Question

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures T1 and T2, with T1 > T2. To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity ε s < 0.5, while the opposite surface has an emissivity of 2ε s. (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity ε s or that of emissivity 2ε s be oriented toward the plate at T1? (b) What orientation will result in the larger value of the shield temperature Ts?

Answer 1

To achieve a larger reduction in heat transfer, the shield with lower emissivity should face the hotter plate. To maximize the shield temperature, the higher emissivity side should face the hotter plate.

Step-by-step explanation:

To reduce the rate of heat transfer by radiation between two large parallel plates with temperatures T1 and T2, where T1 > T2, a thin shield with different emissivities on opposite surfaces is introduced. The correct orientation of the shield is crucial for it to be effective.

• (a) Reducing Heat Transfer

The shield should be oriented so that the surface with the lower emissivity (ε s) faces the hotter plate (T1). This setup will lead to a larger reduction in heat transfer because the lower emissivity surface will emit less radiation towards the hotter plate, compared to the side with an emissivity of 2ε s.

• (b) Maximizing Shield Temperature

If the goal is to attain a higher temperature for the shield itself (ε s), it should be oriented so that the higher emissivity surface (2ε s) faces the hotter plate (T1). The higher emissivity allows the shield to absorb more radiation from the hotter plate, consequently increasing the shield's temperature.

Answer 2

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Step-by-step explanation:

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