1 CHAPTER 2 ONE-DIMENSIONAL STEADY STATE CONDUCTION 2.1 Examples of One-dimensional Conduction: 2.1.1 Plate with Energy Generation and Variable Conductivity.

1

CHAPTER 2

ONE-DIMENSIONAL STEADY STATE

CONDUCTION

2.1 Examples of One-dimensional

Conduction:

2.1.1 Plate with Energy Generation and

Variable Conductivity

2

Example 2.1: Plate with internal energy generationq and a variable k

)1( Tkk o

Find temperature distribution.

L

q C0oC0o

Fig. 2.1

x0

(1) Observations • Variable k

• Symmetry

• Energy generation

• Rectangular system

• Specified temperature at boundaries

3

(2) Origin and Coordinates

Use a rectangular coordinate system

(3) Formulation

(i) Assumptions

• One-dimensional

• Steady

• Isotropic

• Stationary

• Uniform energy generation

4

(ii) Governing Equation

Eq. (1.7):

0

qdx

dTk

dx

d(2.1)

(a) )1( Tkk o

(a) into eq. (2.1)

(1 ) 0o

d dT qTdx dx k

(b)

(iii) Boundary Conditions.

Two BC are needed:

5

(c) 0)0( T

0)( LT (d)

(4) Solution

Integrate (b) twice

2122

22CxCx

k

qTT

o

(e)

BC (c) and (d)

ok

LqC

21

, 02 C (f)

6

(f) into (e)

(g) 0122

L

x

k

xLqTT

o

Take the negative sign

Solving for T

(h)

L

x

k

xLqT

o

111

2

(i)

L

x

k

xLqT

o

111

2

7

(5) Checking

• Dimensional check

• Boundary conditions check

• Limiting check: 0 0, Tq

• Symmetry Check:

(j) )12

)(()1(1

2

1 2

1

2

L

x

k

Lq

L

x

k

Lxq

dx

dT

oo

Setting x = L/2 in (j) gives dT/dx = 0

8

Conservation of energy and symmetry:

(k) 2

)0(ALq

q

2)(

ALqLq

(l)

Fourier’s law at x = 0 and x = L

2

)0()0(1)0(

ALq

dx

dTTkq o

(m)

• Quantitative Check

9

2

)()(1)(

ALq

dx

LdTLTkLq o

(n)

(6) Comments Solution to the special case:

k = constant: Set 0

2.1.2 Radial Conduction in a Composite Cylinder with Interface Friction

10

Example 2.2: Rotating shaft in sleeve, frictional

heat at interface, convection on

outside. Conduction in radial direction.

Determine the temperature

distribution in shaft and sleeve.

(1) Observations

• Composite cylindrical wall

• Cylindrical coordinates

• Radial conduction only

Fig. 2.2

0

sleeve

1T

2T

Th,

sRoR

r

iq

11

• Steady state:

• No heat is conducted through the shaft

Energy generated = heat conducted through the sleeve

• Specified flux at inner radius of sleeve, convection at outer radius

(2) Origin and Coordinates Shown in Fig. 2.2

12

(3) Formulation

(i) Assumptions

• One-dimensional radial conduction

• Steady

• Isotropic

• Constant conductivities

• No energy generation

• Perfect interface contact

• Uniform frictional energy flux

• Stationary

13

(ii) Governing Equation

Shaft temperature is uniform. For sleeve: Eq. (1.11)

(2.2) 01

dr

dTr

dr

d

(iii) Boundary Conditions

:sRSpecified flux at

(a) dr

RdTkq s

i)(1

1

14

Convection at :oR

])([)(

11

1 TRThdr

RdTk o

o (b)

(4) Solution

Integrate eq. (2.2) twice

(c) 211 ln CrCT

BC give C1 and C2

11 k

RqC si (d)

15

and

(d) and (e) into (c)

oo

si

hR

kR

k

RqTC 1

12 ln (e)

o

osi

hR

k

r

R

k

RqTrT 1

11 ln)( (f)

khRo / = Biot number

Shaft temperature T2: Use interface boundary condition

(g) )()()( 122 ss RTRTrT

16

Evaluate (f) at r = Rs and use (g)

os

osi

hR

k

R

R

k

RqTrT 1

12 ln)( (h)

(5) Checking

• Limiting check: 0iq

(6) Comments

• Conductivity of shaft does not play a role

Fig. 2.2

0

sleeve

1T

2T

Th,

sRoR

r

iq

• Dimensional check

• Boundary conditions check

17

• Problem can also be treated formally as a composite cylinder. Need 2 equations and 4 BC.

