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1
CHAPTER 8
APPROXIMATE SOLUTIONS
THE INTEGRAL METHOD
• Obtain approximate solutions when
(a) Exact solution is not available
(b) Form of the exact solution is not convenient to use
• Integral method is used in
(a) fluid flow
(b) heat transfer
(c) mass transfer
2
8.1 Integral Method Applications: Mathematical Simplification
• Differential formulation: Basic laws are satisfied at every point
• Integral formulation: Basic laws are satisfied in an average sense
• Mathematical simplification:
(a) Reduction of the number of independent variables and/or
(b) Reduction of the order of the governing differential equation
3
8.2 Procedure
(1) Integral formulation: Construct the heat-balance integral using one of two methods:
(a) Control volume formulation
• Select a finite control volume
• Apply conservation of energy
(b) Integration of the governing differential equation
• Write the heat equation
• Multiply through by an infinitesimal area dA of the control volume
4
• Integrate each term over the entire control volume
• Make use of Leibnitz’s rule to integrate the term
:dxt
Tb
a
dt
dbttbT
dt
dattaT
dxtxTtd
ddx
t
T tb
ta
tb
ta
]),([]),([
),()(
)(
)(
)(
(8.1)
Each problem has its ownheat-balance integral
5
(2) Assumed temperature profile
• Not unique. Various possibilities: Example: polynomial for Cartesian coordinates
• Must satisfy boundary conditions
• Profile is expressed in terms of unknown parameter or variable
(3) Determination of the unknown parameter or variable
• Use the heat balance integral for the problem
6
8.3 Accuracy of the Integral Method
• Errors are acceptable
• Different assumed profile give different solutions and different errors
• Accuracy is not very sensitive to the form of the assumed profile
• Not possible to identify the most accurate integral solution
7
8.4 Application to Cartesian Systems
• Formulation of the heat-balance integral for each problem by two methods:
(a) Control volume formulation
(b) Integration of the governing differential equation
Example 8.1: Constant Area Fin
x
1.8.Fig
volumecontrol
Th,oTBase temperature = To
Tip is insulated
Ambient Temperature = T
8
x
1.8.Fig
volumecontrol
Th,oT
• Constant area fin • Insulated tip • Convection at surface • Specified base temperature
(2) Origin and Coordinates
(3) Formulation and Solution
(i) Assumptions • Steady state • Fin approximations are valid (Bi 0.1)
(1) Observations
• No energy generation (4) Uniform h and T
9
(ii) Integral Formulation. Use two methods:
(a) Control volume formulation. Conservation of energy applied to control volume
0 outin EE (a)
xd
dTkAE cin
)0( (b)
dxTThCEL
out )(0 (c)
10
(b) and (c) into (a)
Equation (d) is the heat-balance integral for the fin.
(b) Integration of the governing differential equation
ckA
hCm 2 (d)
0)(22
2
TTmxd
Td (2.9)
0)()0(
0
2 L
dxTTmxd
dT(8.2)
11
Multiplying eq. (2.9) by dx and integrating over the length of the fin from x = 0 to x = L
LL
dxTTmdxdx
Td
0
2
0 2
2
0)(
LdxTTm
dx
dT
dx
LdT
0
2 0)()0()( (e)
0)(
dx
LdT (f)
0)()0(
0
2 L
dxTTmxd
dT (8.2)
12
(iii) Assumed Temperature Profile
2210)( xaxaaxT (g)
oTT )0( (h)
Applying (f) and (h) to (g) gives
,0 oTa Laa 2/12
(g) becomes
)2/( 21 LxxaTT o (i)
13
(iv) Determination of the unknown coefficient
a1 is determined by satisfying the heat-balance integral.
