032111 - 5 A volumetric flow of 1" = 0.5 Ina/s of air...

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Unformatted text preview: 5. A volumetric flow of 1:" = 0.5 Ina/s of air at 10°C. 30% RH. enters a heater. The rate of heat transfer to the air is = 15 kW'. Find the exit temperature and relative humidity. ——> $03” L; 0V1“ @ 10°C: 3°25 NJ _> C‘): 152w 4’ spud Tzl¢z 6. The prevfirgprohlel will dem nstrate that simple herdiulfia ld. dry .e. hot air at an even Qfi lower relative humidity. This air can he very uncom- © fortahle as it tends to dry things out. A remedy is to 10.4 25* I I I I I C. add water [in the form of steam or liqurd) to the an 307° 600/ during the heating process. Say we have the same in- s I a let conditions and flow rate as in the previous problem. ,c We now want the exit to he at 25°C and 60% relative g3 IR Ir humidity. To humidity the air during the process. satu- I rated water vapor at 100°C is injected into the system. Find the mass flow rate of the injected steam and the rate of heat transfer to the air. "'. Air at BOQC 70% R.H. is sent into an AC svstenl. The (,3 0 LI) [Q L3 fl ‘ ' v —} o ,\ 5 system] produces 2 1n3I/s of air at 5°C. Calculate the [ I rate of heat transfer from the air and the rate of conden— ‘ , sate removal. l\'Ia.ke Whatever assumptions are necessary L8 (1 I A ‘ Q" andy'or appropriate. 6' . - nQQI f n5 " [meat 37 it); : M46350) (I) '——C72 C2) a ' new.” may: [Lia BO‘C S‘C. 3 767. 3/ Z“ 7 W: C.) I FLUKM‘ F0305 Fl; k5: {080C QJV: (3‘1: HQCDL‘QI) (N.th Sjvr) . , e . r’mar fioi'. hthI wwol now P olml. Id (III O I I/IIIIUII_LII\ IILIjIfibI mots L A». . 1:45 mil-fluca‘ new ...
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