Document EvwQgnkvxEZVpGXe2vYdrEOpV

American Society of Heating and Ventilating Engineers Guide, 1936 c. = 3.46 lb per minute per ton. d. 9000 X 33 qqq = 0.944 hp per ton. e. 200 + 0.944 X 42.5 = 240 Btu per minute per ton. Check:. Heat content leaving compressor = 85.25 + 11.6 = 96.85 (96.85 -- 27.48) 3.46 = 240 Btu per minute per ton. /. Heat Content = 96.85. Pressure = 114.31b per square inch.. From Table 3, Temp. = 146 F. 2 If the velocity of vapor in the suction pipe is'50 fps and there are 10 velocity heads lost between evaporator and compressor, what is the saturation tem perature at the evaporator? h = 10 X. = 388 ft head. Head per degree = (47.28 - 43.16) X (863 + -792^ X ^ = 98.4 ft. Check: Head per degree = 9000 X _47 g* Q* ) = 105 ft. (.Approx.) ooo Temperature = 30 + -gg-j = 34 F. .3 If the refrigerant is sub-cooled to 70 F, what is the'effect on: o. Work per pound: b. Refrigerating effect; c. Horsepower per ton? a. No effect on work per pound. ;85.25 - 23.90) = 61.35. >1.35 -- 57.77 = 3.58 Btu increase. 3 58 gy yy - 6.2 per cent increase. = 3.27 lb per minute per ton 9000 X gl gQQ = 0.891 hp per ton. \ 0.944 -- 0.891 = 0.053 hp decrease. = 5.6 per cent decrease. 4 What is the approximate change in capacity of the following types of systems per degree at 40 F: a. Reciprocating; b. Centrifugal; c: Ejector? a. 2.5 per cent. b. 3.0 per cent. c. 7.5 per cent. 54 Chapter 2--Refrigeration , 5* Which type of system will maintain the most uniform evaporator temperature with change of load? b Which system will maintain the most uniform load with change of evaporator temperature? From Fig. 6. a. Steam ejector. b. Reciprocating and rotary. 6 o. ' What is the velocity of steam expanding from 100 lb per square inch eage " saturated to 0.0178 lb per square inch absolute, corresponding to 50 F if the nozzle has an efficiency of 90 per cent? j What is the velocity of the mixture of this steam with one-third the mass "of entrained steam moving at 300 fps? 0. v = vw>T2i^ro^rai9oT^8i5T) V = 4110 fps. b. Vjnix. 3 X 4110 + 1 X 300 4 = 3158f y 0 a;r entering an open adsorption system at 80 F and 50 per cent relative humidity is dehumidified and cooled to a temperature of 90 F and 12 per cent relative humidity, how much air is cooled per ton of refrigeration and what is the latent heat of the water which is adsorbed? Entering Conditions 66.6 F WB. 30.85 Btu Total, Heat 76.0 Grains per Pound Leaving Conditions 59.1 F WB. 25.59 Btu Total Heat 24.5 Grains per Pound cfm 200 X 13.5 = 513 cfm per ton. (30.85 - 25.59) 513 X (760 - 24.5) X 1044 13.5 X 7000 292.5 Btu. <513 X 0.2415 X 10's Check: = 200 + s 13.5 ) 291.6 Btu. 8 If air entering an open adsorption system at 80 F and 50 per cent relative humidity is cooled to 75 F and 12 per cent relative humidity, how much air is required per ton and what is the latent heat of the water which is absorbed? Entering Conditions 66.6 F WB 30.85 Btu Total Heat 76.0 Grains per Pound Leaving Conditions 50.3 F WB 20.35 Btu Total Heat 14.4 Grains per Pound cfm 200 X 13.5 (30.85 - 20.35) 257 cfm per ton. 257 X (76.0 - 14.4) X 1044 13.5 X 7000 175.2 Btu. Check: L = 200 - (jg7^ 924! 5 X5j = m Q Btu 9 f Why is it possible for a reversed cycle refrigeration unit to show a better performance in heating operation than when operating during the cooling season?