Question

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface is black and has a temperature of \(400 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be $7500 \mathrm{~W} / \mathrm{m}^{2}\( and \)3200 \mathrm{~W} / \mathrm{m}^{2}$, respectively. If the temperature of the side surfaces is \(485 \mathrm{~K}\), the emissivity of the side surfaces is (a) \(0.37\) (b) \(0.55\) (c) \(0.63\) (d) \(0.80\) (e) \(0.89\)

Short answer

Answer: (c) 0.63

Step-by-step solution

Identify the given quantities

Given: Base surface temperature, \(T_b=400\:K\) Radiosity for the top surface, \(J_t=7500\:W/m^2\) Radiosity for the side surfaces, \(J_s=3200\:W/m^2\) Temperature of the side surfaces, \(T_s=485\:K\)

Write the formula to calculate emissivity

Emissivity (\(\epsilon\)) is given by the formula: \(\epsilon=\frac{J}{\sigma T^4}\) where, \(\sigma\) is the Stefan-Boltzmann constant with a value of \(5.67 \times 10^{-8}\: \text{W}/\text{m}^2\text{K}^4\) \(J\) is the radiosity \(T\) is the temperature Our aim is to determine the emissivity of the side surfaces using the given values.

Substitute the given values in the emissivity formula

Now let's substitute the values of \(J_s\), \(\sigma\) and \(T_s\) to find the emissivity of the side surfaces. \(\epsilon=\frac{J_s}{\sigma T_s^4} = \frac{3200}{(5.67 \times 10^{-8})(485)^4}\)

Calculate the emissivity

Now calculate the emissivity by using the substituted values in the formula: \(\epsilon \approx \frac{3200}{(5.67 \times 10^{-8})(485)^4} \approx 0.63\) So, the emissivity of the side surfaces is approximately \(0.63\). The correct answer is (c) \(0.63\).