A marathon runner, treated as a cylinder with diameter 33 cm moving at 19 km h^−1 relative to the surrounding air, has a net radiation load of 300 W m^−2. The air temperature and vapor pressure are 30 °C and 2.40 kPa, respectively. Assuming that the runner’s skin is covered with sweat that is a

Question

A marathon runner, treated as a cylinder with diameter 33 cm moving at 19 km [latex]h^{−1}[/latex] relative to the surrounding air, has a net radiation load of 300 W [latex]m^{−2}[/latex]. The air temperature and vapor pressure are 30 °C and 2.40 kPa, respectively. Assuming that the runner’s skin is covered with sweat that is a saturated salt solution for which the relative humidity of air in contact with the solution is 75%. Estimate the rate of latent heat loss. If all the salt could be washed off when the runner was sprayed with water, what would be the new rate of latent heat loss?

Accepted Answer

The runner’s velocity, 19 km [latex]h^{−1}[/latex], is 5.3 m [latex]s^{−1}[/latex]. Nusselt number [latex] Nu=0.24Re^{0.6} = 0.24 imes (0.33 imes 5.3/16 imes 10^{-6} )^{0.6} = 253 [/latex]. Reistance to sensible heat transfer is [latex] r_{H}= {d}/{kNu} = 0.33/(22.8 imes 10^{-6} imes 253) = 57\; s \; m^{-1} [/latex]. Assume that [latex] r_{V}/r_{H} = 0.93 [/latex]. Saturation deficit δ is (0.75 × 4243 − 2400) = 782 Pa. Other parameters are [latex] \Delta = 244\; Pa\; K^{-1} [/latex] and [latex] \gamma = 66.5 \; Pa \; K^{-1} [/latex]. Applying the Penman-Monteith equation gives λE = [(0.75 × 244 × (300 + 600)) + (1.2 × 10³ × 782/57)]/[(0.75 × 244) + (66.5 × 0.93)] = 739 W [latex]m^{−2}[/latex]. If the salt was washed off, so that the relative humidity at the surface was 100%, but all other terms remained the same, λE = 843 W [latex]m^{−2}[/latex].

Question 13.6: A marathon runner, treated as a cylinder with diameter 33 cm...

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