Question

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\) and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

Answer: (d) 8.3 W/m²K

Step-by-step solution

Calculate the heat transfer rate by radiation

Use the Stefan-Boltzmann Law to find the heat transfer rate due to radiation: \(Q_{rad} = \epsilon \times A \times \sigma \times (T_{person}^4 - T_{room}^4)\) Given values: \(\epsilon = 0.90\), \(A = 1.5 \mathrm{m}^2\), \(T_{person} = 32+273.15 = 305.15 \mathrm{K}\), \(T_{room} = 20+273.15 = 293.15 \mathrm{K}\) To find the value of \(Q_{rad}\), first calculate the difference in temperatures to the power of 4: \((T_{person}^4 - T_{room}^4) = (305.15^4 - 293.15^4) = 156776353.61 \mathrm{K}^4\) Now, multiple this by \(\epsilon\), \(A\), and the Stefan-Boltzmann constant \(\sigma \approx 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\): \(Q_{rad} = 0.90 \times 1.5 \times 5.67 \times 10^{-8} \times 156776353.61 \mathrm{W}\) \(Q_{rad} = 144.03 \mathrm{W}\)

Calculate the heat transfer rate by convection

Since we know that the heat transfer rates by radiation and convection are equal, we can set \(Q_{conv} = Q_{rad} = 144.03 \mathrm{W}\).

Find the combined heat transfer coefficient

Now that we have the heat transfer rates, we can find the combined heat transfer coefficient using the formula for heat transfer by convection: \(Q_{conv} = h \times A \times (T_{person} - T_{room})\) First, find the difference in temperatures: \((T_{person} - T_{room}) = (305.15 - 293.15) = 12\mathrm{K}\) Now, we can rearrange the formula to solve for the combined heat transfer coefficient \(h\): \(h = \frac{Q_{conv}}{A \times (T_{person} - T_{room})} = \frac{144.03 \mathrm{W}}{1.5 \mathrm{m}^2 \times 12 \mathrm{K}}\) Finally, plug in the values: \(h = \frac{144.03}{1.5 \times 12} \approx 8.00 \mathrm{W/m^2K}\) The closest value to our answer is \(8.3 \mathrm{W/m^2K}\), which corresponds to option (d).