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Rocket Man. ME 340 Final Project Bryan Johnson, Devin LeBaron. Introduction. The Problem. How close could a spaceship fly to the sun before melting?. The Sun. Temp = 5800 K Radius = 6.95e8 m Assume blackbody ( ε =1). The Space Ship. Assume Aluminum body Melting Temp = 933 K - PowerPoint PPT Presentation
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ME 340 Final Project
Bryan Johnson, Devin LeBaron
Introduction
How close could a spaceship fly to the sun before melting?
The Problem
Temp = 5800 K
Radius = 6.95e8 m
Assume blackbody (ε=1)
The Sun
Assume Aluminum body
Melting Temp = 933 K
To approximate the shape…
We will use a sphere
The Space Ship
E*As
Energy Balance
αGs*Ac
E*As = αGs*AcεσT^4*As = αGs*AcεσT^4 = αGs/4
Gs = 4εσT^4/α
Ac = πD^2/4As = πD^2
In space, convection and conduction don’t apply
Space Ship
Chart based on graph from Fig 12.17 from the book
Emissivity and AbsorptivityAt Tm = 933 K:F1 = 0.4301; F2 = 0.9628ε(Tm) = 0.2(0.4301) +
0.9(0.9628 - 0.4301) + 0.35(1 – 0.9628)ε(Tm) = 0.58
At Ts = 5800 K:F1 = 0.9904; F2 = 0.9998α(Ts) = 0.2(0.9904) +
0.9(0.9998 - 0.9904) +
0.35(1 - 0.9998)α(Ts) = 0.2066
qs = Es*As
qs = σT^4*As
qs = Gs*Ab
Gs*Ab = σT^4*As
Gs = σT^4*As/Ab
Gs = σT^4*Rs^2/(D+Rs)^2
Radiation of the SunAb
As
Ship
D
Gs = σTs^4*Rs^2/(D+Rs)^2
Gs = 4εσTm^4/α
4εσTm^4/α = σTs^4*Rs^2/(D+Rs)^2
Solving for D:
D = [(αTs^4*Rs^2)/(4εTm^4)]^.5 – Rs
Combine the formulasε = 0.58α = 0.2066Tm = 933 KTs = 5800 KRs = 6.95e8 m
D = 7.32e9 m
The final distance to melting your aluminum space ship would be about:
7.32e9 m
= 4,550,000 miles
The Answer
Mercury is about 5.8e10 m from the sun. Which means you’d be pretty darn close.
What are you doing flying to the sun? Just don’t. But if you must, please use a better material with a higher melting point, better heat resistance and lower absorptivity.
Conclusions and Recommendations
Incropera, Frank P., and Frank P. Incropera. Fundamentals of Heat and Mass Transfer. Hoboken, NJ: John Wiley, 2007. Print.
Rocketman. Perf. Harland Williams, Jessica Lundy, and William Sadler. Disney, 1997.
References