A packed tower is to be used to reduce the ammonia concentration in the gas stream from 4 to 0.3% by volume. A water stream is fed to the top of the tower at a rate of 231 g/s and the gas is fed countercurrently at a rate of 0.2 m³/s. The tower, which is packed with 2.54-cm (1-inch) Raschig rings,

Question

A packed tower is to be used to reduce the ammonia concentration in the gas stream from 4 to $0.3 \%$ by volume. A water stream is fed to the top of the tower at a rate of $231 \mathrm{~g} / \mathrm{s}$ and the gas is fed countercurrently at a rate of $0.2 \mathrm{~m}^{3} / \mathrm{s}$ . The tower, which is packed with $2.54-\mathrm{cm}$ (1-inch) Raschig rings, is operated isothermally at $293 \mathrm{~K}$ and $1.013 \times 10^{5} \mathrm{~Pa}$ .

Determine (a) the composition of the exiting liquid stream; (b) the diameter of the absorption tower if the gas pressure drop is limited to $200 \mathrm{~N} / \mathrm{m}^{2}$ per meter of packing.

Determine (a) the composition of the exiting liquid stream; (b) the diameter of the absorption tower if the gas pressure drop is limited to [latex]200 \mathrm{~N} / \mathrm{m}^{2}[/latex] per meter of packing.

Accepted Answer

The concentration of the three known streams may be expressed on a [latex]\mathrm{NH}_{3}[/latex]-free basis

[latex] \begin{aligned} Y_{\mathrm{NH}_{3}, 1} & =\frac{y_{\mathrm{NH}_{3}, 1}}{1-y_{\mathrm{NH}_{3}, 1}}=\frac{0.04}{0.96}=0.0417 \ Y_{\mathrm{NH}_{3}, 2} & =\frac{y_{\mathrm{NH}_{3}, 2}}{1-y_{\mathrm{NH}_{3}, 2}}=\frac{0.003}{0.997}=0.003 \ X_{\mathrm{NH}_{3}, 2} & =0.0 \end{aligned} [/latex]

The gas enters the tower at plane 1 , and its molar flow rate is

[latex] \begin{aligned} & G_{1}=\frac{\dot{V} P}{R T} \cdot \frac{1}{\mathrm{~A}}=\frac{\left(0.20 \frac{\mathrm{m}^{3}}{\mathrm{~s}} ight)\left(1.013 imes 10^{5} \mathrm{~Pa} ight)}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}} ight)(293 \mathrm{~K})} \cdot \frac{1}{\mathrm{~A}}=8.317 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}} \ & G_{S}=G_{1}\left(1-y_{\mathrm{NH}_{3}, 1} ight)=8.317 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}(0.96)=7.98 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}} \end{aligned} [/latex]

The [latex]\mathrm{NH}_{3}[/latex]-free water flow rate is

[latex]L_{S}=L_{2}\left(1-x_{\mathrm{NH}_{3}, 2} ight)=231 \frac{\mathrm{g}}{\mathrm{s}} \cdot \frac{\mathrm{mol}}{18 \mathrm{~g}} \cdot \frac{1}{\mathrm{~A}}=12.83 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}[/latex]

By an overall balance for ammonia, the concentration of the exiting liquid stream can be established.

[latex] \begin{aligned} L_{S}\left(X_{\mathrm{NH}_{3}, 1}-X_{\mathrm{NH}_{3}, 2} ight) & =G_{S}\left(Y_{\mathrm{NH}_{3}, 1}-Y_{\mathrm{NH}_{3}, 2} ight) \ 12.83 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}\left(X_{\mathrm{NH}_{3}, 1}-0 ight) & =7.98 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}(0.0417-0.003) \ X_{\mathrm{NH}_{3}, 1} & =0.024 \end{aligned} [/latex]

or

[latex]x_{\mathrm{NH}_{3}, 1}=\frac{X_{\mathrm{NH}_{3}, 1}}{1+X_{\mathrm{NH}_{3}, 1}}=\frac{0.024}{1.024}=0.0234[/latex]

The liquid flow rate at end 1 is

[latex]L_{1}=\frac{L_{S}}{1-x_{\mathrm{NH}_{3}, 1}}=\frac{12.83 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}}{1-0.0234}=13.13 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}[/latex]

The maximum mass flow rates for both phases will occur at end 1. Accordingly, we will use these flow rates to determine the diameter of the tower.

