steel balls
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Steel balls. Topic 3. Taiwan Representative. Outline. Question Experimental set up: Vertical collision Materials: Paper, Plastic, and Metal sheets Experimental results: Hole size v.s materials, height… What we learned & discussion Conclusion. Question. - PowerPoint PPT PresentationTRANSCRIPT
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Steel balls
Topic 3
Taiwan Representative
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Outline Question Experimental set up:
Vertical collision Materials: Paper, Plastic, and Metal sheets Experimental results: Hole size v.s materials, height… What we learned & discussion Conclusion
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Question
Colliding two large steel balls with a thin sheet of material (e.g. paper) in between may "burn" a hole in the sheet. Investigate this effect for various materials.
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Experimental Setup: vertical collision
1 m
H
Sheet
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Materials 1. Paper (Burning point: 130o~232oC)
Reference:http://wiki.answers.com/Q/What_is_the_Burning_point_of_paper
2 mm 5 mm
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Inflammable materialsPaper
A thin piece of material
left on the ball
Greatly heated
Burned
O2
Greatly heated
Burned
O2
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Materials 2. Plastic (Melting Point: 106~114 )℃ (Soften temperature: 80~98 )℃
3 mm 4 mm
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Non-flammable materialsPlastic
Compressed
Into thin layerMelted
Wrinkles
Cooled down
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Materials 3. Aluminum foil (Melting point: 660 C)
5 mm
6 mm
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Non-flammable materials
Aluminum foil
Concentric circles wrinkles
Melted
Compressed
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Materials 4.Copper foil (Melting Point: 1083.0 °C)
Reference:www.chemicalelements.com/elements/cu.html
5 mm2mm
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Non-flammable materialsCopper foil
Concentric circles wrinkles
Depends on condition
Compressed
Melted ??
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H
Plastic Paper
= 7.6 cm= 7.6 cm
= 7.6 cm
= 7.6 cm
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Height v.s. Materials
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Height v.s. Materials
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Height v.s. Materials
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Height v.s. Materials
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Estimation of physical parameters during collision
The highest temperature Tmax.= ?
Duration ∆t = ?
Collision area A=?
Average pressure P = ?
Conducted heat ∆Q = ?
Adiabatic process? i.e. ∆W >> ∆Q
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We can melt Al foil . Melting point of Al = 660 .℃
Tmax.= ?
Cooled after melted
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Duration ∆t = ?
Our high speed video: 230 μs/frame
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Material
t=0
Dropping ball
Fixed ball
Edge of pillar
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t=230μs
Material
Dropping ball
Fixed ball
Edge of pillar
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t=230μs x 2
Material
Dropping ball
Fixed ball
Edge of pillar
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t=230μs x 3
Material
Dropping ball
Fixed ball
Edge of pillar
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Duration ∆t = ?
1. Our high speed video: 230 μs/frame
∆t < 230 μs
2. From referenceour mball=287 g
R. Hessel, A. C. Perinotto, R. A. M. Alfaro, and A. A. Freschia, Am. J. Phys. 74 3, 176 (2006).
We can conclude : ∆t < 230 μs
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Collision area A=?
Al foil
Diameter ~ 1.6 mm
Estimate A from the melted Al foil .
Diameter: 1.58± 0.08 mm
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Average pressure P = ?
1F pP
A t A
m=287 g, H=50 cm, g=9.8 m/s2, A< π(0.8mm) 2 , ∆t < 230 μs
2 2 2 2o op m gH p m gH p
)(109.114.30008.0
1
103.2
5.08.92287.02
924 m
NtP
41.9 10P atm
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)(109.3)( 1 JdAPVPW
Work ∆W=?Pressure P > 1.2 x 109 Nt/m2
Collision Area A=πr2 ~ 2x10-6 m2
Thickness of sheet ∆d= 10-4 m
)(108.1~103.21000)102(401 446 JQ
TAt
Q
Conducted heat ∆Q=?∆T < 1000 ℃
A=πr2 ~ 2 x 10-6 m2
σ
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Heat produced
W H m s T
0.39( )W J
Using the lowest production of work,
Increase ofΔT ,
640 25 640 665o oT C T C => Higher than melting point of Al: 660
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Adiabatic process?? Work done during collision ∆W > 3.9 ×10-1 (J) Heat conducted during collision :
∆Q < 2.9 ×10-4 (J)
∆W >> ∆Q
Adiabatic process
ThighTlow
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Conclusion
Height: Diameter of holes increases as height increases
Collision Area: Burning Melting
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Colliding process
P T P T
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Burning
O2O2
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Melting
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Conclusion Height: Diameter of holes increases as
height increases Collision Area:
Melting Burning
Adiabatic Process is the main reason holes are burned because: Δt is small Area is small Pressure is high
high is T
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Conclusion Height: Diameter of holes increases as
height increases Collision Area:
Melting Burning
Adiabatic Process is the main reason holes are burned because: Δt is small Area is small Pressure is high
high is T