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Question

The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h=14.8 V0.69h=14.8 \mathrm{~V}^{0.69} for 0.15<V<1.5 m/s0.15<V<1.5 \mathrm{~m} / \mathrm{s}, where VV is the air velocity. For a person with a body surface area of 1.7 m21.7 \mathrm{~m}^{2} and an average surface temperature of 29C29^{\circ} \mathrm{C}, Find the rate of heat loss from the person in windy air at 10C10^{\circ} \mathrm{C} by convection for air velocities of (a)0.5 m/s(a) 0.5 \mathrm{~m} / \mathrm{s}, (b) 1.0 m/s1.0 \mathrm{~m} / \mathrm{s}, and (c)1.5 m/s(c) 1.5 \mathrm{~m} / \mathrm{s}.

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Answered 2 years ago
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The given quantities are:

h=14.8V0.69 for 0.15<Vlt;1.5 msA=1.7 oCTs=29 oCT=10 oC\begin{align*} h = 14.8V^{0.69}\ \text{for }0.15<V&lt;1.5\ \frac{\text{m}}{\text{s}}\\ A = 1.7\ ^o\text{C}\\ T_s = 29\ ^o\text{C}\\ T_{\infty} = 10\ ^o\text{C}\\ \end{align*}

Since the air is moving, this is the case of forced convection. Thus, hh will be higher than it would be for still air.

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