1-D Transient Conduction: A T T T T T T

1-D TRANSIENT CONDUCTION
The equation to be solved is:-

·
¶T = l ¶²T + ¶²T + ¶²T + Q'''
¶t rcp ( ¶x² ¶y² ¶z² rcp )
¶T = l ¶²T ·
¶t rcp ( ) ¶x²
+ Q'''
rcp
We'll assume initially that there is no internal heat generation,
and divide the 1-D system into equi-spaced planes.
a a
· · ·
n-1 n n+1
At layer n the solution may be written:-
æ Tn-1 - Tn Tn - Tn+1 ö
Tn,1 - Tn, 0 ç - ÷
=ç a a ÷ l where a =
l
Dt ç a ÷ rC P rC P
ç ÷
è ø
This expression can be used to find the temperature at plane 'n'
at time instant '1' from the conditions existing at time instant '0'.
Re-arranging gives:-
aDt
Tn,1 = Tn, 0 + {Tn- 1, 0 - 2Tn, 0 + Tn +1,0 } 2
a
1DTransCond.ppp 1
aDt
Let 2
= F (the non-dimensional grid Fourier Number)
a
ì æ 1ö ü
then Tn,1 = F íTn-1,0 + ç 2 - ÷Tn,0 + Tn +1, 0 ý
î è Fø þ
l @ r
The only unknown in this equation is Tn,1 the temperature at layer
'n' after time interval Dt.
Thus, from a knowledge of the conditions at time zero,
temperatures in successive layers can be found after each time
interval.
The equation gives an EXPLICIT solution for each temperature
in turn and lends itself to tabular, graphical or computer solution.
Examination shows that if the coefficient of T n,0 is negative
(F < 0.5) the solution will become oscillatory and eventually
unstable.
Schmidt Graphical Solution
If F= 0.5 then the equation becomes:-
Tn-1, 0 + Tn +1,0
Tn,1 =
2
This can be solved using graphical construction
1DTransCond.ppp 2
Every other plane can be updated at each time instant.
· Tn+1,0
· Tn,1
Tn-1,0 ·
n-1 n n+1
Example : wall at uniform temperature where one surface

is suddenly raised to a higher temperature.
black t=0
temp
green t=Dt
red t=2Dt
blue t=3Dt
magenta t=4Dt
cyan t=5Dt
1 2 3 4 5 6 7 8
1DTransCond.ppp 3
A convective boundary can be incorporated into the
Schmidt method by equating the convection and
conduction at the boundary at time zero. ie
or
ie, an extra 'conductive' layer of thickness l /h is drawn

outside the wall and used with a false half-layer within the
a a a a black t=0
2 2 green t=Dt
¬ Tf red t=2Dt
blue t=3Dt
¬ l ®
h 1 2 3 4 5
1DTransCond.ppp 4
HEAT TRANSFER into the CONDUCTOR
The total heat transfer into the conductor may be found

either by summing wrt time or distance (layer). ie
all layers
or Q = å m Cp (Tfinal - Tzero)
1DTransCond.ppp 5
CONVECTIVE BOUNDARIES - General
If we go back to our original equation :
ì æ 1ö ü
Tn,1 = F íTn-1,0 + ç 2 - ÷Tn,0 + Tn +1, 0 ý
î è Fø þ
@ l @ r
we can modify it to take account of a convective boundary

condition by energy balance at the surface 'half-layer'.
Tf
w 1 2
For the surface half-layer:

Convection IN/OUT - Conduction OUT/IN = Stored Energy
aDt
As before 2
= F (the grid Fourier No.)
a
ha
and put = Bi (the grid Biot No.)
l
hence:
If the coefficient of Tw,0 is negative (F + F Bi > 0.5) the

solution will become oscillatory and eventually unstable.
NB the expression remains explicit; if we know the
conditions at time t, we can find conditions at time, t + Dt.
1DTransCond.ppp 6
Alternative Finite Difference methods
Because of the instability of the explicit equations they often

call for excessive calculations. This is especially so in 2-D or
3-D where the stability criteria becomes more constraining.
e.g. in 2-D F must be £ 0.25.
The equation:-
is based on 'backward' time differences. ie We found

conditions at plane 'n' at time instant '1' in terms of the
conditions which existed at planes n-1, n, & n+1 at time
instant '0'.
We could equally well have based the equation on 'forward'
time differences. ie We could find conditions at plane 'n' at
time instant '1' in terms of the conditions which will exist at
the planes n-1, n, & n+1 at time instant '1'.
If we do this, the equation becomes:-
The obvious snag is that we don't know conditions at time

instant '1', so therefore we have to write equations for every
plane for a given time and solve them simultaneously.
Alternatively, we may solve them iteratively!
1DTransCond.ppp 7
Writing the equation using Fourier No. (as before) gives:-
We can solve the temperatures by writing the equation in

'residual' form (= R) , and iterating to reduce the residuals to
zero.
An improved value for Tn,1 is given by:-
This method of solution is an 'implicit' technique because we

cannot solve for solutions at each point directly, as we did
using the previous explicit method.
It is found that this method is completely stable and we may

use any value of F which gives the required accuracy.
We may also write the convective boundary condition in

implicit form.
1DTransCond.ppp 8
It is found that compared with the exact solution the explicit
method tends to underestimate the answer, whereas the
implicit method tends to overestimate the answer.
A method using the average of the two above methods (due to

Crank & Nicolson) gives almost correct answers.
(See Bacon, Basic Heat Transfer p. )
Exact Methods
For some regular solids (eg slabs, spheres, cylinders etc.)

exact solutions are available in the form of non-dimensional
charts.
It is also possible to solve some 3-D solutions by

'decomposition' using 1-D solutions.
1DTransCond.ppp 9
LUMPED CAPACITY SYSTEMS
If a solid has a high thermal conductivity, and it loses heat
comparatively slowly, the temperature variations within it are small
and the whole solid can be assumed to be at a constant
temperature throughout.
ha
ie l solid >> h fluid so that (= B i ) is small (< 0.1)
l
Tf ® T
In time interval dt the heat transfer dQ from the body is :-
dQ = hA(T - T f )dt = mC p dT
Putting q = T-Tf, dq = dT, and we may write:
dq hA
= dt
q mC p
æ q ö hA
Integrating gives:- lnçç ÷÷ = (t - t 0 )
q
è 0ø mC p
NB: q0 is the temperature difference at time t0.

t
-
or: q = q0e t* [t0 = 0]
ie an exponential change in temperature with time, where

the time constant, t* = mCp
hA
If heat is generated internally:-
.
-
t [ & Qint = h A (Tinf -Tf) ]
*
q = q inf (1 - e t )
1DTransCond.ppp 10

1-D Transient Conduction: A T T T T T T

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1-D TRANSIENT CONDUCTION

The equation to be solved is:-

Schmidt Graphical Solution

If F= 0.5 then the equation becomes:-

This can be solved using graphical construction

Example : wall at uniform temperature where one surface

ie, an extra 'conductive' layer of thickness l /h is drawn

The total heat transfer into the conductor may be found

we can modify it to take account of a convective boundary

For the surface half-layer:

If the coefficient of Tw,0 is negative (F + F Bi > 0.5) the

Because of the instability of the explicit equations they often

is based on 'backward' time differences. ie We found

The obvious snag is that we don't know conditions at time

Alternatively, we may solve them iteratively!

We can solve the temperatures by writing the equation in

An improved value for Tn,1 is given by:-

This method of solution is an 'implicit' technique because we

It is found that this method is completely stable and we may

We may also write the convective boundary condition in

A method using the average of the two above methods (due to

(See Bacon, Basic Heat Transfer p. )

For some regular solids (eg slabs, spheres, cylinders etc.)

It is also possible to solve some 3-D solutions by

In time interval dt the heat transfer dQ from the body is :-

NB: q0 is the temperature difference at time t0.

ie an exponential change in temperature with time, where

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