On a unit-area basis, the liquid stream at end 1 will contain [latex]12.83 \mathrm{~mol} / \mathrm{s}[/latex] and [latex]\mathrm{H}_{2} \mathrm{O} 13.13-[/latex] [latex]12.83=0.3 \mathrm{~mol} / \mathrm{s} \mathrm{NH}_{3}[/latex]. The total mass flow rate is

[latex]12.83 \frac{\mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}} \cdot\left(\frac{18 \mathrm{~g}}{\mathrm{~mol}} ight)+0.3 \frac{\mathrm{mol} \mathrm{NH}_{3}}{\mathrm{~s}} \cdot \frac{1}{\mathrm{~A}} \cdot\left(\frac{17 \mathrm{~g}}{\mathrm{~mol}} ight)=236.1 \frac{\mathrm{g}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}[/latex]

The average molecular weight of the gas is

[latex]\left(0.04 \frac{\mathrm{mol} \mathrm{NH}_{3}}{\mathrm{~mol}} ight)\left(\frac{17 \mathrm{~g}}{\mathrm{~mol}} ight)+\left(0.96 \frac{\mathrm{mol} \mathrm{air}}{\mathrm{mol}} ight)\left(\frac{29 \mathrm{~g}}{\mathrm{~mol}} ight)=28.5 \frac{\mathrm{g}}{\mathrm{mol}}[/latex]

The total gas-mass flow rate is

[latex]\left(8.317 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}} ight)\left(28.5 \frac{\mathrm{g}}{\mathrm{mol}} ight)=237 \frac{\mathrm{g}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}[/latex]

The ratio of the two mass flow rates is

[latex]\frac{L^{\prime}}{G^{\prime}}=\frac{\left(236.1 \frac{\mathrm{g}}{\mathrm{s}} ight)\left(\frac{1}{\mathrm{~A}} ight)}{\left(237 \frac{\mathrm{g}}{\mathrm{s}} ight)\left(\frac{1}{\mathrm{~A}} ight)}=0.996[/latex]

Note that this ratio can be evaluated without knowing either the diameter or the cross-sectional area. The density of the gas stream entering the tower is

[latex] \begin{aligned} ho_{G}=\frac{n}{V} M_{\mathrm{W}}=\frac{P}{R T} M_{\mathrm{W}} & =\frac{1.013 imes 10^{5} \mathrm{~Pa}}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}} ight)(293 \mathrm{~K})} \cdot 28.5 \frac{\mathrm{g}}{\mathrm{mol}} \ & =1185 \mathrm{~g} / \mathrm{m}^{3}=1.185 \mathrm{~kg} / \mathrm{m}^{3} \end{aligned} [/latex]

The density of the dilute aqueous stream will be essentially that of water at [latex]293 \mathrm{~K}[/latex] that is [latex]998.2 \mathrm{~kg} / \mathrm{m}^{3}[/latex]. The abscissa for Figure 31.25 becomes

[latex]\frac{L^{\prime}}{G^{\prime}}\left(\frac{ ho_{G}}{ ho_{L}- ho_{G}} ight)^{1 / 2}=0.996\left(\frac{1.185}{998.2-1.185} ight)^{1 / 2}=0.034[/latex]

At a pressure drop of [latex]200 \mathrm{~N} / \mathrm{m}^{2}[/latex] per meter of packing, this abscissa value indicates an ordinate value of 0.049 ; consequently

[latex]0.049=\frac{\left(G^{\prime} ight)^{2} c_{f}\left(\mu_{L} ight)^{0.1} J}{ ho_{G}\left( ho_{L}- ho_{G} ight) g_{c}}[/latex]

Upon rearrangement

[latex]\left(G^{\prime} ight)^{2}=\frac{0.049 ho_{G}\left( ho_{L}- ho_{G} ight) g_{c}}{c_{f}\left(\mu_{L} ight)^{0.1} J}[/latex]

From Table 31.2, we evaluate the [latex]c_{f}[/latex] for 1 -in. Raschig rings to be 155 . The viscosity of the aqueous stream at [latex]293 \mathrm{~K}[/latex] is given in Appendix Table I to be [latex]993 imes 10^{-6} \mathrm{~Pa} \cdot \mathrm{s}[/latex].

[latex] \begin{aligned} \left(G^{\prime} ight)^{2} & =\frac{0.049\left(1.185 \frac{\mathrm{kg}}{\mathrm{m}^{3}} ight)\left(998.2-1.185 \frac{\mathrm{kg}}{\mathrm{m}^{3}} ight)(1.0)}{(155)\left(993 imes 10^{-6} \mathrm{~Pa} \cdot \mathrm{s} ight)(1.0)}=0.745 \ G^{\prime} & =0.863 \frac{\mathrm{kg}}{\mathrm{m}^{2} \cdot \mathrm{s}}=863 \frac{\mathrm{g}}{\mathrm{m}^{2} \cdot \mathrm{s}} \end{aligned} [/latex]

As the gas feed rate, [latex]G^{\prime}[/latex], to the tower is equal to [latex]237 \frac{\mathrm{g}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}[/latex] and [latex]863 \frac{\mathrm{g}}{\mathrm{m}^{2} \cdot \mathrm{s}}[/latex], the cross-sectional area of the tower is

[latex]A=\frac{237 \frac{\mathrm{g}}{\mathrm{s}}}{863 \frac{\mathrm{g}}{\mathrm{m}^2 \cdot \mathrm{s}}}=0.275 \mathrm{~m}^2[/latex]

The area is [latex]\frac{\pi D^2}{4}[/latex]; accordingly, the diameter is

[latex]D=\left[\frac{\left(0.275 \mathrm{~m}^2 ight)(4)}{\pi} ight]^{1 / 2}=0.59 \mathrm{~m}[/latex]

Question 31.8: A packed tower is to be used to reduce the ammonia concentra...

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