(i) into (8.2)
0)2/()(0
21
21
L
o dxLxxaTTma
Performing the integration and solving for a1
3/1
)(22
2
1Lm
TTLma o
(j)
14
(j) into (i)
)/)(2/1(3/1
1)(
)(22
22* LxLx
Lm
Lm
TT
TxTxT
o
(8.3)
Tip temperature is T *
(L)
3/1)2/1(1
)()(
22
22*
Lm
Lm
TT
TLTLT
o
(8.4)
Fin heat transfer rate qf : Apply Fourier’s law at x = 0
and use eq. (8.3)
3/1)( 22*
Lm
mL
TThCkA
oc
f
(8.5)
15
(4) Checking
Dimensional check: Each term in eq. (8.3) and eq. (8.5) is dimensionless
Limiting check: If h = 0 no heat leaves the fin. Fin should be at the base temperature. Setting h = m = 0 in
eqs. (8.3) and (8.5) gives T (x) = To and qf = 0(5) Comments. Exact solution:
mLTT
TLTLT
exactoexact cosh
1)()(*
(8.6)
16
mLTThCkA
exactoc
f
exacttanh
)(*
(8.7)
Solutions depend on the parameter mL. Table 8.1 gives the percent error as a function of this parameter.
Table 8.1
Percent Error
mL 0 0.5 1.0 1.5 2.0 3.0 4.0 5.0 10 T
* 0 0.248
3.56
16.0
48.2 225.8 819 2619 5027
q * 0 0.013 1.52 5.27 11.09 24.63 36.8 46.4 70.9 100
17
NOTE
• Heat transfer error is smaller than tip temperature error
• The error increases with increasing mL large mL = large L (long fin)
• As L , T* = -1/2 (exact solution: T* = 0), error
• As L , q* = 0 (exact solution: q* = 1),
error 100
18
Example 8.2: Semi-infinite Region with Time-Dependent Surface Flux
Fig. 8.2
iT)(tqo
)(t
0 x
TSemi-infinite region
Initially at Ti
(1) Observations
• Semi-infinite region
• Time dependent flux at the surface
(2) Origin and Coordinates
)(tqoSurface flux =
19
(3) Formulation and Solution
(i) Assumptions • One-dimensional conduction
• Uniform initial temperature
• Constant properties
(ii) Integral Formulation
(a) Control volume formulation
• (t) = penetration depth or the thermal layer
• At edge of this layer the temperature is Ti
• Control volume: Extends from x = 0 to x =
20
Conservation of energy for control volume
EEE outin (a)
Evaluating each term
)(tqE oin (b)
0outE (c)
)(
0),(
t
ip dxTtxTdt
dcE
(d)
(b)-(d) into (a)
)(
0),()(
t
ipo dxTtxTdt
dctq
(8.8)
21
(b) Integration of the governing differential equation. The heat equation is
t
Tc
x
Tk p
2
2
(e)
Multiplying both sides of (e) by dx and integrating from x = 0 to x = (t)
dxt
Tcdx
x
Tk
t
p
t
)(
0
)(
0 2
2 (f)
Equation (8.8) is the heat-balance integral for this problem.
22
Use Leibnitz’s rule
dxtxTdt
d
dt
dtT
dt
dtTc
x
tTk
x
tTk
t
p
)(
0),(),(
0),0(
),0(),(
(g)
Simplify using B.C.
(2) 0),(
x
tT
(3) iTtT ),(
)(),0(
tqx
tTk o
(1)
23
Substituting into (g)
However
)(
0
tdxT
dt
d
dt
dT
Substituting into (h)
(8.8) )(
0),()(
t
ipo dxTtxTdtd
ctq
(h)
dxtxT
dtd
dtd
Tctqt
ipo
)(
0),()(
24
(iii) Assumed Temperature Profile
2210)( xaxaaxT (8.9)
B.C. give a0, a1 and a2
ktqTa oi 2/)(0
ktqa o /)(1
ktqa o 2/)(2
Equation (8.9) becomes
(8.10) 2])([)(2
)(),( xt
tk
tqTtxT o
i
25
(iv) Determination of the unknown
variable (t) Heat-balance integral gives (t). Substituting (8.10)
into (8.8)
)(
0
2)(
)(2
)()(
to
po dxxttk
tq
dt
dctq
or
)(6
)( 2 tqdtd
k
ctq o
po
(i)
The initial condition on (t) is
0)0( (j)
26
Integrating (i) and using (j)
)()()0()0()()(
)()()(6
222
0
2
0
ttqqttq
ttqddttqc
k
ooo
t
o
t
op
Solving for (t)
2/1
0)(
)(
6)(
t
oo
dttqtq
t (8.11)
27
(5) Checking
Dimensional check: Equations (8.10), (i) and (8.11) are dimensionally consistent.
