An endurance athletic event is conducted when the conditions are extremely stres
ID: 1483324 • Letter: A
Question
An endurance athletic event is conducted when the conditions are extremely stressful, the dry-bulb temperatuer being 45 C. One of the participants is the event, a male, si able to absorb 1.1 liter of water per hour through the gut, although he can lose water through perspiration faster than this if necessary. The latent heat of vaporisation of water at the particpant's skin temperature is 2440 kj/ kg.
a) Determine the maximum rate t which he can reject heat by perspiration without becoming dehydrated.
b)The athlete has a surface area of 1.8 m^2 and his surface heat transfer coefficient, including convection and radiation, is 22 W m^2/ K. During the enurance event his skin temperature is 37 C. Detemine the rate of heat gain or loss due to convention and radiation alone, not including the effect of perspiration.
c) The table above shows the metabolic rate of the athlete doing 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 these conditions?
d) Suppose the event is rescheduled for a cooler day. Again the athlete has sufficient water to perspire and evaporate 1.1 kg/hour and on this occasion he intends to run with a metabolic rate of 800 W. what is the maximum dry-bulb temperature at which he could do this sustainably, assuming his skin temperature is 37 C.
Explanation / 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)