Question

Consider a sealed 20-cm-high electronic box whose base dimensions are $40 \mathrm{~cm} \times 40 \mathrm{~cm}$ placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(100 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

Short answer

Answer: _____________ °C.

Step-by-step solution

Calculate the Surface Area of the Electronic Box

To calculate the amount of power radiated by the box, we first need to find its surface area (excluding the bottom surface, as heat transfer is negligible there). The box has the dimensions 40 cm x 40 cm x 20 cm. To compute the surface area, we will use the formula A = 2lw + 2lh + 2wh Where A is the surface area, l is the length, w is the width, and h is the height. Since we're ignoring the bottom surface, our formula will be A = lw + 2lh + 2wh.

Use Stefan-Boltzmann Law to Relate Power and Temperatures

The Stefan-Boltzmann Law states that the power \(P_{rad}\) radiated from a surface is proportional to the surface area, emissivity, and the fourth power of the temperature: $$ P_{rad} = \epsilon \sigma A (T_s^4 - T_\infty^4) $$ Where \(\epsilon\) is the emissivity (0.95), \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W/m^2 K^4})\), A is the surface area, \(T_s\) is the surface temperature, and \(T_\infty\) is the surrounding temperature. In this problem, we are given the power dissipated by the electronic components inside the box. This power must be equal to the power radiated from the surface of the box to the surroundings by radiation. So we can set up the equation to solve for the surrounding temperature: $$ P = \epsilon \sigma A (T_s^4 - T_\infty^4) $$

Convert Given Temperatures to Kelvin

Before we can substitute the known values into the equation, we must convert the given temperature of 55 degrees Celsius (the maximum allowed surface temperature) to Kelvin. Remember the conversion formula for Celsius to Kelvin: \(T_{K} = T_{\mathrm{C}} + 273.15\). So, \(T_s = 55^{\circ} \mathrm{C} + 273.15 = 328.15 \mathrm{K}\)

Substitute the Values and Solve for Surrounding Temperature

We have all the values required to solve for the surrounding temperature. Our equation now becomes: $$ 100 \mathrm{W} = 0.95 \times 5.67 \times 10^{-8} \mathrm{W/m^2 K^4} \times A \times (328.15^4 - T_\infty^4) $$ Rearrange the equation to isolate \(T_\infty^4\) and then solve for \(T_\infty\). Make sure to first calculate the surface area, A, from Step 1.

Convert the Found Temperature Back to Celsius

Once we've found the value of the surrounding temperature (\(T_\infty\)) in Kelvin, we can convert it back to Celsius using the conversion formula: \(T_{\mathrm{C}} = T_{K} - 273.15\) This will be the temperature at which the surrounding surfaces must be kept if the box is to be cooled by radiation alone, while keeping the box's surface temperature not exceeding \(55^{\circ} \mathrm{C}\).