It is well-known that cold air feels much colder in windy weather than what
the thermometer reading indicates because of the "chilling effect" of the
wind. This effect is due to the increase in the convection heat transfer
coefficient with increasing air velocities. The equivalent wind chill
temperature in \(^{\circ} \mathrm{F}\) is given by \([\mathrm{ASHRAE},\) Handbook
of Fundamentals (Atlanta, GA, \(1993 \text { ), p. } 8.15]\) $$\begin{aligned}
T_{\mathrm{equiv}}=& 91.4-\left(91.4-T_{\text {anbient }}\right) \\
& \times(0.475-0.0203 V+0.304 \sqrt{V})
\end{aligned}$$
where \(V\) is the wind velocity in \(\mathrm{mi} / \mathrm{h}\) and \(T_{\text
{ambicnt }}\) is the ambient air temperature in \(^{\circ} \mathrm{F}\) in calm
air, which is taken to be air with light winds at speeds up to \(4 \mathrm{mi}
/ \mathrm{h}\). The constant \(91.4^{\circ} \mathrm{F}\) in the given equation is
the mean skin temperature of a resting person in a comfortable environment.
Windy air at temperature \(T_{\text {ambient }}\) and velocity \(V\) will feel as
cold as the calm air at temperature \(T_{\text {equiv. }}\) Using proper
conversion factors, obtain an equivalent relation in SI units where \(V\) is the
wind velocity in \(\mathrm{km} / \mathrm{h}\) and \(T_{\text {ambient }}\) is the
ambient air temperature in \(^{\circ} \mathrm{C}\)
