Question 13.4: 30,000 m³ of cool gas (measured at 289 K and 101.3 kN/m² sat...

At 289 K and 101.3 kN/m², the gas is saturated and P_{w0} = 1.8 kN/m².

Thus from equation 13.2,

\mathscr{H} _{o}=\frac{P_{w0}}{P-P_{w0}} \left(\frac{M_{w}}{M_{A}} \right)      (13.2)

\mathscr{H} _{0} = [1.8/(101.3 – 1.8)](18/M_{A}) = (0.3256/M_{A}) kg/kg dry gas, where M_{A} is the molecular mass of the gas.

At 289 K and 340 kN/m² , the gas is in contact with condensed water and therefore still saturated. Thus P_{w0} = 1.8 kN/m² and:

\mathscr{H} _{0} = [1.8/(340 – 1.8)](18/M_{A}) = (0.0958/M_{A}) kg/kg dry gas

At 289 K and 170 kN/m², the humidity is the same, and in equation 13.2:

(0.0958/M_{A}) = [P_{w}/(170 – P_{w})](18/M_{A})

or: P_{w} = 0.90 kN/m²

The percentage humidity is then:

= [( P -P_{w0})/(P – P_w)](100P_{w}/P_{w0})   (equation 13.3)

= [(170 – 1.8)/(170 – 0.90)](100 × 0.90/1.8) = 49.73%