““Very Light” Wing Billet Very Light” Wing Billet Thermal AnalysisThermal Analysis

Brad Allen, Joseph Ashby, Brad Allen, Joseph Ashby,

Aaron WorltonAaron WorltonME 340 Fall 2006ME 340 Fall 2006

Disclaimer:Disclaimer:

This information is not public This information is not public information. The analysis is based information. The analysis is based off a Spectrum Aeronautical airplane off a Spectrum Aeronautical airplane design, and should not be used design, and should not be used without their consent. Thank you.without their consent. Thank you.

The ProblemThe Problem

Find all nodal temperatures during Find all nodal temperatures during the heating of a aluminum wing the heating of a aluminum wing billet.billet.

How many heating elements are How many heating elements are needed to obtain 450 K at the center needed to obtain 450 K at the center of wing billet?of wing billet?

Assumptions:Assumptions:

The billet shape is symmetrical.The billet shape is symmetrical. EGC Enterprises Q Foil Heaters were EGC Enterprises Q Foil Heaters were

used as the heating elements.used as the heating elements. Heaters Produce q’’ = 100 W/in^2 Heaters Produce q’’ = 100 W/in^2

(q_dot = 2.44x10^8 W/m^3)(q_dot = 2.44x10^8 W/m^3) Forced air convectionForced air convection Heaters only cover the top and bottom Heaters only cover the top and bottom

of wing billet.of wing billet.

The Break-DownThe Break-Down

Temperature ProfileTemperature Profile

ConclusionsConclusions

All nodal temperatures were found.All nodal temperatures were found. 450 K was obtained at the 450 K was obtained at the

center of wing billet.center of wing billet. 2880 – 1in^2 heaters needed.2880 – 1in^2 heaters needed.