Humid air at 75°C, 1.1 bar, and 30% relative humidity is fed into a process unit at a rate of 1000 m³ /h.
Determine (1) the molar flow rates of water, dry air, and oxygen entering the process unit, (2) the molal humidity, absolute humidity, and percentage humidity of the air, and (3) the dew point.

Question

Accepted Answer

1. [latex]h_{r}(\%) = 100p_{H_{2}O} / p^{*}_{H_{2}O} (75°C)[/latex]

[latex]\left. \Large{\Downarrow} ight. \begin{matrix} h_{r}= 30\% \ p^{*}_{H_{2}O} (75°C)= 289 mm Hg (from Table B.3) \end{matrix} \ p_{H_{2}O} = (0.3)(289 mm Hg)= 86.7 mm Hg \ \left. \Large{\Downarrow} ight. \begin{matrix} y_{H_{2}O} =p_{H_{2}O}/P \ P= 1.1 bar \Longrightarrow 825 mm Hg \end{matrix} \ y_{H_{2}O} = (86.7 mm Hg)/(825 mm Hg) = 0.105 mol H_{2}O/mol[/latex]

The molar flow rate of wet air is given by the ideal-gas equation of state as

[latex]\dot{n} = P\dot{V} / RT = \begin{array}{c|c|c}1000 m^{3}& 1.1 bar& kmol\cdot K \ \hline h &348 K& 0.0831 m^{3} \cdot bar\end{array} = 38.0 \frac{kmol}{h} [/latex]

Consequently,

[latex]\begin{matrix} \dot{n}_{H_{2}O} = \begin{array}{c|c}38.0 kmol& 0.105 kmol H_{2}O \ \hline h &kmol\end{array} = \boxed{3.99 \frac{kmol H_{2}O}{h}} \\ \dot{n}_{BDA} = \begin{array}{c|c}38.0 kmol& (1 - 0.105) kmol BDA \ \hline h &kmol\end{array} = \boxed{34.0 \frac{kmol BDA}{h}} \\ \dot{n}_{O_{2}} = \begin{array}{c|c}34.0 kmol BDA& 0.21 kmol O_{2} \ \hline h &kmol BDA\end{array} = \boxed{7.14 \frac{kmol O_{2}}{h}}\end{matrix} [/latex]

2. [latex]h_{m}= \frac{p_{H_{2}O}}{P - p_{H_{2}O}} = \frac{86.7 mm Hg}{(825 - 86.7) mm Hg} = \boxed{0.117 \frac{mol H_{2}O}{mol BDA} } [/latex]

The same result could have been obtained from the results of Part 1 as (3.99 kmol H[latex]_{2}[/latex]O/h)=(34.0 kmol BDA/h).

[latex]h_{a}= \begin{array}{c|c|c}0.117 kmol H_{2}O &18.0 kg H_{2}O& 1 kmol BDA \ \hline kmol BDA &kmol H_{2}O &29.0 kg BDA\end{array} = \boxed{0.0726 \frac{kg H_{2}O}{kg BDA}} \ h^{*}_{m} = \frac{p^{*}_{H_{2}O}}{P - p^{*}_{H_{2}O}} = \frac{289 mm Hg}{(825 - 289) mm Hg} = 0.539 \frac{kmol H_{2}O}{kmol BDA}\ h_{p}= 100h_{m}/h^{*}_{m} =(100)( 0.117) /(0.539) = \boxed{21.7\%}[/latex]

[latex]\begin{matrix} \pmb{3.} p_{H_{2}O} = 86.7 mm Hg = p_{H_{2}O}^{*} (T_{dp})\ \left. \Large{\Downarrow} ight. Table B.3 \ \boxed{T_{dp} =48.7°C }\end{matrix} [/latex]

Question 6.3.3: Humid air at 75°C, 1.1 bar, and 30% relative humidity is fed......

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