ENSC 61 Tutoril, Week#9 Non-Recting Mixtures Psychroetrics Applied to Cooling Toer Wter exiting the condenser of poer plnt t 5C enters cooling toer ith ss flo rte of 15000 kg/s. A stre of cooled ter is returned to the condenser fro the cooling toer ith the se flo rte. Mke-up ter is dded in seprte stre t 0C. Atospheric ir enters the cooling toer t 0C ith et bulb teperture of 0C. The oluetric flo rte of oist ir into the cooling toer is 8000 /s. Moist ir exits the toer t 0C nd 90% reltie huidity. Assue n tospheric pressure of 101. kp. Deterine: ) the ss flo rte of dry ir, b) the ss flo rte of ke-up ter, nd c) the teperture of the cooled liquid ter exiting the cooling toer. Step 1: Dr digr to represent the syste Exhust Air T db,=0 C = 90% Wr Condenser Wter,= T =? C,1=15000 kg/s T =5 C Cooled Condenser Wter 1,1 1 Inlet Air V,=8000 /s T db,=0 C T b,=0 C 5 Mke-up Wter =? kg/s T =0 C 5 Step : Prepre property tble H O T (C) (kg/s) 1 (st. liquid) 5 15000 (st. liquid) 15000 5 (st. liquid) 0 h (/kg) M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC 1
Air T db (C) T b (C) (%) (kg /kg ) ( /kg ) h (/kg ) 0 0 0 90 Step : Stte your ssuptions Assuptions: 1) The cooling toer opertes under stedy conditions ) KE, PE 0 ) Cooling toer is rigid nd dibtic W CV, Q CV 0 ) Assue ll liquid ter is sturted 5) The pressure is constnt throughout the cooling toer t 101. kp. Step : Sole Prt ) The ss flo rte of dry ir cn be deterined using the oluetric flo rte of oist ir drn into the cooling toer (gien in the proble s 8000 /s) nd the specific olue of this ir (on per kg dry ir bsis) s shon in Eq1., V (Eq1) The specific olue of the ir entering the cooling toer cn be deterined using the stte point of loction on the psychroetric chrt found ith T db, = 0C nd T b, = 0C. Fro the psychroetric chrt, = 0.87 /kg Substituting this lue nd the gien oluetric flo rte into Eq1 the ss flo rte of dry ir is deterined s shon belo. Prt b) 8000 V, s kg, 916. Anser ), s 0.87 kg 8 To deterine the ss flo rte of the ke-up ter, denoted s, ss blnce cn be perfored on the ter entering/exiting the cooling M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC
toer control olue. At loction 1 there is stre of ter entering the cooling toer fro the condenser, hich ill be denoted s, 1. At loction there is stre of ter exiting the cooling toer (to be returned to the condenser), hich ill be denoted s,. Fro the proble stteent, the ss flo rte of the cooled ter leing the cooling toer (to the condenser) is equl to the ss flo rte of ter entering the cooling toer (fro the condenser). This is expressed in Eq. (Eq), 1, At loction, there is oisture entering the cooling toer control olue crried in by the incoing ir, hich ill be denoted s,. At loction, there is oisture leing the cooling toer control olue crried out by the exiting ir, hich ill be denoted s, cooling toer ter is perfored in Eq.. The ss blnce on the,, 0, (Eq), Note: Eq could he been deeloped ieditely by resoning tht the ount of ter tht needs to be de-up for ill be equl to the ount of oisture tht is picked up in the cooling toer by the ir nd exhusted. The ss flo rte of ter por t nd cn be expressed in ters of the corresponding ss flo rtes of dry ir t loction nd nd their respectie huidity rtios nd s shon in Eq nd Eq5. (Eq),, (Eq5),, Substituting Eq nd Eq5 into Eq, Eq6 is obtined. (Eq6),,,, Since stedy opertion of the cooling toer hs been ssued. The ss flo rte of ir through the toer should rein constnt. This is expressed in Eq7. (Eq7),, M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC
