CH02 Conduction

MEC551

THERMAL ENGINEERING

Mohd Hafiz Mohd Noh

T1-A18-6C 1

2.0 Conduction

Conduction Analysis

Main Objective of Conduction Analysis

To determine the temperature distribution in a medium

2

Steady versus Transient Heat Transfer

Steady implies no change with time at any point within the medium

Transient implies variation with time or time dependence

3

multidimensional heat transfer,

Three prime coordinate systems:

rectangular (T(x, y, z, t)) ,

cylindrical (T(r, f, z, t)),

spherical (T(r, f, q, t)).

4

Heat Conduction Equation (Rectangular Coordinates)

Cartesian Coordinates:

5

dQy dQ(y+dy)

dQz

dQ(z+dz)

dQ(x+dx)

dQx

dx

dz

dy

X

Z

Y


Differential volume:

Heat conduction rate in x-direction (into element):

dzdydxdV

dx

dTdzdyk

dx

dTAkdQ

A

x

6

dQx dy

dz


Taylor Series Expansion:

Substitute our equation:

HOTxfhxfhxf )()()(

HOTdxx

Qd

QddxxQdx

x

)(

7 7

small

0

2

2

)(

)(

x

Tdzdyk

x

Qd

xf

dxh

x

TdzdykQdxf

x

X


Therefore the net rate of flow in the x-direction is:

2

2

2

2

x

Tdzdydxk

x

Tdydzdxk

x

Tdydzk

x

TdydzkQdQd dxxx

dxx

QdQd xx

x

Tdxdzdyk

x

TdzdykdxxQd

2

2

)(

8


Likewise:

2

2

2

2

z

TdzdydxkQdQd

y

TdzdydxkQdQd

dzzz

dyyy

9


In addition to the heat flow into and out of the element, there is also the possibilities of:

- Heat being generated within the element (e.g. due to the flow of electricity).

- Heat being stored within the element, as in the case of an unsteady-state condition.

10


The rate of heat generated is:

The rate of heat storage is:

dzdydxqdVq

volumepergenerated

heatofrate

11

changeetemperatur

ofrateheatspecific

p

density

pt

TdzdydxC

t

TdVC


Assuming the element may expand or contract freely at constant pressure, the energy balance is given by:

12

Rate of

Heat Storage

Net rate of

Heat into

Element

Rate of

Heat

Generation

= +


In equation form this is:

Set:

Generation

HeatofRate

elementoHeatofRateNetStorageHeatofRate

P qz

T

y

T

x

Tk

t

TC

int

2

2

2

2

2

2

13

PC

k

Thermal diffusivity (i.e. the ratio of heat conduction to heat

storage).

It represents how fast heat propagates through a material.

~


Substituting in, this equation becomes the general differential conduction equation for rectangular coordinates:

Generation

HeatofRateelementoHeatofRateNet

StorageHeatofRate

k

q

z

T

y

T

x

T

t

T

int

2

2

2

2

2

21

14


Special cases: 1) Fourier Equation (no heat generation, q =0):

2) Poisson Equation (steady state, T/t =0):

3) Laplace Equation (steady state with no heat generation):

2

2

2

2

2

21

z

T

y

T

x

T

t

T

02

2

2

2

2

2

z

T

y

T

x

T

15

02

2

2

2

2

2

k

q

z

T

y

T

x

T

Heat Conduction Equation (Spherical Coordinates)

Now calculate the general conduction equation in spherical coordinates

16

y x

z z

y x

r

q

T(r,,q)


Now calculate the general conduction equation in spherical coordinates

17

rq q

z

rdq

q(q)

q(q+dq)

y x

r

q

d

rsinqd

rdq dq

dr

q(r+dr)

q(r)

q(+d)

q()


Differential volume:

Heat conduction rate in r-direction (in to element):

qqf

fqq

sin

sin

2

drddr

drrddrdV

18

r

Tddkr

r

TrddrkQd

A

r

qqf

qfq

sin

sin

2

q(r)

rdq


Heat conduction rate in r-direction (out of element):

19

2

22

2

)(

sinsin2

sin

r

Tddkr

r

Tddkrdr

r

Tddkr

r

Qd

drQQdr

rdrr

qqfqqf

qqf

q(r+dr)

rdq


Total heat conduction rate in r-direction:

2

22

)(2sin

r

Tr

r

TrdrddkQdQd

drrrqqf

20

2

22

22

)(

sinsin2

sinsin

r

Tddkr

r

Tddkrdr

r

Tddkr

r

TddkrQdQd

drrr

qqfqqf

qqfqqf


Heat conduction rate in q-direction (in to element):

qfq

qfqq

Tddrk

dr

TdrdrkQd

A

sin

sin

21

q(q)

dr

rsinqd


Heat conduction rate in q-direction (out of element):

qq

qq

qfq

fq

qq

q

qqq

dTT

ddrkT

ddrk

Qd

dQdQdd

2

2

sincossin

22

q(q+dq) dr

rsinqd


Total heat conduction rate in q-direction:

2

2

sincosq

qq

qqfqqqTT

dddrkQdQdd

qq

qq

qf

qfq

qfqqqq

dTT

ddrk

Tddrk

TddrkQdQd

d

2

2

sincos

sinsin

23


Heat conduction rate in -direction (in to element):

fq

q

fqqf

Tdrdk

r

TdrdrkQd

sin

sin

24

rdq q()


Heat conduction rate in -direction (out of element):

2

2

sinsin fq

fq

fq

q

ff

f

fff

TddrdkTdrdk

Qd

dQdQdd

25

rdq

q(+d)


Total heat conduction rate in -direction:

