06. Transient Heat Conduction+SemiInfinite

7/28/2019 06. Transient Heat Conduction+SemiInfinite

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6. ONE-DIMENSIONAL TRANSIENT

CONDUCTION

ForBi> 0.1, lumped capacity is not applicable

Spatial temperature variations must be accounted for

Example: One-dimensional transient conduction

in a plate or in long cylinder or in a

sphere

= heat transfer coefficient

/2 = half thickness of plateTo = initial temperature

T = ambient temperature

a = thermal diffusivity


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Governing Equations

For one-dimensional, no energy generation and

constant conductivity, heat conduction equationbecomes:

(6.1)t

T

a

1

x

T

2

2

=

tTcQ)

zT(

z)

yT(

y)

xT(

x zdr =+

+

+

&


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(c)00

=

=xxT

Initial condition:o

TT(x,0) = (b)

Boundary conditions on both sides:

( ) ( )w0,x0,x TTTTxT

==

=

=(a)

For symmetrical heating or cooling, only one half of the plate

andx=0 placed on the axis of the plate.

x=0 x=/2

( ) ( w2x2x

TTTTxT ==+

==


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Solution to the equation

Analytical

Fourier method of separation of variables.

(6.1)t

T

a

1

x

T2

2

=


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Graphical Representation of Solutions:

Non-dimensional Form (Heislers Charts)

Plates

The form of the equation and boundary conditions

shows that the temperature T(x,t) depends on:

Two variables:xand t

Six physical quantities: , , /2, T ,To and a

= ,

20 ,a,

,Tx, t, TfTT

t

T

a

1

x

T2

2

=

( ) ( )w2x

2x

TTTTx

T ==

+

==

o

TT(x,0) =


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To map the effects of all quantities on T(x,t) can be

tedious (boring) and time consuming

Alternate approach: Express the result in non-

dimensional form (by independent dimensionless

groups) using Buckingham theorem:Required number of dimensionless groups is equal tothe total number of physical quantities n (x,t,,a)minus the number of primary dimensions m requiredto express the dimensional n quantities.

Define the following 4 dimensionless variables

=

,a,,

2,TTt,x,fTT

08 quantities4 primary dimensions

m, s, W, K


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Time: numberFourier

L

tFo

2==

numberBiotLBi ==

Bi)Fo;(X,=

Relation between two variables and six quantities is

replaced by relation between four dimensionlessgroups

Temperature:

=

TT

TT

o

Distance:L

xX=

Characteristic dimensionL

for a plate is /2


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Heisler charts: Equations of the analytical solution

are used to construct charts to determine transienttemperature in plates

Fig. 1: Transient temperature at the center, Tc

Fig. 2:Transient temperature at other locations in

terms ofTc


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Fig. 1: Center-plane transient temperature of a plate of thickness 2L

L

Bi

1=

=

TT

TT

o

c

c

For given time tcalculate Fo

CalculateBiDetermine c temperatureat the center

2L

tFo =

)T(TTT 0cc =


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Fig. 2: Temperature distribution in a plate of thickness 2L

=

TT

TT

c

/L1/Bi=Surfacetemperature

T(TTT c =


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Fig. 3 Center transient temperature of a cylinder


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Fig. 4 Temperature distribution in a Cylinder of radiusor oTterms of

Surface

temperature


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Fig. 5 Center transient temperature of a sphere


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Fig. 6 Temperature distribution of a sphere in terms ofoT


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Multi-dimensional Transient Conduction in

bars, prisms, short cylinders etc.

Superposition of 1D solution product solution

A short cylinder is the intersection of

a long cylinder and a plane wall of the

thickness that equals the height of thecylinder.

Similarly, a long rectangular bar is the

intersection of two plane walls of the

thickness a and b


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Product solution:

cylinderinfinite

wallplane

cylindershort TT

TtrTTT

TtxTTT

TtxrT

=

000

),(.),(),,(

=

TT

TT(0,t)

ocRemind:

Temperature in the center of a finite (short) cylinder:

cylinder

c,infinite

wall

c,plane

cylindershort

0

c .

TT

Tt0,0T =

=

),(

forx, r=0


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Similarly for other locations:

wallplanesurface

cylindercentre1 .

.=

1

wallplane

surface

cylinder

surface .

.

=22

wallplanecenter

cylindercenter .

.

=3

3

wallplanecenter

cylindersurface .

.=4

4


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TRANSIENT CONDUCTION IN SEMI-

INFINITE SOLID

Position of

no temperature

change

Sudden cooling

of surface


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Solution procedure

2

2

xTa

TT

=

Equation for temperature distribution T=f(x,t)

B.C.: One initial condition: at t=0, T=T0, for allx

One boundary condition: at t>0, T=Tw forx=0

x

x=0


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Solution

erf - Gauss error function New variableat

x=

( )

w0

w

TT

Tx,tT

( )

=

=

2

erf

at2

xerf

TT

Tx,tT

w0

w (a)


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Practical impact

/2 = 1,825( )w0

w

TTTx,tT

For /2 > 1,825, i.e.

at3,65x>for

T(x) - Tw > 0.99(To - Tw)

i.e. oTxT )(


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Heat flux transferred on the surface

atTT

xT w0

0

=

=

( )at

TTq ww

0=&Heat flux tq

w

1

&

What about an amount of heat [J]??

Fourier law

0xw

x

Tq

=

=& [W/m2]

After differentiation of (a)( )

=

=

2

erf

at2

xerf

TT

x,t

w0

w

06. Transient Heat Conduction+SemiInfinite

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