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Production of Steel Shot
Josh Ball – [email protected] Calcutt – [email protected] Loney – [email protected]
Introduction
• Background• Process Overview• Calculations• Conclusions
Background
• What is Steel Shot?– Tiny steel balls
• What is it used for?– Cleaning work pieces (Shot Blasting)
• Sand and Scale removal from castings, Surface prep for painting
– Shot Peening– Granite Cutting– Non-Toxic Shotgun ammunition
Process Overview
• Molten Steel flows from a tundish and is made into a spray.
• The droplets (1mm diameter spheres) freefall in a cylindrical chamber containing a gas atmosphere.
• Upon reaching 1000C, they will land in a fluidized bed for further cooling.
Objective
• Determine what gas (He ,Ne, Ar, Kr) in the vessel will result in the fastest solidification time, and therefore the shortest vessel.
• Determine the relationship between air velocity and cooling time in the fluidized bed.
Assumptions
• The steel shot will not deform on impact• The conveyor movement will not impact the
rate of cooling• Density and size of particles does not change
with temperature• 1 Million pounds of shot would be produced
in a 24 hour period
Calculating terminal velocity of the droplets for each atmosphere
vtHe 31.164m
s vtNe 16.058
m
s vtAr 11.827
m
s vtKr 8.111
m
s
ReAr vtAr 2 r vtAr Ar
Ar
vtAr1
4 2 r( ) st Ar 9.8m
s2
3 Ar18.5
2 r Ar
Ar
3
5
5
7
ReAr vtAr1 1.224 103 Re > 500 So using friction factor 2 is incorrect
vtAr
4 2r( ) st Ar 9.81m
s2
3 .44 Ar
1
2
ReAr vtAr 940.997 Friction factor 3 gives an acceptable value for Re (500 < Re < 20000)
Finding the heat transfer coefficient for each process
Finding the heat transfer coefficient, h, for each process
Nu 2 0.6Re
1
2Pr
1
3 Pr Cp
kNu
h 2 rk
hk
2r2 0.6Re
1
2Pr
1
3
PrHe
He CpHe
kHe
hHe
kHe
2 r2 0.6ReHe vtHe
1
2PrHe
1
3
hHe 1.533 103
kg
K s3
hNe 595.92kg
K s3
hAr 293.09kg
K s3
hKr 169.871kg
K s3
Calculating Biot Numbers
Bih r3k
BiHe 6.573 103 BiNe 2.555 10
3 BiAr 1.257 103 BiKr 7.283 10
4
All Biot numbers are below 0.1 = No significant temperature gradients across the solid
Calculating Cooling Time and Chamber Height
Time required to remove superheatt ln
Tmelt Tgas
Ttundish Tgas
1( )r
3
st Cpst
h
tSHHe lnTmelt Tgas
Ttundish Tgas
1( )r
3
st Cpst
hHe
tSHHe 0.071s
tfusionHe
1
3r3
r2 hHe
Hs st Tmelt Tgas
Time required to remove heat of fusion
tfusionHe 0.142s
Time required to cool from melting temp to 1000C
tcoolHe ln1273K Tgas
Tmelt Tgas
1( )r
3
st Cpst
hHe
tcoolHe 0.225sttotalHe tcoolHe tfusionHe tSHHe
ttotalHe 0.437s ttotalNe 1.125s ttotalAr 2.288s ttotalKr 3.948s
ChamberHeightNe vtNe ttotalNe
ChamberHeightHe 13.632m ChamberHeightNe 18.072m ChamberHeightAr 27.064m ChamberHeightKr 32.023m
Void Area fractionVoid Area Fraction of a Close Packed Bed
a2
a2 4 r( )
2
a 8r2
Area a2
8r2 2 8r
2
Area_Spheres 2 r2 2r2
Volume_Voids a2
2r2 r28 2
Void_Area_FracArea_Voids
Area
r28 2
8r2
8 28
8 28
0.215
Bed ThicknessDetermining Thickness of Cooling Bed
DailyProduction 1 106lb
day ConveyorSpeed .05
m
s
BedWidth 0.5m
VolumeFlowRateDailyProduction
st
VolumeFlowRate 6.679 104m3
s
BedThicknessVolumeFlowRate
BedWidth ConveyorSpeed 1
BedThickness 34.017mm
Velocity of AirDetermining Minimum Fluidization Velocity for Steel Shot
Remf 33.673 1134 0.0408Ga( )
1
2 Minimum Fluidization equation
Remf
gas vmf dst
gasGa
dst3 st gas gas 9.81
m
s2
gas2
air 1.202kg
m3
air 1.695105kg
m s
Gaair
2r( )3 st air air 9.81
m
s2
air2
Reair 33.673 1134 0.0408Gaair 1
2
vmfair
Reair air
air 2 r
vmfair 1.211m
s
Terminal velocity was calculated the sameas before
vtair
4 2r( ) st air 9.81m
s2
3 .44 air
1
2
vtair 13.941m
s
vair 1.211m
s1.3
m
s 13.941
m
s
Cooling in Fluidized BedsTcoolair 298K Cpair 1001
J
kg K kair 0.02394
W
m K
Prair
air Cpair
kair
Recoolair vair air vair 2 r
air
hair vair kair
2 r2 0.6Recoolair vair
1
2Prair
1
3
Biair vair hair vair r3 kst
0 5 10 155 10 4
0.001
0.0015
0.002
Biair vair
vair0 5 10 15
100
200
300
400
500
hair vair
vair
Bi is always less than 0.1 so there are notemperature gradients across the solid
Cooling in Fluidized Beds continued
tcoolair vair ln300K Tcoolair
1273K Tcoolair
1( )r
3
st Cpst
hair vair
Gives the amount of time to cool steel shotfrom 1000 degress C to room temperature
0 5 10 1510
15
20
25
30
35
tcoolair vair
vair
Determining length of the bed required
ConveyorLengthvair ConveyorSpeedtcoolair vair
0 5 10 150.5
1
1.5
2
ConveyorLength vair
vair
ConveyorLength 1.211m
s
1.509m
ConveyorLength 13.941m
s
0.558m
The conveyor (when moving at 5 cm/s) must bebetween 0.558 m and 1.509 m depending on the airvelocity. Since we are trying to conserve space,blowing the air at 13.941 m/s would make theconveyor shortest.
Conclusion
• Helium gives the smallest chamber height (13.63m)
• At 0.5m wide the cooling bed is 34mm thick
• The length of the cooling bed depends upon the velocity (0.56m to 1.5m)
Sources
• Dr. Hackney’s wonderful class notes• http://encyclopedia.airliquide.com/
encyclopedia.asp• http://chem.lapeer.org/PhysicsDocs/
Goals2000/Laser1.html