3.044 MATERIALS PROCESSING
LECTURE 3
We will often be comparing heat transfer steps/processes:
When can we neglect one and focus on the other?
Resistance:
10 >
LA
kA
LB
kB
> 0.1
⇒
10 : “B”conducts fast, cannot sustain a gradient
0.1 : “A”conducts fast, cannot sustain a gradient
Reduce Dimensionality:
∂T
= α ∇2 T : T (t, x, y, z)
∂x
1. Steady State
∂T
=0
∂t
2. No Thermal Gradients
∇ T = 0, T = T (t) ONLY
∂T
= ....
∂t
Date: February 15th, 2012.
1
2
LECTURE 3
In general, for solid / “fluid” interfaces: T2 6= Tf
- constant T, B.C. is not appropriate
- fluid cannot always remove heat at the rate it is delivered
How is heat transferred / removed in the fluid?
- conduction: heat moves, atoms sit still
- convection: atoms flow away, carrying heat with them
1. natural convection (T interacts w/ gravity)
2. forced convection (mechanically driven flow)
- radiation: photons carries heat away
What are the proper B.C.?
1. T2 6= Tf
2. @ x = L, specify flux:
heat[ W2 ]
m
z}|{
q
= |{z}
h (T2 − Tf )
heat transfer coeff.[
W
m2 K
⇒
the hotter the material is with respect
to the fluid, the faster heat will flow
]
∂T
∂ 2T
=0=α 2
∂t
∂x
3.044 MATERIALS PROCESSING
Step 1: Solve
T − T1
= xL,
T2 − T1
Θ=χ
where T2 is unknown
Step 2: B.C.
@x = L
qcond = qconv
∂T
−k
= h(T2 − Tf )
∂x
Step 3: Solve for
∂T
∂x
T − T1
x
=
T2 − T1
L
x
(T2 − Tf )
L
∂T
T2 − T1
=
L
∂x
T = T1 +
Plug into: −k ∂T
= h(T2 − Tf )
∂x
T2 − T1
= h(T2 − Tf )
L
kT1
k
+ hTf = h +
L
L
k
Tf
T2 = L k
h+ L
−k
3
4
LECTURE 3
Plug into: T = T1 + Lx (T2 − Tf )
"
#
x Lk T1 + hTf
T = T1 +
− T1
L
h + Lk
"
#
x h(Tf − T1 )
T − T1 =
L
h + Lk
"
#
h Lk
T − T1
x
=
Tf − T1
L 1 + h Lx
#
"
h Lk
Θ=χ
1 + h Lx
hL
h
⇒ k ⇒
k
l
Biot Number:
L
k
1
h
L
where is conductive resistance
k
1
and is convective resistance
h
hL
k
dimensionless, ratio of resistances
3.044 MATERIALS PROCESSING
Three Important Cases:
Generalize:
1. Imperfect interfaces:
qin = qout
= h(T2+ − T2− ),
where
1
= interface resistance
h
2. Geometry:
hL
→ What is L?
k
volume
L≈
, a characteristic dimension
surface area
5
6
LECTURE 3
Examples:
1. plate heated on one side: L = thickness
2. plate heated on both sides: L = half thickness
R
πR2 l
=
3. cylinder: L =
2πRl
2
4
πR3
R
4. sphere (or other 3D shape): L = 3 2 =
4πR
3
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3.044 Materials Processing
Spring 2013
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