Document RjMQ0gXjyxGM0q2Qg4Z5B0pvE

30 CHAPTER 3 1960 Guide for the total (including both phases) mixture and the humidity ratio for saturated air. The shaded solid-liquid vapor region near the lower left of the chart covers nurtures which contain water in all three phases, solid, liquid, and vapor. The temperature is St P throughout this sons. In passing through this region on a line at constant humidity ratio, the change in enthalpy corre sponds to the latent heat involved in converting the liquid water to ice, or vice versa. The fraction of the horizontal dis tance between these boundaries, on a line of constant hu midity ratio, for any given state point, represents the corresponding fractional conversion between liquid and solid. It is possible to obtain two values for the wet-bulb tem perature when this temperature is below 32 P. 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 wetbulb temperature is below 32 F, the temperature indicated by tiie thermometer will drop rapidly and may reach a mini mum below 32 F. If this happens, ice will not be formed 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 tire bulb is frozen, after which it will slowly drop again until equilibrium is reached. The final tem perature may be higher or lower than the firat minimum read ing, or it may be the same depending on the amount of mois ture present in the mixture. In the absence of reliable data on the wet-bulb temperature over subcooled water, the ehart, below 32 F, has been drawn for the equilibrium condition, that is, the values plotted on the ASHRAE Chart are for the condi tion where the rniniwnm temperature is reached with ice on the bulb of the thermometer. USE OF TABLES 2 AND 3 AND THE ASHRAE PSYCHROMETR1C CHART Engineering analysis of air-conditioning problems requires consistently selected data, procedures, and techniques yield ing the desired accuracy, for the application under considera tion. The data and principles contained in this chapter offer the advantage of thermodynamic accuracy, with no limitation upon range of validity, for calculations correctly formulated in terms of enthalpy and humidity ratio. With demonstra tion of principle as the foremost objective here, illustrative examples will be presented to show typical applications of the ASHRAE Psychbometric Cbabt and the tabulated properties. In each of the following it is to be understood that the proc esses in question take place at a constant pressure of 29.921 in. Hg, Le., standard atmospheric pressure. Example t: 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 A*, 24.47 Btu per lb of dry air. Then A at the specified conditions is 19.221 -4- 0.40(24.47) = . 29.01 Btu per lb of dry air.-The term A (Equation 32) is neg ligible for this condition. Solution b: From the ASHRAE Chart. Follow the 80 F drybulb line upward until it intersects the 0.40 degree of satura- Thermodynamics . tion line. From this intersection, follow the line of constant enthalpy to the enthalpy scale and read 29.0 Btu per lb of dry air. (The chart is not.Intended to be read to 0.01 Btu.) Example S: Determine the thermodynamic wet-bulb tem perature of moist air at the conditions of Example 1. A, mation 31 vaporper lb of d^Vnlr7fhe enthalpy of liquidwater at 63.5 F is 31 62 Btu per lb water; this is read from Table 2, which ap plies to liquid water in the presence of air at 1 atmosphere pressure. (Use of Table 3 is wrong here, because it gives data for liquid water in equilibrium with saturated steam at the saturation pressure.) As a second approximation, A* - 29.01 + 0)01257 - 0.00893) (31.62) - 29.15 Btu per lb of dry air. In terpolation in Table 2 gives as the final answer t* *= 63.64 F. Solution 6: From the ASHRAE Chart. At the intersection of the 80 F dry-bulb temperature line and the 0.40 degree of' saturation line, read the thermodynamic wet-bulb temperature. Hg. 8 .... Solution of Example 3 on ASHRAE Psydwometric Chart the final specific volume is 14.65 cu ft per lb of dry air. Substi tuting there values in the energy equation, >9* ~ (20,000/14.65) X (308 - 6.65) - 32,950 Btu per min. + r 4-Os Os w, w2 h, ha 1*12 STEAOY FLOW Hg. 7 .... Mustration of Process of Example 3 Heating of Moist Air at Constant Pressure .With out Addition of Moisture Example 3: Air initially st 20 F, 0.80 degree of saturation, is heated to 120 F. Find the rate of heat supply required to process 20,000 cfm of heated air. The process is diagrammatieally illustrated in Fig. 7. The energy equation for the process U <Ai + >9* =* Qeht 'G*(A, - A0 Soltttion a: From the data of Table 2. The initial humidity ratio, which is the same as the final humidity ratio, is 0-80 X (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; the final volume is 14.611 + 0.02113 (1.905) = 14.651 cu ft per lb of dry air. The terms A and (Equa tions 31 and 32) have oeen neglected. Since 20,000 cfm of heated air are to be processed, the total quantity of heat required per minute is ig, - (20,000/14.651) X 24.111 - 32,914 Btu per min. Solution b: From the ASHRAE Chart. The process is repre sented by the horizontal line 1-2, Fig. 8. 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 (A* -- A0/ (if, p,) <*>. The horisontal line 1-2, Fig. 8, theD 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 the figure. At this condition the final enthalpy is 30.8 Btu per lb of dry air and Cooling of Moist Air at Constant Pressure with Condensation of Water Thisis illustrated in Pig. 9 wheremoistair cooled from State 1 passes through successive states along tire hue, W *= Wj -- constant, until the saturation line is intersected. The tempera ture at this point of intersection is by definition the dew-point temperature for State 1. Further cooling through successive equilibrium states is ac companied by condensation. The succession of states for the total system, moist air, and liquid water, is represented by a continuation of the TF -- Wt Roe into the liquid-vapor region. (Temperatores below 32 F would involve the solid-vapor re gion.) Consider that the final temperature is . The final enthalpy is then A* ; the liquid water formed is (W( -- Wi), where Point 3 is at the intersection of the isotherm through 2 and the saturation cuve; the final humidity ratio of the moist air is Wf; and this final moist ear has dew-point, wet-bulb, and dry-bulb temperatures all equal to U Example + 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 cod the air to 70 F, saturated?