Recently enacted legislation requires that the effectiveness of the absorption system, currently employed to reduce the level of a pollutant in a process off-gas, must be improved to meet new state requirements. Under current operating conditions, 13.6 mol/s· m² of off-gas flows countercurrent to

Question

Recently enacted legislation requires that the effectiveness of the absorption system, currently employed to reduce the level of a pollutant in a process off-gas, must be improved to meet new state requirements. Under current operating conditions, [latex]13.6 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2}[/latex] of off-gas flows countercurrent to [latex]27.2 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2}[/latex] of water in the [latex]12-\mathrm{m}[/latex] tall absorber. The concentration of [latex]A[/latex] is reduced from 2 to [latex]0.5 \mathrm{~mol} \%[/latex].

It has been suggested that the required concentration level in the off-gas may be obtained by doubling the water rate in the existing tower. Its diameter is large enough to permit this higher flow rate. If it is assumed that the process is gas-film controlled and that the overall gas capacity coefficient, [latex]K_{G} a[/latex], is proportional to the mass velocity of the solvent raised to the 0.4 power, determine the concentration of [latex]A[/latex] in the effluent gas when the same [latex]12-\mathrm{m}[/latex] tower and the same gas mass velocity are used with the doubled liquid flow rate.

Equilibrium for the system is defined by a modified Henry's law; [latex]y_{A}^{*}=1.5_{x_{A}}[/latex].

Accepted Answer

As the concentrations are in the dilute range where the equilibrium curve is a straight line, the height of the tower may be evaluated by equation (31-34)

[latex]Z=\frac{G\left(y_{A_{1}}-y_{A_{2}} ight)}{K_{G} a P\left(y_{A}-y_{A}^{*} ight)_{l m}}[/latex]

An overall material balance on the existing system (system I) establishes the composition of its exiting liquid stream.

[latex] \begin{aligned} G\left(y_{A_{1}}-y_{A_{2}} ight) & =L\left(x_{A_{1}}-x_{A_{2}} ight) \ \left(13.6 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)(0.02-0.005) & =\left(27.2 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)\left(x_{A_{1}}-0 ight) \ x_{A_{1}} & =0.0075 \end{aligned} [/latex]

The compositions at each end of the tower for the existing system are

[latex] \begin{aligned} ext { Bottom: } y_{A_{1}} & =0.02 \ x_{A_{1}} & =0.0075 \ y_{A_{1}}^{*} & =1.5 x_{A_{1}}=1.5(0.0075)=0.0113 \ ext { Top: } y_{A_{2}} & =0.005 \ x_{A_{2}} & =0.0 \ y_{A_{2}}^{*} & =1.5 x_{A_{2}}=0.0 \end{aligned} [/latex]

For the existing tower, [latex]\left(y_{A}-y_{A}^{*} ight)_{l m}[/latex] is evaluated by equation (31-35)

[latex] \begin{aligned} & \left(y_{A}-y_{A}^{*} ight)_{l m}=\frac{\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{1}}-\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{2}}}{\ln \left[\frac{\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{1}}}{\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{2}}} ight]} \ & =\frac{(0.02-0.0113)-(0.005-0)}{\ln \frac{(0.02-0.0113)}{(0.005-0.0)}}=0.0067 \end{aligned} [/latex]

Upon substituting the known values into equation (31-34), we obtain

[latex] \begin{aligned} 12 \mathrm{~m} & =\frac{\left(13.6 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)(0.02-0.005)}{K_{G} a_{\mathrm{I}} P(0.0067)} \ 12 \mathrm{~m} & =\frac{\left(13.6 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)(2.24)}{K_{G} a_{\mathrm{I}} P} \end{aligned} [/latex] (31-38)

When we consider the proposed system (system II), we obtain the following relationship with equation (31-34)

[latex]12=\frac{\left(13.6 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)\left(0.02-y_{A_{2}} ight)}{K_{G} a_{\mathrm{II}} P\left(y_{A}-y_{A}^{*} ight)_{l m}}[/latex] (31-39)

