Heat Transfer from a Person Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m^2 and 29°C, respectively,

Question

Heat Transfer from a Person Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m² and 29°C, respectively, and the convection heat transfer coefficient is 6 W/m² · °C (Fig. 2–75).

Accepted Answer

A person is standing in a breezy room. The total rate of heat loss from the person is to be determined.
Assumptions 1 The emissivity and heat transfer coefficient are constant and uniform. 2 Heat conduction through the feet is negligible. 3 Heat loss by evaporation is disregarded.
Analysis The heat transfer between the person and the air in the room will be by convection (instead of conduction) since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body, initiating natural convection currents. It appears that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area (m²) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is, from Eq. 2–53,

[latex] \dot{Q}_{ ext {conv }}=h A\left(T_{s}-T_{f} ight) W[/latex] (2–53)

[latex]\begin{aligned}\dot{Q}_{ ext {conv }} &=h A\left(T_{s}-T_{f} ight) \&=\left(6 W / m ^{2} \cdot{ }^{\circ} C ight)\left(1.6 m ^{2} ight)(29-20){ }^{\circ} C \&=86.4 W\end{aligned}[/latex]

The person will also lose heat by radiation to the surrounding wall surfaces. We take the temperature of the surfaces of the walls, ceiling, and the floor to be equal to the air temperature in this case for simplicity, but we recognize that this does not need to be the case. These surfaces may be at a higher or lower temperature than the average temperature of the room air, depending on the outdoor conditions and the structure of the walls. Considering that air does not intervene with radiation and the person is completely enclosed by the surrounding surfaces, the net rate of radiation heat transfer from the person to the surrounding walls, ceiling, and the floor is, from Eq. 2–57,

[latex] \dot{Q}_{ rad }=\varepsilon \sigma A\left(T_{s}^{4}-T_{ surr }^{4} ight) W[/latex] (2–57)

[latex]\begin{aligned}\dot{Q}_{ rad } &=\varepsilon \sigma A\left(T_{s}^{4}-T_{ sur }^{4} ight) \&=(0.95)\left(5.67 imes 10^{-8} W / m ^{2} \cdot K ^{4} ight)\left(1.6 m ^{2} ight) imes\left[(29+273)^{4}-(20+273)^{4} ight] K ^{4} \&=81.7 W\end{aligned}[/latex]

Note that we must use absolute temperatures in radiation calculations. Also note that we used the emissivity value for the skin and clothing at room temperature since the emissivity is not expected to change significantly at a slightly higher temperature.
Then the rate of total heat transfer from the body is determined by adding
these two quantities to be

[latex] \dot{Q}_{ ext {total }}=\dot{Q}_{ ext {conv }}+\dot{Q}_{ ext {rad }}=86.4+81.7=168.1 W [/latex]

The heat transfer would be much higher if the person were not dressed since the exposed surface temperature would be higher. Thus, an important function of the clothes is to serve as a barrier against heat transfer.
Discussion In the above calculations, heat transfer through the feet to the floor by conduction, which is usually very small, is neglected. Heat transfer from the skin by perspiration, which is the dominant mode of heat transfer in hot environments, is not considered here.

Question 2.19: Heat Transfer from a Person Consider a person standing in a ...

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