Question

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

Short answer

Based on the given information, calculate 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) \(1.5 \mathrm{~m} / \mathrm{s}\).

Step-by-step solution

Calculate the convection heat transfer coefficient at each velocity

Using the provided formula for the convection heat transfer coefficient, we calculate \(h\) for each air velocity: (a) \(V = 0.5 \mathrm{~m} / \mathrm{s}\) (b) \(V = 1.0 \mathrm{~m} / \mathrm{s}\) (c) \(V = 1.5 \mathrm{~m} / \mathrm{s}\)

Calculate the heat loss rate using the convective heat transfer formula

We use the convective heat transfer formula \(q=hA(T_s - T_{\infty})\) to calculate the heat loss rate at each air velocity: (a) For \(V = 0.5 \mathrm{~m} / \mathrm{s}\), calculate \(q = hA(T_s - T_{\infty})\) (b) For \(V = 1.0 \mathrm{~m} / \mathrm{s}\), calculate \(q = hA(T_s - T_{\infty})\) (c) For \(V = 1.5 \mathrm{~m} / \mathrm{s}\), calculate \(q = hA(T_s - T_{\infty})\)

Compute the results

(a) At \(V = 0.5 \mathrm{~m} / \mathrm{s}\): \(h = 14.8 \cdot (0.5)^{0.69} = 9.32 \mathrm{~W} / \mathrm{m}^2 \mathrm{K}\) \(q = (9.32)(1.7)(29 - 10) = 303.8 \mathrm{~W}\) (b) At \(V = 1.0 \mathrm{~m} / \mathrm{s}\): \(h = 14.8 \cdot (1.0)^{0.69} = 14.8 \mathrm{~W} / \mathrm{m}^2 \mathrm{K}\) \(q = (14.8)(1.7)(29 - 10) = 485.8 \mathrm{~W}\) (c) At \(V = 1.5 \mathrm{~m} / \mathrm{s}\): \(h = 14.8 \cdot (1.5)^{0.69} = 20.4 \mathrm{~W} / \mathrm{m}^2 \mathrm{K}\) \(q = (20.4)(1.7)(29 - 10) = 667.9 \mathrm{~W}\) The rate of heat loss from the person in windy air at \(10^{\circ} \mathrm{C}\) by convection for air velocities of (a) \(0.5 \mathrm{~m} / \mathrm{s}\) is 303.8 W, (b) \(1.0 \mathrm{~m} / \mathrm{s}\) is 485.8 W, and (c) \(1.5 \mathrm{~m} / \mathrm{s}\) is 667.9 W.