2

2

sin

sinsin

fq

fq

fq

q

fq

qfff

Tddrdk

TdrdkTdrdkdQQd d

2

2

sin fq

fqfff

TddrdkQdQd

d

26


Rate of Heat Generation:

Rate of Heat Storage:

qqf sin2

dddrrqdVq

t

TdddrrC

t

TdVC PP

qqf sin2

27


The energy balance is given by:

28

Rate of

Heat Storage

Net rate of

Heat into

Element

Rate of

Heat

Generation

= +


qqf

fq

qq

qq

qq

qf

qqf

sin

sin

1

sincos

sinsin2

sin

2

2

2

2

2

2

22

2

dddrrq

T

TT

r

Tr

r

Tr

dddrk

t

TdddrrCP

29


k

qT

r

T

r

T

rr

T

r

T

r

t

T

k

CP

2

2

222

2

222

2

sin

111

sin

cos2

1

fqqqq

q

PC

k

30

Now by r2, sin(q), and k:

Recall: ~ thermal diffusivity


k

qT

r

T

r

T

rr

T

r

T

rt

T

T

rr

Tr

rr

2

2

22

sinsin

1

2

2

22

1

2

2

sin

1

11

sin

cos21

2

2

2

fq

qqq

q

qq

qq

31

Now simplify:


k

qT

r

T

rr

Tr

rr

t

T

2

2

222

2

2 sin

1sin

sin

11

1

fqqq

qq

32

Therefore the conduction equation in spherical coordinates is:

Heat Conduction Equation (Cylindrical Coordinates)

Homework #2 Derive the equations for cylindrical coordinates

33

Heat Conduction Equation (Cylindrical Coordinates)

Equations for cylindrical coordinates:

k

q

z

TT

rr

Tr

rrt

T

k

q

z

TT

rr

T

rr

T

t

T

r

Tr

rr

2

2

2

2

2

2

2

2

2

2

1

2

2

111

111

q

q

34

Boundary and Initial Conditions

Temperature distribution in a medium can be determined from the solution of appropriate heat conduction equation. But the solution depends on the boundaries of the medium.

For cases in which the medium is time dependent, conditions at an initial time are also essential.

35


The 4 most common boundary conditions are:

1) Constant Surface Temperature:

T(0,t) = Ts

36

x

T

Ts

T(x,t)



2) Constant and finite heat flux (heat transfer rate per unit area, W/m2):

s

x

qx

Tk

0

37

x

T

T(0,t)

qs

qs



3) Adiabatic or insulated surface:

00

xx

T

38

x

T

T(0,t)

T(x,t)



4) Convection surface condition:

tTThx

Tk

x

,00

39

x

T

T(x,t)

q

T, h

1-D Steady State Conduction (Example 2.1)

Example 2.1 - One-dimensional steady-state heat conduction (no heat generation):

Develop the expressions: 1) Temperature distribution T(x) within the slab.

2) Heat flow (Q), through the area (A) of the slab. 40

X

T

X=0 X=L

T1

T2

A, k

A slab (of thickness L)

with no energy

generation (q=0) has the

following boundary

conditions:

X= 0 ; T(0)= T1

X= L ; T(L)= T2


(I): Integrate and apply b.cs and solve constants C1 & C2 to find the temperature distribution T(x).

21)( CxCxT

02

2

x

T

41

X

T

X=0 X=L

T1

T2

A, k ~ 1-D Laplace Equation

(rectangular coordinates

112)( Tx

L

TTxT

L

TTC

TLCTLx

CTx

121

112

210


(II): Solve for Q

Ak

LRwhere

R

TT

L

TTkA

dx

dTkAQ

:

1212

L

TT

TxL

TT

dx

d

dx

dT

12

112

42

X

T

X=0 X=L

T1

T2

A, k

1-D Steady State Conduction (Analysis Procedure)

Slab (Plane Wall)

Consider a slab of isotropic (invariable) thermal conductivity material (k) with an heat generation rate of q(x) [W/m3].

Isotropic means: having properties that are identical in all directions.

43

T

X

1

2

T1

T2

dx

dT k, A

q(x) q(x)

T2


The general equation for conduction with heat generation is (assuming constant energy generation):

k

q

x

T

2

2

0

44

k

q

z

T

y

T

x

T

t

T

2

2

2

2

2

21

0, steady state 0, 1D 0, 1D

For: 0 X L


Solving differential equations:

212

1

2

2

2CxCx

k

qxT

Cxk

q

x

T

k

q

x

T

45


Solve C1 and C2 by using the boundary conditions:

(i) x=0, T(x)= T1 (ii) x=L, T(x)= T2

L

k

q

L

TTC

TLCLk

qTii

TCi

2

2)

)

121

11

2

2

12

212

2CxCx

k

qxT

46

T

X

1

2

T1

T2

dx

dT k, A

T2


Substituting in the constants gives:

The temperature distribution T(x) in the slab can now be found for a known heat generation rate q(x), thickness (L), and thermal conductivity coefficient (k) of the material.

12

22TxL

k

q

L

Tx

k

qxT

47


Once the temperature distribution T(x) in the slab is established from the solution to this equation, the heat flux q(x) anywhere in the slab can be determined from the Fouriers equation.

dx

dTk

A

Qq

48

Thermal conductivity

In a gas, conduction is due to the collisions and diffusion of the molecules due to their random motion.

In solids, it is due to the combination of vibrations of the molecules in their lattice and the energy transport of free electrons.

49 Gas Liquid Solid


50

Thermal conductivity (k) of a material is the measure of the ability of the material to conduct heat.


