Question

Consider a person standing in a room at \(23^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and $32^{\circ} \mathrm{C}\(, respectively, and the convection heat transfer coefficient is \)5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

Short answer

Answer: The total rate of heat transfer from the person is 135.83 W.

Step-by-step solution

Convection Heat Transfer Rate

First, we will calculate the convection heat transfer rate using the following formula: \(Q_{conv} = hA(T_{skin} - T_{air})\) Where: \(Q_{conv}\) is the convection heat transfer rate \(h\) is the convection heat transfer coefficient (5 W/m²⋅K) \(A\) is the exposed surface area (1.7 m²) \(T_{skin}\) is the skin temperature (32°C) \(T_{air}\) is the room air temperature (23°C) Plug in the given values: \(Q_{conv} = 5 \mathrm{~W/m}^2\mathrm{~K}\cdot 1.7 \mathrm{~m}^2\cdot (32^{\circ} \mathrm{C} - 23^{\circ} \mathrm{C})\)

Calculate Convection Heat Transfer Rate

Calculate \(Q_{conv}\): \(Q_{conv} = 5 \mathrm{~W/m}^2\mathrm{~K}\cdot 1.7 \mathrm{~m}^2 \cdot (9 \mathrm{~K}) = 76.5 \mathrm{~W}\)

Radiative Heat Transfer Rate

Now, we will calculate the radiative heat transfer rate using the following formula: \(Q_{rad} = \varepsilon\sigma A (T^4_{skin} - T^4_{air})\) Where: \(Q_{rad}\) is the radiative heat transfer rate \(\varepsilon\) is the emissivity of the skin and clothes (0.9) \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8} \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}^4\)) \(T_{skin}\) and \(T_{air}\) are in Kelvin. First, convert Celsius to Kelvin: \(T_{skin(K)} = 32^{\circ} \mathrm{C}+ 273.15 = 305.15 \mathrm{~K}\) \(T_{air(K)} = 23^{\circ}\mathrm{C}+ 273.15 = 296.15\mathrm{~K}\) Plug in the given values: \(Q_{rad} = 0.9\times 5.67\times10^{-8} \mathrm{~W} / \mathrm{m}^2\mathrm{~K}^4 \cdot 1.7\mathrm{~m}^2 (305.15^4\mathrm{~K}^4 - 296.15^4\mathrm{~K}^4)\)

Calculate Radiative Heat Transfer Rate

Calculate \(Q_{rad}\): \(Q_{rad} = 0.9\times 5.67\times 10^{-8}\mathrm{~W} / \mathrm{m}^2\mathrm{~K}^4 \cdot 1.7\mathrm{~m}^2 (8.64\times 10^8\mathrm{~K}^4 - 8.06\times 10^8\mathrm{~K}^4) = 59.33\mathrm{~W}\)

Total Heat Transfer Rate

Now, add the convection and radiative heat transfer rates to get the total heat transfer rate: \(Q_{total} = Q_{conv} + Q_{rad}\) Plug in the calculated values: \(Q_{total} = 76.5\mathrm{~W} + 59.33\mathrm{~W} = 135.83\mathrm{~W}\) The total rate of heat transfer from the person is 135.83 W.