2.1.1 Composite Wall with Energy Generation

Example 2.1: Plate 1 generates heat at .q is sandwiched between two plates.

Outer surfaces of two plates at .oT

Plate 1

Find the temperature distribution in

the three plates.

oT x

q

2L

2L

1L

2k

oT

0

Fig. 2.3

1k

2k

18

oT x

q

2L

2L

1L

2k

oT

0

Fig. 2.3

1k

2k

(1) Observations • Composite wall

• Heat flows normal to plates

• Symmetry and steady state:

Energy generated = Energy conducted out

(2) Origin and Coordinates Shown in Fig. 2.3

• Use rectangular coordinates

• Symmetry: Insulated center plane

19

(3) Formulation (i) Assumptions

• Steady

• One-dimensional

• Isotropic

• Constant conductivities

• Perfect interface contact

• Stationary

(ii) Governing Equations

Two equations:

20

(b) 022

2

dx

Td0

21

2

k

q

dx

Td(a)

(iii) Boundary Conditions

Four BC:

Symmetry:

0)0(1

dx

dT(c)

Interface:

dx

LdTk

dx

LdTk

)2/()2/( 122

111 (d)

21

)2/()2/( 1211 LTLT (e)

Outer surface:

oTLLT )2/( 212 (f)

(4) Solution

Integrate (a) twice

(g) BAxxk

qxT

2

11 2

)(

Integrate (b)

DCxxT )(2 (h)

22

Four BC give 4 constants: Solutions (g) and (h) become

21

2

12

21

1

21

1 4

1

2)(

L

x

Lk

Lk

k

LqTxT o (i)

11

2

2

21

2 2

1

2)(

L

x

L

L

k

LqTxT o (j)

(5) Checking

:2

k

Lq • Dimensional check: units of

23

• Boundary conditions check

• Quantitative check:

1/2 the energy generated in center plate = Heat conducted at 2/1Lx

(k) dx

LdTkq

L )2/(

211

11

CCW/m

mW/m oo

223

)(

)()(

k

Lq

24

(i) into (k)

qL

dx

LdTk

2

)2/( 1111

Similarly, 1/2 the energy generated in center plate= Heat conducted out

(l) dx

LLdTkq

L )2/(

2212

21

(j) into (l) shows that this condition is satisfied.

• Limiting check:

(ii) If 01 L then .)(1 oTxT

,0q then .)()( 21 oTxTxT (i) If

25

(6) Comments Alternate approach: Outer plate with a specified

flux at 2/1Lx and a specified temperature at

.2/ 21 LLx

2.2 Extended Surfaces - Fins

2.2.1 The Function of Fins Newton's law of cooling:

(2.3) )( TThAq sss

26

Options for increasing :sq

• Increase h

• Lower T

• Increase sA

Examples of Extended Surfaces (Fins):

• Thin rods on condenser in back of refrigerator

• Honeycomb surface of a car radiator

• Corrugated surface of a motorcycle engine

• Disks or plates used in baseboard radiators

27

2.2.2 Types of Fins

straight fin constant area (a)

2.5 Fig.

annular fin (d) pin fin (c)

straight fin variable area (b)

28

Terminology and types

• Fin base

• Fin tip

• Straight fin

• Variable cross-sectional area fin

• Spine or pin fin

• Annular or cylindrical fin

2.2.3 Heat Transfer and Temperature Distribution in Fins

• Heat flows axially and laterally (two-dimensional) • Temperature distribution is two-dimensional

29

2.2.4 The Fin Approximation

x

Th ,

T

Tr

6.2.Fig

Neglect lateral temperature variation

)(xTT

Criterion: Biot number = Bi

(2.4) Bi = h /k << 1

cetanresisexternal

cetanresisInternal

h/

k/Bi

1

30

2.2.5 The Fin Heat Equation: Convection at Surface

(1) Objective: Determine fin heat transfer rate.