Upon equating and simplifying equations (31-38) and (31-39), we obtain

[latex]\frac{2.24}{K_{G} a_{\mathrm{I}}}=\left.\frac{\left(0.02-y_{A_{2}} ight)}{K_{G} a_{\mathrm{II}}\left(y_{A}-y_{A}^{*} ight)_{l m}} ight|_{\mathrm{II}}[/latex]

or

[latex]\frac{K_{G} a_{\mathrm{II}}(2.24)}{K_{G} a_{\mathrm{I}}}=\left.\frac{\left(0.02-y_{A_{2}} ight)}{\left(y_{A}-y_{A}^{*} ight)_{l m}} ight|_{\mathrm{II}}[/latex] (31-40)

As the capacity coefficient is proportional to the mass velocity of the solvent raised to the 0.4 power, the ratio of the capacity coefficients is

[latex]\frac{K_{G} a_{\mathrm{II}}}{K_{G} a_{\mathrm{I}}}=\left[\frac{L_{\mathrm{II}}}{L_{\mathrm{I}}} ight]^{0.4}=\left[\frac{\left(54.4 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)(0.018 \mathrm{~kg} / \mathrm{mol})}{\left(27.2 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)(0.018 \mathrm{~kg} / \mathrm{mol})} ight]^{0.4}=1.32[/latex]

Accordingly, equation (31-40) becomes

[latex]\left.\frac{0.02-y_{A_{2}}}{\left(y_{A}-y_{A}^{*} ight)_{l m}} ight|_{\mathrm{II}}=(1.32)(2.24)=2.95[/latex] (31-41)

This equation requires a trial-and-error solution. With a guess of [latex]y_{A_{2}}=0.0021[/latex], we can make an overall balance to establish [latex]x_{A_{1}}[/latex] for the proposed system.

[latex] \begin{aligned} G\left(y_{A_{1}}-y_{A_{2}} ight) & =L\left(x_{A_{1}}-x_{A_{2}} ight) \ \left(13.6 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)(0.02-0.0021) & =\left(54.4 \mathrm{~mol} / \mathrm{s} \cdot \mathrm{m}^{2} ight)\left(x_{A_{1}}-0.0 ight) \ x_{A_{1}} & =0.0045 \end{aligned} [/latex]

The compositions at each end of the proposed tower (system II) are

[latex] \begin{aligned} ext { Bottom: } y_{A_{1}} & =0.02 \ x_{A_{1}} & =0.0045 \ y_{A_{1}}^{*} & =1.5 x_{A_{1}}=1.5(0.0045)=0.0067 \end{aligned} [/latex]

Top: [latex]y_{A_{2}}=0.0021[/latex] (estimated value)

[latex] \begin{aligned} & x_{A_{2}}=0.0 \ & y_{A_{2}}^{*}=1.5_{x_{A_{2}}}=1.5(0.0)=0.0 \end{aligned} [/latex]

The [latex]\left(y_{A}-y_{A}^{*} ight)_{l m}[/latex] for the proposed tower is evaluated by equation (31-35)

[latex] \begin{aligned} & \left(y_{A}-y_{A}^{*} ight)_{l m}=\frac{\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{1}}-\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{2}}}{\ln \left[\frac{\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{1}}}{\left(y_{A}-y_{A}^{*} ight)_{\mathrm{end}_{2}}} ight]} \ & \left(y_{A}-y_{A}^{*} ight)_{l m}=\frac{(0.02-0.0067)-(0.0021-0.0)}{\ln \left[\frac{(0.02-0.0067)}{(0.0021-0.0)} ight]}=0.00607 \ & \frac{0.02-y_{A_{2}}}{\left(y_{A}-y_{A}^{*} ight)_{l m}}=\frac{0.02-0.0021}{0.00607}=2.95 \end{aligned} [/latex]

This satisfies equation (31-41); as a result of the trial-and-error solution, the concentration of [latex]A[/latex] in the effluent gas is [latex]0.21 \%[/latex].

Question 31.6: Recently enacted legislation requires that the effectiveness...

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