Chapter 3 Part 2

One-Dimensional, Steady-State Conduction

Plane Wall with Energy Generation

x

Cold air

Ts

L

hk

Cold air

Ts

T

T0

q

"condq

"convq "

convq

hT

The heat equation for a plane wall with constant uniform energy generation per unit volume and symmetrical boundary conditions can be developed by performing an energy balance around an element of volume ΔV as illustrated below

q

x

Δx

Acc= E In - E Out + E Generated

E In =xdx

dTkA

E Out = xxdx

dTkA

E Generated = xAqVq

0

xAqdxdTkA

dxdTkA

xxx

0

xqdxdTk

dxdTk

xxx

0

q

xdxdTk

dxdTk

xxx

0lim0

qx

dxdTk

dxdTk

xxx

x

02

2

qdx

Tdk

integrate by separation of variables

dx

kq

dxdTd

cxkq

dxdT

at x =0 we know that due to the symmetry so c = 00dxdT

xkq

dxdT

integrate again

'2

2 cxk

qT

xdxkqdT

To determine the constant c’ we can use the boundary condition at x=L, which tells us that the amount of energy generated in half the plane wall as to be transfer to the air since there is no accumulation of energy

TThA LALq

hLqTTL

TThAALq L

'2

2 cLk

qhLqT

2

2' L

kq

hLqTc

22

22L

kq

hLqTx

kqT

If TL is explicitly known than

2

2' L

kqTc L

22

22L

kqTx

kqT L

LTLx

kLqT

2

22

12

Asymmetrical Boundary Conditions

x

Cold air

Ts,2

L

h2

k

Warmer air

Ts,1

2,T

q

"1,convq "

2,convq

h11,T

The same differential equation applies

but the boundary conditions have changed. The general solution of the differential equation is

where c1 and c2 are the constants of integration

if the temperature Ts,1 and Ts,2 are known than

T(-L)= Ts,1 and T(L)= Ts,2

the constant may be evaluated and are of the form

212

2cxcx

kqT

02

2

qdx

Tdk

LTT

c ss

21,2,

1

22

1,2,2

2ss TT

kLqc

In which case the temperature distribution is

and the heat flux at any position x can be determined using

If the surface temperatures are not explicitly known we can use the boundary conditions

and

to determine the integration constants

22

12

1,2,1,2,2

22ssss TT

LxTT

Lx

kLqxT

dx

xdTkxq "

2,2,2

TThdxdTk s

Lx

1,1,2

TThdxdTk s

Lx

Heat Generation in a Wall With One Face Insulated

x

Cold air

Ts

L

h

k

T

T0

q

"condq

"convq

Insu

late

d Su

rfac

e

Solution is as per symmetrical boundary conditions

Example

x

Water

Ts,1

LA

h=1000W m-2 K-1

CT o30

T0

"q

Insu

late

d Su

rfac

e

Ts,2A

B

LB

A plane wall is a composite of two materials, A and B. The wall of material A has uniform heat generation , kA=75 W m-1 K-1 and thickness LA = 500 mm. The inner surface of the material A is well insulated, while the outer surface of material B is cooled by a water stream with and h=1000W m-2 K-1. Sketch the temperature distributuion that exists in the composite under steady-state condition

36105.1 mWq

CT o30

Assumptions:One dimensional wall system Constant conductivity for A and BSteady-StateInner surface A adiabaticNegligible contact resistance

T(x)

xLA LA +LB

Ts,1

T

Ts,2

Determine the temperature T0 of the insulated surface and the temperature Ts,2 of the cooled surface.

Steady State implies that the energy generated per unit area in section A is equal to the heat flux in section B and the heat flux to the water so

TTh

LTT

kLq sB

ssbA 2,2

1,2,

CCKmW

mmWT

hLqT ooA

s 105301000

05.0105.112

36

22,

CCKmW

mmWmT

kLqLT oo

sb

ABs 115105

15005.0105.1

020.0 11

36

2,1,

LTLx

kLqT

2

22

12

In section A

CCKmW

mmWT

kLqxT oo

sA

A 140115150

02.0105.12

0 11

36

1,

2

Hence

Radial System with Energy Generation

Cold Fluid

hT

rr0

q

The heat equation for a cylinder with constant uniform energy generation per unit volume can be developed by performing an energy balance around an element of volume ΔV as illustrated below

q

Acc= E In - E Out + E Generated

E In =rdr

dTrLk2

E Out =

E Generated = rLrqVq 22

rr r

rrdrdTrLk

2

0222 2

rLrqdrdTrLk

drdTrLk

rrr

022

qrr

drdTrLk

drdTkr

rrr

022

lim0

q

rrdrdTrLk

drdTkr

rrr

r

01

q

drdTkr

drd

r

Boundaries conditions :

at r = 0 symmetry

T = Tse at r = r1

d Td r

0

2214

rrkqTT se

Extended SurfacesAim: Increase heat transfer rateMode: Increase surface area

General Conduction Analysis

x xS

q3

A

q1

q2

Assumptions: Steady StateTemperature gradient in fin only in x directionNo energy productionNo radiation

Energy Balance

Accumulation = Energy In – Energy Out + Energy Produced

Energy in = q1

Energy Out = q2 + q3

k Ad Td x

k Ad Td x

h S T Tx x x

0

Divide by and take the limit to 0

0

TT

dxdSh

xdTdkA

dxd

If A and S are independent of x (uniform cross-section)

02

2

TThPdx

TdkA

SolutionBoundary Conditions Temperature Distribution

LxatTThxdTdk

00 xatTT

00 xatTT

00 xatTT

LxatxdTd

0

LxatTT L

o

m L x hm k m L x

m L hm k m L

cosh sinh

cosh sinh

Lm

xLm

o coshcosh

Lm

xLmsmo

L

o sinh

sinhsinh

TT

TT00 kAhPm 2

Heat Transfer RateBoundary Conditions q

LxatTThxdTdk

00 xatTT

00 xatTT

00 xatTT

LxatxdTd

0

LxatTT L

TT

TT00 kAhPm 2

LmkmhLm

LmkmhLm

Mqsinhcosh

coshsinh

AhPkM o

]tanh[mLMq

Lm

LmMq

L

sinh

cosh0

Graphical SolutionFin Efficiency

maxqq fin

f

Ratio of the actual heat transfer rate of the fin over the heat transfer rate if the entire fin temperature was equal to the base temperature.

The fin efficiency is given as graphical function of (figure 3.18 and 3.19)

the fin geometrya characteristic lengtha characteristic areathe conductionthe convective coefficient

When can we assume that the temperature gradient in fin only in x direction:

if we consider a rectangular fin

only if

T TT T

c s

s

01.

y

x Tc

Ts

T

E/2

Energy balance on a slice of the fin

Ts

T

TThyTk s

Ey2

22E

TTkyTk sc

Ey

T TT T

E hk

c s

s

2

E hk2 01 .assumption is valid if Biot Number