ME 340 Final Project Bryan Johnson, Devin LeBaron 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 To approximate the shape… We will use a sphere

Energy Balance Gs = 4εσT^4/α In space, convection and conduction don’t apply 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 E*As Space Ship

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 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 Chart based on graph from Fig 12.17 from the book

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

Combine the formulas D = 7.32e9 m 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

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.

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.

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.