extended surface heat transfer
DESCRIPTION
MODULE 3. Extended Surface Heat Transfer. EXTENDED SURFACES / FINS. Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling law: q = hA(T s -T ). Therefore, to increase the convective heat transfer, one can - PowerPoint PPT PresentationTRANSCRIPT
EXTENDED SURFACES / FINS
Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling law: q = hA(Ts-T). Therefore, to increase the convective heat transfer, one can Increase the temperature difference (Ts-T) between the surface and the fluid.
Increase the convection coefficient h. This can be accomplished by increasing the fluid flow over the surface since h is a function of the flow velocity and the higher the velocity, the higher the h. Example: a cooling fan.
Increase the contact surface area A. Example: a heat sink with fins.
Extended Surface Analysis
x
Tb
q kA dTdxx C q q
dqdx
dxx dx xx
dq h dA T Tconv S ( )( ), where dA is the surface area of the elementS
AC is the cross-sectional area
Energy Balance:
if k, A are all constants.
x
C
q q dq qdqdx
dx hdA T T
kA d Tdx
dx hP T T dx
x dx conv xx
S
C
( )
( ) ,2
2 0
P: the fin perimeterAc: the fin cross-sectional area
Extended Surface Analysis
(contd….) d Tdx
hPkA
T T
x T x T
ddx
m hPkA
D m
x C e C e
C
C
mx mx
2
2
2
22 2 2
1 2
0
0 0
( ) ,
( ) ( ) ,
, , ( )
( ),
A second - order, ordinary differential equation
Define a new variable = so that
where m
Characteristics equation with two real roots: + m & - mThe general solution is of the form
To evaluate the two constants C and C we need to specify two boundary conditions: The first one is obvious: the base temperature is known as T(0) = TThe second condition will depend on the end condition of the tip
2
1 2
b
Extended Surface Analysis (contd...)
For example: assume the tip is insulated and no heat transferd/dx(x=L)=0
The temperature distribution is given by-
The fin heat transfer rate is
These results and other solutions using different end conditions aretabulated in Table 3.4 in HT textbook, p. 118.
T x TT T
m L xmL
q kA dTdx
x hPkA mL M mL
b b
f C C
( ) cosh ( )cosh
( ) tanh tanh
0
Temperature distribution for fins of different configurations
Case Tip Condition Temp. Distribution Fin heat transfer A Convection heat
transfer: h(L)=-k(d/dx)x=L mLmk
hmL
xLmmkhxLm
sinh)(cosh
)(sinh)()(cosh
MmLmk
hmL
mLmkhmL
osinh)(cosh
cosh)(sinh
B Adiabatic (d/dx)x=L=0 mL
xLmcosh
)(cosh mLM tanh0
C Given temperature: (L)=L
mL
xLmxLmb
L
sinh
)(sinh)(sinh)(
mL
mLM b
L
sinh
)(cosh0
D Infinitely long fin (L)=0
mxe M 0
bCbb
C
hPkAMTT
kAhPmTT
,)0(
, 2
Example
An Aluminum pot is used to boil water as shown below. The handle of the pot is 20-cm long, 3-cm wide, and 0.5-cm thick. The pot is exposed to room air at 25C, and the convection coefficient is 5 W/m2 C. Question: can you touch the handle when the water is boiling? (k for aluminum is 237 W/m C)
100 C
T = 25 Ch = 5 W/ m2 C
x
Example (contd...)
We can model the pot handle as an extended surface. Assume that there is no heat transfer at the free end of the handle. The condition matches that specified in the fins Table, case B. h=5 W/ m2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/m C, AC=Wt=0.00015(m2), L=0.2(m)Therefore, m=(hP/kAC)1/2=3.138, M=(hPkAC)(Tb-T)=0.111b=0.111(100-25)=8.325(W)
T x TT T
m L xmL
T x
T x x
b b
( ) cosh ( )cosh
cosh[ . ( . )]cosh( . * . )
,
( ) . * cosh[ . ( . )]
-
25100 25
3138 0 23138 0 2
25 62 32 3138 0 2
Example (contd…)
Plot the temperature distribution along the pot handle
0 0.05 0.1 0.15 0.285
90
95
100
T( )x
x
As shown, temperature drops off very quickly. At the midpointT(0.1)=90.4C. At the end T(0.2)=87.3C.Therefore, it should not be safe to touch the end of the handle
Example (contd...)The total heat transfer through the handle can be calculated also. qf=Mtanh(mL)=8.325*tanh(3.138*0.2)=4.632(W)Very small amount: latent heat of evaporation for water: 2257 kJ/kg. Therefore, the amount of heat loss is just enough to vaporize 0.007 kg of water in one hour.