Substituting Eq7 in Eq6, Eq8 is obtined. ( ) (Eq8) Recll tht the ss flo rte of dry ir s deterined in prt ). The huidity rtio of the ir entering the cooling toer cn be deterined fro stte point on the psychroetric chrt. Fro the psychroetric chrt, = 10.6 g /kg = 0.0106 kg /kg Unfortuntely, stte point (T db, = 0C nd =90%) is off the psychroetric chrt so ill he to be clculted using eqution 1-11b fro Cengel nd Boles s shon belo. Fro Tble A- @ T= 0C, P g = 7.8 kp. 0.6 Pg 0.6(0.9)(7.8) kg 0. 07 P Pg 101. 0.9(7.8) kg Substituting these lues into Eq8, the ss flo rte of the ke-up ter cn be deterined. kg ( ) 916.8 s kg 0. s kg 0.07 0.0106 kg Anser b) Prt c) The teperture of the cooled liquid ter exiting the cooling toer cn be deterined if its enthlpy is knon. Since the liquid exiting the cooling toer is ssued to be sturted liquid, its enthlpy cn be used to interpolte in Tble A- to deterine its teperture. To find the enthlpy of the ter exiting the cooling toer n energy blnce on the cooling toer control olue cn be perfored. At loction 1, the rte of energy entering the control olue crried by the stre of ter coing fro the condenser is h, 1. At loction, the stre of ter leing the cooling toer is crrying y energy t rte of h,. At loction M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC
, the oist ir crries energy into the control olue t rte of h into the control olue. At loction, the oist ir leing the cooling toer control olue crries energy y t rte of h. The ke-up ter crries energy into the control olue t rte of h. Cobining ll of these stteents into one expression to obtin Eq9., 1, h h h h h 0 (Eq9) By rerrnging Eq9, the enthlpy of the ter t loction cn be deterined, s shon in Eq10. h ( h h ), h,1 (Eq10) h nd he preiously been deterined nd is gien in the proble stteent leing h,1, h, h, nd h to be deterined before h, cn be soled for. h,1 Since sturted liquid ter s ssued t loction 1, h,1 cn be deterined fro Tble A- using T 1 = 5C. h, 1 188. 5 kg h Using stte point on the psychroetric chrt h cn be deterined. h 58 kg h As stted preiously, stte point is off the psychroetric chrt so h ust be clculted. Using eqution 1-1 fro Cengel nd Boles, the enthlpy of DRY AIR lone cn be deterined s shon belo. h, c pt 1.005 (0 C) 0. kg C kg Note: the boe clcultion is in fct clcultion of n enthlpy difference ith the enthlpy t 0C being subtrcted fro the enthlpy of interest. The enthlpy t 0C in this cse is zero (becuse 0C is the zero M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC 5
reference point for enthlpy). The enthlpy difference clcultion in full ould pper s shon belo. h h h h,,,, h 0 c c @0 C p p c p T [ C] 7[ K] T [ C] ( T [ K] 7K) 1.005 (0 C) 0. kg C 7K kg The enthlpy of the MOISTURE in the ir cn be deterined fro Tble A- for h g@ T = 0C. h, hg ( T ) 57. kg To cobine the dry ir nd oisture enthlpies t loction into one ter, h, the enthlpy of the oisture ust be conerted to per kg of dry ir bsis, hich is ccoplished by ultiplying it by the huidity rtio,. kg h h, h, 0. 0.07 57. 15. 7 kg kg kg kg h Since sturted liquid ter s ssued t loction 5 h cn be deterined fro Tble A- using T 5 = 0C. h 8. 96 kg Substituting these lues into Eq10, the enthlpy of the ter t loction cn be deterined s shon belo. h, 188.5 kg kg 916.8 s (58 15.7) kg kg 15000 s (0.) kg s (8.96) kg h, 188.5 56.16 1. kg kg kg M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC 6
As stted preiously, since sturted liquid ter s ssued t loction, h, cn be used to find the corresponding teperture in Tble A-. Fro Tble A- it is found tht h, lies in beteen the enthlpies corresponding to tepertures of 0C nd 5C. Interpolting beteen these to lues, the teperture of the ter exiting the cooling toer cn be deterined s shon belo. 1. 15.79 T 0 C 16.68 15.79 5 C 0 C T 1.6[ C] Anser c) Step 5: Sury ) the ss flo rte of dry ir is 916.8 kg/s b) the ss flo rte of ke-up ter is 0. kg/s, nd c) the teperture of the cooled liquid ter exiting the cooling toer is 1.6C. M. Bhri ENSC 61 (S 11) Non-recting ixtures nd HVAC 7