In a hot summer day, the air in Engineering Building is to be maintained at 25 d
ID: 491838 • Letter: I
Question
In a hot summer day, the air in Engineering Building is to be maintained at 25 degree C and 50% relative humidity by passing outside air through an air conditioner. The air enters the conditioner at 32 degree C and 70% relative humidity at a volumetric flow rate of 10000 m^3/min, leaves the conditioner cooled and saturated with water vapor, and is then reheated to 25degree C. Assume ideal gas behavior and assume pressure remains at l atm thought the whole process. a) What is the partial pressure (in atm) of water in the outside air? What is the rate of water (in kg/min) enters the air conditioner? b) Estimate the temperature of the air leaving the air conditioner. c) Calculate the rate of water (kg/min) removed from the conditioner. d) What is the volumetric flow rate (m^3/min) of air entering the building?Explanation / Answer
Vapor pressure of water at 32 deg.c = 35.7 mm Hg
Relative humidity =100* partial pressure of water vapor/vapor pressure of liquid = 70
Partial pressure of water vapor/35.7= 0.7, partial pressure of water vapor =35.7*0.7= 24.99 mm Hg
Under saturated conditions, the partial pressure= vapor pressure = 24.99 mm Hg
At temperature corresponding to this vapor pressure = 26deg.c
Flow rate of air, V = 10000 m3/min =10000*1000 L/min = 107 L/min, T=32+273= 305K, P=1 atm
R= 0.0821 L.atm/mole K, n= PV/RT= 1*107/(0.0821*305)=399353 gmoles/min = 399.353 Kgmoles/min
Molar mass of air =29 ,hence mass of air entering/min = 399.353*29 = 11581.24 kg/min
Mass of water vapor + mass of dry air = 11581.24 kg/min (1)
Entering air
Moles of water vapor/moles of dry air = partial pressure of water vapor/partial pressure of dry air
Moles of water vapor/moles of dry air = 35.7/(760-35.7)= 0.049
Moles of water vapor = 0.049* moles of dry air =0.049*29 kg of dry air =1.421 kg of dry air
mass of water vapor = 1.421*18 kg of dry air =25.6 kg of dry air
From Eq1. 25.6 kg of dry air + kg of dry air = 11581.21
Kg of dry air = 11581.21/26.6 = 435.4 kg of dry air /min
Mass of water vapor = 435.4*25.6 = 11146 kg/min
At out let, partial pressure of water /partial pressure of dry air =moles of water vapor/moles of dry air
24.99/(760-24.99)= moles of water vapor/moles of dry air
Moles of water vapor =0.034* moles of dry air = 0.034*29 kg of dry air =0.986 kg of dry air =0.986*435.4 =429 kg/min
Mass of water vapor = 429*18=7722 kg/min
Mass of water removed =11146-7722 = 3424 kg/min
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