the terminology exact solutions to referto solutions given in terms of fundamental expressions suchas logarithmic trigonometric and exponential Howeveranalytical solutions will be series solutions and in partic-ular those constructed using the DTM The mathematicalmodelling of the problem under consideration is describedin Section 2 A brief discussion on the fundamentals of theDTM is be provided in Section 3 The comparison of theexact and analytical solutions constructed byDTM is given inSection 4 In Section 5 we provide analytical solutions for theheat transfer in longitudinal fins of various profilesHere heattransfer coefficient is given by the power law and we considertwo cases of the thermal conductivity namely the powerlaw and the linear function of temperature Furthermorewe describe the fin efficiency and the heat flux in Section 6Some exciting results are discussed in Section 7 Lastly theconcluding remarks are provided in Section 8
2 Mathematical Models
We consider a longitudinal one dimensional fin of cross-sectional area 119860
119888 The perimeter of the fin is denoted by
119875 and its length by 119871 The fin is attached to a fixed primesurface of temperature 119879
119887and extends to an ambient fluid
of temperature 119879119886 The fin thickness at the prime surface is
given by 120575119887and its profile is given by 119865(119883) Based on the one
dimensional heat conduction the energy balance equation isthen given by (see eg [1])
119860119888
119889
119889119883(120575119887
2119865 (119883)119870 (119879)
119889119879
119889119883) = 119875119867 (119879) (119879 minus 119879
119886)
0 le 119883 le 119871
(1)
where 119870 and 119867 are nonuniform temperature-dependentthermal conductivity and heat transfer coefficients respec-tively119879 is the temperature distribution119865(119883) is the fin profileand119883 is the space variable The length of the fin is measuredfrom the tip to the prime surface as shown in Figure 1Assuming that the fin tip is adiabatic (insulated) and the basetemperature is kept constant then the boundary conditionsare given by
119879 (119871) = 119879119887
119889119879
119889119883
10038161003816100381610038161003816100381610038161003816119883=0
= 0 (2)
Introducing the following dimensionless variables (see eg[1])
119909 =119883
119871 120579 =
119879 minus 119879119886
119879119887minus 119879119886
ℎ =119867
ℎ119887
119896 =119870
119896119886
1198722
=119875ℎ1198871198712
119860119888119896119886
119891 (119909) =120575119887
2119865 (119883)
(3)
119883 = 0 119883 = 119871
Fin tip
Fin profile
Prime surface
120575
119871
119882
Figure 1 Schematic representation of a longitudinal fin of anunspecified profile
with 119896119886being defined as the thermal conductivity at the
ambient temperature and ℎ119887as the heat transfer at the prime
surface (fin base) reduces (1) to
119889
119889119909[119891 (119909) 119896 (120579)
119889120579
119889119909] minus119872
2
120579ℎ (120579) = 0 0 le 119909 le 1 (4)
Here 120579 is the dimensionless temperature 119909 is the dimension-less space variable 119891(119909) is the dimensionless fin profile 119896 isthe dimensionless thermal conductivity ℎ is the dimension-less heat transfer coefficient and119872 is the thermogeometricfin parameter The dimensionless boundary conditions thenbecome
120579 (1) = 1 at the prime surface (5)
119889120579
119889119909
10038161003816100381610038161003816100381610038161003816119909=0
= 0 at the fin tip (6)
We assume that the heat transfer coefficient is given by thepower law
119867(119879) = ℎ119887(119879 minus 119879119886
119879119887minus 119879119886
)
119899
(7)
where the exponent 119899 is a real constant In fact the values of 119899may vary as between minus66 and 5 However in most practicalapplications it lies between minus3 and 3 [27] In dimensionlessvariables we have ℎ(120579) = 120579119899 We consider the two distinctcases of the thermal conductivity as follows
(a) the power law
119870 (119879) = 119896119886(119879 minus 119879119886
119879119887minus 119879119886
)
119898
(8)
with119898 being a constant and(b) the linear function
119870 (119879) = 119896119886[1 + 120574 (119879 minus 119879
119886)] (9)
Mathematical Problems in Engineering 3
The dimensionless thermal conductivity given by the powerlaw and the linear function of temperature is 119896(120579) = 120579119898 and119896(120579) = 1 + 120573120579 respectively Here the thermal conductivitygradient is120573 = 120574(119879
119887minus119879119886) Furthermore we consider a various
fin profiles including the longitudinal rectangular 119891(119909) =1 the longitudinal convex parabolic 119891(119909) = radic119909 and theexponential profile 119891(119909) = e119886119909 with 119886 being the constant (seealso [16])
3 Fundamentals of the DifferentialTransform Method
In this section the basic idea underlying the DTM is brieflyintroduced Let 119910(119905) be an analytic function in a domain DThe Taylor series expansion function of 119910(119905) with the centerlocated at 119905 = 119905
119895is given by [20]
119910 (119905) =
infin
sum
120581=0
(119905 minus 119905119895)120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=119905119895
forall119905 isin D (10)
The particular case of (10) when 119905119895= 0 is referred to as
the Maclaurin series expansion of 119910(119905) and is expressed as
119910 (119905) =
infin
sum
120581=0
119905120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=0
forall119905 isin D (11)
The differential transform of 119910(119905) is defined as follows
119884 (119905) =
infin
sum
120581=0
H120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=0
(12)
where 119910(119905) is the original analytic function and 119884(119905) is thetransformed function The differential spectrum of 119884(119905) isconfined within the interval 119905 isin [0H] where H is aconstant From (11) and (12) the differential inverse transformof 119884(119905) is defined as follows
119910 (119905) =
infin
sum
120581=0
(119905
H)
120581
119884 (120581) (13)
and if 119910(119905) is expressed by a finite series then
119910 (119905) =
119903
sum
120581=0
(119905
H)
120581
119884 (120581) (14)
Some of the useful mathematical operations performedby the differential transform method are listed in Table 1
The delta function 120575(120581 minus 119904) is given by
120575 (120581 minus 119904) = 1 if 120581 = 1199040 if 120581 = 119904
(15)
4 Comparison of Exact andAnalytical Solutions
In this section we consider a model describing temperaturedistribution in a longitudinal rectangular fin with both ther-mal conductivity and heat transfer coefficient being functions
Table 1 Fundamental operations of the differential transformmethod
Original function Transformed function119910(119905) = 119909(119905) plusmn 119911(119905) 119884(119905) = 119883(119905) plusmn 119885(119905)
119910(119905) = 120572119909(119905) 119884(119905) = 120572119883(119905)
119910(119905) =119889119910(119905)
119889119905119884(119905) = (120581 + 1)119884(120581 + 1)
119910(119905) =1198892
119910(119905)
1198891199052119884(119905) = (120581 + 1)(120581 + 2)119884(120581 + 2)
119910(119905) =119889119904
119910(119905)
119889119905119904119884(119905) = (120581 + 1)(120581 + 2) sdot sdot sdot (120581 + 119904)119884(120581 + 119904)
119910(119905) = 119909(119905)119911(119905) 119884(119905) =
120581
sum
119894=0
119883(119894)119885(120581 minus 119894)
119910(119905) = 1 119884(119905) = 120575(120581)
119910(119905) = 119905 119884(119905) = 120575(120581 minus 1)
119910(119905) = 119905119904
119884(119905) = 120575(120581 minus 119904)
119910(119905) = exp(120582119905) 119884(119905) =120582120581
120581
119910(119905) = (1 + 119905)119904
119884(119905) =119904(119904 minus 1) sdot sdot sdot (119904 minus 120581 + 1)
120581
119910(119905) = sin(120596119905 + 120572) 119884(119905) =120596120581
120581sin(120587120581
2+ 120572)
119910(119905) = cos(120596119905 + 120572) 119884(119905) =120596120581
120581cos(120587120581
2+ 120572)
of temperature given by the power law (see eg [2])The exactsolution of (4) when both the power laws are given by thesame exponent is given by [2]
