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

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

\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}

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

\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}}\right)\left(1.013 \times 10^{5} \mathrm{~Pa}\right)}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}}\right)(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}\right)=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}

The \mathrm{NH}_{3}-free water flow rate is

L_{S}=L_{2}\left(1-x_{\mathrm{NH}_{3}, 2}\right)=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}}

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

\begin{aligned} L_{S}\left(X_{\mathrm{NH}_{3}, 1}-X_{\mathrm{NH}_{3}, 2}\right) & =G_{S}\left(Y_{\mathrm{NH}_{3}, 1}-Y_{\mathrm{NH}_{3}, 2}\right) \\ 12.83 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}\left(X_{\mathrm{NH}_{3}, 1}-0\right) & =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}

or

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

The liquid flow rate at end 1 is

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}}

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 12.83 \mathrm{~mol} / \mathrm{s} and \mathrm{H}_{2} \mathrm{O} 13.13- 12.83=0.3 \mathrm{~mol} / \mathrm{s} \mathrm{NH}_{3}. The total mass flow rate is

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}}\right)+0.3 \frac{\mathrm{mol} \mathrm{NH}_{3}}{\mathrm{~s}} \cdot \frac{1}{\mathrm{~A}} \cdot\left(\frac{17 \mathrm{~g}}{\mathrm{~mol}}\right)=236.1 \frac{\mathrm{g}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}

The average molecular weight of the gas is

\left(0.04 \frac{\mathrm{mol} \mathrm{NH}_{3}}{\mathrm{~mol}}\right)\left(\frac{17 \mathrm{~g}}{\mathrm{~mol}}\right)+\left(0.96 \frac{\mathrm{mol} \mathrm{air}}{\mathrm{mol}}\right)\left(\frac{29 \mathrm{~g}}{\mathrm{~mol}}\right)=28.5 \frac{\mathrm{g}}{\mathrm{mol}}

The total gas-mass flow rate is

\left(8.317 \frac{\mathrm{mol}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}\right)\left(28.5 \frac{\mathrm{g}}{\mathrm{mol}}\right)=237 \frac{\mathrm{g}}{\mathrm{s}} \cdot \frac{1}{\mathrm{~A}}

The ratio of the two mass flow rates is

\frac{L^{\prime}}{G^{\prime}}=\frac{\left(236.1 \frac{\mathrm{g}}{\mathrm{s}}\right)\left(\frac{1}{\mathrm{~A}}\right)}{\left(237 \frac{\mathrm{g}}{\mathrm{s}}\right)\left(\frac{1}{\mathrm{~A}}\right)}=0.996

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

\begin{aligned} \rho_{G}=\frac{n}{V} M_{\mathrm{W}}=\frac{P}{R T} M_{\mathrm{W}} & =\frac{1.013 \times 10^{5} \mathrm{~Pa}}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}}\right)(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}

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

\frac{L^{\prime}}{G^{\prime}}\left(\frac{\rho_{G}}{\rho_{L}-\rho_{G}}\right)^{1 / 2}=0.996\left(\frac{1.185}{998.2-1.185}\right)^{1 / 2}=0.034

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

0.049=\frac{\left(G^{\prime}\right)^{2} c_{f}\left(\mu_{L}\right)^{0.1} J}{\rho_{G}\left(\rho_{L}-\rho_{G}\right) g_{c}}

Upon rearrangement

\left(G^{\prime}\right)^{2}=\frac{0.049 \rho_{G}\left(\rho_{L}-\rho_{G}\right) g_{c}}{c_{f}\left(\mu_{L}\right)^{0.1} J}

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

\begin{aligned} \left(G^{\prime}\right)^{2} & =\frac{0.049\left(1.185 \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)\left(998.2-1.185 \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)(1.0)}{(155)\left(993 \times 10^{-6} \mathrm{~Pa} \cdot \mathrm{s}\right)(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}

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

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

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

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