The ammonia–air feed stream described in example 3 is fed cocurrently with an ammonia–free water stream. The ammonia concentration is to be reduced from 3.52 to 1.29 % by volume, using a water stream 1.37 times the minimum. Determine (a) the minimum Ls/Gs ratio, (b) the actual water rate, and (c)

Question

The ammonia–air feed stream described in example 3 is fed cocurrently with an ammonia–free water stream. The ammonia concentration is to be reduced from 3.52 to 1.29 % by volume, using a water stream 1.37 times the minimum. Determine (a) the minimum [latex]L_s / G_s[/latex] ratio, (b) the actual water rate, and (c) the concentration in the exiting aqueous stream.

Accepted Answer

In example 3, the following compositions were evaluated:

[latex] ext { entering } \quad Y_{\mathrm{NH}_3, 1}=0.0365[/latex]

[latex] ext { exiting } \quad Y_{\mathrm{NH}_3, 2}=0.0131[/latex]

[latex] ext { entering } \quad X_{\mathrm{NH}_3, 1}=0.0[/latex]

The moles of G on a solute-free basis were evaluated to be [latex]\frac{0.483}{A} \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}[/latex]. In Figure 31.17, the minimum and actual operating lines are shown. For these operating lines

[latex]\left(\frac{L_s}{G_s} ight)_{\min }=\frac{Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, 2}}{X_{\mathrm{NH}_3, 2}-X_{\mathrm{NH}_3, 2}}=\frac{0.0365-0.0131}{0.01-0}[/latex]

[latex]=2.34 \frac{ ext { moles } \mathrm{NH}_3 ext {-free } L ext { phase }}{ ext { moles } \mathrm{NH}_3 ext {-free } G ext { phase }}[/latex]

and

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {actual }}=1.37\left(\frac{L_S}{G_S} ight)_{\min }=1.37(2.34)=3.21 \frac{ ext { moles } \mathrm{NH}_3 ext {-free } L ext { phase }}{ ext { moles } \mathrm{NH}_3 ext {-free } G ext { phase }}[/latex]

The composition of the exiting stream can be evaluated with the slope of the actual operating line by

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {actual }}=3.21=\frac{Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, 2}}{X_{\mathrm{NH}_3, 2}-X_{\mathrm{NH}_3, 1}}=\frac{0.0365-0.0131}{X_{\mathrm{NH}_3, 2}-0}[/latex]

or

[latex]X_{\mathrm{NH}_3, 2}=\frac{0.0234}{3.21}=0.0073=\frac{\mathrm{mol} \mathrm{NH}_3}{\mathrm{~mol} \mathrm{NH}_3 ext {-free water }}[/latex]

The moles of [latex]\mathrm{NH}_3[/latex]-free water fed to the tower, [latex]L_s[/latex], is also evaluated using the value of

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {actual }}=3.21 \frac{\mathrm{mol} \mathrm{NH}_3 ext { free } L ext { phase }}{\mathrm{mol} \mathrm{NH}_3 ext { free } G ext { phase }}[/latex]

Then

[latex]L_S=3.21 G_S=3.21\left(\frac{0.481}{A} ight)=\frac{1.55}{\mathrm{~A}} \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}[/latex]

Question 31.5: The ammonia–air feed stream described in example 3 is fed co...

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