If a stainless steel handle is used instead, what will happen:For a stainless steel, the thermal conductivity k=15 W/m°C.Use the same parameter as before:
0281.0,47.122/1
C
C
hPkAMkAhPm
Example (contd...)
)](47.12cosh[3.1225)(cosh
)(cosh)(
xLxTmL
xLmTT
TxT
b
0 0.05 0.1 0.15 0.20
25
50
75
100
T x( )
x
Temperature at the handle (x=0.2 m) is only 37.3 °C, not hot at all. This example illustrates the important role played by the thermal conductivity of the material in terms of conductive heat transfer.
Fins-2If the pot from previous lecture is made of other materials other than the aluminum, what will be the temperature distribution? Try stainless steel (k=15 W/m.K) and copper (385 W/m.K).Recall: h=5W/m2C, P=2W+2t=2(0.03+0.005)=0.07(m) AC=Wt=0.00015(m2), L=0.2(m)Therefore, mss=(hP/kAC)1/2=12.47, mcu=2.46Mss=(hPkssAC) (Tb-T)=0.028(100-25)=2.1(W) Mcu= (hPkssAC) b=0.142(100-25)=10.66(W)
For stainless steel, -T x T
T Tm L x
mLT x
T x x
ss
b b
ss
ss
( ) cosh ( )cosh
cosh[ .47( . )]cosh( .47 * . )
,
( ) . * cosh[ .47( . )]
25100 25
12 0 212 0 2
25 12 3 12 0 2
Fins-2 (contd....)
For copper, -T x T
T Tm L x
mLT x
T x x
cu
b b
cu
cu
( ) cosh ( )cosh
cosh[ .46( . )]cosh( .46 * . )
,
( ) . * cosh[ .46( . )]
25100 25
2 0 22 0 2
25 66 76 2 0 2
0 0.04 0.08 0.12 0.16 0.275
80
85
90
95
100
T( )x
T ss( )x
T cu( )x
x
copper
aluminum
stainless steel
Fins-2 (contd...)
Inside the handle of the stainless steel pot, temperature drops
quickly. Temperature at the end of the handle is 37.3C. This is
because the stainless steel has low thermal conductivity and heat
can not penetrate easily into the handle.
Copper has the highest k and, correspondingly, the temperature
inside the copper handle distributes more uniformly. Heat easily
transfers into the copper handle.
Question? Which material is most suitable to be used in a heat
sink?
Fins-2 (contd...) How do we know the adiabatic tip assumption is good? Try using the convection heat transfer condition at the tip (case A in fins table) We will use the aluminum pot as the example.h=5 W/m2.K, k=237 W/m.K, m=3.138, M=8.325W
T x TT T
m L x h mk m L xmL h mk mL
x x
T x x x
b b
( ) cosh[ ( )] ( / )sinh[ ( )]cosh ( / )sinh
cosh[ . ( . )] . sinh[ . ( . )]cosh( . ) . sinh( . )
( ) . {cosh( . . ) . sinh( . . )}
-
3138 0 2 0 00672 3138 0 20 6276 0 00672 0 6276
25 62 09 0 6276 3138 0 00672 0 6276 3138
Long equation
Fins-2 (contd….)
0 0.04 0.08 0.12 0.16 0.285
88.75
92.5
96.25
100
T( )x
T c( )x
x
T: adiabatic tip
Tc: convective tip
T(0.2)=87.32 °CTc(0.2)=87.09 °C
Note 1: Convective tip case has a slightly lower tip temperature as expected since there is additional heat transfer at the tip.Note 2: There is no significant difference between these two solutions, therefore, correct choice of boundary condition is not that important here. However, sometimes correction might be needed to compensate the effect of convective heat transfer at the end. (especially for thick fins)
Fins-2 (contd...)
