thermo A popular method of cooling homes in the southwest US is the evaporative
ID: 1859533 • Letter: T
Question
thermo
A popular method of cooling homes in the southwest US is the evaporative cooler. It cools air entering a home by spraying small drops of water into it; when the drops evaporate they consume energy from the air so the air temperature decreases. Assume you have 1.7 1bm of atmospheric air at 90 F. Into it you spray 0.009 1bm of water at 65 F. Does the water enter as saturated liquid, saturated vapor, compressed liquid, or superheated vapor? (Think of a garden hose or lawn sprinkler). If all the water evaporates to a final temperature of 70 F, what is the final air temperature? Report your answer in F. "Health experts" recommend you "drink eight eight-ounce glasses of water per day" as part of a weight loss regimen. Assume you follow this advice, that the water enters your body as saturated liquid at 60 F, and all of it is exhaled so it exits your body as a saturated vapor at 98 F. How many calories does your body have to supply for this process? [It is useful to know that "food" calories are really kCals, 1000 times the chemists cal, and equal to 4.18 kJ.] How much energy would your body have to supply if all the water exited your body as a saturated liquid at 98 F? What would the value be if half exited as saturated liquid and half as saturated vapor?Explanation / Answer
a)
Water enters as saturated liquid.
b)
Heat balance: m*Cp*dT for air = m*Cp*dT for water + m*L (where m*L is energy associated with phase change from liquid to vapor. L is heat of vaporization = 970.4 Btu/lb)
1.7*0.24*(90 - T) = 0.009*0.47*(70 - 65) + 0.009*970.4
T = 68.5 deg F
2)
8 ounce = 0.5 lb
Energy required to change phase from liquid at 60 F to vapor at 60 F = m*L = 0.5*970.4 = 485.2 Btu
Energy required to raise temperature of vapor from 60 F to 98 F = m*Cp*(T2 - T1) = 0.5*0.47*(98-60) = 8.9 Btu
Total energy = 485.2+8.9 = 494.1 Btu = 124.5 kCals
If all water exits as sat. liquid at 98 F, energy required = m*Cp*(T2 - T1) = 0.5*0.47*(98-60) = 8.9 Btu = 2.243 kCals
If half exits as sat. liq and half as sat. vapor then energy required would be 1/2 *(124.5 + 2.243) = 63.37 kCals
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