A production facility heats various equipment using a network of heated air and

Question

A production facility heats various equipment using a network of heated air and water vapor lines, Aboile Map produces a 200.0 mol/hr stream consisting of 0.510 mol fraction of water and 0.490 mol fraction of air at steam traps 5370.0 mmHg. The air and water vapor cool as the gases move away from the boller, and total of 62.2 collect and remove condensed water from the lines while the remains constant. If and temperature molhr of condensed water is pressure is the composition removed from the air and vapor what pressure of of the air and vapor mixture leaving the steam lines? Use the Antoine equation to find the vapor water at these temperatures. To solve for the requested values, first label the process flow diagram below Label any quantities as "unknown" if they are not given or implied above. Do not leave any blank spaces. Variable are given for reference "BDA tands for bone-dry air, meaning the non-water vapor component of the gas mixture mol/hr mol H2O(vymol mol BDA/mol molUhr Heated Ar and water vapor Lines mmHg mol H20 (v)/rhol mol BDAmol mol H20(lyhr mmHg 34 o 15370 62.2 unknown 2000 490 510

Explanation / Answer

1)

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

0.510=[moles of water vapour]÷200

Moles of water vapour = 102 moles

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

0.490=[air moles]÷200

Air moles=98 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

102 - 62.2 =39.8 mols

Therefore composition of air and water vapour leaving = 39.8+98 = 137.8 Mol's

Mole fraction of water vapor = 39.8 mols /(137.8 Mol's) = 0.29

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 5370 mmHg

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