120579 (119909) = [
cosh (119872radic119899 + 1119909)
cosh (119872radic119899 + 1)]
1(119899+1)
(16)
We use this exact solution as a benchmark or validation ofthe DTM The effectiveness of the DTM is determined bycomparing the exact and the analytical solutionsWe comparethe results for the cases 119899 = 1 and 119899 = 2 with fixed values of119872
41 Case 119899=1 Applying the DTM to (4) with the power lawthermal conductivity 119891(119909) = 1 (rectangular profile) andgivenH = 1 one obtains the following recurrence relation
120581
sum
119894=0
[Θ (119894) (120581 minus 119894 + 1) (120581 minus 119894 + 2)Θ (120581 minus 119894 + 2)
+ (119894 + 1)Θ (119894 + 1) (120581 minus 119894 + 1)Θ (120581 minus 119894 + 1)
minus1198722
Θ (119894)Θ (120581 minus 119894)] = 0
(17)
Exerting the transformation to the boundary condition (6) ata point 119909 = 0
Θ (1) = 0 (18)
The other boundary conditions are considered as follows
Θ (0) = 119888 (19)
4 Mathematical Problems in Engineering
where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows
Θ (2) =1198722
119888
2
(20)
Θ (3) = 0 (21)
Θ (4) = minus1198724
119888
24
(22)
Θ (5) = 0 (23)
Θ (6) =191198726
119888
720
(24)
Θ (7) = 0 (25)
Θ (8) = minus559119872
8
119888
40320
(26)
Θ (9) = 0 (27)
Θ (10) =29161119872
10
119888
3628800
(28)
Θ (11) = 0 (29)
Θ (12) = minus2368081119872
12
119888
479001600
(30)
Θ (13) = 0 (31)
Θ (14) =276580459119872
14
119888
87178291200
(32)
Θ (15) = 0
(33)
These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
241199094
+191198726
119888
7201199096
minus559119872
8
119888
403201199098
+29161119872
10
119888
362880011990910
minus2368081119872
12
119888
47900160011990912
+276580459119872
14
119888
8717829120011990914
+ sdot sdot sdot
(34)
To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have
120579 (1) = 119888 +1198722
119888
2minus1198724
119888
24+191198726
119888
720minus559119872
8
119888
40320
+29161119872
10
119888
3628800minus2368081119872
12
119888
479001600
+276580459119872
14
119888
87178291200+ sdot sdot sdot = 1
(35)
Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07
119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000
Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05
119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000
We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)
42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
81199094
+231198726
119888
2401199096
minus1069119872
8
119888
134401199098
+9643119872
10
119888
13440011990910
minus1211729119872
12
119888
1774080011990912
+217994167119872
14
119888
322882560011990914
+ sdot sdot sdot
(36)
The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)
Mathematical Problems in Engineering 5
08
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(a)
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(b)
Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05
5 Analytical Solutions
It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion
Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product
119891 (119909) sdot 1198911015840
(119909) = 120572 (37)
holds Here 120572 is a real constant
Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes
120572119889120585
119889119909
119889
119889120585[119896 (120579)
119889120579
119889120585] minus119872
2
120579119899+1
= 0 (38)
The most general solution of condition (37) is
119891 (119909) = (2120572119909 + 120574)12
(39)
where 120574 is an integration constant
Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into
119889
119889120585[119896 (120579)
119889120579
119889120585] minus 4119872
2
120585120579119899+1
= 0 (40)
only if 120590 = 12This example implies that the DTM may only be appli-
cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)
51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2
6 Mathematical Problems in Engineering
and applying the DTM one obtains the following recurrencerevelation
120581
sum
119894=0
120581minus119894
sum
119897=0
120581minus119894minus119897
sum
119901=0
[120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)
+ 2120590119901
119901Θ (119897) (119894 + 1)Θ (119894 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
+ 2120590120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
minus1198722
Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0
(41)
One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows
Θ (2) =1198722
1198882
2
Θ (3) = minus1205901198722
1198882
3
Θ (4) =
1198722
1198882
(31205902
minus 21198722
119888)
24
Θ (5) = minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
Θ (6) =
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720
(42)
The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution
120579 (119909) = 119888 +1198722
1198882
21199092
minus1205901198722
1198882
31199093
+
1198722
1198882
(31205902
minus 21198722
119888)
241199094
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
301199095
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
7201199096
+ sdot sdot sdot
(43)
To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is
120579 (1) = 119888 +1198722
1198882
2minus1205901198722
1198882
3+
1198722
1198882
(31205902
minus 21198722
119888)
24
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720+ sdot sdot sdot = 1
(44)
Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution
120579 (119909) = 119888 +1198722
21199092
minus1205901198722
31199093
+
1198722
(1205902
119888 minus 21198722
)
81198881199094
minus
1205901198722
(21205902
119888 minus 211198722
)
601198881199095
+
1198722
(1801198724
minus 1861205902
1198722
119888 + 51205904
1198882
)
7201199096
+ sdot sdot sdot
(45)
Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively
52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +1198722
21199092
minus1198724
41198881199094
+1198726
411988821199096
minus331198728
12211988831199098
+127119872
10
336119888411990910
+ sdot sdot sdot
(46)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +1198882
1198722
21199092
minus1198883
1198724
121199094
+291198884
1198726
3601199096
minus3071198885
1198728
50401199098
+23483119888
6
11987210
45360011990910
+ sdot sdot sdot
(47)
The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering 3
The dimensionless thermal conductivity given by the powerlaw and the linear function of temperature is 119896(120579) = 120579119898 and119896(120579) = 1 + 120573120579 respectively Here the thermal conductivitygradient is120573 = 120574(119879
119887minus119879119886) Furthermore we consider a various
fin profiles including the longitudinal rectangular 119891(119909) =1 the longitudinal convex parabolic 119891(119909) = radic119909 and theexponential profile 119891(119909) = e119886119909 with 119886 being the constant (seealso [16])
3 Fundamentals of the DifferentialTransform Method
In this section the basic idea underlying the DTM is brieflyintroduced Let 119910(119905) be an analytic function in a domain DThe Taylor series expansion function of 119910(119905) with the centerlocated at 119905 = 119905
119895is given by [20]
119910 (119905) =
infin
sum
120581=0
(119905 minus 119905119895)120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=119905119895
forall119905 isin D (10)
The particular case of (10) when 119905119895= 0 is referred to as
the Maclaurin series expansion of 119910(119905) and is expressed as
119910 (119905) =
infin
sum
120581=0
119905120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=0
forall119905 isin D (11)
The differential transform of 119910(119905) is defined as follows
119884 (119905) =
infin
sum
120581=0
H120581
120581[119889120581
119910 (119905)
119889119905120581]
119905=0
(12)
where 119910(119905) is the original analytic function and 119884(119905) is thetransformed function The differential spectrum of 119884(119905) isconfined within the interval 119905 isin [0H] where H is aconstant From (11) and (12) the differential inverse transformof 119884(119905) is defined as follows
119910 (119905) =
infin
sum
120581=0
(119905
H)
120581
119884 (120581) (13)
and if 119910(119905) is expressed by a finite series then
119910 (119905) =
119903
sum
120581=0
(119905
