- #1

- 1

- 0

## Main Question or Discussion Point

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

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