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02. Conduction Heat Equation Boundar Conditions
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Transcript of 02. Conduction Heat Equation Boundar Conditions
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2. THE HEAT CONDUCTION EQUATION
The Need for Temperature Distribution
Examine Fourier's law
x
Tq
x
=& (2.1)
To determinex
q& we needx
T
To determinex
T
we need the temperature distribution
T(x, y, z, t)
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Formulation of the Heat Conduction Equation in
Rectangular Cartesian Coordinates
General case:
Three-dimensional
Unsteady state
Nuclear element
Electric energy dissipation in devices
Metabolic heat production in tissue
Apply conservation of energy to element dx dy dz during
time dt:
Energy generation Eg, examples: z
y
x
1.2.Fig
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=removedEnergy-generatedEnergyaddedEnergyelementhinchange witEnergy
Goal: Express (2.2) in terms ofT.
Energy generation,Eg:
(2.2)akoutgin EEEE =+ [J]
Energy in by conduction,Ein:
dtdzdxqdtdzdyqEyxin&& +=
dtdydxqz&+
(a)[J]
(b)
dtdzdydxQEzdrg
&= [J]
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Energy out by conduction,Eout
(c)
dy)dxdzdty
qq(dx)dydzdt
x
qq(E
y
yx
xout
++
+=
&
&
&
&
dz)dxdydtzqq( z
z
++
&
&
Energy change within the element (accumulation), E
dtt
UE
ak
= (d)
ExpressingEin terms ofT. Neglecting changes inkinetic and potential energy
(e)dxdydzumuU ==
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= density Assume: incompressible material: ccc
pv==
(f)cTu =
dtdzdydx
t
Tcdt
t
UE
=
=
(g)
Substitute (a), (b), (c) and (g) into (2.2) and dividing
through by dx dy dz dt
(h)t
TcQ
z
q
y
q
x
q
zdr
zyx
=+
&
&&&
u = internal energy per unit mass [J/kg]
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Fourier's law (1.5)
(1.5)z
Tq,
y
Tq,
x
Tq
zyx
=
=
= &&&
(1.5) into (h)
(2.4) is the heat conduction equation
Assume: constant
t
T
c
Q+
z
T+
y
T
x
T zdr
2
2
2
2
2
2
=
+
&
(2.5)
t
TcQ)
z
T(
z
)
y
T(
y
)
x
T(
xzdr
=+
+
+
&
(2.4)
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a = thermal diffusivity (souinitel teplotn vodivosti)
defined as
NOTE:(1) (2.5) is the differential formulation of the principle ofconservation of energy. Valid at every point in the
material
(3) Physical significance of each term:
The first three terms = net energy conducted in x,y andzdirections
The fourth term = energy generation
The fifth term = energy storage - accumulation
(2.6)c
a =
storage - accumulation
conduction
(2) Limited to and constant isotropic
t
T
c
Q+
z
T+
y
T
x
T zdr
2
2
2
2
2
2
=
+
&
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(4) Simplifications for special cases:
Steady state: set0
t
T=
One-dimensional: set 02
2
2
2
z
T
y
T =
No energy generation: set 0Qzdr=&
(5) Solutions are simplified for constant a, and zdrQ&
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The Heat Conduction Equation in
Cylindrical and Spherical Coordinates
Cylindrical coordinates zr ,,
(2.7)
t
T
c
q+
z
T+
T
rr
Tr
rr
p
=
+
2
2
2
2
2
1)(
1
Spherical coordinates ,,r
t
T
c
q+)
T
r+
T
rr
Trrr
p
=
+ (sinsin1
sin
1)(
1
22
2
22
2
2
(2.8)
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Boundary Conditions
Boundary conditions are mathematical equations
describing what takes place physically at a boundary.
To write boundary conditions we must:
Select an origin
Select coordinate axes
Identify the physical conditions at the boundaries
Fig. 2.3 shows four typical
boundary conditions:
(B.C. 1) Specified temperature.
Along boundary (0,y) the
temperature is :TTh,
oTW
L
oq x0
convection
insulation
2.3Fig.
,
Q&
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o),0( TyT = (2.9)
Using(B.C. 2a) Specified flux. The heat flux at
boundary (L,y) is .qx&
Fouriers law
Lxx x
Tq
=
=& (2.10)[W/m2]
(2.11)
(B.C. 2b) Insulated boundary. The boundary at (x,W) is
thermally insulated, Fouriers law gives
0==Wyy
T
0.qx=&
Th,oT
W
L
oq x0
convection
insulation
2.3Fig.
,
xQ&
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(B.C. 4) Interface. Two different materials
with a perfect interface contact:
Two types of B.C.:
(i) Equality of temperature:
T1
(0,y) = T2
(0,y) (2.13)
(B.C. 3) Convection. Heat is exchanged at the boundary (x,0)
by convection with a fluid at temperature .TEquating Newton's law with Fourier's law
[ ]0yy
TT(x,0)T
=
= (2.12)
Th,oT
W
L
oq
y
x0convection
insulation
2.3Fig.
,
xQ&
2.4Fig.
x
2k
2T
1k
1T
0
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(2.14)
0x
22
0x
11 x
T
x
T
==
=
Consider the heat equation (2.5)
How many conditions are needed to solve this equation?
(2.15)t
T
a
1
Q+
z
T+
y
T
x
T zdr
2
2
2
2
2
2
=
+
&
(ii) Equality of flux:
2.4Fig.
x
2k
2T
1k
1T
0
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