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54 CHAPTER 3 1952 Guide It is possible to obtain two values for the wet-bulb temperature when this temperature is below 32 F. If the bulb of a thermometer is dipped into water at a temperature slightly above 32 F and held in a stream of air whose wet-bulb temperature is below 32 Fj^the temperature indicated by the thermometer will drop rapidly until a minimum is reached below 32 F. This will be accomplished, without the formation of ice on the bulb, of the thermometer. After reaching this minimum temperature, the reading will jump back to 32 F and remain there until the water on the bulb is frozen, after which it will slowly drop again until equilibrium is reached. The final temperature may be higher or lower than the first -minimum reading, or it may be the same depending on the amount of moisture present in the mixture. In the absence of reliable data on the wet-bulb temperature over sub-cooled water, the chart, below 32 F, has been drawn for the equilibrium condition, that is, the values plotted On the A.S.H.V.E. chart, are for the condition where the minimum temperature is reached with ice on the bulb of the thermometer. USE OF TABLE 2 AND THE A.S.H.V.E. PSYCHROMETRIC CHART The use of Table 2 and the A.S.H.V.E. Psychrometric Chart in'analyzing typical air conditioning problems, is best explained by means of illustrative examples. In each-of the following it is to be understood that the processes in question take place at a constant pressure of 29.921 in. Hg, t.e., standard atmospheric pressure. . Example 1: Determine the enthalpy of moist air at 80 F dry-bulb temperature and 0.40 degree of saturation.' Solution a: From the data of. Table 2, at 80 F, ft. = 19.221 Btu per lb of dry air and A, = 24.47 Btu per lb of dry air. Then ft at the specified conditions is 19.221 + 0.40(24.47) = 29.01. Btu per lb of dry air. Solutionb: From the A.S.H.V.E. Chart. Follow the 80 F dry-bulb line vertically until it intersects the 0.40 degree of saturation line. From this intersection, follow the line of constant enthalpy to the enthalpy scale and read 29.00 Btu per lb of dry air. Example 2: Determine the thermodynamic wet-bulb temperature of moist air at the conditions of Example 1. Solution a: From the data of Table 2. Applying Equation 8, hi = 29.01 Btu per lb of dry air (Example 1). As a first approximation this is A*, the enthalpy at satura tion at the thermodynamic wet-bulb temperature which is, therefore, approximately 63.5 F. W* at 63.5 F is 12.57 X 10"' lb of water vapor per lb of dry air, and W, is 0.02233 X 0.40 = 0.00893 lb of water vapor per lb of dry air. The specific enthalpy of liquid water at 63.5 F is 31.58 Btu per lb of water. As a second approximation, A* = 29.01 + (0.01257 -- 0.00893) (31.58) = 29.12 Btu per lb of dry air. Interpola tion in Table 2 gives as the final answer t* = 63.64 F. ' Solution b: From the A.S.H.V.E. Chart. At the intersection of the 80 F drybulb temperature line and the 0.40 degree of saturation line, read the thermodynamic wet-bulb temperature. Heating of Moist Air at Constant Pressure Without Addition of Moisture Example S: Air initially at 20 F, 0.80 degree of saturation, is heated to 120 F. Find the quantity of heat required to process 20,000 cfm of heated air. The process is diagrammatically illustrated in Fig. 6. The energy equation for the process is GA, + iji = GAj or ,a, = G(As -- Ai) Thermodynamics 55 h.G w(f r G T"~wz 1 ;>/ 7777777 7777/777 i4t Fio. 6.. Illustration of Process of Example 3 Solution a: From the data of Table 2. The initial humidity ratio, which is the same as the final humidity ratio, is 0.80(0.002152) = 0.001722 lb of water vapor per lb of dry air; the.initial enthalpy is 4.804 + 0.80(2.302) = 6.646 Btu per lb of dry air; the final degree :of saturation is 0.001722/0.08149 = 0.02113; the final enthalpy is 28.841 H- 0.02113(90.70) = 30.757 Btu per lb of dry air; the final volume is 14.611 4- 0.02113(1.905) = 114.651 cu ft per lb of dry air. Since 20,000 cfm of heated air are to be processed, the total quantity of heat required is 1 S; ,?( = (20,000/14.651) X 24.111 = 32,914 Btu per min. : Solution b: From ;the- A.S.H.V.E. Chart. The process is represented by the horizontal line 1-2, Fig. 7. The initial enthalpy, at 20 F dry-bulb temperature and 0.80 degree of saturation, is 6.65 Btu per lb of dry air. Since the final humidity ratio is the same as the initial humidity ratio, the ratio (Ai -- Ai)/(1F, -- Wf) = The horizontal line 1-2, Fig. 7, then represents the; condition line for the process, and the final state of the moist air must lie on this line. The final state is located at the fioint at which the 120 F dry-bulb temperature line crosses the condition line, and is abeled point 2 on the figure. At this condition the final enthalpy is 30.8 Btu per lb of dry air and the final specific volume is 14.65 cu ft per lb of dry air. Substituting these values in the energy, equation, . .g,- ^OOD/H^X ,(30.8 - 6.65) = 32,950 Btu per min. - Cooling of Moist Air at Constant Pressure with-Condensation of Water Referring to Fig. 8,mpist air cooled from state i passes through successive states along the line W = Wi = constant, until the saturation line is intersected'. The