In some situations, it might be necessary to include the convective heat transfer at the tip. However, one would like to avoid using the long equation as described in case A, fins table. The alternative is to use case B instead and accounts for the convective heat transfer at the tip by extending the fin length L to LC=L+(t/2).
Original fin length L
With convection
t
Then apply the adiabatic condition at the tip of the extended fin as shown above.
L
LC=L+t/2
t/2
Insulation
Fins-2 (contd...)
T x TT T
m L xmL
T x
T x x
corr
b b
c
c
corr
corr
( ) cosh ( )cosh
cosh[ . ( . )]cosh( . * . )
,
( ) . * cosh[ . ( . )]
-
25100 25
3138 0 20253138 0 2025
25 62 05 3138 0 2025
Use the same example: aluminum pot handle, m=3.138, the
length will need to be corrected to
LC=l+(t/2)=0.2+0.0025=0.2025(m)
0 0.04 0.08 0.12 0.16 0.285
88.75
92.5
96.25
100
T( )x
T c( )x
T corr( )x
x
T(0.2)=87.32 °C Tc(0.2)=87.09 °CTcorr(0.2025)=87.05 °C
slight improvement over the uncorrected solution
Fins-2 (contd...)
Correction Length
The correction length can be determined by using the formula: Lc=L+(Ac/P), where Ac is the cross-sectional area and P is the perimeter of the fin at the tip.
Thin rectangular fin: Ac=Wt, P=2(W+t)2W, since t < W Lc=L+(Ac/P)=L+(Wt/2W)=L+(t/2)
Cylindrical fin: Ac=(/4)D2, P= D, Lc=L+(Ac/P)=L+(D/4)
Square fin: Ac=W2, P=4W, Lc=L+(Ac/P)=L+(W2/4W)=L+(W/4)
Optimal Length of a Fin In general, the longer the fin, the higher the heat transfer. However, a long fin means more material and increased size and cost. Question: how do we determine the optimal fin length? Use the rectangular fin as an example:
tanh , for an adiabatic tip fin
( ) , for an infinitely long fin
Their ratio: R(mL)= tanh( )
f
f
f
f
q M mL
q M
qmL
q
0 1 2 3 40
0.2
0.4
0.6
0.8
1
R( )mL
mL
Note: heat transfer increases with mL as expected. Initially the rate of change is large and slows down drastically when mL> 2.
R(1)=0.762, means any increase beyond mL=1 will increase no more than 23.8% of the fin heat transfer.
Temperature Distribution
For an adiabatic tip fin case:cosh ( )
coshb
T T m L xRT T mL
Use m=5, and L=0.2 as an example:
0 0.05 0.1 0.15 0.20.6
0.8
11
0.648054
R ( )x
0.20 x
High T, good fin heat transfer Low T, poor fin heat transfer
Correction Length for a Fin with a Non-adiabatic Tip
The correction length can be determined by using the formula: Lc=L+(Ac/P), where Ac is the cross-sectional area and P is the perimeter of the fin at the tip.
Thin rectangular fin: Ac=Wt, P=2(W+t)2W, since t < W Lc=L+(Ac/P)=L+(Wt/2W)=L+(t/2)
Cylindrical fin: Ac=(/4)D2, P= D, Lc=L+(Ac/P)=L+(D/4)
Square fin: Ac=W2, P=4W, Lc=L+(Ac/P)=L+(W2/4W)=L+(W/4)
Fin Design
Total heat loss: qf=Mtanh(mL) for an adiabatic fin, or qf=Mtanh(mLC) if there is convective heat transfer at the tip
C C
C,
,
hPwhere = , and M= hPkA hPkA ( )
Use the thermal resistance concept:( )hPkA tanh( )( )
where is the thermal resistance of the fin.
