Question

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Qn.7. Adiabatic mixing 300 m3/min of fresh air at 30 °C DBT and 50% RH is to be mixed with 800 m3/min of recirculated air at 22 °C DBT and 10 °C wet bulb temperature. Determine the (1) enthalpy, (2) specific volume, (3) relative humidity, (4) dry-bulb temperature and (5) dew point temperature of the mixture. Ans. [(1) 38.24 kJ/kg, (2) 0.849 m/kg (3) 29%, (4) 24.2 °C, (5) 5°C]

Answer 1

The process is right as I mentioned below but we are getting different values here but the process is right to find the values of mixture. Kindly go through the process. And if you have any doubts comment me. Thank you
let, given, & = 300 m/min, td, = 30°c, & = 50%, V = 800 m/min, t = 226 tdp = 10c h = enthalpy of the mixture, Now, & drawing the psychrometric chart we can mark the both points 1 and 2. byr

m 20ti12 DRY TERATURE VE SATURATION TEMPERATURE ENTHALPY-O PER KG OF DRY AR dedi & DEW POINT-C 0.- BROWN 17 8 HUMIDITY RATIO.GRAMS OF MISTURE PERILOGRAMOFORY AR 338 WOR PRESSURE MOM OF MERCURY ENTHALPY-KI PERKS OF DRY AIR 39 909

W h , 64.5 KJ/ h cathalpy Rett sot h blom the Phythrometric chart, ol 5k lg consentending to to, y top 6 % - 50% typ = 42 Klug corresponds to to, & 10.8 2 ha-ha loc volume de tapy = 16 & W, = 13.5 31xg of dry ait 22°c 30°c = 1.2 glug of dry air. DET specific volume at 1, now, we know that =0871 mi)ng of dry at 2, V = 0.846 ning (in, tinh = is, h, tinh here in man - h= inh, th, he ingh-h3 How rate mtm, we can find in, & in une, in = = 200 = 342.075 kg/min in = V/v = 800 - 945. 63 149/min in - 5.70 49/8- = 15.76 Mg/s h (5.70464-5)+ (15.76X42 5.70+ 15.76 now, from the chart corresponding to = 47.98 KJ /kg. of dry air, we can find the values of specific volume, humidity, BBT & DPT of mixture Toin the 2-points & the point 3 is lies on the line joining the 2-points as shown in the figure. now, at point 3 from the chart corresponding to h = 47.98 h = 48 165/1g, us, tap & Pe are as follows. Jug = 0.85 m 1kg), (tap = 13c) to = 24.2% (-50%) 0.846 h. - 47.98 KT/kg