# Calculating outlet temperature and heat transfer in an adiabatic spray cooler

Question:

In an adibatic Spray cooler, a fine mist of liquid water is used to cool hot air by evaporating all of the water. The liquid flows in at a rate of 90 g/s and 30°C. The input hot air is at 450°C and 900 torr. The dewpoint of the input air is 47°C and the flow rate is 1362 L/s. Assume no change in pressure occurs for the air stream. Ue the antoine eqn for water.

Relevant eqn's:

Antoine eqn: P*= A-B/(T+c)
Absolte humidity: ha = mass of A in gas/ mass of dry gas
Relative humidity: hr = Pa(T)/P*(T)*100%
Mass balance
Energy Balance
Ideal gas

Progress so far:

I calculated the density of the hot air stream using the ideal gas law and kays rule w/ the assumption that the composition of air is .79 N2 and .21 O2 to be .57379 g/L.

This assumes that the air in was dry air. I suspect that this is where I went wrong. From here I calculated mass flow rate using volumetric flow rate time density.

Mass Balance simplifies to In=Out neglecting the accumulation generation and consumption terms.

I found realtive humidity to be .1152 using mass flow of water / mass flow of air

I found the mol fraction of water in the outlet stream to be
.1557= (90g/s water/18.016 g/mol)/(mol water + (487.93 g/s air)/(28.558 g/mol air)=yw

partial pressure of water is yw*P = .1557*900mmHg = 140.155 mmHg

Using antoine eqn and a= 8.10765 b= 1750.286 c=235 the temperature of water at this partial pressure is 58.66°C.

I don't think I can start using the energy balance w/oknowing the outlet temperature of the air H2O mix. Please help

## Answers and Replies

Chestermiller
Mentor
The outlet temperature is not 58.66 C because the water vapor in the outlet stream is superheated. This problem was not solved correctly. An overall enthalpy balance should have been performed on the device. The only unknown would have been the final temperature of the gas mixture.