1 chapter 4 transient conduction neglect spatial variation: 4.1.1 criterion for neglecting spatial...

1

CHAPTER 4 TRANSIENT CONDUCTION

Neglect spatial variation:

4.1.1 Criterion for Neglecting Spatial Temperature Variation Cooling of wire by surface convection:

(4.1))(tTT

= temperature drop across radius T

4.1 Simplified Model: Lumped-Capacity Method

2

(1) radius (2) conductivity k

(3) heat transfer coefficient h

Define Biot number Bi

Factors affecting T

Neglect for smallTk

h

(4.2)k

hBi

Neglect T if

(4.3)< 0.1 k

hBi

3

• Irregular shape:

Physical significance of Bi:

Determining : • Long cylinder: or • Large plate of thickness L, convection on both sides:

2/L

sA

V= (4.4)

resistanceexternal

resistanceinternalBi

4

• Volume:V

Determine: Transient temperature

Assume:

Conservation of energy:

4.1.2 Lumped-Capacity Analysis• Surface area: sA

• Initial temperature: iT • Convection at surface: Th, • Energy generation: q

)(tTT (a)

(1.6)EEEE outgin

5

(c), (d) and (e) into (b)

EEE outg (b)

Assume that no energy is added )0( inE

Neglect radiation, is by convectionoutE

)( TThAE sout (c)

qVEg (d)

dt

dTVcTThAqV ps )(

dt

dTVcE p (e)

Energy generation gE

Rate of energy change E

6

Separating variables and rearranging

• Equation (4.5) is the lumped-capacity equation for all geometries

• Limited to:

(2) No radiation

(3) Incompressible material

Initial condition:

dtc

q

TTVqhA

dT

ps

1))(/(

(4.5)

(1) Uniform q

Bi < 0.1 (4)

iTT )0( (4.6)

7

Integrate (4.5), use (4.6)

NOTE

(1) Temperature decay is exponential

(2) Solution is independent of k

Integration of (4.5): Assume:

(5) Constant h and T

(6) Constant q

(3) Steady state temperature:set in (4.7):t

)(exp

)(1

TThA

Vqt

Vc

hA

TThA

Vq

TT

TT

isp

s

isi

(4.7)

8

Physical significance of (4.8)shA

VqTT

)( (4.8)

Special Cases:

• Case (i): No energy generation. Set in (4.7)0q

tVc

hA

TT

TT

p

s

i

exp (4.9)

Steady state: TT )(

• Case (ii): No convection. Set h=0 in (4.7)

9

Physical significance: No Steady state

tc

qTT

pi

(4.10)

• Case (iii): Initial and ambient temperatures are thesame. Set in (4.7)TTi

t

Vc

hA

hA

VqTT

p

s

s

exp1 (4.11)

Steady state: setting in (4.11) gives (4.8)t

10

• Bi > 0.1

• Governing equation: PDE

• Solution by separation of variables

• Two HBC and one initial condition

Example 4.1: Plate with Surface Convection Thickness: 2L Initial temperature:

Assume f(x) is symmetrical

L

x

L

0

Fig. 4.1

Th

Th

HBC2

4.2 Transient Conduction in Plates

)()0,( xfxTTi Heat exchange is by convection:0t

Determine ),( txT

11

(1) Observations • Temperature is symmetrical

• Convection BC is NH

(2) Origin and Coordinates

(3) Formulation (i) Assumptions.

(ii) Governing Equations. Let

One-dimensional, constant k, , Th,

TtxTtx ),(),( (a)

tx

12

2

(4.12)

12

(iii) Independent Variable with Two HBC:x-variable

(iv) Boundary and Initial Conditions

L

x

L

0

Fig. 4.1

Th

Th

HBC2 (1) 0),0(

x

t, H

IC (3) Txfx )()0,( , NH

(4) Solution (i) Assumed Product Solution

, H),(),(

tLhx

tLk

(2)

13

(b) into (4.12)

)()(),( txXtx (b)

0)(22

2

xXdx

Xdn (c)

02 ndt

d (d)

(ii) Selecting the Sign of the Terms2n

0)(22

2

xXdx

Xdn (e)

14

(iii) Solutions to the ODE

(f)02 ndt

d

02 n 000

Xyields

xBxAxX nnnnn cossin)( (g)