Need temperature distribution.

(2) Procedure: Formulate the fin heat equation.

Apply conservation of energy.

    • Select an origin and coordinate axis x.

    • Assume ,1.0Bi )(xTT • Stationary material, steady state

31

Th,dx

dy

sy

)(a )(b )(c

ds

CsdA

cdq

xq dxdx

dqq x

x dxx

y

0

7.2.Fig

Conservation of energy for the element dx:

(a) outgin EEE =

(b) xin qE

cx

xout dqdxdx

dqqE (c)

32

(b) and (c) into (a)

(d) cx

g dqdxdx

dqE

Fourier's law and Newton’s law

dx

dTkAq cx (e)

(f) sc dATThdq )( Energy generation

dy

)(b )(c

ds

CsdA

cdq

xq dxdx

dqq x

x dx

33

(e), (f) and (g) into (d)

(2.5a)

( ) ( ) ( ) 0s cc

dTdkA x dx h T T dA q A x dx

dxdx

Assume constant k

0)()()(

12

2

k

q

dx

dATT

xkA

h

dx

dT

dx

dA

xAdx

Td s

c

c

c(2.5b)

• (2.5b) is the heat equation for fins

• Assumptions: (1) Steady state (2) Stationary

dxxAqE cg )( (g)

34

(3) Isotropic

(4) Constant k

(5) No radiation

(6) Bi << 1

2.2.6 Determination of dAs /dxFrom Fig. 2.7b

dsxCdAs )( (a)

= circumference )(xCds = slanted length of the element

• ,/ dxdAc and dxdAs / are determined from

the geometry of fin.

,Ac

35

For a right triangle

(b) 2/122 ][ sdydxds

(b) into (a) 2/12

1)(

dx

dyxC

dx

dA ss (2.6a)

For dxdys / << 1

)(xCdx

dAs (2.6b)

2.2.7 Boundary ConditionsNeed two BC

36

2.2.8 Determination of Fin Heat Transfer Rate :fq

Fig. 2.8

sq

sq

sqx0

)0(q

Th,

Th,

Conservation of energy for :0q

sf qqq )0( (a)

Two methods to determine :fq

37

(2) Convection at the fin surface.

Newton's law applied at the fin surface

(2.8) s

dATxThqqA ssf ])([

• Fin attached at both ends: Modify eq. (2.7) accordingly

• Fin with convection at the tip: Integral in eq. (2.8) includes tip

(1) Conduction at base.

Fourier's law at x = 0

(2.7) dx

dTkAqq cf

)0()0()0(

38

• Convection and radiation at surface: Apply eq. (2.7). Modify eq. (2.8) to include heat exchange by radiation.

2.2.9 Applications: Constant Area Fins with Surface Convection

Fig. 2.9

0 x

CTh,

Th,oT

cA

39

Use eq. (2.5b). Set

(a) 0/ dxdAc

A. Governing Equation

Fig. 2.9

0 x

CTh,

Th,oT

cA = constant sy

0/ dxdys

(a) and (b) into eq. (2.5b)

Eq. (2.6a)

CdxdAs / (b)

0)(2

2

TTkA

hC

dx

Td

c

(2.9)

40

Rewrite eq. (2.9)

TT (c)

ckA

hCm 2 (d)

022

2

m

dx

d(2.10)

Valid for:

(1) Steady state

(2) constant k, cA and T

= constant, (c) and (d) into (2.9)TAssume

41

(3) No energy generation

(4) No radiation

(5) Bi 1

(6) Stationary fin

B. Solution

(2.11a) )exp()exp()( 21 mxAmxAx

Assume: h = constant

mxBmxBx coshsinh)( 21 (2.11b)

