Thermodynamics

734 | ThermodynamicsThen,Water thatleaks outHot, dryairFIGURE 14–26Water in a porous jug left in an open,breezy area cools as a result ofevaporative cooling.COOL,MOISTAIR2'2 1FIGURE 14–27Evaporative cooling.2T wb = ~ const.h = ~ const.Liquidwater1HOT,DRYAIRTherefore, this air-conditioning unit removes moisture and heat from the airat rates of 0.131 kg/min and 511 kJ/min, respectively.Evaporative CoolingConventional cooling systems operate on a refrigeration cycle, and they canbe used in any part of the world. But they have a high initial and operatingcost. In desert (hot and dry) climates, we can avoid the high cost of coolingby using evaporative coolers, also known as swamp coolers.Evaporative cooling is based on a simple principle: As water evaporates,the latent heat of vaporization is absorbed from the water body and the surroundingair. As a result, both the water and the air are cooled during theprocess. This approach has been used for thousands of years to cool water.A porous jug or pitcher filled with water is left in an open, shaded area.A small amount of water leaks out through the porous holes, and the pitcher“sweats.” In a dry environment, this water evaporates and cools the remainingwater in the pitcher (Fig. 14–26).You have probably noticed that on a hot, dry day the air feels a lot coolerwhen the yard is watered. This is because water absorbs heat from the air asit evaporates. An evaporative cooler works on the same principle. The evaporativecooling process is shown schematically and on a psychrometric chartin Fig. 14–27. Hot, dry air at state 1 enters the evaporative cooler, where itis sprayed with liquid water. Part of the water evaporates during this processby absorbing heat from the airstream. As a result, the temperature of theairstream decreases and its humidity increases (state 2). In the limiting case,the air leaves the evaporative cooler saturated at state 2. This is the lowesttemperature that can be achieved by this process.The evaporative cooling process is essentially identical to the adiabatic saturationprocess since the heat transfer between the airstream and the surroundingsis usually negligible. Therefore, the evaporative cooling process follows aline of constant wet-bulb temperature on the psychrometric chart. (Note thatthis will not exactly be the case if the liquid water is supplied at a temperaturedifferent from the exit temperature of the airstream.) Since the constant-wetbulb-temperaturelines almost coincide with the constant-enthalpy lines, theenthalpy of the airstream can also be assumed to remain constant. That is,andm # a V# 1v 1m # w 111.25 kg>min2 10.0216 0.01002 0.131 kg/minQ # out 111.25 kg>min23185.4 39.32 kJ>kg4 10.131 kg>min2158.8 kJ>kg2 511 kJ/min10 m 3 >min 11.25 kg>min0.889 m 3 >kg dry airT wb constanth constant(14–19)(14–20)during an evaporative cooling process. This is a reasonably accurate approximation,and it is commonly used in air-conditioning calculations.