3.044 MATERIALS PROCESSING
LECTURE 4
General Heat Conduction Solutions:
∂T
∂t
= ∇ · k∇T, T (x̄, t)
Trick one: steady state ∇2 T = 0, T (x)
Trick two: low Biot number
∂T
∂t
= α h(Ts − Tf ), T (t)
Low Biot Number Solutions: Newtonian Heating / Cooling
Global Heat Balance:
qconv = qlost
∂T
V
∂t
∂T
−hA
=
∂t
ρcp V
T − Tf
−hA
ln(T − Tf ) =
t+C
ρ cp V
@t = 0, T = Ts
A h(T − Tf ) = −ρ cp
ln(Ts − Tf ) = C
Date: February 21st, 2012.
1
2
LECTURE 4
ln
T − Tf
T s − Tf
=
−hA
t
ρ cp V
−hA
T − Tf
= e ρ cp V
T s − Tf
t
Transient Heat Conduction: depends on position and time
∂T
= α ∇2 T
∂t
You should know:
1) Some common solutions for simple geometries
2) Where to find solutions
3) How to build up complex solutions using simple solutions
Semi-Infinite Solid
- constant T1 at surface
- initially T0 everywhere
x
T − T0
= erfc √
T 1 − T0
2 αt
3.044 MATERIALS PROCESSING
Z
erf(z) =
z
3
2
e−x dx
0
erfc = 1 − erf
T (x) = (T1 − T0 ) erfc
x
√
2 αt
T − T0
x
√
= erfc
T1 − T0
2 αt
x T − T0
√
(−1)
= erfc
(−1)
T1 − T0
2 αt
+ T0
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LECTURE 4
T0 − T
x
√
= erf
− (T − T0 )
T 1 − T0
2 αt
x T − T1
√
= −erf
T1 − T0
2 αt
x T − T1
√
= erf
T0 − T1
2 αt
Semi-Infinite Solid
- convection at surface: qlost = h(T − Tf )
x
hx
x
h√
2
√
√ +
Θ = ERF C
− EXP
+ h k α t · ERF C
αt
k
2 αt
2 αt k
Where to find these solutions:
- Carslaw & Jaeger
- Crank Dimensionless Numbers:
T − T0
x
= erfc √
T1 − T0
2 αt
T − T0
=Θ
T1 − T 0
x
χ=
L
x = Lχ
αt
τ= 2
L
L2 T
t=
α ⎛
⎞
Lχ ⎠
Θ = erfc ⎝ L
2
2 α Lα τ
χ
Θ = erfc √
2 τ
3.044 MATERIALS PROCESSING
5
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3.044 Materials Processing
Spring 2013
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