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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

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Presentation Transcript
slide1

MODULE 2

One Dimensional Steady State Heat Conduction

objectives of conduction analysis

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

slide3

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)

unidirectional heat conduction 1d

&

&

&

&

-

+

=

(

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)

slide5

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

unidirectional heat conduction 1d contd
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
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
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

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

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

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:

thermal resistance electrical analogy
Thermal resistance(electrical analogy)

OHM’s LAW :Flow of Electricity

V=IR elect

Voltage Drop = Current flow×Resistance

thermal analogy to ohm s law
Thermal Analogy to Ohm’s Law :

Temp Drop=Heat Flow×Resistance

1 d heat conduction through a plane wall

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 )

composite walls

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

overall heat transfer coefficient

TA

TB

A

B

Overall Heat transfer Coefficient

Contact Resistance :

slide16

B

AB+AC=AA=AD

A

D

K B

LB=LC

T1

T2

K D

K A

C

K c

Series-Parallel :

series parallel contd

T1

T2

Series-Parallel (contd…)

Assumptions :

  • Face between B and C is insulated.
  • (2) Uniform temperature at any face normal to X.
slide18

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.

slide20

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.

slide21

Set

Critical Insulation Thickness (contd…)

Maximum – Minimum problem

at

Max or Min. ?Take :

slide22

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)

1d conduction in sphere

r2

r1

k

T∞,2

Ts,2

Ts,1

T∞,1

1D Conduction in Sphere

Inside Solid:

slide24

&

E

=

q

&

V

= Energy generation per unit volume

Conduction with Thermal Energy Generation

Applications: * current carrying conductors

* chemically reacting systems

* nuclear reactors

conduction with thermal energy generation

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 :

conduction with thermal energy generation contd

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…)
conduction with thermal energy generation cont
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

cylinder with heat source

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:

cylinder with heat source1

=

=

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.

cylinder with heat source contd
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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