Question

When using radiation shields, if the emissivity’s of all surfaces are equal, then the use of four radiation shields will reduce the radiative heat transfer by? ___ %

the two principal difficulties encountered in the solution of radiation problems are?

Answer 1

We have formula for finding heat transfer with shield and heat transfer with out shield. When all Emissivities are equal then,  

Qshield = Qwithout shield/(N+1)

N is the number of shield = 4

Qshield = Qwithout sheild/(4+1) = 0.2*Qwithout shield

It means heat transfer reduce 20 % of heat transfer with out shield.

Use four shield will reduce radiative heat transfer by 20%.

ans : 20 %

Two principal difficulties for radiation problems solution:

1. First we to aware about the temperature of the matetial and surrounding, always should be taken absolute temperature (K).

2.Second is the heat transfer direction , which surface is heat emit and which surface heat absorbed so that decide view factor of the system.