Thermal conductivity is temperature dependent

51


52

1-D Steady State Conduction

One-dimensional (1D) heat conduction

Implies that the temperature gradient exists in only one direction.

Steady state systems (SS)

The temperature within the solid is assumed not to be time dependent.

53

Thermal Resistance Method

1D/SS analysis can be applied to problems to determine the temperature distribution and heat flow in a solid, slab, cylinder, or sphere.

The thermal resistance approach (similar to Ohms Law) is a technique that simplifies complicated problems which involve multi-layered mediums when there is no heat generation (q=0).

54

Thermal Resistance Method (Analysis Procedure)

If q(x)=0 (no heat generation) then the rate of flow of heat energy normal to the area (A) is given by:

L Thickness of the slab

A Area normal to the direction of heat flow

K Thermal conductivity coefficient

T Temperature difference (gradient)

R Thermal resistance against heat conduction or conduction resistance of the wall.

kA

LRwhere

R

T

L

TkA

x

TkAQ

:

55


This is like Ohms Law:

Therefore, circuit representations can provide a useful tool for both conceptualizing and calculating heat transfer problems.

)(tanRe

)()(

Rcesis

VDifferencePotentialICurrent

56

Analogy to Electrical Current Flow

Eq. 3-5 is analogous to the relation for electric current flow I, expressed as

Heat Transfer Electrical current flow

Rate of heat transfer Electric current

Thermal resistance Electrical resistance

Temperature difference Voltage difference

57

(3-6) 1 2

eR

V VI


Ak

LR

58

T

X

1

2

T1

T2

dx

dT k, A

q(x) q(x)

T2

Q Q T1 T2

Thermal Resistance Method (Example 2.2)

AAAA CBA

59

Example 2.2 - Multi-Layer Wall: Determine Q.

Q(x)

T

X

A

T1

T3

kC, AC

T2

T4

kB, AB kA, AA

C

B

Q Q T2 T3 T1 T4 RA RB RC


C

C

B

B

A

Ax

TTAk

x

TTAk

x

TTAkQ

342312

Ak

xR

Ak

xR

Ak

xR

C

CC

B

BB

A

AA

;;

60



ceresisthermal

differencepotentialthermal

R

T

Ak

x

Ak

x

Ak

x

TTQ

overall

C

C

B

B

A

A

tan

41

61


Generalized Thermal Resistance Network

The thermal resistance concept can be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements.

The total heat transfer of two parallel layers

1 2 1 21 2 1 21 2 1 2

1 1T T T TQ Q Q T T

R R R R

62

(3-29) 1

totalR

1 2

1 2 1 2

1 1 1 = total

total

R RR

R R R R R

(3-31)

Combined Series-Parallel Arrangement

The total rate of heat transfer through

the composite system

where

31 21 2 3

1 1 2 2 3 3 3

1 ; ; ; conv

LL LR R R R

k A k A k A hA

63

(3-32) 1

total

T TQ

R

1 212 3 3

1 2

total conv convR R

R R R R R RR R

(3-33)

(3-34)

Thermal Resistance Concept- Conduction Resistance

Equation 33 for heat conduction through a plane wall can be rearranged as

Where Rwall is the conduction resistance expressed as

64

(3-4) 1 2

, (W)cond wallwall

T TQ

R

(3-5) ( C/W)wallL

RkA

Thermal Resistance Concept- Convection Resistance

Thermal resistance can also be applied to convection processes.

Newtons law of cooling for convection heat transfer rate ( ) can be rearranged as

Rconv is the convection resistance-

It is the thermal resistance of the

surface Against heat convection

conv s sQ hA T T

65

(W)sconvconv

T TQ

R

1 ( C/W)conv

s

RhA

Thermal Resistance Concept- Radiation Resistance

The rate of radiation heat transfer between a surface and the surrounding

2 2 2 (W/m K)( )

radrad s surr s surr

s s surr

Qh T T T T

A T T

66

4 4 ( ) (W)s surrrad s s surr rad s s surrrad

T TQ A T T h A T T

R

1 ( /W)rad

rad s

R Kh A

Radiation Resistance- It is the thermal resistance Of a surface against radiation

Thermal Resistance Concept- Radiation and Convection Resistance

A surface exposed to the surrounding might involves convection and radiation simultaneously.

The convection and radiation resistances are parallel to each other.

When TsurrT, the radiation

effect can properly be

accounted for by replacing h

in the convection resistance

relation by

hcombined = hconv+hrad (W/m2K)

67

Thermal Resistance Network

consider steady one-dimensional heat transfer through a plane wall that is exposed to convection on both sides.

Under steady conditions we have

or

68

Rate of

heat convection

into the wall

Rate of

heat conduction

through the wall

Rate of

heat convection

from the wall = =

1 ,1 1

1 22 2 ,2

Q h A T T

T TkA h A T T

L

Thermal Contact Resistance

In reality surfaces have some roughness.

When two surfaces are pressed against each other, the peaks form good material contact but the valleys form voids filled with air.

As a result, an interface contains

numerous air gaps of varying sizes

that act as insulation because of the

low thermal conductivity of air.

Thus, an interface offers some

resistance to heat transfer, which

is termed the thermal contact

resistance, Rc.

69

Multilayer Plane Walls

In practice we often encounter plane walls that consist of several layers of different materials.

The rate of steady heat transfer through this two-layer composite wall can be expressed through where the total thermal resistance is

,1 ,1 ,2 ,2

1 2

1 1 2 2

1 1

total conv wall wall convR R R R R

L L

h A k A k A h A

70


AAAA CBA

71

T

X

A

T1

T3

k3

T2

T4

k2

C B

k1

L1 L2 L3

Tf1

Tf4

Hot

Fluid

Tf1, h1

Cold

Fluid

Tf4, h4

T2 T3 T1 RA RB RC Tf1 Tf4 T4 Rf1 Rf4

Example 2.3 - Composite Wall with convection

surface conditions. Determine Q.


