Radiation heat loss with a shield can be calculated once we know the temperature of the shield. The radiative flux from the first body to the shield is equal to the radiative flux from the shield to the second body. This is true because no heat is stored in the shield.

Upon rearrangement, T₃ can be found to be

This value of temperature can now be used to compute heat transfer, Q.

Example 5.19: The emissivities of two parallel planes, P₁ and P₂, are 0.75 and 0.38, respectively. They are maintained at steady temperatures of 117 °F and 63 °F respectively. An aluminum shield of emissivity 0.046 is employed. Calculate the radiative heat flux per unit area (a) without the radiation shield, (b) with the radiation shield.

Solution: It can be assumed that all areas are same.

T₁ = 577 °R, T₂ = 523 °R

ε₁ = 0.75, ε₂ = 0.38, ε₃ = 0.046

(a) When there is no radiative shield, heat transfer just occurs between two surfaces, and the radiative heat transfer can be computed as

And the radiative heat transfer is

(b) With a protective shield, the emissivity factor F₁₃ can be found to be

And the emissivity factor F₂₃ can be found to be

Now we can find the temperature of the shield, T₃ as

And Q can be found to be