THE HEAT CONDUCTION EQUATION The Need for Temperature

THE HEAT CONDUCTION EQUATION
The Need for Temperature Distribution
Examine Fourier's law
∂T
q& = − λ
x
∂x
(2.1)
∂T
To determine q& we need
x
∂x
∂T
To determine
we need the temperature distribution
∂x
T(x, y, z, t)
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Based on presentations by professor Latif M. Jiji, The City College of New York
Formulation of the Heat Conduction Equation in
Rectangular Cartesian Coordinates
General case:
• Three-dimensional
• Unsteady state
• Energy generation Eg, examples:
z
• Nuclear element
• Electric energy dissipation in devices
• Metabolic heat production in tissue
y
x
Fig . 2.1
Apply conservation of energy to element dx dy dz during
time dt:
2
Energy added + Energy generated - Energy removed =
Energy change wit hin element
E
in
+E −E
g
out
= ΔE
ak
[J] (2.2)
Express (2.2) in terms of T.
E
in
Energy in by conduction, Ein:
= q& dy dzdt + q& dx dzdt
x
+ q& z dx dydt
y
[J]
(a)
Energy generation, Eg:
E g = Q& zdr dx dy dz dt [J]
(b)
3
Energy out by conduction, Eout
E out = (q& x +
∂ q& x
∂x
+ ( q& +
z
dx)dydzdt + (q& y +
∂ q& z
∂z
∂ q& y
∂y
dy)dxdzdt
(c)
dz)dxdydt
Energy change within the element (accumulation), Δ E
∂U
ΔE =
dt
ak
∂t
(d)
Expressing E in terms of T. Neglecting changes in
kinetic and potential energy
U = mu = u ρ dxdydz
(e)
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u
= internal energy per unit mass
ρ = density
[J/kg]
• Assume: incompressible material: cv = c p = c
u = cT
∂U
∂T
ΔE =
dt = ρc
dx dydz dt
∂t
∂t
(f)
(g)
Substitute (a), (b), (c) and (g) into (2.2) and dividing
through by dx dy dz dt
−
∂q&
x −
∂x
∂q& y
∂q& z
∂T
&
−
+ Qzdr = ρc
∂y
∂z
∂t
(h)
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Fourier's law (1.5)
∂T
∂T
∂T
, q& = − λ
q& = − λ
, q& = − λ
x
∂x y
∂y z
∂z
(1.5)
(1.5) into (h)
∂ ∂T
∂ ∂T
∂ ∂T
∂T
&
(λ
) + (λ
) + (λ
) + Q = ρc
zdr
∂x ∂x
∂y ∂y
∂z ∂z
∂t
(2.4)
• (2.4) is the heat conduction equation
• Assume: constant λ
&
⎡ ∂ 2T
ρc ∂ T
∂ 2T
∂ 2 T ⎤ Q zdr
+
+
=
+
⎢
2
2
2 ⎥
λ
λ ∂t
∂y
∂z ⎦
⎣ ∂x
(2.5)
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a = thermal diffusivity (součinitel teplotní vodivosti)
defined as
conduction
λ
(2.6)
a =
ρc
• NOTE:
storage - accumulation
(1) (2.5) is the differential formulation of the principle of
conservation of energy. Valid at every point in the
material
(2) Limited to isotropic and constant λ
(3) Physical significance of each term:
• The first three terms = net energy conducted in x,
y and z directions
Q& zdr
⎡ ∂ 2T
∂ 2T
∂ 2T ⎤
ρc ∂ T
+
+
+
=
⎢
2
2
2 ⎥
λ
λ ∂t
∂y
∂z ⎦
⎣ ∂x
• The fourth term = energy generation
• The fifth term = energy storage - accumulation7
(4) Simplifications for special cases:
∂T
=0
• Steady state: set
∂t
∂ 2T ∂ 2 T
=
=
0
• One-dimensional: set
∂ y2 ∂z2
• No energy generation: set
Q& zdr = 0
&
(5) Solutions are simplified for constant a, λ and Qzdr
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The Heat Conduction Equation in
Cylindrical and Spherical Coordinates
• Cylindrical coordinates r , φ , z
⎡1 ∂
1 ∂ 2T ∂ 2T ⎤
∂T
q ′′′
∂T
(r
)+
α⎢
=
+
⎥+
2
2
2
r ∂φ
∂ z ⎦ ρ cp ∂ t
⎣r ∂ r ∂ r
(2.7)
• Spherical coordinates r ,θ , φ
⎡ 1 ∂ 2 ∂T
∂ T ⎤ q′′′ ∂ T
∂ 2T
∂
1
1
+
)⎥ +
α⎢
=
(r
)+
(sinθ
2∂r
2
2
2
2
∂ r r sin θ ∂ φ r sinθ ∂ θ
∂ θ ⎥⎦ ρ c p ∂ t
⎢⎣r
(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
y
Fig. 2.3 shows four typical
boundary conditions:
(B.C. 1) Specified temperature.
Along boundary (0, y) the
temperature is Tο :
To
insulation
Qq& ′′
W
ox
L
0
convection α,h, T∞
Fig. 2.3
x
10
y
T (0, y ) = To
insulation
(2.9)
To
W
0
L
convection α,h, T∞
Fig. 2.3
(B.C. 2a) Specified flux. The heat flux at
boundary (L, y) is q& . Using
x
Fourier’s law
∂T
2]
q& = − λ
[W/m
(2.10)
x
∂x x = L
(B.C. 2b) Insulated boundary. The boundary at (x,W) is
thermally insulated, q& = 0. Fourier’s law gives
x
∂T
∂y
=0
(2.11)
y =W
11
Qq& ′′
ox
x
(B.C. 3) Convection. Heat is exchanged at the boundary (x,0)
by convection with a fluid at temperature T ∞ .
Equating Newton's law with Fourier's law
y
insulation
∂T
(2.12)
α T − T(x,0) = − λ
∞
∂y y =0
Qq& ′′
[
To
W
0
L
h,T∞
convection α,
]
ox
x
Fig. 2.3
(B.C. 4) Interface. Two different materials
with a perfect interface contact:
Two types of B.C.:
T1
T2
kλ1
λk22
1
x
0
(i) Equality of temperature:
T1 (0,y) = T2 (0,y)
(2.13)
Fig. 2.4
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(ii) Equality of flux:
−λ
1
∂T1
∂x
= −λ
x =0
2
∂T2
∂x
(2.14)
x =0
Consider the heat equation (2.5)
T1
T2
kλ1
λk22
1
x
0
Fig. 2.4
⎡ ∂ 2 T ∂ 2 T ∂ 2 T ⎤ Q& zdr
1 ∂T
=
+
+
⎢ 2 +
2
2 ⎥
λ
a ∂t
∂y
∂z ⎦
⎣ ∂x
(2.15)
How many conditions are needed to solve this equation?
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