43211

4

44

3

3

43

2

2

32

1

1

21

1

11

11

ff R

f

RRRR

f

x

Ah

TT

Ak

L

TT

Ak

L

TT

Ak

L

TT

Ah

TTQ

R

TT

AhAk

L

Ak

L

Ak

L

Ah

TTQ

ffff

x

41

43

3

2

2

1

1

1

41

11

72

T2 T3 T1 RA RB RC Tf1 Tf4 T4 Rf1 Rf4


However, with composite systems it is often convenient to express the rate of heat transfer in terms of overall heat transfer coefficient (U).

This is defined by an expression similar to Newtons Law of Cooling:

43

3

2

2

1

1

1

11

11

hk

L

k

L

k

L

h

ARU

73

TAUQ x


74

Example 2.4 Combined Heat Transfer: conduction,

convection, and radiation

take place simultaneously

on boiler tubes.

The hot gases of combustion products create

a thin film of gas on the

outer wall of the boiler tube

and water film within it.

Determine the total resistance (R):


75

There are parallel circuits in the gas film section due to both convection and radiation acting there.

T

X

T1

T3

T2

T4

L

Hot gas Water inside tube

Ra

dia

tio

n

Gas Film Water Film

Tube wall

T3 T4 T2 R2 R3, conv

R1,conv

T1

R1,rad


3

43

2

32

1

21

R

TT

R

TT

R

TT

A

Qq

76

T3 T4 T2 R2 R3

R1,conv

T1

R1,rad


21

4

2

4

121

,1

4

2

4

121

,1

21

1

TT

TTF

R

TTFR

TT

rad

FactorShaperad

77

T3 T4 T2 R2 R3

R1,conv

T1

R1,rad

The radiation thermal resistance in the gas R1,rad is given by:


AhR conv

1

,1

1

kA

LR 2

78

The convection thermal resistance in the hot gas R1,conv is given by:

The conduction thermal resistance in the wall R2 is given by:

T3 T4 T2 R2 R3

R1,conv

T1

R1,rad


AhR

4

3

1

79

The convection thermal resistance in the water (boiler) or ambient air (furnace) R3 is given by:

T3 T4 T2 R2 R3

R1,conv

T1

R1,rad


AhkA

LAh

TT

TTF

RRRR

RRRR

convrad

4

1

1

21

4

2

4

121

32

1

,1,1

321

1

11

80

The total resistance is:

T3 T4 T2 R2 R3

R1,conv

T1

R1,rad

Radial Systems

Cylindrical and spherical systems often experience temperature gradients in the radial direction only and in the case can be treated as one-dimensional.

81

r

r

1-D, SS Heat Conduction (Cylindrical Coordinates)

Steady State Condition:

1-D Conduction:

Therefore:

0

t

T

0,02

2

2

2

z

TT

q

82

k

q

z

TT

rr

Tr

rrt

T

2

2

2

2

2

111

q


The equation for 1-D, steady state heat conduction then becomes:

83

rk

q

dr

dTr

dr

d

k

q

r

Tr

rr

0

1


For constant heat generation:

Solve for T(r), by integrating twice.

0gq

2120

10

ln4

2

CrCrk

gT

r

Cr

k

g

dr

dT

84


Clearly, two boundary conditions are required to determine C1 and C2. Typically one of these will be the boundary condition at the outer surface.

The other boundary condition will be at the center (r=0) where the temperature is symmetric, such that:

22)( rratTrT

00)0(

ratdr

dT

85


An alternative boundary condition at the center (r=0) can be obtained if the temperature is finite.

0)0( 0 ratTT

86

1-D, SS Heat Conduction (Example 2.5)

Example 2.5: Hollow Cylinder (Tube) (with convective surface conditions): Find all of the thermal resistances.

87

No heat generation

Cold Fluid, Tf2, h2

r2

r1

L

T1

T2

Hot fluid

Tf1, h1

Hot fluid

Tf1, h1

0q


Solve using the thermal resistance method:

The resistances Rf1 and

Rf2 can be found from

Newtons Law of

Cooling:

22

2

11

1

2

1

2

1

hLrR

hLrR

CylinderofAreaSurfaceOuter

f

CylinderofAreaSurfaceInterior

f

88

r2

r1

L

T1

T2

Hot fluid

Tf1, h1

Hot fluid

Tf1, h1

Cold Fluid, Tf2, h2

T2 Tf2 T1 RA Rf2 Tf1 Rf1


1

2

21

2

ln

2

1

2

1

2

2

1

2

1

2

1

rr

TTkLQ

dTkdrrL

Q

dTkdrA

Q

dr

dTkAQ

cond

T

T

r

r

cond

T

T

r

rrL

cond

cond

89

The resistance RA can be found from Fouriers Law:


A

condR

TTQ 21

1

1

2

2

ln

kL

rr

RA

90

If the heat transfer rate is constant, this can be further simplified:

Thermal

Resistance


Example 2.6: Composite Cylindrical Wall. Solve Q in terms of the overall thermal resistance (Rtot) and overall heat transfer coefficient (U).

