amayaroberson6727 amayaroberson6727
  • 21-08-2019
  • Engineering
contestada

Calculate the heat dissipation by radiation through a 0.2m^2 opening of a furnace at 1000K into an ambient at 300K. Assume that both the furnace and the ambient blackbodies.

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Netta00 Netta00
  • 02-09-2019

Answer:

Heat dissipation is 11.24 KW.

Explanation:

Given that

Temperature of furnace= 1000 K

Temperature of surrounding = 300 K

Also given that both bodies are black bodies so ∈=1

We know heat radiation for black bodies is given as follows

[tex]Q=\sigma A(T_1^4-T_2^4)[/tex]

Now by putting the values in the above formula

[tex]Q=5.67\times 10^{-8} \times 0.2(1000^4-300^4)[/tex]

Q=11.24 KW

So heat dissipation is 11.24 KW.

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