Document JrVvNw7koY4Lvnp4ZeMZovrXO

HEATING VENTILATING AIR CONDITIONING GUIDE 1942 liminary calculation shows that the final mixture contains liquid. The final weight of water per pound of dry air is determined from 33.96 + (W -- 0.01574) X 38.0 - 17.98 ,,r,, ,, W - 0.00331 1156 3 where the specific enthalpy of the injected water is 1156.3 Btu per pound. The answer is W = 0.01718 lb water per pound dry air. Therefore, the weight of water added is 0.01718 -- 0.00331 = 0.01387 lb per pound dry air as shown in Fig. 7. Cooling Load In the calculation of the cooling load for an air conditioned space, the problem usually reduces to determining the quantity of inside air that must be withdrawn and the condition to which it must be brought by cooling, separating and possibly reheating so that return of the conditioned air will have the net effect of removing given amounts of energy and water from the air conditioned space. Fig. 8. Diagram Illustrating Example 19 Let m denote the weight of dry air withdrawn per hour. With it will be withdrawn energy of amount mh\ Btu per hour and water of amount mWj pounds per hour, where hi and Wi denote enthalpy and humidity ratio, respectively, of inside air. The weight of dry air returned per hour will be the same as that withdrawn but with it must be returned a smaller amount of energy, mh Btu per hour, and a smaller quantity of water, mW pounds per hour, where h and W denote enthalpy and humidity ratio of conditioned air. With this understanding, the requirements of the cooling load problem are, mh = mh\ -- AQ mW = mW - AW where AQ and A IF are the given amounts of energy and water, respec tively, to be removed. Eliminating m from these two equations, h -- h, W - W, AQ AW which says that all possible states for the conditioned air lie on a straight line, on the Mollier Chart, which passes through the state point of the in 32 CHAPTER 1. THERMODYNAMICS OF AIR AND WATER MIXTURES side air with a slope determined by the ratio of the quantity of energy to be removed to the quantity of water to be removed. This straight line is called the condition-line for the given problem. The border scale facilitates the graphical solution of this problem. In practice the point at which the condition line crosses the saturation . curve may dictate an excessive number of air changes for the particular space to be conditioned. If so it might be necessary to cool to a lower temperature; but if the requirements of the problem are to be exactly met both as regards the removal of water and the removal of energy, the mixture returned to the conditioned space must contain a certain amount of liquid. In other words, its state point must lie on the condition line; otherwise excessive dehumidification will result. Example 19. In order to maintain a condition of 80 F dry-bulb, 67 F wet-bulb in a certain store, it is found necessary to remove 115,060 Btu of energy per hour and 15.97 lb of water per hour. Analyze the problem illustrated in Fig. 8. Solution. The state point of the inside air is easily located on the Mollier Chart. Through it draw a line having the slope 115,060 4- 15.97 = 7205 Btu per pound water as determined from the border scale. This line crosses the saturation curve at 58.02 F. Hence a possible conditioning process is to cool some of the inside air to 58.02 F, separate the liquid thus formed, and return the resulting saturated mixture to the store. The thermodynamic properties entering the calculation are: Inside Air After Cooling After Separating t.......................... 80.0 F................................58.02 F. ...................... 58.02 F W.......................... 0.01115............................. 0.01115......... .................. .0.01027 h......................... 31.41. ........................... 25.09................................. 25.07 The weight of dry air to be withdrawn is 115,060 4- (31.41 -- 25.07) = 18,130 lb per hour. Adiabatic. Saturation Any case of adiabatic mixing in which the resulting mixture is saturated may properly be called adiabatic saturation. For example, if enough water at 352 F be sprayed into dry air at 80 F to produce a saturated mixture, the-resulting enthalpy will.be hs = 19.19 + (Ws -- 0) 324; and since hB and Wa are functions of the same temperature, this temperature is determined by the equation to be 53.0 F. Thus, adiabatic saturation of dry air at 80 F by injecting liquid water at 352 F results in a tempera ture of 53.0 F when saturation is reached. But in practice, much more is usually read into the term adiabatic saturation, it being generally understood that saturation is to be pro duced by injecting liquid water at such a temperature as will coincide with that at which the saturation curve is reached. With this under standing it may be said that thermodynamic wet-bulb temperature is the result of adiabatic saturation. Thus, if liquid water at 48126 F instead of 352 F be injected into dry air at 80 F a saturated mixture at 48.26 F instead of 53.0 F will be produced. Therefore, 48.26 F is the thermo dynamic wet-bulb temperature of dry air at 80 F. It is possible to produce adiabatic saturation, interpreting the term literally, by mixing two air streams neither of which is itself saturated. In order for this to. be possible, the straight line connecting the repre sentative points on the Mollier diagram must cut the saturation curve twice. 33