Radiation/convection heat transfer problem. (Thermodynamics)

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SUMMARY

The discussion centers on calculating the total heat loss from a 1.5-m² black surface at 120°C, losing heat to surrounding air at 30°C through convection and radiation. The convection heat transfer coefficient is 18 W/(m²·°C), resulting in a convection heat loss of 2430 Watts. The radiation heat loss calculation requires the emissivity value, which is assumed to be 1 for a black surface, as it is treated as an ideal blackbody. The total heat loss combines both convection and radiation losses.

PREREQUISITES
  • Understanding of heat transfer principles, specifically convection and radiation.
  • Familiarity with the Stefan-Boltzmann law for radiation heat transfer.
  • Knowledge of emissivity and its significance in thermal calculations.
  • Ability to perform calculations involving temperature conversions between Celsius and Kelvin.
NEXT STEPS
  • Study the Stefan-Boltzmann constant and its application in thermal radiation calculations.
  • Learn about emissivity values for different materials and their impact on heat transfer.
  • Explore advanced heat transfer topics, including transient heat transfer and thermal resistance.
  • Review practical applications of convection and radiation in engineering systems.
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Students in thermodynamics, engineers working on heat transfer problems, and anyone involved in thermal management of systems.

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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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