3.044 MATERIALS PROCESSING

LECTURE 8

Radiation:

∂T−k =∂x

−εσ[T 4surf − T 4

source

]L

M = εσkT 3surf

∂T= α

∂t∇2T

q = εσ(T 4surf − T 4

source)

⇒ Very few analytical solutions, some charts

Date: March 5th, 2012.1

2 LECTURE 8

CVD: Chemical Vapor Deposition

At steady state the thermocouple outputs a temperature reading TTC

TC is heated by gas (convection)

in︷ ︸︸ ︷−h(T − Tf )A =

out︷ ︸︸εσ(T 4

surf − T 4source)A

εσ

︷TTC = Tf − (T 4

h surf − T 4source)

TTC �= Tf

Tf ≈ 1000◦C, Twall ≈ 500◦C,

Wε = 0.1, h = 100

m2K

TTC ≈ 830◦C, ΔT ≈ 200◦C

Conclusions:

1. Objects that “see” cold surroundings are colder than you think2. If an object “sees” a hot source it can be unexpectedly hot3. “In vacuum” indicates no convection → radiation must be important

Topics Covered So Far:

Heat Beat Mixheat transfer solid mechanics diffusion

fluid mechanics phase transition

3.044 MATERIALS PROCESSING 3

Next step is to discuss heat transfer combined⇒with diffusion and phase transformations

Solidification: Heat transfer plus phase transition, single component solidification

What are the B.C at the solid/liquid interface?

1. T = Tm

2. heat balance:qin︷ ︸︸ ︷

−ks∂T

∂x

∣∣∣∣s

=

qout︷ ︸︸ ︷−kl

∂T

∂s

∣∣∣∣l

3. heat of fusion

Look closely:

∂Tqin = −kl

∂x

∣∣∣∣x=s,l

4 LECTURE 8

∂Tqout = −ks

∂x

∣∣∣∣x=s,s

Fusion: −Hf

[kJ

kg

]

In time Δt the interface moved Δs ⇒ Volume transformed = AΔs(∂T−kl∂x

)A

︸ ︷︷∂T

+

(ks

in︸ ∂x

)A

︸ ︷︷ ︸out

−Hfρ

(Δs

Δt

)A

︸ ︷︷ ︸Heat of Fusion

= 0

Δswhere

Δt=

∂s= interface velocity

∂t

=−k ∂T

s ∂x

∣∣s− kl

∂T∂x

∣∣l

Hfρ

k=

∂T

Hfρ

[∂x

∣∣∣∣s

− ∂T

∂x

∣∣∣∣l

](within factor of two)

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3.044 Materials ProcessingSpring 2013

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