)exp()( 2 tCtnn

(h)

(iv) Application of Boundary and Initial Conditions

BC (1): An = 0

xBxX nnn cos)( (i)

15

where

Substituting into (b)

IC (3):

BC (2): gives n

BiLL nn tan (4.13)

khLBi / (j)

xeaTtxTn

n

t

nn

0

cos ),(2

(4.14)

0ncos)(

nn xaTxf (k)

16

(v) Orthogonality

Eq. (3.6):

in equation (k) are solutions to (e). Comparing (e) with eq. (3.5b)

xncos

021 aa and 13 a

p = w = 1 and q = 0

BC at x = 0 and x = L are homogeneous.

w(x) = 1. Multiply both sides of (k) by integrate and apply orthogonality.

Therefore, xncos are orthogonal with respect to ,cos dxx

m

))(cos(sin

cos)(2

nnn

0 nn

LLL

xTxfa

L

n

(4.15)

17

Special case: Uniform initial temperature iT

(5) Checking Dimensional check

: Limiting check Plate is initially at T Differential Equation

Boundary and initial conditions

(6) Comments

(ii) Initial temperature f(x) need not be symmetrical

))(cos(sin

))(sin(2

nnn

n

LLL

LTTa i

n

(4.16)

(i) Both separated equations have term2n

18

Example 4.2: Plate with Energy Generation,Specified Surface Temperature

Thickness = L

4.3 Non-homogeneous Equations and Boundary Conditions

Initial temperature: iTintroduce

:0t q At

Surfaces at and 1T 2T

Determine: ),( txTFig. 4.2

2T1TL

q x0

19

(1) Observations • Asymmetry

• x-variable has two NHBC

• The heat equation is non-homogeneous


(3) Formulation (i) Assumptions.

(ii) Governing Equations: eq. (1.8)

One-dimensional, constant qk ,,

t

T

k

q

x

T

1

2

2

(4.17)

20

(iii) Independent Variable with Two HBC: x-variable


Fig. 4.2

2T1TL

q x0

(1) 1),0( TtT

(2) 2),( TtLT

(3) iTxT )0,(

(4) Solution)(),(),( xtxtxT (a)

21

Split (b). Let

(a) into BC (1)

(a) Into eq. (4.17)

tk

q

dx

d

x

12

2

2

2

(b)

(c)tx

12

2

(d)02

2

k

q

dx

d

1)0(),0( Tt

22

0),0( t (c-1)

1)0( T (d-1)

(a) into BC (2)

IC (3)

Solution to (d)

0),( tL (c-2)

2)( TL

)()0,( xTx i (c-3)

(e)212

2)( CxCx

k

qx

23

(i) Assumed Product Solution

(f) into (c), separating variables

)()(),( txXtx (f)

022

2

nn

n Xdx

Xd (g)

(ii) Selecting the Sign of the Two Terms2n

0)(22

2

xXdx

Xdnn

n (i)

(h)02 nn

n

dt

d

24

(iii) Solutions to the Ordinary Differential Equations

(j)02 nn

n

dt

d

for ,02 n

000 X

xBxAxX nnnnn cossin)( (k)

)(exp)( 2 tCtnnn

(l)


BC (c-1) 0n

B

25

BC (c-2)

Substituting into (f) and summing

...3,2,1,,0sin nnLL nn (m)

1

2 sin )][exp(),(n

nnn xtatx (n)

Solution to :)(x

xxLk

q

L

xTTTx )(

2)()( 121

(o)

NH initial condition (c-3):

Return to BC (d-1) and (d-2) give and ),(x .2

C1

C

26

(v) Orthogonality

(p)

1

sin)(n

nni xaxT

in (p) are solutions (i). Compare (i) with eq. (3.5a): Sturm- Liouville equation with

xnsin.1)( xw

Multiply both sides of (p) by apply orthogonality, gives

,sin dxxm

integrate and

na

L

n

Lni

ndx

dxxxTa

0

2

0

sin

sin)]([

(q)

27

(o) into (q), evaluate integrals and use (m)