H)
120581
119884 (120581) (14)
Some of the useful mathematical operations performedby the differential transform method are listed in Table 1
The delta function 120575(120581 minus 119904) is given by
120575 (120581 minus 119904) = 1 if 120581 = 1199040 if 120581 = 119904
(15)
4 Comparison of Exact andAnalytical Solutions
In this section we consider a model describing temperaturedistribution in a longitudinal rectangular fin with both ther-mal conductivity and heat transfer coefficient being functions
Table 1 Fundamental operations of the differential transformmethod
Original function Transformed function119910(119905) = 119909(119905) plusmn 119911(119905) 119884(119905) = 119883(119905) plusmn 119885(119905)
119910(119905) = 120572119909(119905) 119884(119905) = 120572119883(119905)
119910(119905) =119889119910(119905)
119889119905119884(119905) = (120581 + 1)119884(120581 + 1)
119910(119905) =1198892
119910(119905)
1198891199052119884(119905) = (120581 + 1)(120581 + 2)119884(120581 + 2)
119910(119905) =119889119904
119910(119905)
119889119905119904119884(119905) = (120581 + 1)(120581 + 2) sdot sdot sdot (120581 + 119904)119884(120581 + 119904)
119910(119905) = 119909(119905)119911(119905) 119884(119905) =
120581
sum
119894=0
119883(119894)119885(120581 minus 119894)
119910(119905) = 1 119884(119905) = 120575(120581)
119910(119905) = 119905 119884(119905) = 120575(120581 minus 1)
119910(119905) = 119905119904
119884(119905) = 120575(120581 minus 119904)
119910(119905) = exp(120582119905) 119884(119905) =120582120581
120581
119910(119905) = (1 + 119905)119904
119884(119905) =119904(119904 minus 1) sdot sdot sdot (119904 minus 120581 + 1)
120581
119910(119905) = sin(120596119905 + 120572) 119884(119905) =120596120581
120581sin(120587120581
2+ 120572)
119910(119905) = cos(120596119905 + 120572) 119884(119905) =120596120581
120581cos(120587120581
2+ 120572)
of temperature given by the power law (see eg [2])The exactsolution of (4) when both the power laws are given by thesame exponent is given by [2]
120579 (119909) = [
cosh (119872radic119899 + 1119909)
cosh (119872radic119899 + 1)]
1(119899+1)
(16)
We use this exact solution as a benchmark or validation ofthe DTM The effectiveness of the DTM is determined bycomparing the exact and the analytical solutionsWe comparethe results for the cases 119899 = 1 and 119899 = 2 with fixed values of119872
41 Case 119899=1 Applying the DTM to (4) with the power lawthermal conductivity 119891(119909) = 1 (rectangular profile) andgivenH = 1 one obtains the following recurrence relation
120581
sum
119894=0
[Θ (119894) (120581 minus 119894 + 1) (120581 minus 119894 + 2)Θ (120581 minus 119894 + 2)
+ (119894 + 1)Θ (119894 + 1) (120581 minus 119894 + 1)Θ (120581 minus 119894 + 1)
minus1198722
Θ (119894)Θ (120581 minus 119894)] = 0
(17)
Exerting the transformation to the boundary condition (6) ata point 119909 = 0
Θ (1) = 0 (18)
The other boundary conditions are considered as follows
Θ (0) = 119888 (19)
4 Mathematical Problems in Engineering
where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows
Θ (2) =1198722
119888
2
(20)
Θ (3) = 0 (21)
Θ (4) = minus1198724
119888
24
(22)
Θ (5) = 0 (23)
Θ (6) =191198726
119888
720
(24)
Θ (7) = 0 (25)
Θ (8) = minus559119872
8
119888
40320
(26)
Θ (9) = 0 (27)
Θ (10) =29161119872
10
119888
3628800
(28)
Θ (11) = 0 (29)
Θ (12) = minus2368081119872
12
119888
479001600
(30)
Θ (13) = 0 (31)
Θ (14) =276580459119872
14
119888
87178291200
(32)
Θ (15) = 0
(33)
These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
241199094
+191198726
119888
7201199096
minus559119872
8
119888
403201199098
+29161119872
10
119888
362880011990910
minus2368081119872
12
119888
47900160011990912
+276580459119872
14
119888
8717829120011990914
+ sdot sdot sdot
(34)
To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have
120579 (1) = 119888 +1198722
119888
2minus1198724
119888
24+191198726
119888
720minus559119872
8
119888
40320
+29161119872
10
119888
3628800minus2368081119872
12
119888
479001600
+276580459119872
14
119888
87178291200+ sdot sdot sdot = 1
(35)
Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07
119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000
Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05
119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000
We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)
42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
81199094
+231198726
119888
2401199096
minus1069119872
8
119888
134401199098
+9643119872
10
119888
13440011990910
minus1211729119872
12
119888
1774080011990912
+217994167119872
14
119888
322882560011990914
+ sdot sdot sdot
(36)
The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)
Mathematical Problems in Engineering 5
08
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(a)
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(b)
Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05
5 Analytical Solutions
It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion
Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product
119891 (119909) sdot 1198911015840
(119909) = 120572 (37)
holds Here 120572 is a real constant
Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes
120572119889120585
119889119909
119889
119889120585[119896 (120579)
119889120579
119889120585] minus119872
2
120579119899+1
= 0 (38)
The most general solution of condition (37) is
119891 (119909) = (2120572119909 + 120574)12
(39)
where 120574 is an integration constant
Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into
119889
119889120585[119896 (120579)
119889120579
119889120585] minus 4119872
2
120585120579119899+1
= 0 (40)
only if 120590 = 12This example implies that the DTM may only be appli-
cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)
51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2
6 Mathematical Problems in Engineering
and applying the DTM one obtains the following recurrencerevelation
120581
sum
119894=0
120581minus119894
sum
119897=0
120581minus119894minus119897
sum
119901=0
[120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)
+ 2120590119901
119901Θ (119897) (119894 + 1)Θ (119894 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
+ 2120590120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
minus1198722
Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0
(41)
One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows
Θ (2) =1198722
1198882
2
Θ (3) = minus1205901198722
1198882
3
Θ (4) =
1198722
1198882
(31205902
minus 21198722
119888)
24
Θ (5) = minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
Θ (6) =
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720
(42)
The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution
120579 (119909) = 119888 +1198722
1198882
21199092
minus1205901198722
1198882
31199093
+
1198722
1198882
(31205902
minus 21198722
119888)
241199094
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
301199095
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
7201199096
+ sdot sdot sdot
(43)
To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is
120579 (1) = 119888 +1198722
1198882
2minus1205901198722
1198882
3+
1198722
1198882
(31205902
minus 21198722
119888)
24
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720+ sdot sdot sdot = 1
(44)
Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution
120579 (119909) = 119888 +1198722
21199092
minus1205901198722
31199093
+
1198722
(1205902
119888 minus 21198722
)
81198881199094
minus
1205901198722
(21205902
119888 minus 211198722
)
601198881199095
+
1198722
(1801198724
minus 1861205902
1198722
119888 + 51205904
1198882
)
7201199096
+ sdot sdot sdot
(45)
Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively
52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +1198722
21199092
minus1198724
41198881199094
+1198726
411988821199096
minus331198728
12211988831199098
+127119872
10
336119888411990910
+ sdot sdot sdot
(46)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +1198882
1198722
21199092
minus1198883
1198724
121199094
+291198884
1198726
3601199096
minus3071198885
1198728
50401199098
+23483119888
6
11987210
45360011990910
+ sdot sdot sdot
(47)
