Question

An electronic box that consumes \(200 \mathrm{~W}\) of power is cooled by a fan blowing air into the box enclosure. The dimensions of the electronic box are \(15 \mathrm{~cm} \times 50 \mathrm{~cm} \times 50 \mathrm{~cm}\), and all surfaces of the box are exposed to the ambient environment except the base surface. Temperature measurements indicate that the box is at an average temperature of \(32^{\circ} \mathrm{C}\) when the ambient temperature and the temperature of the surrounding walls are \(25^{\circ} \mathrm{C}\). If the emissivity of the outer surface of the box is \(0.75\), determine the fraction of the heat lost from the outer surfaces of the electronic box.

Short answer

Answer: The fraction of heat lost from the outer surfaces of the electronic box is approximately 43.95%.

Step-by-step solution

Compute the Surface Area of the Electronic Box

First, we need to calculate the surface area of the exposed sides of the electronic box. The dimensions of the electronic box are 15 cm x 50 cm x 50 cm. We can convert these dimensions to meters: 0.15 m x 0.50 m x 0.50 m. All surfaces are exposed except the base surface, so we need to sum up the surface areas of the exposed sides: top, left, right, front, and back. Surface area = Top + Left + Right + Front + Back Surface area = (0.15 m * 0.50 m) + (0.50 m * 0.50 m) + (0.50 m * 0.50 m) + (0.15 m * 0.50 m) + (0.15 m * 0.50 m) Surface area = 0.075 m² + 0.25 m² + 0.25 m² + 0.075 m² + 0.075 m² Surface area = 0.725 m²

Find the Heat Transfer by Radiation

Now, we can use the Stefan-Boltzmann law for radiation heat transfer to find the heat transfer rate by radiation. The Stefan-Boltzmann law is given by the equation: \(q = \epsilon \sigma A (T_{box}^{4} - T_{ambient}^{4})\) where: \(q\) is the heat transfer rate by radiation, \(\epsilon\) is the emissivity of the outer surface of the box (given as 0.75), \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8} \mathrm{W/m^2 K^4}\)), \(A\) is the surface area of the exposed sides of the box, \(T_{box}\) is the average temperature of the box in Kelvin, and \(T_{ambient}\) is the ambient temperature in Kelvin. First, we need to convert the temperature values from Celsius to Kelvin: \(T_{box} = 32^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 305.15 \mathrm{K}\) \(T_{ambient} = 25^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 298.15 \mathrm{K}\) Now, we can substitute the given values into the Stefan-Boltzmann law: \(q = 0.75 \times 5.67\times10^{-8} \mathrm{W/m^2 K^4} \times 0.725 \mathrm{m^2} \times (305.15 \mathrm{K}^{4} - 298.15 \mathrm{K}^{4})\) \(q \approx 87.89 \mathrm{~W}\)

Determine the Fraction of Heat Lost from the Outer Surfaces

Finally, we need to find the fraction of heat lost from the outer surfaces of the electronic box. Divide the heat transfer rate by radiation (\(q\)) by the power consumed by the electronic box, and multiply by 100 to get the percentage: Fraction of heat lost = (\(87.89 \mathrm{~W}\) / \(200 \mathrm{~W}\)) * 100 Fraction of heat lost = 43.95 % The fraction of heat lost from the outer surfaces of the electronic box is approximately 43.95%.