Objectives of conduction analysis

To determine the temperature field, T(x,y,z,t), in a body(i.e. how temperature varies with position within the body) T(x,y,z,t) depends on:

- boundary conditions - initial condition

- material properties (k, cp, …) - geometry of the body (shape, size)

Why we need T(x,y,z,t) ? - to compute heat flux at any location (using Fourier’s eqn.) - compute thermal stresses, expansion, deflection due to temp. etc. - design insulation thickness - chip temperature calculation - heat treatment of metals

T(x,y,z)

0Area =

A xx x+x

= Internal heat generation per unit vol. (W/m3)

qx qx+xA

Solid bar, insulated on all long sides (1D heat conduction)

Unidirectional heat conduction (1D)

Unidirectional heat conduction (1D)

First Law (energy balance)

t

EqxAqq xxx

)(

stgenoutin EEEE )(

t

TxcA

t

uxA

t

E

uxAE

)(

xx

qqq

x

TkAq

xxxx

x

If k is a constantt

T

t

T

k

c

k

q

x

T

1

2

2

Unidirectional heat conduction (1D)(contd…)

t

T cq

x

Tk

x

t

TxAcqxAx

x

Tk

xA

x

TkA

x

TkA

Longitudinal conduction

Thermal inertiaInternal heat generation

Unidirectional heat conduction (1D)(contd…)

For T to rise, LHS must be positive (heat input is positive)

For a fixed heat input, T rises faster for higher In this special case, heat flow is 1D. If sides were not

insulated, heat flow could be 2D, 3D.

Boundary and Initial conditions:

The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body.

We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).

Boundary and Initial conditions (contd…)

How many BC’s and IC’s ?

- Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate.

* 1D problem: 2 BC in x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction

* 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir.

- Heat equation is first order in time. Hence one IC needed

1- Dimensional Heat Conduction

The Plane Wall :

.. .. . . … . . . . . . .. . . . . . . . . . .

. . . . . .

.. . . . . . . . . . .. . . . . . . . . . . . . .

…. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . .

. .

k

T∞,2

Ts,1Ts,2

x=0 x=LHot fluid

Cold fluid

0

dx

dTk

dx

d

kAL

TTTT

L

kA

dx

dTkAq

ssssx

/

2,1,2,1,

Const. K; solution is:

Thermal resistance(electrical analogy)

OHM’s LAW :Flow of Electricity

V=IR elect

Voltage Drop = Current flow×Resistance

Thermal Analogy to Ohm’s Law :

thermqRT

Temp Drop=Heat Flow×Resistance

1 D Heat Conduction through a Plane Wall

qx

T∞,1 Ts,1 Ts,2 T∞,2

Ak

LAh2

1

Ah1

1

.. .. . . … . . . . . . .. . . . . . . . . . .

. . . . . .

.. . . . . . . . . . .. . . . . . . . . . . . . .

…. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . .

. .

k

T∞,2

Ts,1Ts,2

x=0 x=LHot fluid

Cold fluid

T∞,1

AhkA

L

AhRt

21

11(Thermal Resistance )

Resistance expressions

THERMAL RESISTANCES

Conduction Rcond = x/kA

Convection Rconv = (hA)-1

Fins Rfin = (h

Radiation(aprox) Rrad = [4AF(T1T2)

1.5]-1

Composite Walls :

A B CT∞,1

T∞,2

h1

h2K A K B K C

L A L B L C

q x

T∞,1 T∞,2

Ah1

1Ah2

1Ak

L

A

A

Ak

L

B

B

Ak

L

C

C

TUA

Ahk

L

kL

kL

Ah

TT

R

TTq

C

C

B

B

A

Atx

21

2,1,2,1,

11

AR

Uwheretot

1, Overall heat transfer coefficient

Overall Heat transfer Coefficient

21

total

h1

kL

h1

1

AR

1U

Contact Resistance :

A B

TA

TB T

xc,t

q

TR

21

111

hk

L

k

L

k

L

h

U

C

C

B

B

A

A

Series-Parallel :

AB

D

CT1 T2

K c

K DK A

K B

AB+AC=AA=AD

LB=LC

Series-Parallel (contd…)

Assumptions :

(1) Face between B and C is insulated.