The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
where 119888 is a constant Equation (17) is an iterative formula ofconstructing the power series solution as follows
Θ (2) =1198722
119888
2
(20)
Θ (3) = 0 (21)
Θ (4) = minus1198724
119888
24
(22)
Θ (5) = 0 (23)
Θ (6) =191198726
119888
720
(24)
Θ (7) = 0 (25)
Θ (8) = minus559119872
8
119888
40320
(26)
Θ (9) = 0 (27)
Θ (10) =29161119872
10
119888
3628800
(28)
Θ (11) = 0 (29)
Θ (12) = minus2368081119872
12
119888
479001600
(30)
Θ (13) = 0 (31)
Θ (14) =276580459119872
14
119888
87178291200
(32)
Θ (15) = 0
(33)
These terms may be taken as far as desired Substituting(18) to (32) into (13) we obtain the following analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
241199094
+191198726
119888
7201199096
minus559119872
8
119888
403201199098
+29161119872
10
119888
362880011990910
minus2368081119872
12
119888
47900160011990912
+276580459119872
14
119888
8717829120011990914
+ sdot sdot sdot
(34)
To obtain the value of 119888 we substitute the boundarycondition (5) into (34) at the point 119909 = 1 Thus we have
120579 (1) = 119888 +1198722
119888
2minus1198724
119888
24+191198726
119888
720minus559119872
8
119888
40320
+29161119872
10
119888
3628800minus2368081119872
12
119888
479001600
+276580459119872
14
119888
87178291200+ sdot sdot sdot = 1
(35)
Table 2 Results of the DTM and exact solutions for 119899 = 1119872 = 07
119909 DTM Exact Error0 0808093014 080809644 000000342601 0810072036 081007547 000000343402 0815999549 081600301 000000345903 0825847770 082585127 000000350004 0839573173 083957673 000000355905 0857120287 085712392 000000363306 0878426299 087843002 000000372207 0903426003 090342982 000000381408 0932056648 093206047 000000382009 0964262558 096426582 000000326310 1000000000 100000000 0000000000
Table 3 Results of theDTMand Exact Solutions for 119899 = 2119872 = 05
119909 DTM Exact Error0 0894109126 0894109793 000000066501 0895226066 0895226732 000000066602 0898568581 0898569249 000000066803 0904112232 0904112905 000000067204 0911817796 0911818474 000000067805 0921633352 0921634038 000000068506 0933496900 0933497594 000000069407 0947339244 0947339946 000000070208 0963086933 0963087627 000000069409 0980665073 0980665659 000000058610 1000000000 1000000000 0000000000
We omit presenting the tedious process of finding 119888 Howeverone may obtain the expression for 120579(119909) upon substituting theobtained value of 119888 into (34)
42 Case 119899=2 Following a similar approach given inSection 41 and given 119899 = 2 one obtains the analyticalsolution
120579 (119909) = 119888 +1198722
119888
21199092
minus1198724
119888
81199094
+231198726
119888
2401199096
minus1069119872
8
119888
134401199098
+9643119872
10
119888
13440011990910
minus1211729119872
12
119888
1774080011990912
+217994167119872
14
119888
322882560011990914
+ sdot sdot sdot
(36)
The constant 119888may be obtained using the boundary conditionat the fin base The comparison of the DTM and the exactsolutions are reflected in Tables 2 and 3 for different valuesof 119899 Furthermore the comparison of the exact and theanalytical solutions is depicted in Figures 2(a) and 2(b)
Mathematical Problems in Engineering 5
08
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(a)
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(b)
Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05
5 Analytical Solutions
It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion
Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product
119891 (119909) sdot 1198911015840
(119909) = 120572 (37)
holds Here 120572 is a real constant
Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes
120572119889120585
119889119909
119889
119889120585[119896 (120579)
119889120579
119889120585] minus119872
2
120579119899+1
= 0 (38)
The most general solution of condition (37) is
119891 (119909) = (2120572119909 + 120574)12
(39)
where 120574 is an integration constant
Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into
119889
119889120585[119896 (120579)
119889120579
119889120585] minus 4119872
2
120585120579119899+1
= 0 (40)
only if 120590 = 12This example implies that the DTM may only be appli-
cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)
51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2
6 Mathematical Problems in Engineering
and applying the DTM one obtains the following recurrencerevelation
120581
sum
119894=0
120581minus119894
sum
119897=0
120581minus119894minus119897
sum
119901=0
[120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)
+ 2120590119901
119901Θ (119897) (119894 + 1)Θ (119894 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
+ 2120590120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
minus1198722
Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0
(41)
One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows
Θ (2) =1198722
1198882
2
Θ (3) = minus1205901198722
1198882
3
Θ (4) =
1198722
1198882
(31205902
minus 21198722
119888)
24
Θ (5) = minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
Θ (6) =
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720
(42)
The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution
120579 (119909) = 119888 +1198722
1198882
21199092
minus1205901198722
1198882
31199093
+
1198722
1198882
(31205902
minus 21198722
119888)
241199094
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
301199095
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
7201199096
+ sdot sdot sdot
(43)
To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is
120579 (1) = 119888 +1198722
1198882
2minus1205901198722
1198882
3+
1198722
1198882
(31205902
minus 21198722
119888)
24
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720+ sdot sdot sdot = 1
(44)
Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution
120579 (119909) = 119888 +1198722
21199092
minus1205901198722
31199093
+
1198722
(1205902
119888 minus 21198722
)
81198881199094
minus
1205901198722
(21205902
119888 minus 211198722
)
601198881199095
+
1198722
(1801198724
minus 1861205902
1198722
119888 + 51205904
1198882
)
7201199096
+ sdot sdot sdot
(45)
Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively
52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +1198722
21199092
minus1198724
41198881199094
+1198726
411988821199096
minus331198728
12211988831199098
+127119872
10
336119888411990910
+ sdot sdot sdot
(46)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +1198882
1198722
21199092
minus1198883
1198724
121199094
+291198884
1198726
3601199096
minus3071198885
1198728
50401199098
+23483119888
6
11987210
45360011990910
+ sdot sdot sdot
(47)
The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 5
08
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(a)
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
DTMExact
(b)
Figure 2 Comparison of analytical and exact solutions In (a) 119899 = 1119872 = 07 and (b) 119899 = 2119872 = 05
5 Analytical Solutions
It is well known that exact solutions forODEs such as (4) existonly when thermal conductivity and the term containingheat transfer coefficient are connected by differentiation(or simply if the ODE such as (4) is linearizable) [4] Inthis section we determine the analytical solutions for thenonlinearizable (4) firstlywhen thermal conductivity is givenby the power law and secondly as a linear function oftemperature In both cases and throughout this paper theheat transfer coefficient is assumed to be a power law functionof temperature These assumptions of the thermal propertiesare physical realistic We have noticed that DTM runs intodifficulty when the exponent of the power law of the thermalconductivity is given by fractional values and also when thefunction 119891(119909) is given in terms of fractional power law Onemay followMoradi and Ahmadikia [16] by introducing a newvariable to deal with fractional powers of 119891(119909) and on theother hand it is possible to remove the fractional exponent ofthe heat transfer coefficient by fundamental laws of exponentand binomial expansion
Proposition 1 A nonlinear ODE such as (4) may admit theDTM solution if 119891(119909) is a constant or exponential functionHowever if 119891(119909) is a power law then such an equation admitsa DTM solution if the product