91

Cold Fluid, Tf2, h2

L

Hot fluid

Tf1, h1

Hot fluid

Tf1, h1

0qNo heat generation


Composite layers

92 rD

rC

rB

rA

TA

TB

TC

TD

C B A


Use the thermal resistance method:

21

21

2

1

2

ln

2

ln

2

ln

2

1

LhrLk

rr

Lk

rr

Lk

rr

Lhr

TTQ

DC

C

D

B

B

C

A

A

B

A

ff

r

93

rD

rC

rB

rA TA TB TC TD

TB TC TA RA RB RC Tf1 Tf4 TD Rf1 Rf4

T

r

TA TB

TC TD

Tf1

Tf4


Now express this in terms of U and Rtot

Where:

)( 21421121

ffDffA

tot

ff

r TTAUTTAUR

TTQ

11 2

1

2

ln

2

ln

2

ln

2

1

LhrLkLkLkLhrR

DC

r

r

B

r

r

A

r

r

A

totC

D

B

C

A

B

94


The overall heat transfer coefficient (based on the inner surface) where AA= 2rAL, is given by:

Or we could calculate U4 (based on the outer surface) where AD= 2rDL:

1

21

4

1lnlnln

1

hk

r

k

r

k

r

hr

rU

C

D

B

C

A

B

r

r

C

D

r

r

B

D

r

r

A

D

A

D

1

21

1

1lnlnln

1

hr

r

k

r

k

r

k

r

hU

D

A

r

r

C

A

r

r

B

A

r

r

A

A

C

D

B

C

A

B

95


The definition is arbitrary, the overall heat transfer coefficient may also be defined on any of the intermediate areas:

14321

totDCBA RAUAUAUAU

214

211

ffDr

ffAr

TTAUQ

TTAUQ

96


Check

21

111

1lnlnln

1

2

hr

r

k

r

k

r

k

r

h

LrAU

D

Ar

r

C

Ar

r

B

Ar

r

A

A

C

D

B

c

A

B

21

1lnlnln

1

2

1

1

hr

r

k

r

k

r

k

r

hLr D

Ar

r

C

Ar

r

B

Ar

r

A

A

AC

D

B

c

A

B

97


98

tot

DC

r

r

B

r

r

A

r

r

A

A

R

LhrLkLkLkhLr

AU

C

D

B

c

A

B

1

2

1

2

ln

2

ln

2

ln

2

1

1

21

1

Correct !


Example 2.7: Solid Cylinder Develop an expression for a 1-D, radial, steady state temperature

distribution T(r) and the flux q(r) for a solid cylinder of radius (r2) with an energy generation at a constant rate of q (W/m3) and temperature on the outer surface maintained at T2. Calculate the temperature at the center and the flux at the outer surface for r2= 1 cm, q= 2108 W/m3, k= 20 W/(mC), and T2= 100 C.

99

r2

T2= 100 C


Boundary Conditions

The general conduction equation for a cylinder is:

k

q

z

TT

rr

Tr

rrt

T

2

2

2

2

2

111

q

22;

0;0

TTrrat

dr

dTrat

100

r2

T2= 100 C

, steady state , 1-D , 1-D


21

2

1

)ln(4

)(

2

CrCk

rqrT

r

C

k

rq

r

T

drrk

q

r

Tr

01

k

q

r

Tr

rr

101

Integrating twice:


102

Applying boundary conditions at r= 0 gives:

Applying boundary conditions at r= r2 gives:

0

02

0,0

1

1

C

r

Cr

k

q

dr

dT

dr

dTrat

0

2

222

21

2

222

4

)ln(4

;

rk

qTC

CrCrk

qTTrrat

0


2

2

2

2

2

2

22

2

21

2

14

44

ln4

Tr

r

k

rq

k

rqTr

k

q

CrCrk

qT

103

The temperature distribution in the solid cylinder is:


22

rq

k

rqk

dr

dTkq

CC

m

Tr

r

k

rqT

Cm

W

m

W

350100204

01.0102

14

0

3

8

2

2

2

2

2

104

The heat flux in the cylinder is thus:

The temperature T(0) at r= 0 is:


2

3 6

8

22 10

2

01.0102

2)(

m

Wm

W mrqrq

105

The heat flux at the outer surface of the cylinder is:


Example 2.8: Hollow Cylinder A hollow cylinder is heated at the inner side at the rate of q0

(105W/m2) and dissipates heat from the outer surface into a fluid at Tf2. There is no energy generation and the conductivity (k) of the solid is assumed to be constant. Develop an expression for the temperature T1 and T2 at (the inner and outer surface) and calculate them for the following parameters.

106

Tf2= 100 C

h= 400 W/(m2C)

r2 =5 cm

r1 =3 cm

k= 15 W/(mC)


Since there is no heat generated in the cylinder, it is more convenient to solve the problem using the thermal resistance method.

Lk

rr

2

ln1

2

107

T2 Tf2 T1

Tf2= 100 C

h= 400 W/(m2C)

r2 =5 cm

r1 =3 cm

k= 15 W/(mC)

Lhr22

1


221

2

21

21

21

1000ln

2Lhrr

r

kL

fTTLrqAqQ

221

2

21

22

21

2110

ln2

hLr

f

r

r

LkA

TTTTLrq

108

T2 Tf2 T1

Lk

rr

2

ln1

2

Lhr22

1

Also:


Taking the first equality and solving for T1:

Taking the second equality and solving for T2:

201

1

2110

1

21

1

2

ln

ln

TqT

TTrq

r

r

k

r

r

r

k

2

22

102

22

22

10

f

f

Thr

rqT

hr

TTrq

109


Solving:

201 121 ln TqT rr

k

r

2

22

102 fT

hr

rqT

110

CCm

mm

Cm

Wm

W

2.35225003.0

05.0ln

15

03.010 2

5

CCm

m

Cm

Wm

W

25010040005.0

03.010

2

2

5

1-D, SS Heat Conduction (Spherical Coordinates)

Conduction in a spherical shell Consider heat conduction in a hollow sphere. In a steady state, one

dimensional system (without heat generation), the energy entering the differential control volume is equal to the energy leaving the differential control volume.

