Question

A thin aluminum sheet with an emissivity of \(0.12\) on both sides is placed between two very large parallel plates maintained at uniform temperatures of \(T_{1}=750 \mathrm{~K}\) and \(T_{2}=400 \mathrm{~K}\). The emissivities of the plates are \(\varepsilon_{1}=0.8\) and \(\varepsilon_{2}=0.7\). Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and the temperature of the radiation shield in steady operation.

Short answer

Question: Calculate the net heat transfer per unit surface area between two large parallel plates with temperatures of 750K and 400K, and emissivities of 0.8 and 0.7, respectively, separated by an aluminum radiation shield with emissivity 0.12 on both sides. Answer: First, find the view factors for each surface in contact using the given emissivities. Next, calculate the temperature of the radiation shield using the equality of heat transfers between the shield and both plates. Finally, find the net heat transfer per unit surface area between the two plates using the heat transfer equation.

Step-by-step solution

Finding the view factors

To determine the view factors of the two plates, we first look at the emissivity of the radiation shield, which is given as 0.12 for both sides. Since the aluminum sheet is a radiation shield, both plates and the radiation shield will have view factors close to 1. Therefore, we can assume \(F_{1-shield} = 0.12\) and \(F_{2-shield} = 0.12\). To find the view factors between the plates \(F_{12}\), we can use the relationship \(F_{12} = 1 - F_{1-shield} - F_{2-shield}\).

Calculating the temperature of the radiation shield

In steady operation, the heat transfer between plate 1 and the shield and between plate 2 and the shield should be equal. So, we can represent these two heat transfers using the following equations: *q_{1-shield} = σ * ε_1 * ε_{shield} * (T_1^4 - T_{shield}^4) * F_{1-shield} = σ * ε_2 * ε_{shield} * (T_{shield}^4 - T_2^4) * F_{2-shield}$ We can plug in the given values for the emissivities (\(ε_1 = 0.8\), \(ε_2 = 0.7\), and \(ε_{shield} = 0.12\)) and temperatures (\(T_1 = 750K\) and \(T_2 = 400K\)) into this equation, and solve for the temperature of the radiation shield, \(T_{shield}\).

Calculating the net heat transfer per unit surface area

Now that we have the temperature of the radiation shield, we can find the net heat transfer per unit surface area between the two plates using the total heat transfer equation: *q_{total} = σ * ε_1 * ε_2 * (T_1^4 - T_2^4) * F_{12}$ We know all the variables in the equation, so we can plug them in and solve for the net heat transfer.

Finalizing the solution

In conclusion, first, we found the view factors for each surface in contact. Then, we calculated the temperature of the radiation shield using the equality of heat transfers between the shield and both plates. Finally, we found the net heat transfer per unit surface area between the two plates.