119891 (119909) sdot 1198911015840
(119909) = 120572 (37)
holds Here 120572 is a real constant
Proof Introducing the new variable 120585 = 119891(119909) it follows fromchain rule that (4) becomes
120572119889120585
119889119909
119889
119889120585[119896 (120579)
119889120579
119889120585] minus119872
2
120579119899+1
= 0 (38)
The most general solution of condition (37) is
119891 (119909) = (2120572119909 + 120574)12
(39)
where 120574 is an integration constant
Example 2 (a) If 119891(119909) = 119909120590 then 120585 = 119909120590 transforms (4) into
119889
119889120585[119896 (120579)
119889120579
119889120585] minus 4119872
2
120585120579119899+1
= 0 (40)
only if 120590 = 12This example implies that the DTM may only be appli-
cable to problems describing heat transfer in fins withconvex parabolic profile In the next subsections we presentanalytical solutions for (4) with various functions 119891(119909) and119896(120579)
51 The Exponential Profile and Power Law Thermal Con-ductivity In this section we present solutions for equationdescribing heat transfer in a fin with exponential profile andpower law thermal conductivity and heat transfer coefficientThat is given (4) with 119891(119909) = e120590119909 and both heat transfercoefficient and thermal conductivity being power law func-tions of temperature we construct analytical solutions In ouranalysis we consider 119899 = 2 and 3 indicating the fin subject tonucleate boiling and radiation into free space at zero absolutetemperature respectively Firstly given 119899 = 3 and 119898 = 2
6 Mathematical Problems in Engineering
and applying the DTM one obtains the following recurrencerevelation
120581
sum
119894=0
120581minus119894
sum
119897=0
120581minus119894minus119897
sum
119901=0
[120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)
+ 2120590119901
119901Θ (119897) (119894 + 1)Θ (119894 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
+ 2120590120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
minus1198722
Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0
(41)
One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows
Θ (2) =1198722
1198882
2
Θ (3) = minus1205901198722
1198882
3
Θ (4) =
1198722
1198882
(31205902
minus 21198722
119888)
24
Θ (5) = minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
Θ (6) =
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720
(42)
The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution
120579 (119909) = 119888 +1198722
1198882
21199092
minus1205901198722
1198882
31199093
+
1198722
1198882
(31205902
minus 21198722
119888)
241199094
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
301199095
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
7201199096
+ sdot sdot sdot
(43)
To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is
120579 (1) = 119888 +1198722
1198882
2minus1205901198722
1198882
3+
1198722
1198882
(31205902
minus 21198722
119888)
24
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720+ sdot sdot sdot = 1
(44)
Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution
120579 (119909) = 119888 +1198722
21199092
minus1205901198722
31199093
+
1198722
(1205902
119888 minus 21198722
)
81198881199094
minus
1205901198722
(21205902
119888 minus 211198722
)
601198881199095
+
1198722
(1801198724
minus 1861205902
1198722
119888 + 51205904
1198882
)
7201199096
+ sdot sdot sdot
(45)
Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively
52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +1198722
21199092
minus1198724
41198881199094
+1198726
411988821199096
minus331198728
12211988831199098
+127119872
10
336119888411990910
+ sdot sdot sdot
(46)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +1198882
1198722
21199092
minus1198883
1198724
121199094
+291198884
1198726
3601199096
minus3071198885
1198728
50401199098
+23483119888
6
11987210
45360011990910
+ sdot sdot sdot
(47)
The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Mathematical Problems in Engineering
and applying the DTM one obtains the following recurrencerevelation
120581
sum
119894=0
120581minus119894
sum
119897=0
120581minus119894minus119897
sum
119901=0
[120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 2)Θ (120581 minus 119894 minus 119897 minus 119901 + 2)
+ 2120590119901
119901Θ (119897) (119894 + 1)Θ (119894 + 1)
times (120581 minus 119894 minus 119897 minus 119901 + 1)Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
+ 2120590120590119901
119901Θ (119897) Θ (119894) (120581 minus 119894 minus 119897 minus 119901 + 1)
times Θ (120581 minus 119894 minus 119897 minus 119901 + 1)
minus1198722
Θ(119901)Θ (119897) Θ (119894) Θ (120581 minus 119894 minus 119897 minus 119901) ] = 0
(41)
One may recall the transformed prescribed boundaryconditions (18) and (19) Equation (41) is an iterative formulaof constructing the power series solution as follows
Θ (2) =1198722
1198882
2
Θ (3) = minus1205901198722
1198882
3
Θ (4) =
1198722
1198882
(31205902
minus 21198722
119888)
24
Θ (5) = minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
Θ (6) =
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720
(42)
The pervious process is continuous and onemay consideras many terms as desired (but bearing in mind that DTMconverges quite fast) Substituting (18) to (19) and (42) into(13) we obtain the following closed form of the solution
120579 (119909) = 119888 +1198722
1198882
21199092
minus1205901198722
1198882
31199093
+
1198722
1198882
(31205902
minus 21198722
119888)
241199094
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
301199095
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
7201199096
+ sdot sdot sdot
(43)
To obtain the value of 119888 we substitute the boundarycondition (5) into (43) at the point 119909 = 1 That is
120579 (1) = 119888 +1198722
1198882
2minus1205901198722
1198882
3+
1198722
1198882
(31205902
minus 21198722
119888)
24
minus
1198722
1198882
(1205903
minus 41205901198722
119888)
30
+
1198722
1198882
(51205904
minus 781205902
1198722
119888 + 581198724
1198882
)
720+ sdot sdot sdot = 1
(44)
Substituting this value of 119888 into (43) one finds theexpression for 120579(119909) On the other hand given (119899119898) = (2 3)one obtains the solution
120579 (119909) = 119888 +1198722
21199092
minus1205901198722
31199093
+
1198722
(1205902
119888 minus 21198722
)
81198881199094
minus
1205901198722
(21205902
119888 minus 211198722
)
601198881199095
+
1198722
(1801198724
minus 1861205902
1198722
119888 + 51205904
1198882
)
7201199096
+ sdot sdot sdot
(45)
Here the constant 119888 maybe obtained by evaluating theboundary condition 120579(1) = 1 The solutions (43) and (45) aredepicted in Figures 3(b) and 3(a) respectively
52 The Rectangular Profile and Power Law Thermal Con-ductivity In this section we provide a detailed constructionof analytical solutions for the heat transfer in a longitudinalrectangular finwith a power law thermal conductivity that iswe consider (4) with 119891(119909) = 1 and 119896(120579) = 120579119898 The analyticalsolutions are given in the following expressions
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +1198722
21199092
minus1198724
41198881199094
+1198726
411988821199096
minus331198728
12211988831199098
+127119872
10
336119888411990910
+ sdot sdot sdot
(46)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +1198882
1198722
21199092
minus1198883
1198724
121199094
+291198884
1198726
3601199096
minus3071198885
1198728
50401199098
+23483119888
6
11987210
45360011990910
+ sdot sdot sdot
(47)
The constant 119888 is obtained by solving the appropriate 120579(119909)at the fin base boundary condition The analytical solutionsin (46) and (47) are depicted in Figures 4(a) and 4(b)respectively
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 7
0 02 04 06 08 1094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(a)
0 02 04 06 08 1093
094
095
096
097
098
099
1
119909
120579
119872 = 02
119872 = 03
119872 = 04
119872 = 05
(b)
Figure 3 Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
082
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
(a)
084
086
