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28 CHAPTER 3 1959 Guide point at which the break in the isotherm through the point in question occurs, is the weight of water per pound of dry air in the vapor phase.' Consequently, the difference between the two ordinates -is the weight of condensed moisture per pound of dry air in the vapor phase. - The shaded -solid-liquid-vapor region at 32 F is an iso thermal three-phase zone, and separates the liquid-vapor zone from the solid-vapor zone. The temperature is 32 F throughout the shaded area. It is possible to obtain two values for the wet-bulb tem perature 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 jus whose wetbulb temperature is below 32 F, 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 reach ing 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 subcooled-water, the chart', below 32 F, has Been drawn for the equilibrium condition, that is, the values plotted on'the ASHAE chart are for the condition where the minimum tem perature is reached with ice on the bulb of the thermometer. USE OF TABLE 2 AND THE ASHAE PSYCHROMETRIC CHART The use of Table 2 and the ASHAE Pstchbometric 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, i.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, A. = 19.221 Btu per lb of dry air and h,, = 24.47 Btu per lb of dry air. Then A at the specified conditions is 19.221 -f 0.40(24.47) = 29.01 Btu per lb of dry air. Solution b: From the ASHAE Chart. Follow the 80 F drybulb line upward until it intersects the 0.40 degree of satura tion 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 t: Determine the thermodynamic wet-bulb tem- ' perature of moist air at the conditions of Example l. Solution a: From the data of Table 2. Applying Equation 8, hi =* 29.01 Btu per lb of dry air (Example I). As a first approxi mation this is A,* the enthalpy at saturation at the thermo dynamic wet-bulb temperature which is, therefore, approxi mately 63 A F. IF* at 63.5 F is 12.57 X 10"* lb of water vapor per lb of dry air, and IF, 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 approxima tion, A* - 29.01 + (0.01257 - 0.00893) (31.58) - 29.12 Btu Thermodynamics per lb of dry air. Interpolation in Table 2 gives as the final answer I* " 63.64 F. Solution b: From the ASHAE Chart. At the intersection of the 80 F dry-bulb temperature line and the 0.40 degree of satu ration line, read the thermodynamic wet-buib temperature. Heating of Moist Air at Constant Pressure With out 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 proc ess 20,000 cfm of heated air. The process is diagrammatically illustrated in Fig. 6. The energy equation for the process is .* GA, ij* TM G(A, -- Ax) 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 + 0.02113(90.70) 30.757 Btu per lb of dry air; tne final volume is 14.611 + 0.02113 (1.905) = 14.651 cu ft per lb of dry air. Since 20,000 cfm of heated air are to be processed, the total quantity of heat re quired is . . i?t - (20iO<j6/14.651) X 24.111 32,914 Btu per min. `Hg. 6 . Illustration of Process of Example 3 - Solution b:'-From the ASHAE Chart. The process.is repre sented by the horizontal line 1-2, Fig. 7. The initial enthalpy, at 20 Fldry-bulb temperature ana 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. (A, -- A,)/ (IF, -- IF,) =* <. The horizontal line 1-2, Fig. 7, then repre sents 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 point at which the 120 F dry-bulb temperature line crosses the condition line,.and is labeled Point 2 on tbe 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. Substi tuting these values in the energy equation, ,q, - (20,000/14.65) 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, moist air cooled from State 1 passes through'successive.states along the line, IF = IF, = constant, until the saturation line is intersected. The temperature at this point of intersection is by definition the dew-point tem perature-for State-1'. Further cooling through successive equilibrium states is accompanied by condensation. :The succession of states for the total system, moistair, and liquid water, is represented by a continuation of the IF ** IF, line into the liquid-vapor region. (Temperatures below 32 F would involve the solid-vapor re gion.) Consider that the final temperature is I, . The final enthalpy is then A, ; the liquid water formed is (FF, -- IF,), 29 where Point 3 is at the intersection of tbe isotherm through 2 and the saturation curve; the final humidity ratio of the moist sir is W, ; and this final moist air has dew-point, wet-bulb, and dry-buib temperatures all equal to (, . Example 4: How much beat must be removed from 20,000 cfm of air at 95 F dry-bulb temperature and 0.50 degree of saturation to cool the air to 70 F, saturated? Solution a: From the data of Table 2. The initial humidity ratio is 0.50(0.03673) = 0.0IS37 lb of water vapor per lb of dry air; tbe initial enthalpy is 22.827 + 0.50(40.49) TM 43.072 Btu per lb of dry air; the humidity ratio at saturation at the final temperature is 0.01582 lb of water vapor per lb of dry air; the quantity of liquid formed- is 0.01837 -- 0.01582 = 0-00255 lb of water vapor per lb of dry air; A,, at 70 F is 38.11 Btu per lb of water; tne initial specific volume is 13.980 + 0.50(0.822) = 14.391 cu ft per lb of ary air. Fig. 9 illustrates the process diagrammatically. The energy equation for the process is Gkt * GA, + <7(IF, - W,)h** + iqt - G[A, - A, - (IF, - IF,)A^1 = X (43.072 - 34.09 - 0.00255 X 38.07) -- '12,350 Btu per min. Solution b: From the ASHAE Chart. Two methods may be used to solve the problem by use of the psychrometric chart. The simpler is to use the region to the left of the saturation line (Fig. 8). From Point l draw a horizontal line on the chart Hg. 8 .... Cooling of Air at Constant Pressure Shown on ASHAE Psychrometric Chart