Consider radiative heat transfer between two large parallel planes of surface emissivities 0.8. How many thin radiation shields of emissivity 0.05 be placed between the surfaces is to reduce the radiation heat transfer by a factor of 75?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Problem Statement:

Radiative heat transfer between two large parallel planes of surface emissivities 0.8. How many thin radiation shields of emissivity 0.05 be placed between the surfaces is to reduce the radiation heat transfer by a factor of 75?

Solution:

The reduction in radiation heat transfer is given by,

$$\frac{q_{1}}{q_{2}}=\frac{1}{75}$$

where, $q_{1}$ is the heat transfer rate without any shields and $q_{2}$ is the heat transfer rate with n number of shields.

The heat transfer rate can be given by,

$$q=\frac{\sigma A}{1/\epsilon_{1}+n/\epsilon_{i}+1/\epsilon_{2}}(T_{1}^{4}-T_{2}^{4})$$

where, $\sigma$ is the Stefan-Boltzmann constant, A is the area of the plane, $\epsilon_{1}$ and $\epsilon_{2}$ are the surface emissivities of the planes, $\epsilon_{i}$ is the emissivity of the shield, T1 and T2 are the temperatures of the planes.

Substituting the values, we get,

$$\frac{1}{75}=\frac{\frac{\sigma A}{1/0.8+n/0.05+1/0.8}(T_{1}^{4}-T_{2}^{4})}{\frac{\sigma A}{1/0.8+1/0.8}(T_{1}^{4}-T_{2}^{4})}$$

Simplifying the above equation, we get,

$$n=3$$

Therefore, the number of thin radiation shields required to reduce the radiation heat transfer by a factor of 75 is 3.

Answer: Option C (3)

Similar Chemical Engineering Doubts

Two large parallel planes with emissivity 0.4 are maintained at different temperatures and exchange heat only by radiation. What percentage change in net radiative heat transfer would occur if two equally large radiation shields with surface emissivity 0.04 are introduced in parallel to the plates?

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