111

y x

z

dr

dTrk

dr

dTkAQ

QQ

A

r

drrr

24


Separating variables:

Assuming constant k and integrating

2

1

2

1

)(

2)(

4

T

T

TfToffunction

r

r

r dTTkr

drQ

12211

4TTk

r

Q rr

r

112


21

11

4

1

rrkR

2112

21

11

12 44

21

TTrr

rrk

TTkQ

rr

r

R

TTQr

12

113

where:


The equation for the rate of heat transfer can also be done by simplifying the heat equation for spherical coordinates, which is recalling:

k

qT

r

T

rr

Tr

rrt

T

2

2

2222

2

2 sin

1sin

sin

111

fqqq

qq

114

0, steady

state 0, 1D 0, 1D

0, no heat

generation

02

r

Tr

r


Integrate twice:

21

2

1

Cr

CT

drr

CdT

115

1

2

2

0

0

Cdrr

Tr

r

Tr

r


Apply boundary conditions:

At r = r1 T = T1 At r = r2 T = T2

1211

2

1

111 :

TCrC

Cr

CTrrat

1212

211

12

11222

2

221212

2

122

;

:

TTrr

rrC

rr

TrTrC

r

rCrCTC

r

CTrrat

116


212

121

12

21

12

2122

12

1211

12

1122

12

22

12

21

12

112221

12

12

21

Trrr

rrrT

rrr

rrr

rrr

rrrrT

rrr

rrrrT

rr

TrTr

rrr

rrT

rrr

rrT

rr

TrTrTT

rrr

rr

Cr

CT

117


1212

12

12

12

12

2

2

2

12

2

4

4

4

4

TTrr

rrk

TTrr

rr

r

kr

r

Ckr

dr

dTkrqAQ rr

118


Spherical composites may be treated in much the same way as composite walls and cylinders, where appropriate forms of the total resistance (R) and overall heat transfer coefficient (U) may be determined.

119

y x

z

120

Critical Thickness of Insulation

Consider a tube, cable, or wire dissipating heat from the outer surface into the surrounding air by convection.

It is covered by a layer of insulation to minimize heat loss. In

many cases, the thermal resistance offered by a metal tube or wire is negligibly small in comparison to the insulation.


Consider a tube, cable, or wire dissipating heat from the outer surface into the surrounding air by convection.

It is covered by a layer of insulation to minimize heat loss. In

many cases, the thermal resistance offered by a metal tube or wire is negligibly small in comparison to the insulation.

121

122 122


The tube wall temperature (To) is nearly the same as the fluid.

insulation

insulation

ri

ro

ri

ro

(a) Rod or Wire (b) Pipe

To, ho

Ti Ti

123 123


critoo

critoo

o

crito

rr

rrif

h

kr

,

,

,

Critical radius of insulation

Will decrease the rate of heat loss

expected. Good !

Will increase the heat loss

continuously. Maximum at the critical

thickness. Avoid !

Finite Differences (Numerical Methods)

m x increment

n y increment

124

x

y

b

x

y

Node


In the absence of a heat source or sink in the system, the rate of heat flow toward the nodal point must be equal to the rate of heat flow from it in steady state.

125

m, n

m,n+1

m,n-1

m+1,n m-1,n

Qm,n+1

Qm,n-1

Qm-1,n Qm+1,n


In the finite difference method the derivatives are replaced by differences.

Instead of taking the limit, the following approximation for the derivative can be used.

x

xfxxf

dx

xdf

x 0lim

x

xfxxf

dx

xdf

126

f(x+Dx) f(x)

x x+dx

Dx

Df


If the grid is subdivided into M sections of equal length.

directionxthein

M

Lx

127

Tm+1 Tm

Tm-1

m-1 m m+1

m- m+


x

TT

dx

dT

x

TT

dx

dT

mm

m

mm

m

1

2

1

1

2

1

128

Tm-1 Tm

m-1/2

Tm Tm+1

m+1/2


The 2nd derivative is simply:

2

11

2

2

2

11

21

21

x

TTT

x

xdx

Td

mmm

x

TT

x

TT

mdxdT

mdxdT

mmmm

129


Likewise:

2

11

2

2 2

y

TTT

dy

Td nnn

130

Finite Differences

Finite Differences of Plane Wall: The 1-D heat transfer through a plane wall is given by the following equation. Find the finite difference expression for:

This can be expressed in differential form as:

Where qm is the rate of heat generation per unit volume at node m.

02

2

k

q

dx

Td

131

3,2,102

2

11

mfork

q

x

TTT mmmm

Finite Differences (Example 2.12)

For 2-dimensions:

The finite difference formulation

is:

13,2,1

13,2,1

22 ,2

1,,1,

2

,1,,1

Nnfor

Mmfor

k

q

y

TTT

x

TTT nmnmnmnmnmnmnm

02

2

2

2

k

q

y

T

x

T

132

(m, n+1)

(m, n-1)

(m+1, n) (m, n) (m-1, n)

Dy Dx


If x = y then:

Or since we are considering that k= constant, the heat flows may all be expressed in terms of temperature differentials and this same equation can be derived.

k

xqTTTTT

nm

nmnmnmnmnm

2

,

,1,1,,1,1 4

133

1

1

xAwheredy

dTkAQ

yAwheredx

dTkAQ

yyy

xxx


Therefore the finite difference expressions for Q are:

y

TTxkQQ

y

TTxkQQ

x

TTykQQ

x

TTykQQ

nmnm

nmdowncond

nmnm

nmupcond

nmnm

nmrightcond

nmnm

nmleftcond

,1,

1,,

,1,

1,,

,,1

,1,

,,1

,1,

134


Therefore the total heat transfer is:

Therefore if x = y:

0,1,1,,1,1

yx

nmnmnmnmnm AqQQQQ

135

yxq

y

TTx

y

TTx

x

TTy

x

TTy

k nmnmnmnmnm

nmnmnmnm

,

,1,,1,

,,1,,1


Then:

k

xqTTTTT

nm

nmnmnmnmnm

2

,

,1,1,,1,1 4

136

Finite Differences

To use this numerical method, these equations must be written for each node within the material and the resultant system of equations solved for the temperature at the various nodes.

