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) qT(x,y,z,t) depends on:

­ boundary conditions ­ initial condition ­ material properties (k, c p , ρ …) ­ geometry of the body (shape, size)

qWhy 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)

0 Area = A x

x x+∆x

q &= Internal heat generation per unit vol. (W/m 3 )

q x q x+∆x A

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

Unidirectional heat conduction (1D)

Unidirectional heat conduction (1D)

First Law (energy balance)

t E q x A q q x x x ∂

∂ = ∆ + − ∆ + & ) (

st gen out in E E E E & & & & = + − ) (

t T

xc A t u

x A t E

u x A E

∂

∂ ∆ =

∂

∂ ∆ =

∂

∂

∆ =

ρ ρ

ρ ) (

x x q

q q

x T

kA q

x x x x

x

∆ ∂

∂ + =

∂

∂ − =

∆ +

If k is a constant t T

t T

k c

k q

x T

∂ ∂

= ∂ ∂

= + ∂ ∂

α ρ 1

2

2 &

Unidirectional heat conduction (1D)(contd…)

t T c q

x T k

x

t T x Ac q x A x

x T k

x A

x T kA

x T kA

∂ ∂

= +

∂ ∂

∂ ∂

∂ ∂

∆ = ∆ + ∆

∂ ∂

∂ ∂

+ ∂ ∂

+ ∂ ∂

−

ρ

ρ

&

&

Longitudinal conduction

Thermal inertia Internal heat generation

Unidirectional heat conduction (1D)(contd…)

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

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

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

Boundary and Initial conditions:

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

q 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 x­direction * 2D problem: 2 BC in x­direction, 2 in y­direction * 3D problem: 2 in x­dir., 2 in y­dir., and 2 in z­dir.

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

1­ Dimensional Heat Conduction The Plane Wall :

.. .. . . … . .

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

.

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

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

. .

k

T ∞,2

T s,1 T s,2

x=0 x=L Hot fluid

Cold fluid

0 =

dx dT k

dx d

( ) kA L T T T T

L kA

dx dT kA q s s

s s x /

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 :

therm qR T = ∆

Temp Drop=Heat Flow×Resistance

1 D Heat Conduction through a Plane Wall

q x

T ∞,1 T s,1 T s,2 T ∞,2

A k L

A h 2 1

A h 1 1

.. .. . . … . .

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

.

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

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

. .

k

T ∞,2

T s,1 T s,2

x=0 x=L Hot fluid

Cold fluid

T ∞,1

∑ + + = A h kA

L A h

R t 2 1

1 1 (Thermal Resistance )

Resistance expressions

THERMAL RESISTANCES

• Conduction Rcond = ∆x/kA

• Convection Rconv = (hA) ­1

• Fins Rfin = (hηΑ) −1

• Radiation(aprox) Rrad = [4AσF(T1T2) 1.5 ] ­1

Composite Walls :

A B C T ∞,1

T ∞,2

h 1 h 2 K A K B K C

L A L B L C

q x

T ∞,1 T ∞,2

A h1 1

A h2 1

A k L

A

A

A k L

B

B

A k L

C

C

T UA

A h k L

k L

k L

A h

T T R T T

q

C

C

B

B

A

A t x ∆ =

+ + + +

− =

− = ∞ ∞ ∞ ∞

∑ 2 1

2 , 1 , 2 , 1 ,

1 1

= = A R

U where tot

1 , Overall heat transfer coefficient

Overall Heat transfer Coefficient

2 1

total

h 1

k L

h 1

1 A R

1 U + Σ +

= =

Contact Resistance :

A B

T A T B T ∆

x c , t

q T R

∆ =

2 1

1 1 1

h k L

k L

k L

h

U

C

C

B

B

A

A + + + + =

Series­Parallel :

A B

D

C T 1 T 2

K c

K D K A

K B

A B +A C =A A =A D

L B =L C

Series­Parallel (contd…)

Assumptions :

(1) Face between B and C is insulated.