42

C. Special Case (i):

• Finite length

• Specified temperature at base, convection at tip

Boundary conditions:

(e) o)0( TT

])([)(

TLThdx

LdTk t

(f)

o)0( (h)

Fig. 2.10cA

C

th0

oT

Th,

xTh,

43

)()(

Lhdx

Ldk t

(i)

Two BC give B1 and B2

fqEq. (2.7) gives

mLmkhmL

mLmkhmLTThCAkq

t

tcf sinh )/( cosh

] cosh)/()[sinh(][ o2/1

(2.13)

TT

TxTx

o o

)()(

(2.12)

mLmkhmL

xLmmkhxLm

t

t

sinhcosh

sinhcosh

44

C. Special Case (ii):

• Finite length

• Specified temperature at base, insulated tip

BC at tip:

0)(

dx

Ld(j)

0thSet eq. (2.12)

mL

xLm

TT

TxTx

cosh

)-( cosh)()(

oo

(2.14)

0thSet eq. (2.13)

(2.15) mLTTChkAq ocf tanh21

45

2.2.10 Corrected Length Lc

• Insulated tip: simpler solution

• Simplified model: Assume insulated tip, compensate by increasing length by cL

• The corrected length is cL

cc LLL (2.16)

• The correction increment Lc depends on the

geometry of the fin:

Circular fin:

cLrr o2

o 2

cLIncrease in surface area due to = tip area

46

2/oc rL

4/tLc

= total surface areasA

Square bar of side t

f2.2.11 Fin EfficiencyDefinition

maxq

q ff (2.17)

TThAq osmax

47

Eq. (2.17) becomes

(2.18) )( o

TTAh

q

s

ff

2.1.12 Moving Fins

Examples:

• Extrusion of plastics

• Drawing of wires and sheets

• Flow of liquids

Udx

Th,

oT x

surT

(a)

dx

hm ˆ )(ˆ

ˆ dxdx

hdhm

dxdx

dqq x

x xq

cdqrdq

Fig. 2.11

(b)

48

Heat equation:

• Steady state• Constant area• Constant velocity U• Surface convection and radiation

Conservation of energy for element dx

rcx

xx dqdqdx

hdmhmdx

dx

dqqhmq

ˆˆˆ

(a)

cUAm (b)

dx

hm ˆ )(ˆ

ˆ dxdx

hdhm

dxdx

dqq x

x xq

cdqrdq

Fig. 2.11(b)

Assume:

49

dTchd pˆ (c)

Fourier’s and Newton’s laws

dx

dTkAq cx (d)

sc dATThdq )( (e)

dxCdAs (f)

ssurr dATTdq )( 44 (g)

(b)-(g) into (a) assume constant k

0)()( 442

2

surc

p TTk

TTkA

hC

dx

dT

k

Uc

dx

Td

(2.19)

50

Assumptions leading to eq. (2.19): (1) Steady state

(3) Isotropic

(4) Gray body

(5) Small surface enclosed by a much larger surface and

,(2) Constant U, k, P and

(6) Bi << 1

51

Example 2.4

insulated bottomfurnace

Fig. 2.12

t

oT

xTh,

WU

2.2.13 Application of Moving Fins

Determine the temperature distribution in the sheet.

(2) Bi < 0.1(3) No radiation(4) No heat transfer from bottom

Plastic sheet leaves furnace at .oT

Sheet is cooled at top by convection.