Complete solution

(r)

k

Lq

n

nkLq

TTn

TTn

a

n

n

i

n

n

2

3

222

121

)(

2)1)(2()2/(

)()1(2

)()1(1

2

)/sin()/exp(

)(2

)(),(

1

222

121

LxnLtna

xxLk

q

L

xTTTtxT

nn

(s)

28


Limiting check: Temperature at steady state

(6) Comments

(i) Non-uniform initial temperature

(ii) Non-homogeneous boundary conditions

29

4.4 Transient Conduction in Cylinders

Example 4.3: Cylinder with Energy Generation

Long cylinder

Surface cooling by convectionTh,

Th,r

0q q

Fig. 4.3

2 HBC

Initial temperature i

T

Energy generation :0t q

Determine ),( trT

(1) Observations • Symmetry

30

• Heat equation is non-homogeneous


(3) Formulation (i) Assumptions

(4) Negligible end effect

(ii) Governing Equations

makes convection BC homogeneous • , TT

(1) One-dimensional,

(2) Uniform h and ,T (3) Constant ,,k

31

TtrTtr ),(),(

Heat equation (1.11)

(iii) Independent Variable with 2 HBC: r-variable

(iv) Boundary Conditions

tk

q

rrr

11

2

2

(4.18)

or (1) 0),0(

r

t

finite),0( t Th,

Th,r

0q q

Fig. 4.3

2 HBC

32

(4) Solution

(2) ),(),(

trhr

trk o

o

IC (3) TTr i)0,(

)(),(),( rtrtr (a)

(a) into eq. (4.18)

tk

q

rd

d

rdr

d

rrr

111

2

2

2

2 (b)

Split (b)

33

(a) into BC (1)

Let

(c)trrr

11

2

2

01

2

2

k

q

rd

d

rdr

d (d)

0)0(),0(

dr

d

r

t

),0( t (c-1)0),0(

r

tor finite

34

BC (2):

Initial condition

Integrating (d)

(c-2)),(),(

trhr

trk o

o

0)0(

dr

d(d-1)

)()(

oo rh

rd

rdk

(d-2)

)()()0,( rTTr i (c-3)

(e)212 ln

4)( CrCr

k

qr

35

(f) into (c), separating variables

(i) Assumed Product Solution

(f))()(),( trRtr

01 2

2

2

kkkk R

dr

dR

rdr

Rd (g)

02 kkk

dt

d (h)

(ii) Selecting the Sign of the Terms2k

01 2

2

2

kkkk R

dr

dR

rdr

Rd (i)

36


(j)02 kkk

dt

d

For :02 k 000 R

)()()( 00 rYBrJArR kkkkk (k)

)exp()( 2 tCt kkk (l)


BC (c-1) and (c-2)0kB

37

(k) and (l) into (f), summing

Initial condition (c-3)

and

(m))()()( 10 okokok rJrrJBi

)()exp(),(1

02 rJtatr k

kkk

(n)

Return to solution (e) for BC (d-1) and (d-2) give

1C and 2C),(r

h

rqrr

k

qr o

o 2)(

4)( 22

(o)

38

00 )()()(

kkki rJarTT (p)

(v) Orthogonality

Eq. (3.6) gives

Functions in (p) are solutions to (i). Comparing

(i) with eq. (3.5a) shows that it is a Sturm-Liouville equation with

)(0 rJ k

and0,/1 21 ara 13 a

HBC at and therefore are orthogonal with respect to

0r ,o

rr )(0 rJ k.)( rrw

and rwp 0q

39

Multiply both sides of (p) by integrate and

apply orthogonality

,)(0 drrrJ i

(q)

o

o

r

k

r

kikk

rdrrJ

rdrrJrTTa

0

20

0 0

)(

)()()(2

(o) into (q) and evaluate the integrals

)()(

1

2

1

)(1

)()(

)())((2

0

2

22

2

22221

0

oki

o

ok

i

o

okok

okokik

rJTTk

rq

rBi

TTk

rq

rJrBi

rJrTTa

(r)

40

Complete solution

)/1()(4

),( 222

oi

o

irr

TTk

rq

TT

TtrT

)()exp()(

1

)(21

02

2

rJtaTTBiTTk

rqk

kkk

ii

o

(4.19)


Limiting check:

