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