Limiting check:
(i) If ,0)( tqo the temperature remains at Ti.
Setting 0)( tqo in eq. (8.10) gives T (x,t) = Ti .
(ii) If , the penetration depth . Setting = in eq. (8.11) gives = .
28
(6) Comments
Special case: Constant heat flux: oo qtq )( = constant,
eq. (8.11) becomes
tt 6)( (k)
(k) into (8.10)
26
62),( xt
tkq
TtxT oi
(l)
Surface temperature: set x = 0 in (l)
tk
qTtT o
i 62
),0(
29
or
225.12
3
/)(
),0(
kttq
TtT
o
i
Exact solution:
128.14
/)(
),0(
exacto
i
kttq
TtT
Error = 8.6%.
30
8.5 Application to Cylindrical Coordinates
3.8.Fig
Th,w 0 r
Th,
Base temperature = To
Inner radius = ri
Outer radius = ro
Thickness = w
(1) Observations • Constant fin thickness
• Specified temperature at base
• Insulated tip
Example 8.3: Cylindrical Fin
31
(2) Origin and Coordinates
(3) Formulation and Solution
(i) Assumptions • Steady state
• Fin approximations are valid (Bi < 0.1)
• Uniform h and T
(ii) Integral Formulation
(a) Control volume formulation
Control volume: Entire fin.
32
Conservation of energy:
0 outin EE (a)
Fourier’s law:
rdrdT
wkrE iiin
)(2 (b)
Newton’s law:
drrTThEo
r
irout )(4 (c)
(b) and (c) into (a)
0)(2)(
o
r
iri
i drrTTwrk
h
rd
rdT(8.12)
33
(b) Integration of the governing differential equation: Fin eq. (2.24)
0)(21
2
2
TTwk
h
rd
dT
rrd
Td(2.24)
Rewrite
0)(21
TT
wk
h
rd
dTr
rd
d
r(d)
Multiplying (d) by 2r dr and integrating from r = ro
to r = ri
02)(2
21 )(
or
ir
or
ir
drrTTwk
hdrr
rd
dTr
rd
d
r
34
or
0)(2
)(
or
ir
or
irdrrTT
wk
h
rd
dTrd (e)
or
0)(2)()(
or
ir
ii
oo drrTT
wk
h
dr
rdTr
dr
rdTr
Insulated tip
0)(
rd
rdT o (f)
(e) becomes
0)(2)(
or
iri
i drrTTrwk
h
dr
rdT(8.12)
35
(iii) Assumed Temperature Profile2
210)( raraarT (g)
additional B.C.
oi TrT )( (h)
(f) and (h) into (g)
1
2
2)1
2()( a
r
rr
r
rrTrT
oo
iio
(i)
Unknown is a1
36
(iv) Determination of the unknown coefficient
Use the heat-balance integral. Substituting (i) into (8.12).