(ii) Steady state

(i) and TTi 0q

41

(6) Comments

Initial temperature (i) )()0,( rfrT (ii) Two parameters: the Biot number Bi and

)(/)( 2 TTkrq io

4.5 Transient Conduction in Spheres

Example 4.4: Sphere with Surface Convection

Determine: transient temperature

Fig. 4.4

HBC2Th,

ror0

Initial temperature i

T Convection at surface :0t

42

(1) Observations


• Convection BC is non-homogeneous

• One-dimensional, transient ),( trT

• Define TT to give r-variable 2HBC


(1) One-dimensional

(2) Constant ,k

(ii) Governing Equations

43

Eq. (1.13) gives

(iii) Independent Variable with 2 HBC: r-variable


TtrTtr ),(),( (a)

trr

rr

11 2

2 (4.20)

or (1) 0),0(

r

t

finite),0( tFig. 4.4

HBC2Th,

ror0

44

(4) Solution (i) Assumed Product Solution

(b) into eq. (4.20), separate variables

(2) ),(),(

trhr

trk o

o

IC (3) TTr i)0,(

(b))()(),( trRtr

02 222

22 kk

kk Rrdr

dRr

dr

Rdr (c)

02 kkk

dt

d (d)

45


Use eqs. (2.32) and (2.33)

(ii) Selecting the Sign of the Terms2k

(e)02 222

22 kk

kk Rrdr

dRr

dr

Rdr

02 kkk

dt

d (f)

gives02 k

000 R

)()()( 2/12/12/1 rJBrJArrR kkkkk

)cossin(1

)( rBrAr

rR kkkkk (g)

46

Solution to (f)


BC (1)

BC (2)

(g) and (h) into (b)

(h))exp()( 2 tCt kkk

0kB

,tan)1(okok

rrBi k = 1, 2, … (i)

khrBi o /

47

1

2 sin)exp(),(),(

k

kkk r

rtaTtrTtr

(4.21)

IC (3):

(j)

1

sin

k

kki r

raTT

(v) Orthogonality

,/21 ra ,02 a 13 a

,2rp ,0q 2rw

The characteristic functions in (j) are

it is a Sturm-Liouville problem with

)/1( r rksin solutions to (e). Comparing (e) with eq. (3.5a) shows that

48

are orthogonal with respect to

Multiply (j) by

orthogonality

)/1( r rksin .)( 2rrw ,))(sin/1( 2drrrr i integrate and invoke

(5) Checking

Dimensional check

(k)

]cossin[

cossin)(2

sin

sin)(

0

2

0

okokokk

okokoki

r

k

r

ki

k

rrr

rrrTT

drr

drrrTTa

o

o

49

(ii) Steady state

(6) Comments

Limiting check:

(i) Initial temperature = T

(i) Non-uniform initial temperature, )()0,( rfrT (ii) Solution is expressed in terms of a single

parameter: Biot number Bi

50

Examples:

• Solar heating• Reentry aerodynamic heating

• Reciprocating surface friction

• Periodic oscillation of temperature or flux

• Limitation: Linear equations

4.6 Temperature Dependent Boundary Conditions: Duhamel’s Superposition Integral

51

Duhamel’s superposition integralUse solution of an auxiliary problem with constantboundary condition to construct the solution to the

same problem with time dependent condition

4.6.1 Formulation of Duhamel’s Integral

Initial temperature: 0iT:0

it boundary is at 01T

:i

t boundary is at 02T01T

02TT

1

0102 TT

t11.4.Fig

52

SolutionDecompose into two problems: One starts and the

second at Each problem has a constant BC

0t.

1t

),( txT Solution to auxiliary problem, constant surfaceTemperature of magnitude unity.Thus

(b)),(),( 011 txTTtxT

),(),(),( 21 txTtxTtxT (a)

Second problem starts at with surface temperature

Solution to 1

).(0102

TT :),(2

txT

53

Adding (b) and (c)

• Generalize to arbitrarily boundary condition

(c)),()(),( 101022 txTTTtxT

),()(),(),( 1010201 txTTTtxTTtxT (d)

= surface temperature, or)(tF

= ambient temperature, or

= heat flux

to many problems, each havinga small step change, in B.C.