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

T 1 T 2

A k L

A

A

A k L

D

D A k

L

B

B

A k L

C

C

Consider a composite plane wall as shown:

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

Example:

k I = 20 W/mk

A I = 1 m 2 , L = 1m

k II = 10 W/mk

A II = 1 m 2 , L = 1m

T 1 = 0°C T f = 100°C

h = 1000 W/ m 2 k

q x

1 D Conduction(Radial conduction in a composite

cylinder) h 1

T ∞,1

k 1

r 1

r 2

k 2 T ∞,2 r 3

h 2

T ∞,1 T ∞,2

) 2 )( ( 1

1 1 L r h π ) 2 )( ( 1

2 2 L r h π

1 2 ln

2

1

Lk r r

π 2 2

ln 3

2

Lk r r

π

∑ ∞ ∞ −

= t

r R T T

q 1 , 2 ,

Critical Insulation Thickness :

r 0 r i T ∞

T i h

Insulation Thickness : r o ­r i

h L r kL R i r

r

tot ) 2 ( 1

2 ) ln(

0

0

π π + =

Objective : decrease q , increases

Vary r 0 ; as r 0 increases ,first term increases, second term decreases.

tot R

Critical Insulation Thickness (contd…)

Set 0 0

= dr dR tot

0 2

1 2

1 0

2 0

= − hLr L kr π π

h k r = 0

Max or Min. ? Take : 0 0

2

2

= dr R d tot at

h k r = 0

h k r

tot

hL r L kr dr R d

=

+ −

= 0

0 2

0 2

0 2

2 1 2

1 π π

0 2 3

2

Lk h

π =

Maximum – Minimum problem

Minimum q at r 0 =(k/h)=r c r (critical radius)

good for steam pipes etc. good for electrical cables

R c r =k/h

r 0

R t o t

Critical Insulation Thickness (contd…)

1D Conduction in Sphere 1D Conduction in Sphere

r 1

r 2

k

T s,1 T s,2

T ∞,1

T ∞,2

( ) ( )

( ) ( )

k r r R

r r T T k

dr dT kA q

T T T r T

dr dT kr

dr d

r

cond t

s s r

r r r r s s s

π

π

4 / 1 / 1

/ 1 / 1 4

) (

0 1

2 1 ,

2 1

2 , 1 ,

/ 1 / 1 2 , 1 , 1 ,

2

2

2 1

1

− = →

− −

= − = →

− = →

=

− −

−

Inside Solid:

Applications: * current carrying conductors

* chemically reacting systems

* nuclear reactors

= Energy generation per unit volume V E q &

&=

Conduction with Thermal Energy Generation

Conduction with Thermal Energy Generation

The Plane Wall :

k T s,1

T s,2

x=0 x=+L

Hot fluid

Cold fluid

T ∞,2 T ∞,1

x= ­L

q& Assumptions:

1D, steady state, constant k,

uniform q &

Conduction With Thermal Energy Generation (contd…)

C x C x k

q T Solution

T T L x

T T L x cond Boundary

k

q

dx

T d

s

s

+ + − =

= + =

= − =

= +

2 :

,

, : .

0

2 1

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

dx dT k q flux Heat

T T L x T T

L x

k L q T solution Final

C and C find to conditions boundary Use

x

s s s s

− = ′ ′

+ +

− +

− =

:

2 2 1

2 : 1 , 2 , 1 , 2 ,

2

2 2

2 1

&

Conduction with Thermal Energy Generation (cont..)

Cylinder with heat source

Assumptions: 1D, steady state, constant k, uniform

Start with 1D heat equation in cylindrical co­ordinates:

k q

dr dT r

dr d

r = +

0 1 &

r o r

T s

T ∞ h

q&

q&

r k

T r cond

= =

0 4

, : .

s

s

T r r q r T Solution

dr dT r

T r Boundary

+

− =

= =

2

2 2

0

0

1 ) ( :

0 , 0

&

T s may not be known. Instead, T ∝ and h may be specified.

Exercise: Eliminate T s , 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 Ω. The length of the wire is 1m. The wire is submerged in a liquid at 100°C, and the heat transfer coefficient is 10W/m 2 K. Calculate the centre temperature of the wire at steady state condition.