Document vVy14LM1mMpJdEpEnEe3ErL1w

American Society of Heating and Ventilating Engineers Guide, 1937 PROBLEMS IN PRACTICE X Dichlorodifluoromethane (Fn) at a saturated temperature of 30 F but superheated 25 F to 55 F is compressed to a saturation temperature of 90 F with a compressor having an overall efficiency of 70 per cent. It leaves the con denser sub-cooled 5 F. What is the: o. Work per pound; b. Refrigerating effect; c. Pounds per minute per ton; d. Horsepower per ton; e. Heat rejected to condenser neglecting radiation;/ Equivalent discharge temperature? a. X 43.16 X 144 X 0.939 X ~ 0.151 490 (By Pressure--Volume Method). -1]xok = 9000ft-lb; Check: (93.35 85.25) X Q-y = 11.6 Btu per pound = 9000 ft-lb. (By Heal Content Method). b. 95.25 - 27.48 = 57.77 Btu per pound. c.. = 3.46 lb per minute per ton. d. 9000 X 3.46 = 0.944 hp per ton. 33,000 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 = 114.3 lb per square inch. From Table 3, Temp. = 146 F. 4 2 0 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= V, 2g h = 10 X = 388 ft head. 2g) Head per. degree = (47.28 -- 43.16) X ( 0.863 + 0.792\ j Xy = 98.4 ft.' : Check: Head per degree = 9000 X 0.7 = 105 ft. (Approx.) (90 - 30) 388 Temperature = 30 + = 34 F. 98.4 3 If the refrigerant is sub-cooled to 70 F, what is the effect on: a. Work per pound: b. Refrigerating effect; c. Horsepower per ton? o. No effect on work per pound. b. (85.25 -.23.90) .= 61.35. 61.35 -- 57.77 = 3.58 Btu increase. 3.58 57.77 6.2 per cent increase. 52 Chapter 2--Refrigeration 200 = 3 27 lb per minute per ton. Cm 61.35 9000 X 3^60 = 0 891 hP Pef t0"' 0.944 -- 0.891 = 0.053 hp decrease. 0 053 = 5 6 per cent decrease. 0.944 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. 3 a. 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 centrifugal. 6 0 a. What is the velocity of steam expanding from 100 lb per square inch gage, saturated to 0.0178 lb per square inch absolute, corresponding to 50 F if the nozzle has an efficiency of 90 per cent? b. What is the velocity of the mixture of this steam with one-third the mass of entrained steam moving at 300 fps? a. V. = V 778 X 2g X 0.9 X (1189.0 - 810.3) V = 4130 fps. ,, 3 X 4130 + 1 X 300 b. rmix---------------------------7 '.= 3172 fps. 7 If air 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 = = 513 cfm per ton. (30.85 - 25.59) 7 _ 513 X (760 - 24.5) X 1044 ,,,,,, ,, ---------------------13.5 X 7000' " " 292'5 BtU` Check: L = 200 + f513 X 0.2415 X 10> l 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 Leasing Conditions 50.3 F WB 20.35 Btu Total Heat 14.4 Grains per Pound 53