30,000 m³ of cool gas (measured at 289 K and 101.3 kN/m² saturated with water vapour) is compressed to 340 kN/m² pressure, cooled to 289 K and the condensed water is drained off. Subsequently the pressure is reduced to 170 kN/m³ and the gas is distributed at this pressure and 289 K. What is the

Question

30,000 m³ of cool gas (measured at 289 K and 101.3 kN/m² saturated with water vapour) is compressed to 340 kN/m² pressure, cooled to 289 K and the condensed water is drained off. Subsequently the pressure is reduced to 170 kN/m³ and the gas is distributed at this pressure and 289 K. What is the percentage humidity after this treatment? The vapour pressure of water at 289 K is 1.8 kN/m².

Accepted Answer

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

Thus from equation 13.2,

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

[latex]\mathscr{H} _{0}[/latex] = [1.8/(101.3 - 1.8)](18/[latex]M_{A}[/latex]) = (0.3256/[latex]M_{A}[/latex]) kg/kg dry gas, where [latex]M_{A}[/latex] 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 [latex]P_{w0}[/latex] = 1.8 kN/m² and:

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

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

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

or: [latex]P_{w}[/latex] = 0.90 kN/m²

The percentage humidity is then:

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

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

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

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