Question

A circular grill of diameter \(0.25 \mathrm{~m}\) has an emissivity of \(0.8\). If the surface temperature is maintained at \(150^{\circ} \mathrm{C}\), determine the required electrical power when the room air and surroundings are at \(30^{\circ} \mathrm{C}\).

Short answer

Answer: The required electrical power to maintain the surface temperature is 57.684 W.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the temperatures given in Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperature: Grill temperature: \(T_1 = 150^{\circ}\mathrm{C} + 273.15 = 423.15 \mathrm{K}\) Surrounding temperature: \(T_2 = 30^{\circ}\mathrm{C} + 273.15 = 303.15 \mathrm{K}\)

Calculate the surface area of the circular grill

As we know, the surface area (A) of a circle can be calculated using the formula \(A = \pi r^2\), where \(r\) is the radius of the circle. We are given the diameter as \(0.25\mathrm{~m}\), and the radius can be calculated as half of the diameter. So, the radius is \(0.125\mathrm{~m}\). Calculating the surface area, we get: \(A = \pi (0.125)^2 = 0.04909 \mathrm{m^2}\)

Calculate the net power radiated by the grill using Stefan-Boltzmann Law

The Stefan-Boltzmann Law can be written as \(P = \varepsilon A \sigma (T_1^4 - T_2^4)\), where \(P\) is the net power radiated by the grill, \(\varepsilon\) is the emissivity, \(A\) is the surface area, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\mathrm{Wm^{-2}K^{-4}}\)), \(T_1\) is the grill temperature, and \(T_2\) is the surrounding temperature. Substituting the values, we find the net power radiated by the grill: \(P = 0.8 \times 0.04909 \times 5.67 \times 10^{-8} \times (423.15^4 - 303.15^4) = 57.684 \mathrm{W}\)

Determine the required electrical power

The power radiated by the grill in the form of thermal radiation is equal to the electrical power required to maintain the temperature. Therefore, the required electrical power is: \(P_{required} = P = 57.684 \mathrm{W}\)