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American Society of Heating and Ventilating Engineers Guide, 1937 Pa _ A p P, ~ TV * 2VP, r, (96 + 460) (29.92 X 0.491 X 144) (78 + 460) (0.491 X 144) 30.90 in. Hg. 6 Given saturated air at a temperature of 75 F and a barometric pressure of 29.92 in. Hg; determine the heat content of the mixture per pound of dry air, including the heat content of the liquid above 32 F. From Equation 12, A = 0.24 (1 - 0) + WK inhere As = 1059.2 + 0.45< (Empirical equation derived from Keenan's Steam Tables). /. = 75 F. W = 0.01873 lb of water vapor (Table 6). A = 0.24 (75 - 0) 4- 0.01873 (1059.2 + 0.45 X 75). A = 38.46 Btu per pound of dry air. 7 A building requires 50,000 cu ft of air per hour to be raised from --10 F dry-bulb and 75 per cent relative humidity to 72 F dry-bulb and 30 per cent, relative humidity. Determine the amount of heat and the weight of water which it is necessary to supply per hour if the temperature of the supply water is 50 F and the barometric pressure is 28.75 in. Hg. Assume air volume to be dry air at 70 F. Weight of air = 0.075 X 50,000 = 3750 lb per hour. From Table 6, Pressure of vapor in the mixture, outside air = 0.75 X 0.0221 = 0.0166 in. Hg. Specific humidity, outside air = 0.622 ^ 28 75 ^ (T0166 ) = -^589 lb. From Table 6, Pressure of vapor in the mixture, inside air = 0.30 X 0.7906 = 0.2372 in. Hg. (28Specific humidity, inside air = 0.622 75~--^237fs) = ^-005174 lb. Water to be added = 3750 (0.005174 - 0.0003589 ) = 18.06 lb per hour. Heat content, inside air = 0.24 X 72 + 0.005174 (1059.2 + 0.45 X 72) = 22.925 Btu per pound. Heat content, outside air = 0.24 X (-10) + 0.0003589 (1059.2 + (0.45 X -10) = --2.021 Btu per pound. Btu added incident to the water per pound of dry air. * (0.005174 - 0.0003589) (50 - 32) = 0.0867 Btu per pound. Heat requirement per hour = (22.925 -- (--2.021 4- 0.0867) X 3750 = 93,221 Btu. 8 Determine the amount of heat and water that must be extracted to cool 3750 lb of air (weighed dry) from 95 F and 60 per cent relative humidity to 50 Fand 100 per cent relative humidity with a barometric pressure of 28.75 in. Hg. Pressure of vapor in the mixture, outside air = 0.6 X 1.659 = 0.995 in. Hg. Specific humidity, outside air = 0.622 ( j-995 \ = 0.0223 lb. \Z8.75 -- 0.995/ ` Specific humidity, inside air = 0.007626 lb. Weight of water to be extracted per hour = (0.0223 -- 0.007626) X 3750 = 55.03 lb. Heat content, outside air = 0.24 X 95 4- 0.0223 (1059.2 -f- 0.45 X 95) = 47.37 Btu per pound. Heat content, inside air = 0.24 X 50 + 0.00764 (1059.2 + 0.45 X 50) = 20.26 Btu per pound. Heat to be extracted = (47.37 -- 20.26) X 3750 = 101,662 Btu. `' 34 Chapter 2 REFRIGERATION Classification of Systems, Refrigerants, Mechanical Compression Systems, Theoretical Cycle, Refrigerating Effect per Pound, Coefficient of Performance, Steam Ejector System, System Characteristics, Closed Absorption Systems, Open Adsorption Systems, Reverse Cycle THE various types of refrigeration systems most commonly used for air conditioning purposes may be classified fundamentally as follows: 1. Compression Systems. o. Mechanical--Reciprocating, Rotary, and Centrifugal. b. Ejector. 2. Absorption Systems. a. Closed. b. Open. Of these, the mechanical systems are the most extensively used at the present time and will be given complete consideration in the following discussion. REFRIGERANTS The common refrigerants are volatile liquids which produce refrigera tion by their evaporation under reduced pressure. Factors usually in fluencing the choice of refrigerant are safety, chemical stability, operating pressures and adaptability for the type of system to be used. Of the six refrigerants whose properties are listed in Tables 1 to 6, ammonia, carbon dioxide, dichlorodifluoromethane (Fu) and methyl chloride are used in reciprocating and rotary mechanical compression systems. Monofluorotrichloromethane (Fu) and water are used in cen trifugal compression systems. Water is used almost to the exclusion of other refrigerants in ejector systems. Closed absorption systems may use ammonia, methyl chloride, water, or others as the refrigerant. MECHANICAL REFRIGERATION SYSTEMS While the mechanical refrigeration systems differ in the methods used for compression of the refrigerant vapor, they are all fundamentally similar. Refrigerant vapor, usually saturated or slightly superheated, is drawn into a compressor (see Fig. 1). It is then compressed and dis charged at a higher pressure to a condenser. The vapor is condensed as it contacts a heat transfer surface over which is flowing a cooling medium such as water, air or a combination of the two. The liquid refrigerant 35 A