)1)(4/1()1)(3/2(
)1)(2)(2/1()1(2
/)1)((
43
222
222
1
RRRR
RRrmR
rRTTrma
o
ioo
(k)
Performing the integration
(j)
02
12
)(2
1
1
2
1
rdrar
r
irr
r
rrTT
rwk
h
ar
r
oo
ii
r
ir
oi
o
i
o
37
where
kw
hm
22 (l)
o
i
r
rR (m)
Substituting into (i)
)1)(4/1()1)(3/2(
)1)(2)(2/1()1(2
)1()2/()/()2)(2/1()(
43
222
2222
RRRR
RRrmR
RrrrrrRrm
TTTrT
o
oiio
o
o
(8.13)
38
Heat transfer rate: Apply Fourier’s law at r = ri
rd
rdTwrkq i
i)(
2 (n)
Substituting (8.13) into (n) and introducing the
dimensionless fin heat transfer rate q*
)1)(4/1()1)(3/2(
)1)(2)(2/1()1(2
)1(2
))((2*
43
222
22
RRRR
RRrmR
R
TTrrh
o
oio
(8.14)
39
(4) Checking
Dimensional check: Equations (8.13) and (8.14) are dimensionally correct.
Limiting check: For ro = , the dimensionless heat
transfer rate q* should vanish. Setting ro = in eq.
(8.14) gives q* = 0.
(5) Comments
• The fin equation (2.24) can be solved exactly. q* depends on two parameters: R and m ro.
Table 8.2 compares the integral solution with the exact solution for R = 0.2
40
• Integral solution becomes increasingly less accurate as m ro is increased
Table 8.2
Percent error in q* for ri / ro = 0.2
m ro 0.2 0.5 1 2 3 4
% Error 1.7 3.5 16.8 33.0 43.4 51.2
41
8.6 Non-linear Problems
Example 8.4: Semi-infinite Region with Temperature Dependent Properties
A semi-infinite region
Initial temperature = Ti
Surface flux = qo
Conductivity, density and specific heat depend on temperature
T
)(tiT
x0
oq
Fig. 8.4
42
T
)(tiT
x0
oq
Fig. 8.4
)1()( 1TkTk o (a)
)1()( 2TcTc pop (b)
)1()( 3TT o (c)
(1) Observations
• Semi-infinite region
• Temperature dependent properties
• Transient one-dimensional conduction
43
(2) Origin and Coordinates
(3) Formulation and Solution
(i) Assumptions
• One-dimensional transient conduction
• Uniform initial temperature
(ii) Integral Formulation
Integrating the differential equation:
Variable properties heat equation:
tT
cxT
kx p
)( (8.15)
44
B.C.
(1) oqx
tTk
),0(
(2) iTtT ),(
(t) = penetration depth
Simplify (8.15), use Kirchhoff transformation
dTTcTc
T pT
poo)()(
1)(
0
(8.16)
(3) 0)(
xt,T
45
Differentiating (8.16)
poo
p
c
c
dT
d
(d)
Thus
,x
T
dT
d
x
or xc
c
x
T
p
poo
,t
T
dT
d
t
or tc
c
t
T
p
poo
Substituting into (8.15)
txT
x
][ )( (8.17)
46
where
)()(
)()(
TcT
TkT
p (8.18)
Left term of eq. (8.17) is non-linear
Transformation of B.C.B.C. (1) becomes
op
poo qx
t
c
kc
),0(
or
poo
os c
q
x
t
),0(
(e)
S = diffusivity at surface temperature (0,t)
47
BC (2) becomes
ipT
poodTTcT
ct
i
)()(1
),(0
(f)
BC (3) becomes
0),(
x
t (g)
Solve eq. (8.17) for (x,t) using the integral method. Once (x,t) is determined, the temperature distribution T(x,t) can be obtained from eq. (8.16) using (b) and (c)
dTTcTc
T poT
opoo
)1()1(1
)( 30
2
48
Carrying out the integration gives
332232
32)( TTTT
(8.19)
Heat-balance integral: Integrating eq. (8.17). Multiply eq. (8.17) by dx and integrate from x = 0 to x= (t)
Use Leibnitz’s rule
(h) dxt
dxxx
tt
)(
0
)(
0)(
(i)
dxtxdtd
dtd
tdtd
tx
tx
t
t
si
)(
0),(
),(0
),0(),0(),(
49
Simplify (i) using 3 B.C.
