O sl of work with must he lose dur Strenuous exercis curr of activity Table 22.1
ID: 1770241 • Letter: O
Question
O sl of work with must he lose dur Strenuous exercis curr of activity Table 22.1 Metabolic rates associated with various leve ork at an average does not lose or 22.9 loses heat sa nutritional in- when the con nx a ander22.9 An endurance athletic event is conduct An endura lb temperature be 45 °C. One of the participants in the event, a maie sorb 1.1 litre of water per hour through the t to the ditions are extremely stressful, the dry-bulb t male, is able to ab. lose water through perspiration faster than this if n eneration rate 170W by con- latent heat of vaporisation of water at the participants . The isture will she perature is 2440 kJkg-1. match lasting tennis player skin tem. (a) Determine the maximum rate at which he can reject heas perspiration without becoming dehydrated ation of water (b) The athlete has a surface area of 1.8m2 and his surface hea transfer coefficient, including convection and radiation is es to go for a 4MIkg ile running. n tempera- ture is 37°C. Determine the rate of heat gain (or loss) due to convection and radiation alone, not including the effect of perspiratiorn ical power () The table above shows the metabolic rate of the athlete do- ing different activities. Assume this is the same as the rate of production of thermal energy in his body. Determine the maximum sustainable metabolic rate for the athlete during the endurance event without any temperature rise. What is the most vigorous activity that could be sustained under n transfer erature is atements ect) these conditions? d) Suppose the event is rescheduled for a cooler day. Again by a few the athlete has sufficient water to perspire and evapora 1.1 kg per hour and on this occasion he intends to run w a metabolic rate of 800 W. What is the maximum dry air onto te by con- temperature at which he could do this sustainably, assum ing his skin temperature is 37 C? ry-bulbExplanation / Answer
a)
Athlete can perspire at a rate 1.1 l/hr. Taking density of water as 1kg/l, the mass of water lost is 1.1 kg/hr
Heat given out Q = mL for vaporization
Q = 1.1 * 2440 = 2684 kJ/hr = 2684000/3600 W = 745.5 W
b)
Rate of heat gained = surface heat transfer coefficient * area * temperature difference
= 22 * 1.8 * (45-37) = 316.8 W
c)
For no temperature rise heat gained = heat lost
heat due to metabolism + heat convection/radiation = heat lost by perpiration
heat due to metabolism = 745.5 - 316.8 = 428.7 W (Ans)
d)
heat due to metabolism + heat convection/radiation = heat lost by perpiration
800 + 22 * 1.8 * ( T -37) = 745.5
T = 35.6 deg C (Ans)
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