088
09
092
094
096
098
1
120579
0 02 04 06 08 1
119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(b)
Figure 4 Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity In (a) the exponents are(119899119898) = (2 3) and (b) (119899 119898) = (3 2)
53 The Convex Parabolic Profile and Power Law ThermalConductivity In this section we present solutions for theequation describing the heat transfer in a fin with con-vex parabolic profile and power law thermal conductivityEquation (40) is considered Here we consider the values(119899119898) = (2 3) (3 2) The final analytical solution is givenby
(a) Case (119899119898) = (2 3)
120579 (119909) = 119888 +21198722
311990932
minus21198724
51198881199093
+231198726
45119888211990992
minus1909119872
8
247511988831199096
+329222119872
10
2598751198884119909152
+ sdot sdot sdot
(48)
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Mathematical Problems in Engineering
075
08
085
09
095
1
119872 = 025
119872 = 04
119872 = 05
119872 = 065
120579
0 02 04 06 08 1119909
(a)
119872 = 025
119872 = 04
119872 = 05
119872 = 065
0 02 04 06 08 1119909
082
084
086
088
09
092
094
096
098
120579
1
(b)
Figure 5 Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity In (a) the exponentsare (119899119898) = (2 3) and (b) (119899119898) = (3 2)
(b) Case (119899119898) = (3 2)
120579 (119909) = 119888 +21198882
1198722
311990932
minus41198883
1198724
451199093
+41198884
1198726
2711990992
minus9921198885
1198728
74251199096
+30064119888
6
11987210
212625119909152
+ sdot sdot sdot
(49)
The constant 119888 is obtained by evaluating the appropriate 120579(119909)at the fin base boundary condition The solutions in (48) and(49) are depicted in Figures 5(a) and 5(b) respectively
54 The Rectangular Profile and LinearThermal ConductivityIn this section we present solutions for the equation repre-senting the heat transfer in a fin with rectangular profile andthe thermal conductivity depending linearly on temperatureThat is we consider (4) with 119891(119909) = 1 and 119896(120579) = 1 + 120573120579 Theanalytical solutions for this problem for different values of 119899are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1198724
119888 (minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
1198726
119888 (1 minus 16120573119888 + 281205732
1198882
)
720(1 + 120573119888)51199096
minus
1198728
119888 (minus1 + 78120573119888 minus 6001205732
1198882
+ 8961205733
1198883
)
40320(1 + 120573119888)7
1199098
+
11987210
119888 (1 minus 332120573119888 + 78121205732
1198882
minus 398961205733
1198883
+ 511841205734
1198884
)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(50)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
11987241198883(minus1 + 2120573119888)
24(1 + 120573119888)31199094
+
11987261198884(10 minus 16120573119888 + 19120573
21198882)
720(1 + 120573119888)5
1199096
minus
11987281198885(minus80 + 342120573119888 minus 594120573
21198882+ 559120573
31198883)
40320(1 + 120573119888)7
1199098
+
119872101198886(1000 minus 7820120573119888 + 24336120573
21198882minus 36908120573
31198883+ 29161120573
41198884)
3628800(1 + 120573119888)9
11990910
+ sdot sdot sdot
(51)
(c) Case 119899 = 2
120579 (119909)
= 119888 +
1198722
1198883
2 (1 + 120573119888)
1199092
+
1198724
1198885
8(1 + 120573119888)31199094
+
1198726
1198887
(3 + 21205732
1198882
)
80(1 + 120573119888)51199096
+
1198728
1198889
(49 minus 20120573119888 + 661205732
1198882
minus 401205733
1198883
)
4480(1 + 120573119888)7
1199098
+
11987210
11988811
(427 minus 440120573119888 + 11161205732
1198882
minus 10201205733
1198883
+ 6721205734
1198884
)
134400(1 + 120573119888)9
11990910
+ sdot sdot sdot
(52)
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 9
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092+
11987241198887(4 + 120573119888)
24(1 + 120573119888)31199094
+
119872611988810(52 + 32120573119888 + 25120573
21198882)
720(1 + 120573119888)5
1199096
+
119872811988813(1288 + 1020120573119888 + 1212120573
21198882minus 95120573
31198883)
40320(1 + 120573119888)7
1199098
+
1198721011988816(52024+45688120573119888+77184120573
21198882minus680120573
31198883+ 15025120573
41198884)
3628800(1 + 120573119888)9
11990910
+
(53)
The constant 119888may be obtained from the boundary conditionon the appropriate solution The solutions in (50) (51) (52)and (53) are depicted in Figure 6
55The Convex Parabolic Profile and LinearThermal Conduc-tivity In this section we present solutions for the equationdescribing the heat transfer in a fin with convex parabolicprofile and the thermal conductivity depending linearly ontemperature That is we consider (40) with 119896(120579) = 1 + 120573120579The analytical solution for this problem for different values of119899 is given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +21198722
119888
3 (1 + 120573119888)11990932
minus21198724
119888 (minus2 + 3120573119888)
45(1 + 120573119888)3
1199093
+
1198726
119888 (2 minus 25120573119888 + 331205732
1198882
)
405(1 + 120573119888)5
11990992
minus
1198728
119888 (minus10 + 599120573119888 minus 35821205732
1198882
+ 40591205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(54)
(b) Case 119899 = 1
120579 (119909)
= 119888 +21198722
1198882
3 (1 + 120573119888)11990932
minus21198724
1198883
(minus4 + 120573119888)
45(1 + 120573119888)3
1199093
+
21198726
1198884
(9 minus 11120573119888 + 101205732
1198882
)
405(1 + 120573119888)5
11990992
minus
21198728
1198885
(minus330 + 1118120573119888 minus 15841205732
1198882
+ 10931205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(55)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
1
095
09
085
08
075
07
065
06
055
0 02 04 06 08 1
119909
120579
Figure 6 Temperature profile in a longitudinal rectangular fin withlinear thermal conductivity and varying values of 119899 Here 120573 = 05and119872 = 15 are fixed
(c) Case 119899 = 2
120579 (119909)
= 119888 +21198722
1198883
3 (1 + 120573119888)11990932
+21198724
1198885
(6 + 120573119888)
45(1 + 120573119888)31199093
+
1198726
1198887
(16 + 3120573119888 + 71205732
1198882
)
135(1 + 120573119888)5
11990992
+
1198728
1198889
(1160 minus 107120573119888 + 11161205732
1198882
minus 3671205733
1198883
)
22275(1 + 120573119888)7
1199096
+ sdot sdot sdot
(56)
(d) Case 119899 = 3
120579 (119909)
= 119888 +21198722
1198884
3 (1 + 120573119888)11990932
+21198724
1198887
(8 + 3120573119888)
45(1 + 120573119888)31199093
+
41198726
11988810
(23 + 17120573119888 + 91205732
1198882
)
405(1 + 120573119888)5
11990992
+
21198728
11988813
(5000+4868120573119888 + 43561205732
1198882
+ 3631205733
1198883
)
66825(1 + 120573119888)7
1199096
+ sdot sdot sdot
(57)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (54) (55)(56) and (57) are depicted in Figure 7
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Mathematical Problems in Engineering
1
09
08
07
06
05
120579
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
Figure 7 Temperature profile in a longitudinal convex parabolic finwith linear thermal conductivity and varying values of 119899 Here 120573 =05 and119872 = 15 are fixed
56 The Exponential Profile and LinearThermal ConductivityIn this section we present solutions for equation heat transferin a fin with convex parabolic profile and the thermalconductivity depending linearly on temperature That is weconsider (4) with 119896(120579) = 1 + 120573120579 and 119891(119909) = e120590119909 Here 120590is a constant The analytical solutions for this problem fordifferent values of 119899 are given by
(a) Case 119899 = 0
120579 (119909)
= 119888 +
1198722
119888
2 (1 + 120573119888)
1199092
minus
1205901198722
119888 (2 + 120573119888 + 1205732
119888)
6(1 + 120573119888)21199093
+
1198722
119888 (1198722
(1 minus 21205732
119888) + 1205902
(3 +3120573119888+1205733
1198882
+1205734
1198882
+1205732
119888 (3 + 119888)))
24(1 + 120573119888)4
1199094
+ sdot sdot sdot
(58)
(b) Case 119899 = 1
120579 (119909)
= 119888 +
11987221198882
2 (1 + 120573119888)
1199092minus
12059011987221198882(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198882(1198722119888 (2+120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
1199094