dxtxdt
d
dt
d
c
q t
ipoo
o )(
0),(
(j)
However
)(
0
tdx
dt
d
dt
d
Substituting into (j)
)(
0),(
t
ipoo
o dxtxdt
d
c
q
(8.20)
where i = diffusivity at initial temperature i
50
(iii) Assumed Temperature Profile
2210)( xaxaax (k)
B.C. (1), (2) and (3) give a0 , a1 and a2
spoooi cqa 2/0
spooo cqa /1
spooo cqa 2/2
Equation (k) becomes
)(2/2/)(),( 2 txxtcq
txspoo
oi
(8.21)
51
(v) Determination of the unknown variable (t)
Use the heat-balance integral, (8.20). Eq. (8.21) into (8.20)
Initial condition on (t)
0)0( (m)
Integrate (l) and use (m)
tt s 6)( (8.22)
)(
0
2 )(2/2/)(t
spoo
o
poo
o dxtxxtcq
dtd
cq
or (l) )(6
11 2 t
dt
d
s
52
S depends on (0,t). Setting x = 0 in (8.21) gives (0,t).
2/)(),0( tc
qt
spoo
oi
Eliminating (t) by using eq. (8.22)
spoo
oi
tc
qt
23
),0(
(8.23)
Eq.(8.18) and (8.19) give S in terms of (0,t). Using this result with eq. (8.23) gives S as a function of time.
Equation (8.22) is then used to obtain (t) which when substituted into eq. (8.21) gives (x,t). The
transformation eq. (8.19) gives the temperature distribution T (x,t).
53
(4) Checking
Dimensional check: Equations (8.21) and (8.22) are dimensionally correct.
Limiting check: For the special case of constant properties the solution agrees with the result of Example 8.2.
(5) Comments
Improving accuracy: Using a cubic polynomial of the form
33
2210)( xaxaxaax (n)
54
Need a fourth B.C.
0),(
2
2
x
t (o)
(n) becomes
3)/1(3
),( xc
qtx
poos
oi
(8.24)
Corresponding solution is
tt s 12)( (8.25)
55
Example 8.5: Conduction with Phase Change
T
5.8.Fig
x)(txi
0
oT
fTsolidliquid
Semi-infinite region
Initially solid at the fusion
temperature Tf
Surface at x = 0 is TO Tf
(1) Observations
• Semi-infinite region
• Solid phase remains at Tf
(2) Origin and Coordinates
56
(3) Formulation and Solution
(i) Assumptions • One-dimensional transient conduction • Constant properties • Uniform initial temperature
(ii) Integral Formulation
The heat equation is
tT
x
T
2
2
(a)
Multiplying by dx and integrating from x = 0 to x = xi (t)
(b) dxtT
dxx
ixix
00 2
2
57
Integrate and use Leibnitz’s rule
dxtxTdtd
dt
dxtxT
dtd
tT
xtT
x
txT
ixi
i
i
0),(),(
0),0(
),0(),( (c)
Simplify using B.C.
(1) oTtT ),0(
(2) fi TtxT ),(
(3)dt
dxL
xtxT
k ii
),(
58
Use B.C. (2) and (3)
Equation (d) is the heat-balance integral for this problem.
(iii) Assumed Temperature Profile 2
221 )()()( ii xxaxxaax (e)
• BC (3) is not suitable. It leads to 2nd order DE for xi
(d)
dxtxTdtd
dtdx
Tx
tTdt
dxkL ix
if
i
0
),(),0(
59
• Alternate approach: Combine (2) and (3). Total derivative of T (xi,t) in condition (2) is
0),(),(
dtt
txTdx
x
txTdT ii
f at x = xi (f)
Setting dx = dxi in (f)
0),(),(
t
txT
td
ds
x
txT ii (g)
B.C. (3) into (g)
t
Tcx
txT
p
i
L2
),( (h)
60
Using B.C. (1), (2) and (h) give the coefficients a1 , a2 and a3
fTa 1 (i)
2/12 )1(1
ixk
aL (j)
22
3)(
i
foi
x
TTxaa
(k)
where
)(2
fop TT
c
L (l)
61
s (t) is unknown. Use the heat-balance integral.
