Document ZJmLQkyk9NGVEyjyj6dexY73Y

44 CHAPTER 3 1965 Guide And Data -Book Psychrometries - 9BT 45 REFRISCIUNT . Rg. 6 .... Adiabatic Mixing of Two Streams of Moist Air Example 4: Moist air, saturated at35 F, eaters a hwtimg coil at a rate of 20,000 cfm. Air leaves the coil at 100 F. Find thereauired rate of heat addition in Btu per hr.' j - Solution: Big. 3 schematically shows the solution. State 1 is located on the saturation curve at 35 F. Thus, A, - 13.01 Btu/lb dry air, IT, = 0.00428 U> water/lb dry air, and p, - 12.55 cu ft/lb "y ^ staA? 2 ny be located at the intersection of t - 100 F d IT* ** Wt " 0.00428 lb water/lb dry air. Thus, 5, -- 28.77 Btu/lb dry air. The rnaan flow of drjr air is (20,000) (60) 12.55 95,620 lb dry.air/hr. . From Equation 25 ' ' , ... .1?* " (95,620)(28.77 - 13.01) - 1,507,000 Btu/hr. Cooling of Moist Air Ifmoist air is'cooled to atemperature below its ihitial'dew point, separation of.moisture will occur. Pig. 4 shows a thematic cooling coil where it is assumed that moist air is Uniformly processed."'Although water may be separated at various temperatures,'tanging from the initial dew point to tiie final saturation temperature, it is assumed that the ooni densed water is cooled to the final air temperature fa'before H drains from the system.' For the system of Fig. 4, the steady-flow energy .and ma terial balance equations are ` 1 P*A.." mji* + ,9* + mwhmx ..... rn.IT* + Thus m.(FT, 1 Wt) (26) "vl(Ai - AO - Xwi JTS)A-*1 . " (27) Example 6air at 85 F dry-bulb temperature and 50 per cent relative humidity enters a cooling coil at a rate of 10,000 cfm. Ine an is processed;to a final condition of saturation at,50 F. Bind the tons of refrigeration required. " Solution: Fig. 5 shows the schematic solution. State 1 is located intersection of f = 85 F and * - 50 percent. Thus, . At - 34.62 Btu/lb dry air, IT, -- 0.01292 lb water/lb dry air, pd " 14.01 cu ft/lb.dry air. State 2 is located on the saturato curve at 50 F. Thus,. 5, = 20.30 Btu/lb dry air and W, - 0.00766 lb water/lb dry air. From Table 1, 1L* - 18.11 Btu/lb water. The mass flow of dry air is - 10,000 14.01. ' 713.8 lb dry air/min From Equation 27 iff*- (713.8)1(34.62 - 20.30) ' ' --(0.01292 -- 0.00766) (18.11)] = 10,150 Btu/min ffinc^one ton of refrigeration equals a heat withdrawal rate of 200 Btu/min, the required refrigerating capacity,is 50.75 tnn., Adiabatic Mixing of Two Streams of Moist Air A common process involved in air-conditioning systems is the adiabatic mixing of two streams of moist air. Fig. 6 schematically 1. B .... Schematic injection of Water into Moist Air Rg. i(nO : Rg; 10____ Schematic Problem of Air Conditioning a Space ^ o,e problem. If the miring in Babtic, it must be governed by the three equations ` m.iA, + )>uA - fltii *+ el** = tnj - ny,ITi +,m^fT* " Elimination of T7tn gives A* --.A*. Wt -- Wt . m*, ht -- A, *: Wt -- Wi . . ms. . s, (28) according to which:.on the'ASHRAJS chart the etato~poini of the resulting mixtureJieson the..straight'line connecting the state- points of the two streams.being mixed and divides theline into hoo segments which are in the same ratio as are the.masses of dry air in the too streams.[ . v, / - - Example ff.'Astream of 5000 cfm of outdoor air at 40 F dry- bulb temperature and 35 F thermodynamic.wet-bulb temperature is adiabaticaQy with 15,000 cfm of recirculated air at 75 F dry-bulb temperature and 50 percent relative humidity.-Find the dry-bulb temperature and thermodynamic wet-bulb temperature of the resulting mixture. 1 V- , Solution:-Fig.7 shows the schematic solution. Statea l' and 2 may.' be, located'on the. ASHRAE. chart. Thus,- d, = 12.65 .cu ft/fij dry.air, and 13.68,cu ft/lb dry air.-Thusm =: /-.. i ,.5000.', : 12.6& --395 lb .dry air/mn 15,000 1 ------ = 1096 lb dry air/min . 13.68 " According to Equation 28 !Iine'3-2 'pw :"'Line 1-3 m* 1096 ^ 1 :.Iine.l-3--. m* .. - Line 1-2 n* !1491 . Thus, the length of line segment i--3 is 0.735 times the length of the entire' line' 1-2. Through use of a ruler,.State'3 may be lo cated, the values fa = 65.9 F and fa* " 56.6 F found. Adiabatic Mixingof Moist Air with Injected Water . A frequent air-conditioning process is the injection of steam or' liquid water into a: moist^tir stream' in order to. raise' the humidity,ratio of the moist air. Fig. 8 schematically shows the problem; If the miring is adiabatic, the .following equations apply ; ' ;* ' T . .tnjii + mihm " m*A Thus *nlT + m, - m.TT* W* - W> (2?) according to which: on the ASHRAB chart the final state-posnti cf the moist air must lie on a straight line whose direction is fixed by the specific enthalpy of the injected water and which drawn through the initial state-point <f the moist air , , Example 7 . Moist air at 70 F dry-bulb temperature mid 45 F fo>rnKMfyn*mift wet-bulb temperature is to be processed to a final dew-point temperature of 55 r by injection of satorated steam at 230 t. Tbe^ate of dry air flow;is 200 lb per min..Find