A cyclist with a total skin area of 1.5 m2 is riding a bicycle on a 20*C day....?

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kat_nasty

New member
A cyclist with a total skin area of 1.5 m2 is riding a bicycle on a 20*C day. Her skin temperature is 34*C. She generates 400 W of excess heat from riding. Estimate how much water must evaporate from her skin each hour to remove this heat. Assume the skin has an emissivity of 0.70.
 
Emitted power intensity I = σE(T^4-Ta^4) = 60.14 w/m^2 (ref.)
Emitted power = I*area = 90.2 w
This leaves 309.8 w to dissipate through evaporation.
The heat of vaporization of water is 2.26E6 J/kg (there are many variations on this; I've also seen 2.3E6, 2.6E6 and 2.256E6 on one google page)
Then in 1 hour, water consumed = 309.8*3600/2.26E6 = 0.4935 kg.
 
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