Document YjjZgerNkG6qr79mBR846K7B8

American Society of Heating and Ventilating Engineers Guide, 1935 Chapter 9--Central Air Conditioning Systems Solution. 199,736 X 55.2 ieolo , 60 X 12 ~ 15-313cfra 1146 lb per minute. If a system similar to the one shown in Fig. 1 is used, 1146 lb per minute will be | capacity of the dehumidifier as well as of the fan equipment. ` '* Example 8. If in addition to the 199,736 Btu per hour sensible heat load, then.--, ditioned space has a moisture gain of 384,000 grains per hour, calculate the appals dew point required to give maintained conditions of 80 F dry-bulb and 66 F wet-! ~ with a corresponding 563^ F dew point. Solution. With 384,000 grains of moisture per hour to be picked up, the entering point temperature should be low enough so that the addition of this moisture will increase the dew point above 56T F. Grains per pound of air saturated at 56H F (Table 5, Chapter 1) Less: Grains per pound to be picked up, 384,000 1146 X 60' J l Grains per pound allowable iri entering air r*T TO BE REMOVED BY COOLING AND DEHUMIDIFYING HEAT iv APPARATUS , Assume the same data as for Example 3. If the amount of outside air, at Example 4- yg p wet-bulb, required for ventilation has been found to be 169 lb 95 F <!ry',bU determine the refrigeration capacity required, pff minute. As the total weight of the air introduced per minute is 1146 lb, and 41 per Solution ^ tf,e dehumidifier, the total work to be done may''be computed ijent of it g0^ 11 5 as follows: M passing`de a.sfor ventilation.................................................................. Less: Return air...................................... .............................................................................................. The refrigeration required for the return air is: Tntal heat per pound at 65 F........................ Total heat per pound at 51.2 F..................................... .... KS9 lb 301 lb 29.65 Btu 20.85 Btu Reouirement for cooling 1 lb of return air.......................................... 8.80 Btu 301 lb X 8.80 Btu = 2649 Btu per minute required to cool the return air. This corresponds to an apparatus dew-point temperature of 54.17 F. Example 3. Illustration of the by-pass system. (See Fig. 2.) Assume the same data as for Example 2. Instead of passing all of the air through | dehumidifier for cooling and dehumidifying, a portion may be passed through and th balance be mixed with the conditioned air at the leaving end of the dehumidifier, & 5; mixture being proportioned so that the resultant conditions will be those required! give proper conditions in the area considered. | Solution. The quantity of air to be dehumidified, the quantity to be by-passed, ii the apparatus dew-point temperature may be calculated as follows: Let The refrigeration required for the outside air is: ' Total heat per pound of outside air............................................. ........ 37.81 Btu Less: Total heat per pound at 51.2 F................................................. 20.85 Btu Requirement to cool 1 lb of outside air................................. ............. 16.96 Btu > 169 lb X 16.96 Btu = 2866 Btu per minute required to cool the . outside air. Thus, the total refrigeration required is: 2649 Btu + 2866 Btu = 5515 Btu per minute, which is equivalent to a load of 27.6 tons of refrigeration. X = percentage of air to be by-passed. K = percentage of air to be passed through the dehumidifier. Id = apparatus dew-point temperature, degrees Fahrenheit. The quantity X of 80-F air must mix with the quantity Y of dehumidified airi produce air with a resultant 66 F wet-bulb temperature. Also, X quantity of airi 56H F dew point must be mixed with Y quantity of dehumidified air to give a results apparatus dew-point temperature of 54.17 F. It is assumed that the air passing thrt^ the dehumidifier is saturated. Solving simultaneous equations, 80.0AT + Kid = 68.00 56.5* + Kid = 54.17 23.5X +0 = 13.83 SIZE OF REHEATERS A.properly designed air-conditioning system will have reheaters of sufficient capacity to heat the conditioned air from the apparatus dew1;:point temperature to the outlet delivery temperature.. If winter heating is to be accomplished, consult Chapter 22. The following general formula may be used to determine the amount of heat-necessary to reheat a given quantity of air: I:!# sohere'' Hi = 0.24 (ly - Id) M (5) , beat to be supplied to reheater coil, Btu per hour. X = 13.83 X 23.5 100 = 59 per cent, air by-passed. = 100 - X = 41 per cent, air passed through waste The second step is to determine the apparatus dew-point temperature. Substitute; in either Equation 3 or Equation 4, and solve for Id: 80 X 0.59 + Id X 0.41 = 68 68-47 Id = 0.41 = 51.2 F, the apparatus dew point. ^ i^uhedfk^ Assume the same data as for Example 1, and find the amount of reheating .SiJution. Hi = 0.24 (68 - 54.17) 1146 X 60 = 228,200 Btu per hour. SURFACE COOLINC PROBLEM iSdam0Unt co'l sur^ace required for a given amount of work is ^surfaecmesS'Pmt audPeonafvaaciltaobrsleprbeyvidouiffselyrelinsttemd.anOufbavcitouurselrys, the various types of will have different 160 161