Radiation/convection heat transfer problem. (Thermodynamics)

AI Thread Summary
A 1.5-m² black surface at 120°C loses heat to surrounding air at 30°C through convection and radiation. The convection heat loss is calculated to be 2430 Watts using the convection heat transfer coefficient of 18 W/(m²°C). The radiation heat loss requires the emissivity value, which is not provided, but a black surface typically has an emissivity of 1. The user is uncertain whether to assume this value for calculations or disregard it due to lack of information. Clarification on the emissivity assumption is sought before contacting the instructor for further guidance.
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Homework Statement



A 1.5-m^2 black surface at 120 degrees C is losing heat to the surrounding air at 30 degrees C by convection with a convection heat transfer coefficient of 18 W/(m^2 degree C), and by radiation to the surrounding surfaces at 10 degrees C. The total rate of heat loss from the surface is : ?

Homework Equations



Q (convection) = h * A *( delta T)

Q (radiation) = \epsilon * \sigma * A * (T^4 (surface) - T^4 (surroundings))

The Attempt at a Solution



I calculate the heat lost by convection to be 2430 Watts. However, I need to find the total heat loss from the surface, which includes the loss from radiation.

I have the surface Area, the sigma value (Stefan-Boltzmann constant) , and the temperatures (393 K for the emitting surface and 283 K for the surrounding surface) but no epsilon value is given. (emissivity). All it says is "black surface" and not the material. Should I assume the surface is ideal (blackbody radiation) and put a value of 1? Should I disregard the term altogether?

I wanted to post this here before I emailed my instructor, because I unless I'm doing something wrong I feel I lack enough information to complete the problem.
 
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"Black" implies an emissivity of 1.
 
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