For a fin with an adiabatic tip, the fin
b bc
bf b
t f
t f
m T TkA
T Tq mL T TR
R
,C
resistance can be expressed as( ) 1
hPkA [tanh( )]b
t ff
T TRq mL
Tb
T
Fin EffectivenessHow effective a fin can enhance heat transfer is characterized by the fin effectiveness f: Ratio of fin heat transfer and the heat transfer without the fin. For an adiabatic fin:
ChPkA tanh( )tanh( )
( )If the fin is long enough, mL>2, tanh(mL) 1, it can be considered an infinite fin (case D of table3.4)
In order to enhance heat tra
f ff
C b C C
fC C
q q mL kP mLq hA T T hA hA
kP k PhA h A
nsfer, 1.
However, 2 will be considered justifiable
If <1 then we have an insulator instead of a heat fin
f
f
f
Fin Effectiveness (contd...)
To increase f, the fin’s material should have higher thermal
conductivity, k.
It seems to be counterintuitive that the lower convection
coefficient, h, the higher f. But it is not because if h is very high,
it is not necessary to enhance heat transfer by adding heat fins.
Therefore, heat fins are more effective if h is low. Observation: If
fins are to be used on surfaces separating gas and liquid. Fins are
usually placed on the gas side. (Why?)
fC C
kP k PhA h A
P/AC should be as high as possible. Use a square fin with
a dimension of W by W as an example: P=4W, AC=W2,
P/AC=(4/W). The smaller W, the higher the P/AC, and
the higher f.
Conclusion: It is preferred to use thin and closely spaced
(to increase the total number) fins.
Fin Effectiveness (contd...)
Fin Effectiveness (contd...)
, ,
, ,
The effectiveness of a fin can also be characterized as( ) /
( ) ( ) /
It is a ratio of the thermal resistance due to convection to the thermal resistance of a fin
f f b t f t hf
C b b t h t f
q q T T R Rq hA T T T T R R
. In order to enhance heat transfer,the fin's resistance should be lower than that of the resistancedue only to convection.
Fin Efficiency
Define Fin efficiency:
where represents an idealized situation such that the fin is made upof material with infinite thermal conductivity. Therefore, the fin shouldbe at the same temperature as the temperature of the base.
f
q
q hA T T
f
f b
max
max
max ( )
T(x)<Tb for heat transferto take place
Total fin heat transfer qf
Real situation Ideal situation
For infinite kT(x)=Tb, the heat transferis maximum
Ideal heat transfer qmax
Tb x x
Fin Efficiency (contd…)
Fin Efficiency (cont.)
Use an adiabatic rectangular fin as an example:
max
f
max
,,
( ) tanhtanh( ) ( )
tanh tanh (see Table 3.5 for of common fins)
The fin heat transfer: ( )
, where 1/( )
f c bf
f b b
c
f f f f b
b bf t f
f f t f
q hPkA T T mLM mLq hA T T hPL T T
mL mLmLhP L
kAq q hA T T
T T T Tq RhA R
,
, , b
1
Thermal resistance for a single fin.1As compared to convective heat transfer:
In order to have a lower resistance as that is required to enhance heat transfer: or A
f f
t bb
t b t f f f
hA
RhA
R R A
Overall Fin Efficiency
Overall fin efficiency for an array of fins:
Define terms: Ab: base area exposed to coolant
Af: surface area of a single fin
At: total area including base area and total
finned surface, At=Ab+NAf
N: total number of fins
qb
qf
Overall Fin Efficiency (contd…)
( ) ( )
[( ) ]( ) [ (1 )]( )
[1 (1 )]( ) ( )
Define overall fin efficiency: 1 (1 )
t b f b b f f b
t f f f b t f f b
ft f b O t b
t
fO f
t
q q Nq hA T T N hA T T
h A NA N A T T h A NA T T
NAhA T T hA T T
ANAA
Heat Transfer from a Fin Array
,,
,
,
1( ) where
Compare to heat transfer without fins1( ) ( )( )
where is the base area (unexposed) for the fin
To enhance heat transfer Th
bt t O b t O
t O t O
b b b f b
b f
t O
T Tq hA T T RR hA
q hA T T h A NA T ThA
A
A A
Oat is, to increase the effective area .tA
Thermal Resistance Concept
T1T
TbT2
T1 TTbT2
L1 t
R1=L1/(k1A)
Rb=t/(kbA)
)/(1, OtOt hAR
1 1
1 ,b t O
T T T TqR R R R
A=Ab+NAb,f