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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