+ sdot sdot sdot
(59)
119899 = 0
119899 = 1
119899 = 2
119899 = 3
0 02 04 06 08 1
119909
1
095
09
085
08
075
07
120579
Figure 8 Temperature profile in a longitudinal fin of exponentialprofile with linear thermal conductivity and varying values of 119899Here 120573 = 05 and119872 = 15 are fixed
(c) Case 119899 = 2120579 (119909)
= 119888 +
11987221198883
2 (1 + 120573119888)
1199092minus
12059011987221198883(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198883(11987221198882(3+2120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(60)
(d) Case 119899 = 3120579 (119909)
= 119888 +
11987221198884
2 (1 + 120573119888)
1199092minus
12059011987221198884(2 + 120573119888 + 120573
2119888)
6(1 + 120573119888)21199093
+
11987221198884(11987221198883(4+3120573119888minus2120573
2119888)+1205902(3+3120573119888+120573
31198882+12057341198882+1205732119888 (3+119888)))
24(1 + 120573119888)3
times1199094+ sdot sdot sdot
(61)
The constant 119888 may be obtained from the boundarycondition on the appropriate 120579(119909)The solutions in (58) (59)(60) and (61) are depicted in Figure 8
6 Fin Efficiency and Heat Flux
61 Fin Efficiency Theheat transfer rate from a fin is given byNewtonrsquos second law of cooling
119876 = int
119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119909 (62)
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 11
Fin efficiency is defined as the ratio of the fin heat transferrate to the rate that would be if the entire fin were at the basetemperature and is given by (see eg [1])
120578=
119876
119876ideal=
int119871
0
119875119867 (119879) (119879 minus 119879119886) 119889119883
119875ℎ119887119871 (119879119887minus 119879119886)
(63)
In dimensionless variables we have
120578 = int
1
0
120579119899+1
119889119909 (64)
We consider the solutions (50) (51) (52) and (53) and depictthe fin efficiency (63) in Figure 9
62 Heat Flux The fin base heat flux is given by the Fourierrsquoslaw
119902119887= 119860119888119870 (119879)
119889119879
119889119909 (65)
The total heat flux of the fin is given by [1]
119902 =119902119887
119860119888119867(119879) (119879
119887minus 119879119886) (66)
Introducing the dimensionless variable as described inSection 2 implies
119902 =1
119861119894
119896 (120579)
ℎ (120579)
119889120579
119889119909 (67)
where the dimensionless parameter 119861119894 = ℎ119887119871119896119886is the Biot
number We consider a number of cases for the thermalconductivity and the heat transfer coefficient
621 Linear Thermal Conductivity and Power Law HeatTransfer Coefficient In this case (67) becomes
119902 =1
119861119894(1 + 120573120579) 120579
minus119899119889120579
119889119909 (68)
The heat flux in (68) at the base of the fin is plotted inFigure 10
622 Power LawThermal Conductivity and Heat TransferCoefficient In this case (67) becomes
119902 =1
119861119894120579119898minus119899
119889120579
119889119909 (69)
Not surprisingly heat flux in one-dimensional fins is highergiven values 119861119894 ≪ 1 The heat flux in (69) is plotted inFigures 11(a) 11(b) and 11(c)
7 Some Discussions
The DTM has resulted in some interesting observations andstudy We have observed in Figures 2(a) and 2(b) an excellentagreement between the analytical solutions generated byDTM and the exact solution obtained in [2] In particular we
119872
1
095
09
085
08
075
07
065
06
055
120578
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 9 Fin efficiency of a longitudinal rectangular fin Here 120573 =075
119872
25
20
15
10
5
0
119902
2151050
119899 = 0
119899 = 1
119899 = 2
119899 = 3
Figure 10 Base heat flux in a longitudinal rectangular finwith linearthermal conductivity Here 120573 = 01
considered a fin problem in which both thermal conductivityand heat transfer coefficient are given by the same power lawFurthermore we notice from Table 2 that an absolute errorof approximately 35119890 minus 005 is produced by DTM of order119874(15) In Table 3 an absolute error of approximately 65119890 minus006 is produced for the same order This confirms that theDTM converges faster and can provide accurate results witha minimum computation As such a tremendous confidencein the DTM in terms of the accuracy and effectiveness wasbuilt and thus we used this method to solve other problemsfor which exact solutions are harder to construct
In Figures 3(a) 3(b) 4(a) 4(b) 5(a) and 5(b) weobserve that the fin temperature increaseswith the decreasing
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Mathematical Problems in Engineering
values of the thermogeometric fin parameter Here thevalues of the exponents are fixed Also we observe that fintemperature is higher when 119899 minus 119898 gt 0 that is when heattransfer coefficient is higher than the thermal conductivityWe observe in Figures 6 7 and 8 that the fin temperatureincreases with the increasing values of 119899 Furthermoreit appears that the fin with exponential profile performsthe least in transferring the heat from the base since thetemperature in such a fin is much higher than that ofthe rectangular and the convex parabolic profiles In otherwords heat dissipation to the fluid surrounding the extendedsurface is much faster in longitudinal fins of rectangular andconvex parabolic profiles In Figure 9 fin efficiency decreaseswith increasing thermogeometric fin parameter Also finefficiency increases with increasing values of 119899 It is easyto show that the thermogeometric fin parameter is directlyproportional to the aspect ratio (extension factor) with squareroot of the Biot number being the proportionality constantAs such shorter fins are more efficient than longer onesElse the increased Biot number results in less efficient finwhenever the space is confined that is where the length ofthe fin cannot be increased Figure 10 depicts the heat fluxat the fin base The amount of heat energy dissipated fromthe fin base is of immense interest in engineering [28] Weobserve in Figure 10 that the base heat flux increases withthe thermogeometric fin parameter for considered values ofthe exponent 119899 (see also [28]) Figures 11(a) 11(b) and 11(c)display the heat flux across the fin length We note that theheat flux across the fin length increases with increasing valuesof the thermogeometric fin parameter
8 Concluding Remarks
In this study we have successfully applied the DTM to highlynonlinear problems arising in heat transfer through longitu-dinal fins of various profiles Both thermal conductivity andheat transfer coefficient are given as functions of temperatureThe DTM agreed well with exact solutions when the thermalconductivity and heat transfer coefficient are given by thesame power law A rapid convergence to the exact solutionwas observed Following the confidence in DTM built by theresults mentioned we then solved various exciting problemsThe exotic results have been shown in tables and figures listedin this paper
The results obtained in this paper are significant improve-ments on the known results In particular both the heattransfer coefficient and the thermal conductivity are allowedto be given by the power law functions of temperature andalsowe considered a number of finprofilesWenote that exactsolutions are difficult if not impossible to construct when theexponents of these properties are distinct
Perhaps the notable advantage of the DTM is the general-ization of the Taylor method to problems involving unusualderivative procedures such as fractional fuzzy or q-derivative[22] Some generalizations have been made by Odibat et al[24] and they referred to their new method as the General-ized Differential Transform Method (GDTM) This showedgreat improvement compared to the Fractional Differential
0 02 04 06 08 1119909
45
40
35
30
25
20
15
10
5
0
minus5
119902
(a)
0 02 04 06 08 1119909
30
25
20
15
10
5
0
119902
(b)
35
30
25
20
15
10
5
0
119902
0 02 04 06 08 1119909
119872 = 025
119872 = 04
119872 = 05
119872 = 065
(c)
Figure 11 Heat flux across a longitudinal rectangular fin with linearthermal conductivity (a) 119899minus119898 lt 0 (b) 119899minus119898 gt 0 and (c) 119899minus119898 = 0Here 119861119894 = 001
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 13
