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One Dimensional Steady State Heat Conduction






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MODULE 2. One Dimensional Steady State Heat Conduction . T(x,y,z). 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
One Dimensional Steady State Heat Conduction

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Slide 1

MODULE 2

One Dimensional Steady State Heat Conduction

Slide 2

T(x,y,z)

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

Slide 3

Unidirectional heat conduction (1D)

Area = A

x

x+x

0

x

A

qx

qx+x

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

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

Slide 4

&

&

&

&

-

+

=

(

E

E

)

E

E

in

out

gen

st

E

-

+

D

=

q

q

A

(

x

)

q

&

+

D

x

x

x

t

=

r

D

E

(

A

x

)

u

T

=

-

q

kA

x

x

E

u

T

=

r

D

=

r

D

q

A

x

A

xc

=

+

D

x

q

q

x

t

t

t

+

D

x

x

x

x

Unidirectional heat conduction (1D)

First Law (energy balance)

Slide 5

æ

ö

T

T

T

T

-

+

+

D

+

D

=

r

D

ç

÷

kA

kA

A

k

x

A

x

q

Ac

x

&

x

x

x

x

t

è

ø

æ

ö

T

T

+

=

r

ç

÷

k

q

c

&

x

x

t

è

ø

Thermal inertia

Internal heat generation

Longitudinal conduction

r

2

T

q

c

T

1

T

&

+

=

=

a

2

x

k

k

t

t

Unidirectional heat conduction

(1D)(contd…)

If k is a constant

Slide 6

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.

Slide 7

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).

Slide 8

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

Slide 9

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

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

Cold fluid

k

Ts,1

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

Ts,2

T∞,2

Hot fluid

x=L

x=0

1- Dimensional Heat Conduction

The Plane Wall :

Const. K; solution is:

Slide 10

Thermal resistance(electrical analogy)

OHM’s LAW :Flow of Electricity

V=IR elect

Voltage Drop = Current flow×Resistance

Slide 11

Thermal Analogy to Ohm’s Law :

Temp Drop=Heat Flow×Resistance

Slide 12

T∞,1

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

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

Cold fluid

k

Ts,1

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

Ts,2

T∞,2

Hot fluid

x=L

x=0

1

L

1

A

A

A

h

h

k

1

2

1 D Heat Conduction through a Plane Wall

T∞,1

Ts,1

Ts,2

T∞,2

qx

(Thermal Resistance )

Slide 13

Resistance expressions

Slide 14

T∞,1

A

B

C

h1

K A

K B

K C

h2

T∞,2

T∞,1

T∞,2

L A

L B

L C

q x

1

1

A

A

h

h

1

2

Composite Walls :

Overall heat transfer coefficient

Slide 15

TA

TB

A

B

Overall Heat transfer Coefficient

Contact Resistance :

Slide 16

B

AB+AC=AA=AD

A

D

K B

LB=LC

T1

T2

K D

K A

C

K c

Series-Parallel :

Slide 17

T1

T2

Series-Parallel (contd…)

Assumptions :

  • Face between B and C is insulated.

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

Slide 18

kI = 20 W/mk

AI = 1 m2, L = 1m

qx

Tf = 100°C

h = 1000 W/ m2 k

T1 = 0°C

kII = 10 W/mk

AII = 1 m2, L = 1m

Example:

Consider a composite plane wall as shown:

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

Slide 19

h1

r1

T∞,1

r2

h2

T∞,2

k1

r3

k2

T∞,1

T∞,2

1 D Conduction(Radial conduction in a composite cylinder)

Slide 20

T∞

r0

ri

h

Ti

Critical Insulation Thickness :

Insulation Thickness : r o-r i

Objective :decrease q , increases

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

Slide 21

Set

Critical Insulation Thickness (contd…)

Maximum – Minimum problem

at

Max or Min. ?Take :

Slide 22

R t o t

good for electrical cables

good for steam pipes etc.

R c r=k/h

r0

Critical Insulation Thickness (contd…)

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

Slide 23

r2

r1

k

T∞,2

Ts,2

Ts,1

T∞,1

1D Conduction in Sphere

Inside Solid:

Slide 24

&

E

=

q

&

V

= Energy generation per unit volume

Conduction with Thermal Energy Generation

Applications: * current carrying conductors

* chemically reacting systems

* nuclear reactors

Slide 25

k

Ts,1

Ts,2

Assumptions:

1D, steady state, constant k,

uniform

q

&

T∞,1

T∞,2

Hot fluid

Cold fluid

q

&

x= -L

x=+L

x=0

Conduction with Thermal Energy Generation

The Plane Wall :

Slide 26

2

d

T

q

&

+

=

0

2

dx

k

=

-

=

Boundary

cond

.

:

x

L

,

T

T

,

1

s

=

+

=

x

L

,

T

T

,

2

s

q

&

=

-

+

+

2

Solution

:

T

x

C

x

C

1

2

2

k

Conduction With Thermal Energy Generation (contd…)

Slide 27

Conduction with Thermal Energy Generation (cont..)

Use

boundary

conditions

to

find

C

and

C

1

2

-

+

æ

ö

T

T

T

T

2

2

q

L

x

x

&

ç

÷

=

-

+

+

,

2

,

1

,

2

,

1

s

s

s

s

Final

solution

:

T

1

ç

÷

2

2

k

L

2

L

2

è

ø

Not linear any more

dT

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

¢

¢

=

-

Heat

flux

:

q

k

x

dx

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

Slide 28

T∞ h

ro

q

&

r

æ

ö

1

d

dT

q

&

+

=

ç

÷

r

0

r

dr

è

dr

ø

k

Ts

q

&

Cylinder with heat source

Assumptions:

1D, steady state, constant k, uniform

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

Slide 29

=

=

Boundary

cond

.

:

r

r

,

T

T

0

s

dT

=

=

r

0

,

0

dr

æ

ö

2

q

r

&

ç

÷

2

=

-

+

Solution

:

T

(

r

)

r

1

T

ç

÷

0

s

2

4

k

r

è

ø

0

Cylinder With Heat Source

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

Exercise: Eliminate Ts, using T and h.

Slide 30

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.


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