Cooling Tower

Main Question or Discussion Point

A natural draught cooling tower is designed to have a 3% loss of the mass of the water entering the tower by evaporation into the atmosphere. Atmospheric air enters the tower at a volume flow rate of 4 m^3/s a temp of 10C and a relative humidity of 50%. Air leaves the tower in a saturated condition at a temp of 34C. The atm pressure is 0.995 bar. The water enters the tower at a temperature of 36C. Find the mass flow rate of water entering the tower.

The problem is I don't know how to find the h for the water at the outlet, any suggestions? Thx~

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Have you considered using CpΔT instead of enthalpy? This should be close enough if you use the average heat capacity between temperatures.

The point is if i don't know the outlet temp for water how will i be able to do CpT

I guess you need to assume standard cooling tower outlet conditions - 25 or 30 C maybe. Are there any design parameters to go by?

SteamKing
Staff Emeritus
Homework Helper
You know the inlet and outlet conditions for the air. You know the inlet conditions of the water, and that a certain amount of the water is lost due to evaporation. Don't you think you can determine how much water enters the tower based on the amount of energy required to heat the air leaving the cooling tower? The problem does not ask you to find the temperature of the cooled water, only how much enters initially.

But there's one equation and 2 unknowns. So the duty the air supplies, Q = water flow*Cp*(T1-T2). We don't have T2 and mass flow of water. We need to take a different path. We have to use the 3% loss in water mass that leaves the tower as steam.

i agree with Mrmiller, but how can I actually use the 3% and find H.... I have been stuck here

russ_watters
Mentor
You're not trying to find energy here, so I'm not sure what direction you guys are going with the problem. You have the inlet and outlet conditions of the air, so just read from a table or psych chart the moisture content at each, subtract, then convert to mass (then divide by 3%).