In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit
Answer: hello your question lacks some data below is the missing 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
Answer :
a) 34.98 lit/min
b) 1432.53 m^3/min
Explanation:
a) Calculate how much water is produced
density of water = 1 kg/liter
First we will determine the mass of condensed water using the relation below
inlet mass - outlet vapor mass = 0.0339508 * n * 18/1000 ----- ( 1 )
where : n = 57241.57
hence equation 1 = 34.98 Kg/min
∴ volume of water produced = mass of condensed water / density of water
= 34.98 Kg/min / 1 kg/liter
= 34.98 lit/min
b) calculate the Volumetric flow rate of air entering the unit
applying the relation below
Pv = nRT
101325 *V = 57241.57 * 8.314 * 305
∴ V = 1432.53 m^3/min