## Related questions with answers

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

Solution

VerifiedThe given quantities are:

$\begin{align*} h = 14.8V^{0.69}\ \text{for }0.15<V<1.5\ \frac{\text{m}}{\text{s}}\\ A = 1.7\ ^o\text{C}\\ T_s = 29\ ^o\text{C}\\ T_{\infty} = 10\ ^o\text{C}\\ \end{align*}$

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

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