(2) Uniform temperature at any face normal to X.

T1T2

Ak

L

A

A

Ak

L

D

DAk

L

B

B

Ak

L

C

C

Consider a composite plane wall as shown:

Develop an approximate solution for the rate of heat transfer through the wall.

Example:

kI = 20 W/mk

AI = 1 m2, L = 1m

kII = 10 W/mk

AII = 1 m2, L = 1m

T1 = 0°C Tf = 100°C

h = 1000 W/ m2 k

qx

1 D Conduction(Radial conduction in a composite

cylinder)h1

T∞,1

k1

r1

r2

k2T∞,2 r3

h2

T∞,1 T∞,2

)2)((

1

11 Lrh )2)((

1

22 Lrh

12

ln2

1

Lkrr

22

ln3

2

Lkrr

t

r R

TTq 1,2,

Critical Insulation Thickness :

r0ri

T∞

Tih

Insulation Thickness : r o-r i

hLrkLR ir

r

tot )2(

1

2

)ln(

0

0

Objective : decrease q , increases

Vary r0 ; as r0 increases ,first term increases, second term decreases.

totR

Critical Insulation Thickness (contd…)

Set 00

dr

dRtot

02

1

2

1

02

0

hLrLkr

h

kr 0

Max or Min. ? Take : 00

2

2

dr

Rd tot ath

kr 0

h

kr

tot

hLrLkrdr

Rd

0

02

02

02

2 1

2

1

02 3

2

Lk

h

Maximum – Minimum problem

Minimum q at r0 =(k/h)=r c r (critical radius)

good for steam pipes etc.good for electrical

cablesR c r=k/h

r0

R t o t

Critical Insulation Thickness (contd…)

1D Conduction in Sphere1D Conduction in Sphere

r1

r2

k

Ts,1

Ts,2

T∞,1

T∞,2

k

rrR

rr

TTk

dr

dTkAq

TTTrT

dr

dTkr

dr

d

r

condt

ssr

rr

rrsss

4

/1/1

/1/1

4

)(

01

21,

21

2,1,

/1

/12,1,1,

2

2

21

1

Inside Solid:

Applications: * current carrying conductors

* chemically reacting systems

* nuclear reactors

= Energy generation per unit volumeV

Eq

Conduction with Thermal Energy Generation

Conduction with Thermal Energy Generation

The Plane Wall :

k

Ts,1

Ts,2

x=0 x=+L

Hot fluid

Cold fluid

T∞,2T∞,1

x= -L

q Assumptions:

1D, steady state, constant k,

uniform q

Conduction With Thermal Energy Generation (contd…)

CxCxk

qTSolution

TTLx

TTLxcondBoundary

k

q

dx

Td

s

s

2:

,

,:.

0

21

2

2,

1,

2

2

Not linear any more

Derive the expression and show that it is not independent of x any more

Hence thermal resistance concept is not correct to use when there is internal heat generation

dxdT

kqfluxHeat

TT

LxTT

Lx

kLq

TsolutionFinal

CandCfindtoconditionsboundaryUse

x

ssss

:

221

2: 1,2,1,2,

2

22

21

Conduction with Thermal Energy Generation (cont..)

Cylinder with heat source

Assumptions:

1D, steady state, constant k, uniform

Start with 1D heat equation in cylindrical co-ordinates:

kq

drdT

rdrd

r

01

ro

r

Ts

T∞ h

q

q

rk

Trcond

04

,:.

s

s

Tr

rq

rTSolution

dr

dTr

TrBoundary

2

22

0

0

1)(:

0,0

Ts may not be known. Instead, T and h may be specified.

Exercise: Eliminate Ts, using T and h.

Cylinder With Heat Source

Cylinder with heat source(contd…)

Example:

A current of 100A is passed through a stainless steel wire having a thermal conductivity K=25W/mK, diameter 3mm, and electrical resistivity R = 2.0 . The length of the wire is 1m. The wire is submerged in a liquid at 100°C, and the heat transfer coefficient is 10W/m2K. Calculate the centre temperature of the wire at steady state condition.