Conduction of Heat and Radiation problem

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SUMMARY

The discussion focuses on a heat conduction and radiation problem involving a small sphere with an emissivity of 0.90 and a spherical asbestos shell. The inner sphere maintains a temperature of 800.0°C, while the inner surface of the shell is at 600.0°C. The relationship between the inner and outer radii is given as r2/r1 = 10.0, and the thermal conductivity of asbestos is specified as k(asbestos) = 0.090 J/s·m·°C. The solution requires applying conservation of energy principles, equating the heat transfer by conduction and radiation to find the temperature of the outer surface of the shell.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction and radiation.
  • Familiarity with the Stefan-Boltzmann law for radiation: Q = eσT^4A.
  • Knowledge of thermal conductivity and its units, particularly in the context of materials like asbestos.
  • Ability to manipulate equations involving area and temperature gradients in spherical coordinates.
NEXT STEPS
  • Study the conservation of energy in thermal systems to understand heat transfer dynamics.
  • Learn about the Stefan-Boltzmann law and its application in thermal radiation problems.
  • Explore the derivation and application of Fourier's law of heat conduction.
  • Investigate the impact of emissivity on thermal radiation and its calculation in practical scenarios.
USEFUL FOR

Students in thermodynamics, engineers working with thermal systems, and anyone involved in heat transfer analysis or thermal management of materials.

Lexxian
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Homework Statement


The question - A small sphere (emissivity = 0.90, radius r1) is located at the center of a spherical asbestos shell (thickness=1.0cm, outer radius r2). The thickness of the shell is small compared to the inner and outer radii of the shell. The temperature of the small sphere is 800.0 degree C, while the temperature of the inner surface of the shell is 600.0 degree C, both temps remain constant. Assuming that r2/r1 = 10.0 and ignoring any air inside the shell, find the temperature of the outer surface of the shell.

k(asbestos)=0.090 J/s x m x degrees Celsius


Homework Equations


Q=(kA deltaT)t/L (minus the t)
Q= e sigma T^4^At (minus the t)
A= 4 pi r^2^

The Attempt at a Solution


I know that I will be using both equations. I am unsure if I set them equal to each other (similar to conservation of energy equations) which I feel I was lead to that thought process by the statement of ignoring any air inside the shell. Also, in order to find the radius thus giving me the Area I need to solve the equations listed. Unless...the r2/r1 = 10 will actually provide me with...nah...because the Area of the smaller sphere is smaller than the area of the whole sphere. Please help.
 
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Hi Lexxian, welcome to PF. Yes, it would be a good idea to apply conservation of energy to this problem. At steady state, all the energy leaving the inner sphere as radiation will be conducted through the outer sphere. Does this help you get started?

(Also, check your equations - it looks like there are a few typos.)
 

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