Assume:

(1) Steady state

52

Solution

(1) Observations

• Constant area fin

• Temperature is one-dimensional

• Convection at surface

• Specified furnace temperature

• Fin is semi-infinite

• Constant velocity

insulated bottomfurnace

Fig. 2.12

t

oT

xTh,

WU

(2) Origin and Coordinates

53

(3) Formulation

(i) Assumptions (1) One-dimensional

(2) Steady state

(3) Isotropic

(4) Constant pressure

0)(2

2

TTkA

hC

dx

dT

k

Uc

dx

Td

c

p(2.20)

, (5) Constant U, k, P and (6) Negligible radiation

(ii) Governing Equation

Eq. (2.19)

54

WtAc (a)

tWC 2 (b)

(a) and (b) into eq. (2.20)

cTmdx

dTb

dx

Td 2

2

2

2 (c)

where

,2k

Ucb p

,)2(2

kWt

tWhm

T

tWk

tWhc

)2(

(d)

(iii) Boundary Conditions

oTT )0( (e)

)(T finite (f)

55

Eq. (c) is:

• Linear• Second order• Constant coefficients

(4) Solution:

Eq. (A-6b), Appendix A

01 C (g)

222

2

221

)exp(

)exp(

m

cxmbbxC

xmbbxCT

(2.21)

B.C. (f)

56

B.C. (e)

202m

cTC (h)

(d), (g) and (h) into (2.21)

xtWk

tWh

k

Uc

k

Uc

TT

TxT pp

o

)2(2

22exp

)( )(

(2.22)

(5) Checking

Dimensional check: Each term in the exponential in

eq. (2.22) is dimensionless

57

Boundary conditions check: Eq.(2.22) satisfies (e) and (f).

Limiting checks:

:0h(i) If

:U(ii) If

(6) Comments

(i) Temperature decays exponentially

(ii) Motion slows decay

oTxT )(

oTxT )(

58

2.2.14 Variable Area Fins

)(xAA cc

Example: Cylindrical or annular fin

Governing equation:

Usually has variable coefficients

Case (i) : The Annular Fin

0)()()(

12

2

dr

dATT

rkA

h

dr

dT

dr

dA

rAdr

Td s

c

c

c (2.23)

2.13 Fig.

Th,

drr

Th,

59

trrAc 2)( (a)

tdrdAc 2/ (b)

Eq.(2.6a) gives drdAs /

2/12

1)(

dr

dyrC

dr

dA ss

0/ drdys

)2(2)( rrC (c)

For

sy = constant:

2.13 Fig.

Th,

drr

Th,

60

Eq. (2.6a):

)2(2 rdr

dAs (d)

(a), (b) and (d) into eq. (2.23)

0))(/2(1

2

2

TTkthdr

dT

rdr

Td (2.24)

Case (ii): Triangular straight fin

Fin equation: Constant k,eq. (2.5b)

2.14 Fig.

Th,

L

dx

xsy

0

y

tTh,

)(2 xyWA sc (e)

61

(f) )2/)(/( tLxys

xLtWAc )/( (g)

LtWdx

dAc / (h)

Eq. (2.6a):

2/122/12 ])2/(1[2])/(1[2 LtWdxdyWdx

dAs

s (i)

2.14 Fig.

Th,

L

dx

xsy

0

y

tTh,

62

(g), (h) and (i) into eq. (2.5b)

0)(/12/1/21 2/12

2

2

TTxLtkthLdxdT

xdx

Td

(2.25)

Equations (2.24) and (2.25) are:

•Linear

•Second order

•Variable coefficients

63

(2.26) 0)21(

2)21(

22222222

22

22

ynCAxABxBxDC

dxdy

xBxAdx

ydx

C

2.3 Bessel Differential Equations and Bessel Functions

2.3.1 General form of Bessel Equations

Note the following:

(1) Eq. (2.26) is linear, second order with variable coefficients

(2) A, B, C, D, and n are constants

64

(3) n is called the order of the differential equation

(4) D can be real or imaginary

2.3.2 Solution: Bessel Functions

• Form: Infinite power series solutions

• General solution: depends on the constants n and D

(1) n is zero or integer, D is real

)()()exp()( 21C

nC

nA xDYCxDJCxBxxy

(2.27) where

21 ,CC = constant of integration

65

)( Cn DxJ = Bessel function of order n of the first kind

)( Cn DxY = Bessel function of order n of the second kind

Note the following:

(i) The term )( CxD is the argument of the Bessel

function

(ii) Values of Bessel functions are tabulated

(2) n is neither zero nor a positive integer, D is real

(2.28)

)()()exp()( 21C

nC

nA xDJCxDJCxBxxy

66

(3) n is zero or integer, D is imaginary

(2.29)

)()()exp()( 21C

nC

nA xpKCxpICxBxxy

where

,i

Dp i is imaginary = 1

= modified Bessel function of order n of the first kind nI

nK = modified Bessel function of order n of the second

kind

67

(4) n is neither zero nor a positive integer, D is imaginary

)()()exp()( 21C

nC

nA xpICxpICxBxxy

(2.30)

2.3.3 Form of Bessel Functions

nnnnn IIJYJ ,,,, :nKand

• Symbols for infinite power series

• The form of each series depends on n

(2.31)

0 !!k 2

)2/()1()(

22

2 kk

xxJ

kk

Example: ,2n )(2 xJBessel function

68

2.3.4 Special Closed-form Bessel Functions:

2

integer oddn

For 2/1n

(2.32) xx

xJ sin2

)(2/1

and

xx

xJ cos2

)(2/1 (2.33)

69

For n = 3/2, 5/2, 7/2, … use Eq. (2.32) or eq. (2.33) and the recurrence equation:

)()(12

)( 3/22/12/1 xJxJx

kxJ kkk

k = 1, 2, 3, … (2.34)

For the modified Bessel functions, n = 1/2

xx

xI sinh2

)(2/1 (2.35)

and

xx

xI cosh2

)(2/1 (2.36)

70

For n = 3/2, 5/2, 7/2, …. use eq. (2.35) or eq. (2.36) and the recurrence equation

(2.37)

)()(12

)( 3/22/12/1 xIxIx

kxI kkk

k = 1, 2, 3, …

2.3.5 Special Relations for n = 0, 1, 2, …

)()1()( xJxJ nn

n (2.38a)

)()1()( xYxY nn

n (2.38b)

)()( xIxI nn (2.38c)

)()( xKxK nn (2.38d)

71

2.3.6 Derivatives and Integrals of Bessel Functions

(2.39)

mxZmx

mxZmxmxZx

dx

d

nn

nn

nn

1

1 IYJZ ,,

KZ (2.40)

(2.41)

mxZmx

mxZmxmxZx

dx

d

nn

nn

nn

1

1 KYJZ ,,

IZ (2.42)

72

(2.45)

mxZx

nmxmZ

mxZx

nmxmZ

mxZdx

d

nn

nn

n

1

1 KYJZ ,,

IZ (2.46)

(2.43)

mxZx

nmxmZ

mxZx

nmxmZ

mxZdx

d

nn

nn

n

1

1 IYJZ ,,

KZ (2.44)

IYJZmxZxmdxmxZx nn

nn ,,)()/1()(1

(2.47)

73

2.3.7 Tabulation and Graphical Representation of Selected Bessel Functions

(2.48)

mxZxmdxmxZx nn

nn

11

KYJZ ,,

Table 2.1

x J0(x) Jn(x) I0(x) In(x) Yn(x) Kn(x)

0 1 0 1 0 - 0 0 0 0

74

Fig. 2.15 Graphs of selected Bessel functions

75

2.4 Equidimensional (Euler) Equation

(2.49) 0012

22 ya

xdyd

xaxd

ydx

Solution:Depends on roots r1 and r2

2

4)1()1( 02

112,1

aaar

(2.50)

Three possibilities:

(1) Roots are distinct

2121)( rr xCxCxy (2.51)

76

(2) Roots are imaginary

)logsin()logcos()( 21 xbCxbCxxy a (2.52)

(3) One root only

)log()(21

xCCxxy r (2.53)

2.5 Graphically Presented Fin Solutions to Fin

Heat Transfer Rate fq

Fin efficiency :f

)( o

TTAh

q

s

ff (2.18)

77

Fig. 2.16 Fin efficiency of three types of straight fins [5]

78

Fig. 2.17 Fin efficiency of annular fins of constant thickness [5]