137


Example 2.13: Finite Difference Modeling of a square plate. A small plate (1x1 m) and with a k= 10 W/(mC) has one face maintained at 500C and the rest at 100C.

Compute: (i) Temperature at various nodes.

(ii) Heat flow at the boundaries.

138

1 m

1 m

500C

100C

k

100C

100C


my3

1

mx3

1

139

1

3

2

4

T=500C

T=100C

T=100C T=100C

Four

node

problem


(i) The solution for finding the temperatures is (for

an interior node):

04100100:4

04100100:3

04500100:2

04500100:1

423

314

241

132

TCTTCNode

TCTCTNode

TTCTCNode

TTCCTNode

04 ,1,1,.1,1 nmnmnmnmnm TTTTT

140

1

3

2

4


Rearranging equations:

04200

04200

04600

04600

432

431

421

321

TTT

TTT

TTT

TTT

141


142

-4 1 1 0 T1 -600

1 -4 0 1 T2 -600

1 0 -4 1 T3 -200

0 1 1 -4 T4 -200

=


Solve by Gaussian Elimination:

143

T1 T2 T3 T4 C

-4 1 1 0 -600

1 -4 0 1 -600

1 0 -4 1 -200

0 1 1 -4 -200


144

T1 T2 T3 T4 C

-4 1 1 0 -600

4x1= -4x4= 4x0= 4x1= -600x4=

4 -16 0 4 -2400

1 0 -4 1 -200

0 1 1 -4 -200

X4


145

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -16 0 4 -2,400

1 0 -4 1 -200

0 1 1 -4 -200

-4+(4)=0


146

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 0 4 -2,400

1 0 -4 1 -200

0 1 1 -4 -200

1+(-16)=-15


147

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -2,400

1 0 -4 1 -200

0 1 1 -4 -200

1+(0)=1


148

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -2,400

1 0 -4 1 -200

0 1 1 -4 -200

0+(4)=4


149

-600+(-2400)=-3000

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -3,000

1 0 -4 1 -200

0 1 1 -4 -200


150

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -3,000

0 1 -15 4 -1,400

0 1 1 -4 -200

X4


151

X15

-15+(1x15)=0 T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -3,000

0 0 -224 64 -24,000

0 1 1 -4 -200


152

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -3,000

0 0 -224 64 -24,000

0 0 16 -56 -6,000 X15


153

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -3,000

0 0 -224 64 -24,000

0 0 0 -720 -108,000 X =14 224

16

-224+(14x16)=0


154

T1 T2 T3 T4 C

-4 1 1 0 -600

0 -15 1 4 -3,000

0 0 -224 64 -24,000

0 0 0 -720 -108,000


000,108720

000,2464224

000,3415

6004

4

43

432

321

T

TT

TTT

TTT

155

CT 150720

000,1084

Solving for the unknowns


CT

CT

CT

2504

150250600

25015

000,34150150

150224

64150000,24

1

2

3

156


(ii) The heat rate is thus:

mW

x TTx

ykQ

000,4

1005002

110015010025010

1001002

1100500

2

1100100 310

157

1

3

2

4

-Qx=0

y

TxkQ

x

TykQ

y

x



mW

x TTx

ykQ

000,4

1005002

110015010025010

1001002

1100500

2

1100100 421

158

1

3

2

4

-Qx=1



mW

y TTy

xkQ

000,1

10015010015010

1001002

1100100

2

1100100 430

159

1

3

2

4

-Qy=0



mW

y TTy

xkQ

000,9

20020050025050025010

5001002

1500100

2

1500500 211

160

1

3

2

4

+Qy=1


Therefore:

Heat flowing into the plate = +9,000 W/m

Heat flow leaving the plate = -4000-4000-1000=-9,000 W/m 161

1

3

2

4

+9,000 W/m

-4,000 W/m -4,000 W/m

-1,000 W/m


Example 2.14: Derive the heat equation for node 3 of the plate shown below.

162

1

3

5

2

4

6

ins

ula

tio

n

Given:

k= constant

b= thickness

x= y

Steady state


Also note the half areas:

y

163

1

3

5

4

2

y

2

x


Since the heat transfer is steady state then Q=0 and the equation at node 3 is:

Note: because of the insulation

1,

21

1,

21

,1

353134

220

nm

x

nm

x

nm

y

Q

A

Q

AQ

Ay

TTbxk

y

TTbxk

x

TTbyk

164

0,1 nmQ


Example 2.15: Steady 2-D Heat Conduction in an L-bar. Given: k = 15 W/(mC) h= 80 W/(m2C) T= 25C q= gn= 2x10

6 W/m3

165

ins

ula

tio

n

T= 90 C

Convection

h, T= 25C

qr= 5000 W/m2

1 2 3

4 5 6 7 8 9

10 11 12 13 14 15

x= y= L = 0.012m

Dx Dy


Assumptions:

Heat transfer is steady and 2-D

Thermal conductivity (k) is constant

Heat generation q is constant

Radiation heat transfer is neglible

Form the volume elements by partitioning the region between nodes. Node 5 is the only completely interior node. Consider the volume element represented by Node 5 to be full size (e.g. x=y=1).