(e) into (d)
2/1
2/1
)1(5
)1(16
td
xdx i
i (m)
Solve for xi (t) and use interface initial condition, xi (0) = 0
ttxi 2)( (n)
where
2/1
2/1
2/1
)1(5
)1(13
(o)
62
Dimensional check: Eqs. (h), (m) and (n) are dimensionally consistent
(5) Comments Problem is identical to Stefan’s problem.
Exact solution is
2
)(erf
2
L
fop TTce (p)
(4) Checking
63
Table 8.3 compares the two solutions
Table 8.3
xi / xie
0 1.000
0.4 1.026
0.8 1.042
1.2 1.052
1.6 1.059
2.0 1.064
2.4 1.068
2.8 1.070
3.2 1.072
3.4 1.073
4.0 1.073
64
8.7 Energy Generation
Example 8.6: Semi-infinite Region with Energy Generation
Semi-infinite region
Initial Temperature Ti = 0
Surface at x = 0 is at T (0,t) = 0 Fig. 8.6
T
0 x
)(tq 3t
2t
1t
Fig. 8.6(1) Observations
• Semi-infinite region
• Transient conduction
• Time dependent energy generation
)(tq At t > 0 apply
65
(2) Origin and Coordinates
(3) Formulation and Solution
(i) Assumptions
• One-dimensional transient conduction
• Constant properties
• Uniform initial temperature
(ii) Integral Formulation
Integrating the differential equation:
tT
ctq
x
T
p
)(
2
2(a)
66
Multiplying by dx and integrating from x = 0 to x = (t)
Use Leibnitz’s rule
(b) dxtT
dxctq
dxx
tt t
p
)(
0
)(
0
)(
02
2 )(
(c)
dxtxTdtd
dtd
tTdtd
tT
tc
qx
tTx
tT
t
p
)(
0),(),(
0),0(
)(),0(),(
67
Simplify using B.C.
(2) 0),(
x
tT
Substituting into (c)
Equation (d) is the heat-balance integral for this problem.
(d)
dxtxTdtd
dtd
tTtc
qx
tT
t
p
)(
0),(
),()(),0(
0),0( tT(1)
68
(iii) Assumed Temperature Profile 3
32
210)( xaxaxaax (e)
Two additional B.C.
A fourth condition is obtained by evaluating differential equation (a) at x = . Using boundary condition (3), equation (a) becomes
Integrate
)(
0)(
1),(
t
pdttq
ctT
(f)
0),(
2
2
x
tT (3)
t
tT
c
tq
p
),()(
69
Define
(g) into (f) gives the fourth B.C.
The assumed profile becomes
][ 3)/1(1)(
),(
xctQ
txTp
(h)
Unknown is (t). Use the heat-integral equation.
(g) t
dttqtQ0
)()(
pc
tQtT
)(
),( (4)
70
Substitute B.C. (4) and the assumed profile (h) into (d)
td
dtQ
td
dQtqtQ
)(3)(4)(12 22 (i)
Differentiate (g)
td
dQtq )( (j)
Substitute into (i)
)(12)( 222 tQtd
dQ
td
dQtQ
or
)(24)( 222 tQtQtd
d (k)
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Integration of (k)
tt
tdtQQ0
2
0
22 )(24 (l)
The initial condition on (t) is
0)( t
Substitute into (l)
)(
)(24)(
2/1
0
2
tQ
tdtQt
t
(m)
(m) into (h) gives the transient temperature distribution
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(4) Checking
Dimensional check: Equations (h) and (m) are dimensionally consistent.
Limiting check: If ,0q the temperature remains at the initial value. Setting 0)( tq in (g) gives Q = 0.
When this is substituted into (h) gives T (x,t) = 0
(5) Comments
Special case: Constant energy generation rate, (g) gives
tqtQ )(
73
Substitute into (k)
tt 22)(
Equation (h) becomes
3)( 22/11),( tx
ctq
txTp
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