1 / 12

Rocket Man

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

dmccowan
Download Presentation

Rocket Man

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Rocket Man ME 340 Final Project Bryan Johnson, Devin LeBaron

  2. Introduction

  3. The Problem • How close could a spaceship fly to the sun before melting?

  4. The Sun • Temp = 5800 K • Radius = 6.95e8 m • Assume blackbody (ε=1)

  5. The Space Ship • Assume Aluminum body • Melting Temp = 933 K • To approximate the shape… • We will use a sphere

  6. Energy Balance In space, convection and conduction don’t apply • E*As αGs*Ac Ac = πD^2/4 As = πD^2 E*As = αGs*Ac εσT^4*As = αGs*Ac εσT^4 = αGs/4 Gs = 4εσT^4/α Space Ship

  7. Emissivity and Absorptivity At Tm = 933 K: F1 = 0.4301; F2 = 0.9628 ε(Tm) = 0.2(0.4301) + 0.9(0.9628 - 0.4301) + 0.35(1 – 0.9628) ε(Tm) = 0.58 • Chart based on graph from Fig 12.17 from the book At Ts = 5800 K: F1 = 0.9904; F2 = 0.9998 α(Ts) = 0.2(0.9904) + 0.9(0.9998 - 0.9904) + 0.35(1 - 0.9998) α(Ts) = 0.2066

  8. Radiation of the Sun Ab • 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 Ship D As

  9. Combine the formulas • 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 ε = 0.58 α = 0.2066 Tm = 933 K Ts = 5800 K Rs = 6.95e8 m D = 7.32e9 m

  10. The Answer • The final distance to melting your aluminum space ship would be about: • 7.32e9 m = 4,550,000 miles Mercury is about 5.8e10 m from the sun. Which means you’d be pretty darn close.

  11. Conclusions and Recommendations • 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.

  12. References • 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.

More Related