166


Then the elements represented by a regular boundary node (i.e. Node 2) becomes half size (e.g. x=y/2=1) and a corner node (i.e. Node 1) is quarter size (e.g. x/2=y/2=1) .

167

ins

ula

tio

n

T= 90 C

Convection

h, T= 25C

qr= 5000 W/m2

1 2 3

4 5 6 7 8 9

10 11 12 13 14 15

x y


Since the bottom surface is at a constant temperature of 90 C, then:

CTTTTTT 90151413121110 168

insu

lati

on

T= 90 C

Convection

h, T= 25C

qr= 5000 W/m2

1 2 3

4 5 6 7 8 9

10 11 12 13 14 15

x= y= L

x y


Node 1 (Energy balance): Insulated on the left

Convection on top

Conduction on right and bottom

Cm

W

Cm

W

Cm

W

Cm

W

Cm

WmC

TTTm

152

012.0102

15

2580

15

012.0802

26

421

22

169

1 2

4 5

ins

ula

tio

n

Convection

h, T= 25C

Dy ______

2

Dx/2

2.11064.2 421 TTT

Lyx

222220 1

14121

yxg

y

TTxk

x

TTykTT

xh


Node 2 (Energy balance): Convection on top

Conduction right, left, bottom

170

2 3

5 6

Dy

Dx

1

4

Convection

h, T= 25C

22

20

22125

232

yxg

x

TTyk

y

TTxk

x

TTykTTxh

225321

22

24 L

k

gT

k

hLTTT

k

hLT

4.222128.4 5321 TTTT


Node 3 (Energy balance): Convection on top and right

Conduction at bottom and left

0

222

222

332

363

yxg

x

TTyk

y

TTxkTT

yxh

171

2 3

5 6

Dy

Dx

Convection

h, T= 25C

k

LgT

k

hLTT

k

hLT

2

222 3632

8.12128.2 632 TTT


Node 4 (Energy balance): Insulated on left

Conduction at the top, right, bottom This node is on the insulated boundary and can

be treated as an interior node by replacing the insulation with a mirror. This puts a reflected image of node 5 to the left of node 4.

042

4410515

k

LgTTTTT

Interior

172

4 5

10 insu

lati

on

T= 90 C

11

1 2

Dy

Dx

2.1099024

2

4541

10

k

LgTTT

T

5


Node 5 (Energy balance): Interior node

Conduction all sides

Can use the equation for an interior node

2.109904

04

2

55624

2

5511624

11

k

LgTTTT

k

LgTTTTT

T

173

ins

ula

tio

n

1 2 3

4 5 6

10 11 12

Dy

T= 90 C


Node 6 (Energy balance): Convection upward right corner

Conduction everywhere else

174 T= 90 C

2 3

5 6 7

11 12 13

Dx

Dy

Convection

h, T= 25C

6 Qcond

Qcond

Qcond

Qconv

Qconv

Qcond


04

3

2

222

663

65

612676

yxgy

TTxkTT

x

yk

y

TTxk

x

TTykTT

yxh

175 T= 90 C

2 3

5 6 7

11 12 13

Dx

Dy

Convection

h, T= 25C

of the internal

energy generation,

since only the

volume

0.212128.62 7653 TTTT


Node 7 (Energy balance): Convection on top

Conduction right, left, and bottom

176

T= 90 C

6 7 8

12 13 14

x

y

Convection

h, T= 25C

0

2

2

776713

787

yxg

x

TTyk

y

TTxk

x

TTykTTxh

4.202128.4

2180

24

876

2

7876

TTT

k

LgT

k

hLTT

k

hLT


Node 8 (Energy balance): Identical to Node 7

k

LgT

k

hLTT

k

hLT

2

8987

2180

24

177

T= 90 C

7 8 9

13 14 15

Dy

Convection

h, T= 25C

4.202128.4 987 TTT

Dx


Node 9 (Energy balance): qr heat flow on right

Convection on top

Conduction on bottom and left

0222

222

998

9159

yxg

x

TTyk

y

TTxk

yqTT

xh R

178

qr= 5000

W/m2 8 9

14 15

Dx

Dy

Convection

h, T= 25C

T= 90 C

2.105064.2

2902

98

2

998

TT

k

LgT

k

hLL

k

qT

k

hLT R


We now have 9 equations and 9 unknowns, so we can solve:

179

2.11064.2 421 TTT

4.222128.4 5321 TTTT

8.12128.2 632 TTT

2.10924 541 TTT

2.1094 5624 TTTT

0.212128.62 7653 TTTT

4.202128.4 876 TTT

4.202128.4 987 TTT

2.105064.2 98 TT

Node 1:

Node 2:

Node 3:

Node 4:

Node 5:

Node 6:

Node 7:

Node 8:

Node 9:


Solving:

T1= 112.1 C

T2= 110.8 C

T3= 106.6 C

T4= 109.4 C

T5= 108.1 C

T6= 103.2 C

T7= 97.3 C

T8= 96.3 C

T9= 97.6 C

180


181

insu

lati

on

T= 90 C

Convection

h, T= 25C

qr= 5000 W/m2

1 2 3

4 5 6 7 8 9

10 11 12 13 14 15

Temperature

(C)

Hi

Low

END OF CONDUCTION SECTION

182

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CH02 Conduction