Question

Two very large parallel plates are maintained at uniform temperatures of \(T_{1}=1100 \mathrm{~K}\) and \(T_{2}=700 \mathrm{~K}\) and have emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.5\), respectively. It is desired to reduce the net rate of radiation heat transfer between the two plates to one-fifth by placing thin aluminum sheets with an emissivity of \(0.1\) on both sides between the plates. Determine the number of sheets that need to be inserted.

Short answer

Based on the information given, calculate the number of aluminum sheets needed to be inserted between the parallel plates to reduce the net radiation heat transfer to one-fifth of its initial value. Use the given emissivities and temperatures, as well as the Stefan-Boltzmann constant to determine the initial heat transfer rate, the desired heat transfer rate, and finally the number of aluminum sheets required to achieve the heat transfer reduction.

Step-by-step solution

Understand the initial heat transfer between the plates.

Using the equation of radiative exchange between parallel surfaces, the initial heat transfer can be calculated, given by: $$q_{1,2} = A\frac{\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}-1}$$ Here, \(A\) is the surface area of the parallel plates, \(\sigma\) is the Stefan-Boltzmann constant, and temperatures are in Kelvin.

Calculate the initial heat transfer rate.

First, let's calculate the initial heat transfer rate between the two given plates, using the emissivities and temperatures: $$q_{1,2} = A\frac{(5.67\times10^{-8}\mathrm{W}/\mathrm{m}^2\mathrm{K}^4)(1100^4 - 700^4)}{\frac{1}{0.5}+\frac{1}{0.5}-1}$$

Calculate the desired heat transfer rate

Now, as we want to reduce the heat transfer rate to one-fifth of the initial value, the desired heat transfer rate would be: $$q_{\text{desired}} = \frac{1}{5}q_{1,2}$$

Use the method of view factors.

To determine the number of aluminum sheets required, let's use the method of view factors \(F_{1,i}\) for parallel surfaces. We take the view factor from plate 1 to the i-th aluminum sheet as \(F_{1, i}\). The view factor from plate 1 to plate 2 is \(1\) since they are parallel surfaces. We can write the view factors in terms of the emissivities of aluminum sheets and the number of sheets, as follows: $$1 = \sum_{i=1}^n\varepsilon_iF_{1,i} + F_{1,2}$$, where \(n\) is the number of aluminum sheets.

Determine the heat transfer rate with aluminum sheets.

With the aluminum sheets between the plates, the heat transfer rate can be written as: $$q_{\text{desired}} = A\sum_{i=1}^n\frac{\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_{1}}+\frac{1}{\varepsilon_{i}}-1}F_{1,i}$$ Now, we want to find the smallest integer \(n\) such that the desired heat transfer rate is one-fifth of the initial heat transfer rate.

Calculate the number of aluminum sheets.

Now, we need to find the number of aluminum sheets \(n\) necessary for the desired heat transfer rate: $$n = \left\lceil\frac{q_{1,2}}{5\left(\frac{\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_{1}}+\frac{1}{0.1}-1}\right)}\right\rceil$$ We use the ceiling function \(\lceil \cdot \rceil\) to find the smallest integer value for \(n\). After calculating the values, we find the number of aluminum sheets required to reduce the heat transfer rate to one-fifth of its initial value.