Document MG2vQpRdjadDKN3YbMRLDwwJL

30 CHAPTER 3 1959 Guide Energy balance for the process, <?iAi + {7A* -- Gth Maas balance for the water vapor, (?ifF, + <7,tF* = G\Wt Eliminating Gt and combining the three equations yields the equation, ht -- ht W, - W, G, hs-ht^Wt-Wt^Gt (34) until it intersects the constant temperature line in the liquidvapor region corresponding to the final temperature 70 F. This is shown as Point 2 on the diagram. Then, i$ = (?(/i - At) The initial enthalpy is 43 Btu per lb of dry sir; the initial spe cific volume is 14.4 cu ft per lb of dry air; and the final enthalpy is 34.2 Btu per lb of dry air. The solution of the problem is 'tfi - X <43 -- 34.2) = 12,200 Btu per min. The other method is to use an energy balance, i<> - <7[A, - A, - k*Wt - W*)] The initial humidity ratio is 0.01S3 lb of water vapor per lb of dry air, and the final humidity ratio is 0.0158 lb of water vapor per lb of dry air. Therefore, the beat to be removed is ifr - X <43 - 34.1 - 0.0025 X 38.07) ** 12,130 Btu per min. AdiaboHc Mixing of Two Steady Flow Air Streams at Constant Pressure The process is diagrammed in Fig. 10. By applying the prin ciples of the conservation of mass and energy, three equations may be written: Mass balance for the dry air, (7, -f- Gt - <?, Example 6: Outdoor air at 0 F dry-bulb temperature and 0.80 degree of saturation is to be mixed adiabatic&lly with recircu lated indoor air at 70 F dry-bulb temperature and 0.20 degree of saturation, in the ratio of one pound of dry air in the former to four in the latter. Find the temperature and degree of satu ration in the resulting mixture. Solution a: From the data of Table.2. The only unknown properties are the humidity ratio TP and the enthalpy A* of the resulting mixture. These may be determined from Equation 34. Thus, 20.270 - A 0,003164 - IF, 1 A, - 0.668 " W* - 0.000630 " 4 from which At = 16.350 and IP* K 0.002657. The enthalpy of the final mixture may also be expressed by Equation 28: At = A, + f(Aaa Since n by definition is Wt/W,, Equation 28 may be rewritten as 16.350 - K + (0.002657/fF.) X A.. At 56 F the right side of the equation is 16.332, and at 57 F it is 16.582. Interpolation gives as the final dry-bulb temperature of the mixture 56.07 F. At this temperature the humidity ratio at saturation is 0.00960 lb of water vapor per lb of dry air. Therefore, the final degree of saturation is ft - 0.002657/0.00960 - 0.277 Solution b: From the ASHAE Chart. Equation 34 indicates that the state point of the resulting mixture lies on a straight line connecting the state points of the two streams being mixed, and divides this line into two segments whose respective lengths are inversely proportional to the rates of dry air flow in the corresponding streams. This is illustrated in Fig. 11. Points 1 and 2 are located and connected by a straight line. Hie state of the final mixture is set so that (?' _ Dt-t 1 Gt Dt-i 4 Rg. 10.... Illustration of Mixing of Two Steady Row Streams at Constant Pressure Rg. If -- . Solution of Example 5 on ASHAE Psydirometric Chart Thermodynamics Sealing the distances on the chart, the required solution to Esampk 5 is 56 F dry-bulb temperature and 0.28 degree of saturatioo- Addition of Moisture to an Adiabatic Stream Consider a stream of moist air flowing adiabatic&lly between two sections, 1 and 2, as in fig. 12} with moisture addition at the rate Gi{Wt -- and the moisture having the enthalpy A. Btu per pound of moisture. An energy balance yields <7,A, + tfiCR't - iP,)A - <7,As (35) Example 6: Liquid water chilled to 40 F is injected into an air stream initially at 95 F dry-bulb temperature and 80 F thermodynamic wet-bulb temperature. At what temperature will saturation be reached? How much water must be evapo rated to reach saturation? Solution a; From the data of Table 2. The solution of Equa tion 35 for As yields. A, = Ai + (VP, - H\)A. The initial enthalpy of the moist air Ai must be found from Equation 8, A, * A* -- (IP* - 1P,)A.* 22.827 + *40.49 - 43.69 - (0.02233. - 0.03673*) (48.05) from which * = 0.511. Hence, A, - 22.827 + 0.511(40.49) = 43.52 Btu per lb of dry air. and W, => 0.03673(0.511) -- 0.01877 lb per lb of dry air. The solution of Equation 35 is A, - 43.52 + (IV, - 0.01877)(8-09) By trial and error, this equation will be satisfied at the tem perature 79.87 F. At this temperature the humidity ratio Wt is 0.02223. The weight of water evaporated is therefore 0.02223 - 0.01877 - 0.00346 lb per lb of dry air. Solution 6: From the ASHAE Chart. Solution of Equation 35 for the ratio (A* -- A*)/(IP* -- fPi) yields A, - At (36) The slope of the condition line is therefore determined by 31 the enthalpy of the water which is supplied. This slope is es tablished on the chart by connecting the center of the pro tractor on the psyebrometric chart with the value of A on the protractor. Draw a line parallel to this reference line through the initial Point 1 (Fig. 13). The second line is the condition line for the process. Since the conditions of the problem re quire the final point to lie on the saturation line, the intersec tion of the condition line with the saturation line gives the desired solution. Adiabatic Saturation Adiabatic saturation is the designation given any process in winch the state of moist air is changed from some initial unaaturated condition to a saturated one without the addition or removal of beat. According to this definition, the addition of moisture to an adiabatic stream may become an adiabatic saturation process. Example 6 is an illustration. A type of adiabatic saturation of further practical interest is the use of continually recirculated spray water in a saturat ing air washer. Here the spray water will ultimately come to the same temperature as the saturated leaving air; thia tem perature is, by definition, the thermodynamic wet-bulb tem perature. Hence, adiabatic saturation in this manner will have the final state.point on the saturation curve, with the same thermodynamic wet-bulb temperature as the original state point. la e process such as this, the moist air enthalpy changes very slightly. The moisture added and temperature change may be obtained from the psychrometric chart as sketched in Fig. 14. Example 7: Moist air at 75 F dry-bulb temperature and O.GO degree of saturation, is saturated adiabatic&lly with recircu- X AT ENTHALPY h Rg. 12___ Illustration of Addition of Moisture to an Adiabatic Stream Rg. 14___ Moisture Added and Temperature Change