Ammonia is to be absorbed from an air mixture at 293 K and 1.013 × 10^5 Pa pressure in a countercurrent packed tower, using water at 293 K as the absorbent. An inlet gas rate of 1.21 × 10^-2m^3/s and an ammonia-free water rate of 9.46 × 10^-3 kg/s will be used. If the ammonia, NH3, concentration is

Question

Ammonia is to be absorbed from an air mixture at 293 K and [latex]1.013 imes 10^5 \mathrm{~Pa}[/latex] pressure in a countercurrent packed tower, using water at 293 K as the absorbent. An inlet gas rate of [latex]1.21 imes 10^{-2} \mathrm{~m}^3 / \mathrm{s}[/latex] and an ammonia-free water rate of [latex]9.46 imes 10^{-3} \mathrm{~kg} / \mathrm{s}[/latex] will be used. If the ammonia, [latex]\mathrm{NH}_3[/latex], concentration is reduced from 3.52 to 1.29 % by volume, determine the ratio of [latex]\left(L_S / G_S ight)_{ ext {actual }} /\left(L_S / G_S ight)_{\min }[/latex]. Equilibrium data for the system at 293 K and [latex].013 imes 10^5 . \mathrm{Pa}[/latex] are as follows:

[latex]\mathrm{X} \frac{\mathrm{kg} \mathrm{mol} \mathrm{NH}_3}{\mathrm{~kg} \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}[/latex]	0.0164	0.0252	0.0349	0.0455	0.0722
[latex]\mathrm{Y} \frac{\mathrm{kg} \mathrm{mol} \mathrm{NH}_3}{\mathrm{~kg} \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}[/latex]	0.021	0.032	0.042	0.053	0.08

Accepted Answer

The total moles of gas entering the tower per hour may be evaluated using the ideal gas law.

[latex]G_1 A= ext { moles entering gas } / \mathrm{s}=\frac{P \dot{V}}{R T}=\frac{\left(1.013 imes 10^5 \mathrm{~Pa} ight)\left(1.21 imes 10^2 \mathrm{~m}^2 / \mathrm{s} ight)}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^3}{\mathrm{~mol} \cdot \mathrm{K}} ight)(293 \mathrm{~K})}[/latex]

[latex]=0.503 \mathrm{~mol} / \mathrm{s}[/latex]

The gas enters the tower with a mole fraction of ammonia, [latex]\mathrm{Y}_{A_1}[/latex], equal to 0.0352. Choosing a cross-sectional area for the tower of A m², we may evaluate the moles of G on a solute-free basis as

[latex]G_S=G_1\left(1-y_{A 1} ight)=\left(0.503 \frac{\mathrm{mol}}{\mathrm{s}} ight)\left(\frac{0.9648}{A \mathrm{~m}^2} ight)=\frac{0.483}{A} \frac{\mathrm{mol}}{\mathrm{s} \cdot \mathrm{m}^2}[/latex]

The moles of phase L on a solute-free basis are

[latex]L_S=\left(9.46 imes 10^{-3} \frac{\mathrm{kg}}{\mathrm{s}} ight)\left(\frac{\mathrm{kg} \mathrm{mol}}{18 \mathrm{~kg}} ight)\left(\frac{1}{A \mathrm{~m}^2} ight)=\frac{5.26 imes 10^{-4}}{A} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{s} \cdot \mathrm{m}^2}=\frac{0.526}{A} \frac{\mathrm{mol}}{\mathrm{s} \cdot \mathrm{m}^2}[/latex]

The ratio of the actual [latex]L_S[/latex] to [latex]G_S[/latex] is evaluated as

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {actual }}=\frac{0.526}{A} \frac{A}{0.483}=1.09 \frac{ ext { mole } \mathrm{NH}_{3 ext {-free }} L ext { phase }}{ ext { mole } \mathrm{NH}_{3 ext {-free }} G ext { phase }}[/latex]

The composition of the known streams, [latex]G_1, G_2,[/latex] and [latex]L_2[/latex] on a solute-free basis, are evaluated from the known mole fraction as

[latex]Y_{\mathrm{NH}_3, 1}=\frac{y_{\mathrm{NH}_3, 1}}{1-y_{\mathrm{NH}_3, 1}}=\frac{0.0352}{0.9648}=0.0365[/latex]

[latex]Y_{\mathrm{NH}_3, 2}=\frac{y_{\mathrm{NH}_3, 2}}{1-y_{\mathrm{NH}_3, 2}}=\frac{0.0129}{0.9871}=0.0131[/latex]

[latex]X_{\mathrm{NH}_3, 2}=\frac{x_{\mathrm{NH}_3, 2}}{1-x_{\mathrm{NH}_3, 2}}=0[/latex]

The exiting composition, [latex]X_{\mathrm{NH}_3, 1},[/latex] can be evaluated by

[latex]G_S\left(Y_{A_1}-Y_{A_2} ight)=L_S\left(X_{A_1}-X_{A_2} ight)[/latex]

[latex]\frac{0.483}{A}(0.0365-0.0131)=\frac{0.526}{A}\left(X_{A_1}-0 ight)[/latex]

[latex]X_{A_1}=0.0215[/latex]

The actual and minimum operating lines are shown in Figure 31.12. The composition of the solution in equilibrium with [latex]Y_{\mathrm{NH}_3, 1}=0.0365[/latex] is obtained from the equilibrium curve as [latex]X_{\mathrm{NH}_{3, ext { equll }}}=0.0296[/latex]. The slope of the minimum operating line is

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {minimum }}=\frac{\Delta Y}{\Delta X}=\frac{0.0365-0.0131}{0.0296-0}=0.79 \frac{ ext { mole } \mathrm{NH}_{3 ext {-free }} ext { liquid phase }}{ ext { mole } \mathrm{NH}_{3 ext {-free }} ext { gas phase }}[/latex]

The desired ratio, [latex]\left(L_S / G_S ight)_{ ext {actual }} /\left(L_S / G_S ight)_{ ext {minimum }}[/latex] is then a ratio of the two values, 1.09/0.79 or 1.38.

Question 31.3: Ammonia is to be absorbed from an air mixture at 293 K and 1...

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