Question

Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) 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 \(120 \mathrm{~W}\) 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: The surrounding surfaces must be kept at approximately 11.25°C if the box is to be cooled by radiation alone.

Step-by-step solution

Calculate surface area of the box

To find the surface area, we need to know the total area of all the walls of the box exposed to the surroundings except the base, which is assumed to have negligible heat transfer. Therefore, the area is given by the sum of the areas of four vertical walls which can radiate heat. \(A_{\mathrm{surface}} = 4 \times (\mathrm{height} \times \mathrm{width}) = 4 \times ( 0.2 \, \mathrm{m}\times 0.5 \, \mathrm{m}) = 0.4 \, \mathrm{m^2}\)

Arrange the radiative heat transfer formula

We need to find the surrounding temperature (\(T_{\mathrm{surrounding}}\)), so we will rearrange the radiative heat transfer formula to isolate this variable. \(\displaystyle T_{\mathrm{surrounding}}^4 = \frac{q - \epsilon \sigma A_{\mathrm{surface}}T_{0}^4}{\epsilon \sigma A_{\mathrm{surface}}}\)

Insert given values and solve for surrounding temperature

Now we will plug in the values of \(\epsilon\), \(\sigma\), \(A_{\mathrm{surface}}\), \(q\), and \(T_{0}\) to find \(T_{\mathrm{surrounding}}\). \(\epsilon = 0.95\) \(\sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4}\) \(A_{\mathrm{surface}} = 0.4 \, \mathrm{m^2}\) \(q = 120 \, \mathrm{W}\) \(T_{0} = 55 + 273.15 = 328.15 \, \mathrm{K}\) \(\displaystyle T_{\mathrm{surrounding}}^4 = \frac{120 - 0.95 \cdot 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \cdot 0.4 \, \mathrm{m^2} \cdot (328.15\,\mathrm{K})^4}{0.95 \cdot 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \cdot 0.4 \, \mathrm{m^2}}\) After calculating, we get: \(T_{\mathrm{surrounding}}^4 = 1.921\times 10^{11} \, \mathrm{K^4}\) Taking the fourth root of both sides, we get: \(T_{\mathrm{surrounding}} = 284.4\, \mathrm{K}\)

Convert the result back to Celsius

We will now convert the surrounding temperature back to Celsius. \(T_{\mathrm{surrounding}}^{\circ\mathrm{C}} = 284.4 - 273.15 = 11.25^{\circ}\mathrm{C}\) The surrounding surfaces must be kept at approximately \(11.25^{\circ} \mathrm{C}\) if the box is to be cooled by radiation alone.

Key concepts

Heat Dissipation in Electronics

Electronics generate heat when operating, which can affect performance and longevity if not managed properly. Heat dissipation is crucial in maintaining the operational integrity of electronic devices. Most electronics have a thermal management system, which can include passive or active cooling methods. Passive methods, like heat sinks, enhance heat transfer through increased surface area and materials with high thermal conductivity, whereas active methods can include the use of fans or liquid cooling systems.

In the given exercise, an electronic box dissipates heat solely through radiation, a form of passive cooling. Radiation allows heat to be transferred without direct contact or an intervening medium, ideal for a vacuum chamber. The challenge is to ensure that the temperature of the electronic box does not exceed a set limit, requiring a precise calculation of radiative heat transfer and control of the surrounding temperature.

Surface Area Calculation

Calculating the surface area is essential when dealing with heat transfer in devices. A larger surface area usually means more heat can be dissipated, leading to better cooling.

The problem provided requires calculating the area that radiates heat. This is done by summing the areas of all radiating surfaces. For box-shaped objects, like the given electronic box, the total area can be found by adding the individual areas of the sides that contribute to heat transfer. In the solution, the base's heat transfer is negligible, so it's excluded from the calculation.

Students must remember that accurate measurement and using the correct units are crucial to ensure the efficacy of the calculation, especially when applying formulas that involve area, like the Stefan-Boltzmann law.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a fundamental principle in thermodynamics that relates the power radiated from a black body to the fourth power of its temperature. It's expressed as:

\( P = \epsilon \sigma A T^4 \)
where \( P \) is the power radiated per unit surface area, \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \mathrm{W/m^2K^4}) \), \( A \) is the surface area, and \( T \) is the absolute temperature in kelvins.

In the context of our problem, the law allows us to calculate the required temperature of surrounding surfaces to dissipate the heat from the electronic box. By rearranging the formula to solve for the surrounding temperature, we can keep the electronic components within a safe operating temperature. Understanding the Stefan-Boltzmann law is crucial for professionals dealing with thermal management in various fields, including electronics, mechanical engineering, and aerospace.