• Solution: superimpose solutions

F

t

)(tF

F

tFig. 4.12

54

solution

• BC for first problem: F(0) starts at time t = 0,

t

)(tF

F

tFig. 4.12

),()0(),(0 txTFtxT (e)

• Contribution of the ith problem,

BC is )( iF

),()(),( iii txTFtxT

Adding all solutions

(f)

n

iii txTFtxTFtxT

1

),()(),()0(),(

55

or

i

n

ii

i

i txTF

txTFtxT

1

),()(

),()0(),(

,n , di

,)()(

d

dFF

i

i

Integration by parts with 0)0,( xT

(4.29)

t

dt

txTFtxT

0

),()(),(

t

dtxTdt

dFtxTFtxT

0),(

)(),()0(),(

(4.28)

56

NOTE:

• Duhamel’s method applies to linear equations

• Equations (4.28) and (4.29) apply to any coordinate system

• 0iT

Define • If :0i

T iTT

equal to unity • Auxiliary solution: ),( txT is based on a constant BC

• Method applies to:

(1) )(tq (2) Lumped-capacity models )(tT

• Integrals in (4.28) and (4.29) are with respect to the dummy variable

57

4.6.2 Extension to Discontinuous Boundary Conditions

and tat

)(1 tF

)(2 tF)(tF

Fig. 4.13

If is discontinuous, modify Duhamel’s integral. )(tF

atttFtF 0)()( 1

atttFtF )()( 2

• :a

tt use )()(1

tFtF in (4.28) & (4.29)

• :a

tt modify (4.28) and (4.29)

and with discontinuity)(1 tF )(2

tF at :at

58

Result:

For (4.28) add another solution, :),( txTa

),()]()([),( 12 aaaa ttxTtFtFtxT

Integration by parts with gives0)0,( xT

dtxTd

dFdtxT

d

dF

ttxTtFtF

txTFtxT

t

at

at

aaa

),()(

),()(

),()]()([

),()0(),(

2

0

1

12

1

(4.30)

59

dt

txTF

dttxT

F

ttxTtFtFtxT

t

t

taaa

a

a

),()(

),()(

),()()(),(

2

01

12

(4.31)

• Extension to several discontinuities

60

4.6.3 ApplicationsExample 4.5: Plate with Time Dependent

Surface TemperatureBC and IC:

(1) 00 ),( tT

(2) AttftLT )(),(

(3) )0,(xT 0iT

Determine ),( txT

(1) Observations

0 x

xAT 0T 0)0,( xT

L

Fig. 4.14

• One BC depends on time

0iT •

61



(1) 1-D

(ii) Governing equation

(iii) Boundary and Initial conditions

t

T

x

T

1

2

2 (a)

(1) 0),0( tT

(2) constant ,,k

62

0 x

xAT 0T 0)0,( xT

L

Fig. 4.14

(2) tAtLT ),(

(3) 0)0,( xT

(4) Solution

Eq.(a) is linear, apply Duhamel’s integral, eq .(4.28)

tAtF )(

),( txT solution to problem with BC (2) replaced by1),( tLT

t

dtxTdt

dFtxTFtxT

0),(

)(),()0(),(

(4.28)

63

Use result of Example 4.2: set

)/sin(])/(exp[)1(2

),(

1

2 LxntLnnL

x

txT

n

n

(b)

where

nn n

a )1(2 (c)

Find and )0(F ddF /)(

,)( tAF 0)0(, FAd

dF

(d)

0,01

i

TTq 12

Tand

64

Substituting (b) and (d) into eq. (4.28)

dtLnLxnn

A

dL

xAtxT

n

tn

t

)]()/(exp[)/sin()1(2

),(

10

2

0

(e)

Evaluating the integrals in (e)

)]/exp(1[)/sin()1(2

),(

222

133

2LtnLxn

n

LA

tL

xAtxT

n

n

(f)

65

(5) Checking

(i) A = 0(ii) t

Dimensional check

Limiting check:

Example 4.6: Lumped-Capacity Method with Time Dependent Ambient Temperature

Metal foil, 1.0Bi

Initial temperature iT

Oven temperature starts at and changes with time asiT

66

(1) Observations

io TtTtT )]exp(1[)( (a)

