Question

A production facility heats various equipment using a network of heated air and water vapor lines. A boiler produces a 195.0 mol/hr stream consisting of 0.590 mol fraction of water and 0.410 mol fraction of air at 131.0°C and 5200.0 mmHg. The air and water vapor cool as the gases move away from the boiler, and steam traps collect and remove condensed water from the lines while the pressure remains constant.

a) If a total of 65.6 mol/hr of condensed water is removed from the air and vapor lines, what is the composition of the air and vapor mixture leaving the steam lines?

b) Use the Antoine equation for water valid between 60°C and 150ºC to find the temperature of the air and vapor mixture leaving the heating lines.

Answer 1

1)we can find the no.of moles of water vapour using the mole fraction given

0.590=[moles of water vapour]÷195

Moles of water vapour=115.05 moles

In the same way we can find the no.of air moles

0.410=[air moles]÷195

Air moles=79.95 mols

Assuming that only water vapour is getting condensed,total composition of water vapour leaving can be found by deducting the moles of water condensed from total water vapour

115.05-65.6=49.45mols

Therefore composition of air and water vapour leaving =49.45+79.95=129.4 Mol's

2)Antoine equation

P=10^A-B/C+T

The equation is used using temp in °C and pressure in mmHg

Temp =165.12 °C at 5200 mmHg