3.044 MATERIALS PROCESSING
LECTURE 15
Pilkington Sheet Glass Process
Pour molten glass on a liquid, let it settle/flatten and then let it cool
· higher ρ than glass → ρglass = 2500 mkg3
· liquid temperature → between Trigid ∼ 550◦ C and Tworking ∼ 900◦ C
· immiscible
Answer: Liquid Metal
Sn (tin) → ρSn = 7000 mkg3
The velocity of the sheet is determined by economic input:
⇒ need 750 tonnes/day to compete with block casting and grinding
Vsheet = 10
Date: April 9th, 2012.
m
min
1
2
LECTURE 15
Two Physical Goals
1) Fluid Flow → flat, stable film, level glass
2) Heat Transfer → cool to 550◦ C from 900◦ C
Fluid Flow:
kg
m3
2
= 10 Pa · s
kg
= 7000 3
m
= 3 × 10− 3 Pa · s
ρglass = 2500
μglass
ρSn
μSn
3.044 MATERIALS PROCESSING
3
How long does it take to remove the transient? How long does it take to
get to steady-state?
In Diffusion:
τ
=
dimensionless
time
αt
=1
L2
kinematic
viscosity
ν
t
τ=
2
L
2
L
tss = μ
ρ
For glass:
L = 0.007 m,
μ = 102 Pa · s,
ρ = 2500
⇒ tss = 0.0001 s
kg
m3
For tin:
μ = 3 × 10− 3 Pa · s,
m
⇒ tss = 114 s, Vsheet = 10
min
L = 0.007 m,
Length of travel to achieve steady state: ≈ 20 m
ρ = 7000
kg
m3
4
LECTURE 15
Heat Transfer:
W
m2 K
W
Glass: k = 1.7
,
mK
hL
Bi:
= 0.04
k
Vapor/glass interface: h = 10
L = 0.007 m
Bi is Low! conduction is rapid
in glass and convection is slow
Tin: k = 35
W
mK
Compare the resistances of glass and Sn:
resistance of Sn is low → Sn is a great heat sink
How long does it take to cool?
3.044 MATERIALS PROCESSING
τ =1
t ≈ 15 − 30 sec
Summary of Fluid Flow
∂v̄
F¯ ∇P
= ν ∇2 v̄ + −
ρ
∂t
ρ
2
∂vx
∂ vx Fx 1 ∂P
=ν 2 +
−
ρ ∂x
∂t
∂y
ρ
Navier-Stokes Equation: solve this with boundary conditions → laminar flow
Fluids can transfer momentum to solids
5
6
LECTURE 15
Fluid Flow in a Pipe:
Pipe:
Vx =
At the wall:
τyx
ΔP 2
R − r2
4Lμ
∂Vz = −μ
∂r r=R
Stress on the Inner Pipe Wall:
τ0 =
μΔP
ΔP · R
2R =
4Lμ
2L
Kinetic Force: (drag force, total force exerted by the fluid on the solid)
Fk =
τ dA
A
ΔP R
=
2πRL
2L
Fk = ΔP πR2
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3.044 Materials Processing
Spring 2013
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