Document 3ekqK6Q3GqyamyVajp4omeggy

62 CHAPTER 3 -\1946i Guide. passes through the state point of the Inside Air and intersects the saturation curve at the apparatus dew-point. .- According to. Equation 11 tlie enthalpy he and humidity ratio Ws at the. apparatus dew-point must satisfy the equation , 7l48.8^a - hs = 7148:8 X 0.01122 - 31.514 = 48.681 At 58 F the left-hand member has the value 48.513; at 59 F its value is 50.641; by inter polation the apparatus dew-point is 58.08 F. It would be a mistake to assume that the refrigeration to be supplied is equal to the net energy to be removed; for in general water is to.be removed simultaneously and unless this is removed as liquid at 32 F. it will automatically take some energy with it. Thus, unless the water, is reftioved as solid (ice) the refrigeration to be supplied will be somewhat less than the net energy to be removed. . Example IS. Referring to the cooling load problem of Example 12, suppose that the conditioning process consists of cooling a portion of the inside axr to the apparatus dew point temperature, separating out the liquid thus formed; and returning the resulting saturated mixture to the conditioned space. Find the quantity of inside air that must be processed in this manner and the corresponding quantity of refrigeration required. 'ThermodYnaniics ,-_______________________ _____ !___________ ;63 formed-in the.cooling.operation .is represented by line BC whose projection on.the ordinate axis is the quantity of liquid so separated per pound of dry air. Point C is the apparatus dew-point and lies on.the condition line as required. . / In. practice it may not .be feasible to choose the apparatus .dew-pointas the point on the condition line to which to condition the inside air because to do so would require an excessive number of air changes in the given space. Or it may be that the condition/ line does not cross the saturation curve at all so that the apparatus dew-point as defined does not exist. Finally, it is rarely possible to obtain complete saturation in conventional air conditioning apparatus. Nevertheless the requirements of the cooling load problem can be exactly .met if the conditioned air is brought to any point on the condition line of the problem. Heating Load \ The condition line is also useful in the analysis of heating load problems as may best be illustrated by means of an illustrative example. Fig. 9; Illustration of Use of Mollier Diagram in Solution of Example 12 Solution. During the cooling operation the enthalpy of the Inside Air is reduced to the value, , ' . ; . A. =* 25.17..-}- (0^01122 --.0^01033)' >/26.20 = 25.193 V where 25.17 and 0.01033 are the values of enthalpy and humidity ratio at saturation at the apparatus dew-point temperature and 26.20 is the specific'enthalpy of liquid water at that temperature.. It follows that the quantity of.refrigeration reQuired is 31.514 -- 25.193 = 6.321 Btu per pound of dry air. * , The inside air being processedleaves% the store with an enthalpy of 31.514 arid is returned with an enthalpy of 25.17;'it therefore removes energy of axhount 6.344'Btu. per pound of dry air; This means that the weight of dry air involved in the-process is 114,510/6.344. = 18,050 lb"per hour and that the total refrigeration to'be suDolied' is 18,050 X 6.321 = 114,090 Btu per hour, 6r 9.508 tons.. . ' \ /: The quantity of liquid separated out. during the conditioning process is 18,050 X .. (0.01122 -- 0.01033) .= 16.018.1b per hour as required. . In leaving;.the apparatus it takes with it energy of amount 16.018 X 26.20 = 420 Btu per'hour/ This plus, the refrigeration accounts for the total energy.removal of 114,510 Btu per Hour as required. On'.the'Mollier'Diagrain, Fig: 40, the'cooling operation is represented by line AB whoser length?is thVquantity bf refrigeration per pound of dry air; the separation of the liquid Fig. 10. Illustration of Use of Mollier Diagram in Solution of Example 13 Example 14* A certain space is to be maintained at 70 F and 50. per cent saturation with outside conditions at 0 F and 80 per cent saturation. The normal heat trans mission through walls, partitions, floor, roof, glass and doors is estimated at 75,000 Btu per hour. Energy gained from lights and appliances is estimated at 15,000 Btu per hour. Energy and water gains from occupants are to be disregarded in the calculations. Double * doors-and windows are used, so that infiltration is negligible! The ventilation require ment is -30,000 cu ft per hour of outside, air. The requirements of the problem are to be met in the following manner; preheat the. ventilating air; mix it adiabatically with recirculated inside air; saturate the mixture adiabatically with recirculated spray water; heat the resulting mixture to 105 F and return it to the conditioned space as supply dir. * Analysis., Every pound of dry,air admitted to the system (air conditioned space plus air-conditioning apparatus) with the ventilating air displaces a* pound of dry'air /rom the systeiri with inside air.' Since the ventilating air is not admitted, directly to. the space, then for every pound of dry air withdrawn with inside air there is a pound of dry air returned with supply air. This has to have, the net effect of adding energy of amount 60,000 Btu per hour and water of amount zero pounds per hour. Thus the ratio.q deter mining the direction of the condition line is infinite, which means that the condition line is horizontal as indicated by'the protractor on the Mollier'Diagram. ; . The properties of inside air are: h = 25.451, W = 0.007910. Since the state point of the supply air must be on the condition line at 105 F, its properties are:: h = 33.986, /^,