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Homework Statement
A light bulb has an internal filament with a total surface area of 2.5 X 10-3 ft2 that operates at a temperature of 2240 F. The filament has an emissivity of 1.0 and radiates through a vacuum to the glass wall of the bulb. Only 10 percent of the total energy emitted as radiation from the filament is absorbed by the surrounding glass bulb. Heat that is absorbed by the glass bulb is subsequently lost from the outer surface of the glass bulb (ε = 1.0) by both radiation and natural convection to the surroundings (surrounding temperature = 80 F). The heat transfer coefficient for natural convection under these conditions is 1.17 BTU / hr ft2 F.
What is the surface temperature of the bulb assuming the bulb can be modeled as a sphere 3 inches in diameter?
Homework Equations
The Attempt at a Solution
I calculated the convection term and radiation from the glass to the surroundings, and set it equal to 0.1*energy from filament to glass. I am not sure if my expressions are correct for all these various forms of heat transfer.