1 CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD Obtain approximate solutions when (a) Exact...

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:



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.


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

qq

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

qq

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


)(tqoSurface flux =

19


(i) Assumptions • One-dimensional conduction

• Uniform initial temperature

• Constant properties

(ii) Integral 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


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



(i) Assumptions • Steady state

• Fin approximations are valid (Bi < 0.1)

• Uniform h and T



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

qq

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


• Temperature dependent properties

• Transient one-dimensional conduction

43



(i) Assumptions

• One-dimensional transient conduction



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)


(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


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


• Solid phase remains at Tf


56


(i) Assumptions • One-dimensional transient conduction • Constant properties • Uniform initial temperature


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


• Transient conduction

• Time dependent energy generation

)(tq At t > 0 apply

65



(i) Assumptions

• One-dimensional transient conduction

• Constant properties



Integrating the differential equation:

tT

ctq

x

T

p

)(

2

2(a)

66

Multiplying by dx and integrating from x = 0 to x = (t)


(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)

71

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

72

(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

1 CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD Obtain approximate solutions when (a) Exact...

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Transcript of 1 CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD Obtain approximate solutions when (a) Exact...