Question

B5. (15%) (Show all your calculation steps.) Moist air at 20°C, 1 atm, and 43% relative humidity and a volumetric flow rate of 900 m3/hrs. enters a control volume at steady state and flow along a surface maintained at 65° C, through which heat transfer occurs. Liquid water at 20° C is injected at a rate of 5 kg/hr and evaporates into flowing stream. For the control volume, Wex=0, and kinetic and potential energy effects are negligible. Moist air exits at 32°C, 1 atm. (a) draw the system schematic with given data, (b) determine the rate of heat transfer in kW and (c) calculate the relative humidity at exit in % (d) use the attached psychrometric chart to compare your answers in (b) and (c) by drawing the process on the chart.

use engineering methodology and show units calculations please

Answer 1

Date Page Liquid water uma Tz = 20°c 5kg/he Can T3: 32°c Parlatm T-20°c Pi= 43% LAV), 900 m m3/h P1 = 1 atm To= 33QK From Psychrometry chart & $1 = 43% at Ti= 20°c air volume grow rate humidity ratio 21=0.83 m3/kg of dry kg/kg of dry 0.00 62 air mass how air of dry mal = (Auli Vi 900 0-83 - 1084 : 33 = 0.3012 kg I see 33 kg / he umi= wimal 6.722 kg/h hr ima imitime 11.722 kg the

CIASSMATE Date Page W3 - M3 e uma 0.01081 rair kg/kg of dry mar mai ( ima, - indi ) at T3= 32°C and W3 = 0.01081 regikg of any vair I3326 from Psychrometry chart humidity relative at enit = 103 = 37 1. CC) By energy consewationi hit (W3-Wilhfa tqv=h3 gov = h3-hi - (W3 - wil hfz Cpl T3-TI) +w3 - [ 2500 + 1.88 T3) Wi[2500 + 1-88 T1 ]-|W3-wilhfz steam table at T2 = 20°C, hf2 = 84.2043 kJ/kg by rong You = 12.06+ 27.67-15.73 -0.308 = 23.61 kJ / kg of dry air qu= ma q av = 7.1119 kW b

Page (a) uusing pohycometric chart $1 = 43 % = لا 0.0062 20°C 37% (37% W3=0.0/08 43% Loi=0.0062 20c T3: 3200 W3 = 0.0108 neglky 93= 37%.
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