Question

In a device to produce drinking water, humid air at 32C, 90% relative humidity and 1 atm is cooled to 5C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air.

Data: Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol

H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.

H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg

What is the flow of humid air entering the system in mol/min?

What is the flow of humid air entering the system in m^3?

How much drinking water is produced by the system in lit/min?

Answer 1

1) Relative humidity = partial pressure of water/vapor pressure of water*100

vapor pressure of water at 32 DegC = 4.75925 kPa, and 1 atm = 101.325 kPa, partial pressure = totla pressure*mole fraction of water

0.9 = 101.325*mole fraction of water/4.75925

so mole fraction of water in entering humid air = 0.04227

now assuming water vapor in equilibirium finally,

mole fraction of water in outgoing humid air = water vapor pressure at 5 DegC/total pressure

=0.872575/101.325 = 0.0086116

Now lets assume n moles of humid air are entering to system

water mass enetering = mole*molar mass = n*0.04227*18/1000 Kg

mole of dry air entering = n(1-0.04227) = 0.95773n moles

moles of dry air in outlet vapor stream = same = 0.95773n(since dry air will not condense)

lets assume moles of water in outlet humid air = y

then mole of water = y/(y+0.95773n) = 0.0086116 so y = 0.0083192n moles

or mass of water vapor in outlet air = n*0.0083192*18/1000 Kg

so water mass condensed = inlet mass - outlet vapor mass =

n*0.04227*18/1000 Kg - n*0.0083192*18/1000 Kg = 0.0339508n*18/1000 Kg

applying enthalpy balance:

(Enthalpy in) - (enthalpy going out) = enthalpy change

[0.95773n*0.204 +n*0.04227*18/1000*2560] - [0.95773n*(-0.576)+n*0.0083192*18/1000*2510.8 + 0.0339508n*18/1000*21] = 2200

solving for n we get, n = 954 mol/s

mol/min of humid air entering = 954*60 = 57241.57

2) applying PV = nRT for humid air

101325*V = 57241.57*8.314*(273+32) or V = 1432.53 m3/min

3) As calculated in part 1, water condensed = 0.0339508n*18/1000 = 0.0339508*57241.57*18/1000

= 34.98 Kg/min

density of water = 1 kg/liter,

so liter /min = 34.98 lit/min