Question

Chemical Engineering question.

Air at 38.0°C and 99.0% relative humidity is to be cooled to 18.0°C and fed into a plant area at a rate of 710.0 m³/min. You may assume that the air pressure is 1 atm in all stages of the process.
A.) Calculate the rate at which water condenses in kg/min .

B.) Calculate the cooling requirement in tons (1 ton of cooling = 12,000 Btu/h), assuming that the enthalpy of water vapor is that of saturated steam at the same temperature and the enthalpy of dry air is given by the expression H∧ (kJ/mol) = 0.0291[T(°C) – 25].

Answer 1

A)Vapor pressure data (mm Hg)   : 38 deg.c= 49.7 and 18 deg.c = 15.5 mm Hg

Flow rate of air water vapor mixture= 710m3/min=710*1000L/min= 7.1*10⁵ L/min

Relative humidity = 100*partial pressure of water vapor/vapor pressure of liquid

Hence at 38 deg.c, 0.99= partial pressure of water vapor/49.7

Partial pressure of water vapor= 49.7*0.99=49.2 mm Hg

Total moles of air vapor mixture from gas law, n= PV/RT, where P= 1atm, R=0.0821 L.atm/mole.K, T= 38 deg.c= 38+273=311K

Hence no of moles of air water vapor mixture = 1*7.1*10⁵/(0.0821*311)= 27807 moles/min

At inlet conditions of 38 deg.c

Moles of water vapor/total moles = partial pressure of water vapor/total pressure

Moles of water/ 27807= 49.2/760= 0.065

Moles of water vapor at inlet= 27807*0.065=1807 moles/min

Moles of dry air at inlet= 27807-1807=26000 moles/min. This remains the same through out the cooling process and is considered as tie substance.

At outlet, the air is assumed to be saturated with water vapor. Hence at the exit temperature of 18 deg.c, partial pressure of water vapor= vapor pressure of water =15.5 mm Hg

Hence partial pressure of water vapor/ partial pressure of dry air= moles of water vapor/ moles of dry air

15.5/(760-15.5)= moles of water vapor/26000

Moles of water vapor at the exit/min= 541 moles/min

Moles of water condensed/min= molar flow rate of water vapor at the inlet- molar flow of water vapor at the exit= 1807-541=1266 moles/min

Mass flow rate of water condense= molar flow rate of water condensed*molar mass = 1266*18 g/min=22788g/min= 22.788 kg/min

Enthalpy of mixture at the inlet at 38deg.c= enthalpy of dry air + enthalpy of water vapor

Enthalpy of mixture at the outlet at 18 deg.c= enthalpy of dry air+ enthalpy of water vapor

Enthalpy change= Enthalpy of dry air at inlet- enthalpy of dry air at outlet+ enthalpy of water

Condensed=

latent heat of vaporization of water= 40.7 KJ/moe

enthalpy change = cooling requirement {0.0291*(38-25)- 0.0291*(18-25)}26000Kj+40.7*1266= 66658.2Kj/min=1111 Kj/hr

1Btu= 1.05 KJ

Hence 1111 Btu/hr= 1111*1.05 Kj/hr =1166.55 Btu/hr

1Ton=12000 Btu/hr

Hence Tonnage= 1166.55/12000 =0.097 Tons