Determine: )(tT

• Oven temperature is time dependent

• Use Duhamel’s method if lumped-capacity equation is linear

appears in DE • T

• 0iT


(3) Formulation

67

(ii) Governing Equation

Define

(i) Assumptions

(1) Bi < 0.1

(2) constant hck p ,,,

Set in (4.5)0q

(a)dtVc

hA

tTT

dT

p

s

)(

iTtTt )()( (b)

Oven temperature

iTtTt )()( (c)

68

Substitute into (a)

Initial condition

(4) Solution

Apply Duhamel’s integral

dtVc

hA

tt

d

p

s

)()(

(d)

0)0( (e)

t

dtxTdt

dFtxTFtxT

0),(

)(),()0(),(

(4.28)

69

Constant oven temperature equal to unity:

Rewrite (a)

= solution to auxiliary problem of constantoven temperature equal to unity and zeroinitial temperature

)(t

(f))]exp(1[)()()( tTTtTttF oi

1 iTT (g)

dtVc

hAd

p

s

1

(h)

= time dependent oven temperature)(tF

70

Initial condition

0)0( (i)

Integrate and use IC

tVc

hAt

p

s

exp1)( (j)

Determine ddF /)(

(k))exp(

oTd

dF

Substituting (j) and (k) into eq. (4.28), use 0)0( F

71

(l)

dtVc

hATt

t

p

so

0)(exp1)exp()(

Performing the integration

(5) Checking

Dimensional check

)/exp()exp()/(

)exp()(

VcthAtVchA

T

tTTTtT

psps

o

ooi

(m)

Limiting check:

t

72

Differential equation check

Initial condition check: 0t

4.7 Conduction in Semi-infinite Regions: The Similarity Method

• Limitations of separation of variables: 2 HBC separated by a finite distance

• Method fails for semi-infinite regions

• Alternate approach: Similarity method

Combine two independent variables into one andtransform PDE to ODE

• Basic idea:

73

(i) Semi-infinite region

• Limitations:

(ii) Limitations on BC and IC

Fig. 4.15

oT

0 x

iT

Example: Semi-infinite plate, uniform initial Surface is suddenly temperature .

iT

maintained at temperature .o

T

t

T

x

T

1

2

2

(a)

BC (1) oTtT ),0(

(2) iTtT ),(

IC (3) iTxT )0,(

74

Define:

io

i

TT

TT

(b)

Substitute into (a)

BC and IC become

tx

12

2

(c)

Fig. 4.15

oT0 x

iT

(1) 1),0( t

(2) 0),( t(3) 0)0,( x

75

Assume that x and t can be combined into a singlevariable ),( tx

)(),( tx (d)

Try:

Conditions on Must transform PDE, BC and IC:),( txin terms of and eliminate x and t.

t

x

4 (4.32)

Using (4.32), construct the derivatives in (c)

td

d

xd

d

x

4

1

76

2

2

2

2

4

1

4

1

xd

d

txtd

d

d

d

x

dd

tdd

ttx

tx

dd

tdd

t

21

42

422/3

(c) becomes

022

2

d

d

d

d (4.33)

77

NOTE:

(1) x and t are eliminated and replaced by(2) Solution to PDE (c) must satisfy 3 conditions.

(3) Solution to ODE (4.33) can only satisfy two conditions.

Transformation of BC and IC:

Using t

x

4

ttx ,0 transforms to 0ttx , transforms to 0, txx transforms to

78

BC and IC transform to:

NOTE: One BC and IC coalesce. Fig. 4.15

oT0 x

iT

(1) 1)0( (2) 0)( (3) 0)(

Solution to (4.33): Separating variables

ddd

ddd2

/

)/(

Integrating

Ad

dlnln 2

79

or

Integrating and using first BC

dAd e2

dAd e

2

01

dA e

2

01 (e)

Define

de

2

0

2erf (f)

Where00erf

80

and

Equation (e) becomes

BC (2) gives B = -1.

(g) becomes

1erf

erf1)( B (g)

erf1)( (4.33a)

tx

tx

4

erf1),( (4.33b)

Derivative of :erf 22

)(erf

ed

d (4.34)

1 chapter 4 transient conduction neglect spatial variation: 4.1.1 criterion for neglecting spatial...

Documents

initial conditions bc

initial condition example

solution b

coordinates convection

summing nh initial condition

solution split

symmetrical convection

observations temperature