Transform Method (FDTM) introduced by Arikoglu andOzkol [23]
We have shown with the help of an example that DTMmay only be applied to fin problems involving heat transferthrough fins with convex parabolic profile Note that given anODE such as (4) with a power law heat transfer coefficient ofa fractional exponent then one can easily remove the fractionby basic exponential rules and employment of the Binomialexpansion However using the DTM one runs into difficultyif the power law thermal conductivity in the same equation isgiven by the fractional exponent We do not know whetherthese observations call for the ldquomodifiedrdquo DTM to solveproblems arising in heat flow through fins with other profilessuch as longitudinal triangular and concave parabolic andalso with fractional power law thermal conductivity
The main results obtained in this paper give insightinto heat transfer in boiling liquids where the heat transfercoefficient is temperature dependent and may be given by apower law The thermal conductivity of some materials suchas gallium nitride (GaN) and Aluminium Nitride (AlN) maybe modeled by power law temperature dependency [29 30]Thus the solutions constructed here give a better comparisonof heat transfer in terms of material used since in many engi-neering applications thermal conductivity is given as a linearfunction of temperature Furthermore a good study in termsof performance and efficiency of fin with different profiles isundertaken These finding could help in the design of fins Itis claimed in [14] that DTM results are more accurate thanthose constructed by Variational Iteration Methods (VIM)and Homotopy Perturbation Methods (HPM) However itwould be risky to use the DTM approximate solutions asbenchmarks for the numerical schemes Nevertheless wehave also shown that DTM converges rapidly in just fifteenterms to the exact solution
Acknowledgments
The authors wish to thank the National Research Foundationof South Africa and the University of the Witwatersrandfor the generous financial support The authors are gratefulto Professor D P Mason and the anonymous reviewers fortheir invaluable comments which improved this paper signif-icantly
References
[1] A D Kraus A Aziz and J Welty Extended Surface Heat Trans-fer Wiley New York NY USA 2001
[2] R JMoitsheki T Hayat andM YMalik ldquoSome exact solutionsof the fin problem with a power law temperature-dependentthermal conductivityrdquo Nonlinear Analysis Real World Applica-tions vol 11 no 5 pp 3287ndash3294 2010
[3] R J Moitsheki ldquoSteady one-dimensional heat flow in a lon-gitudinal triangular and parabolic finrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 10 pp3971ndash3980 2011
[4] R J Moitsheki ldquoSteady heat transfer through a radial fin withrectangular and hyperbolic profilesrdquo Nonlinear Analysis RealWorld Applications vol 12 no 2 pp 867ndash874 2011
[5] Mo Miansari D D Ganji and Me Miansari ldquoApplication ofHersquos variational iteration method to nonlinear heat transferequationsrdquo Physics Letters A vol 372 no 6 pp 779ndash785 2008
[6] C H Chiu and C K Chen ldquoA decomposition method forsolving the convective longitudinal fins with variable thermalconductivityrdquo International Journal of Heat and Mass Transfervol 45 no 10 pp 2067ndash2075 2002
[7] A Rajabi ldquoHomotopy perturbation method for fin efficiencyof convective straight fins with temperature-dependent thermalconductivityrdquo Physics Letters A vol 364 no 1 pp 33ndash37 2007
[8] G Domairry and M Fazeli ldquoHomotopy analysis method todetermine the fin efficiency of convective straight fins withtemperature-dependent thermal conductivityrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no2 pp 489ndash499 2009
[9] A Campo andR J Spaulding ldquoCoupling of themethods of suc-cessive approximations and undetermined coefficients for theprediction of the thermal behaviour of uniform circumferentialfinsrdquo Heat and Mass Transfer vol 34 no 6 pp 461ndash468 1999
[10] A Campo and F Rodrfguez ldquoApproximate analytic tempera-ture solution for uniform annular fins by adapting the powerseriesmethodrdquo International Communications inHeat andMassTransfer vol 25 no 6 pp 809ndash818 1998
[11] R Chiba ldquoApplication of differential transformmethod to ther-moelastic problem for annular disks of variable thickness withtemperature-dependent parametersrdquo International Journal ofThermophysics vol 33 pp 363ndash380 2012
[12] S Sadri M R Raveshi and S Amiri ldquoEfficiency analysis ofstraight fin with variable heat transfer coefficient and thermalconductivityrdquo Journal ofMechanical Science andTechnology vol26 no 4 pp 1283ndash1290 2012
[13] M Torabi H Yaghoori and A Aziz ldquoAnalytical solution forconvective-radiative continously moving fin with temperaturedependent thermal conductivityrdquo International Journal of Ther-mophysics vol 33 pp 924ndash941 2012
[14] M Torabi andH Yaghoobi ldquoTwo dominant analytical methodsfor thermal analysis of convective step fin with variable thermalconductivityrdquoThermal Science In press
[15] AMoradi ldquoAnalytical solutions for finwith temperature depen-dant heat transfer coefficientrdquo International Journal of Engineer-ing and Applied Sciences vol 3 no 2 pp 1ndash12 2011
[16] A Moradi and H Ahmadikia ldquoAnalytical solution for differentprofiles of fin with temperature-dependent thermal conductiv-ityrdquoMathematical Problems in Engineering vol 2010 Article ID568263 15 pages 2010
[17] H Yaghoobi and M Torabi ldquoThe application of differentialtransformation method to nonlinear equations arising in heattransferrdquo International Communications in Heat and MassTransfer vol 38 no 6 pp 815ndash820 2011
[18] S Ghafoori M Motevalli M G Nejad F Shakeri D D GanjiandM Jalaal ldquoEfficiency of differential transformation methodfor nonlinear oscillation Comparison with HPM and VIMrdquoCurrent Applied Physics vol 11 no 4 pp 965ndash971 2011
[19] A A Joneidi D D Ganji andM Babaelahi ldquoDifferential trans-formation method to determine fin efficiency of convectivestraight fins with temperature dependent thermal conductivityrdquoInternational Communications in Heat and Mass Transfer vol36 no 7 pp 757ndash762 2009
[20] J K Zhou Differential Transform Method and Its Applicationsfor Electric circuIts Huazhong University Press Wuhan China1986
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Mathematical Problems in Engineering
[21] S Mukherjee ldquoReply to comment on lsquosolutions of the duffing-van der pol oscillator equation by the differential transformmethodrsquordquo Physica Scripta vol 84 Article ID 037003 2 pages2011
[22] C Bervillier ldquoStatus of the differential transformationmethodrdquoApplied Mathematics and Computation vol 218 no 20 pp10158ndash10170 2012
[23] A Arikoglu and I Ozkol ldquoSolution of fractional differentialequations by using differential transform methodrdquo ChaosSolitons and Fractals vol 34 no 5 pp 1473ndash1481 2007
[24] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[25] CW Bert ldquoApplication of differential transformmethod to heatconduction in tapered finsrdquo Journal of Heat Transfer vol 124no 1 pp 208ndash209 2002
[26] F Khani and A Aziz ldquoThermal analysis of a longitudinal trape-zoidal fin with temperature-dependent thermal conductivityand heat transfer coefficientrdquo Communications in NonlinearScience and Numerical Simulation vol 15 no 3 pp 590ndash6012010
[27] H C Unal ldquoAn anlytical study of boiling heat transfer from afinrdquo International Journal of Heat and Mass Transfer vol 31 no7 pp 1483ndash1496 1988
[28] M H Chang ldquoA decomposition solution for fins with tempera-ture dependent surface heat fluxrdquo International Journal of Heatand Mass Transfer vol 48 no 9 pp 1819ndash1824 2005
[29] A Jezowski B A Danilchenko M Bockowski et al ldquoThermalconductivity of GaN crystals in 42ndash300 K rangerdquo Solid StateCommunications vol 128 no 2-3 pp 69ndash73 2003
[30] S Vitanov V Palankovski S Maroldt and R Quay ldquoHigh-temperature modeling of AlGaNGaN HEMTsrdquo Solid-StateElectronics vol 54 no 10 pp 1105ndash1112 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of