Question 31.3: Ammonia (NH3) will be absorbed from an air mixture at 293 K ......

The analysis begins with a process material balance. First, the carrier gas flow molar rate is determined by

$A G_{S}=A G_{1}\left(1-y_{A_{1}}\right)=5.03\times10^{-4}\,\mathrm{kgmole}/\mathrm{s}(1-0.0352)=4.85\times10^{-4}\,\mathrm{kgmole}/\mathrm{s}$

where A refers to the cross-sectional area of the tower, and the subscript A refers to the transferring solute. The total molar flow rate of the solvent entering the top of the tower is

$A L_{\mathrm{{S}}}=A L_{\mathrm{{2}}}(1-x_{\mathrm{{A_{2}}}})=A L_{\mathrm{{2}}}(1-0)=A L_{\mathrm{{2}}}=\left(9.46\times10^{-3}{\frac{\mathrm{{Kg}}}{s}}\right)\left({\frac{1\,{\mathrm{kgmole}}}{18\,{\mathrm{kg}}}}\right)=5.26\times10^{-4}\,{\mathrm{kgmole/s}}$

The mole ratio compositions of the streams of known mole fraction are

$Y_{A_{1}}={\frac{y_{A_{1}}}{1-y_{A_{1}}}}={\frac{0.0352}{1-0.0352}}=0.0365$

$Y_{A_{2}}={\frac{y_{A_{2}}}{1-y_{A_{2}}}}={\frac{0.0129}{1-0.0129}}=0.0131$

$X_{A_{2}}={\frac{x_{A_{2}}}{1-x_{A_{2}}}}={\frac{0}{1-0}}=0$

The mole ratio for solute A in the liquid exiting the bottom of the tower, $X_{A_{1}}$, is determined by a material balance on solute A around the terminal streams, given by

$A G_{S}Y_{A_{1}}+A L_{S}X_{A_{2}}=A G_{S}Y_{A_{2}}+A L_{S}X_{A_{1}}$

or

$X_{A_{1}}={\frac{A G_{S}(Y_{A_{1}}-Y_{A_{2}})+A L_{S}X_{A_{2}}}{A L_{S}}}={\frac{4.85\times10^{-4}(0.0365-0.0131)+0}{5.26\times10^{-4}}}=0.0216$

With the terminal stream compositions now known, the equilibrium and operating lines in mole ratio coordinates are shown in Figure 31.12. Given that the shape equilibrium line is nearly linear in the composition range of interest, the minimum solvent flow rate will occur when the composition of the solute in the liquid exiting the tower is in equilibrium with the composition of the solute A in the gas entering the tower. Consequently, $X_{A_{1,}{\mathrm{min}}} = 0.0453\ {\mathrm{at}}\,Y_{A_{1}}=0.0365.{\mathrm{With}}\,X_{A_{1,}{\mathrm{min}}}$ known, the minimum solvent flow rate is estimated by material balance

$A G_{S}Y_{A_{1}}+A L_{S,\mathrm{min}}X_{A_{2}}=A G_{S}Y_{A_{2}}+A L_{S,\mathrm{min}}X_{A,\mathrm{min}}$

or

$A L_{ S,\mathrm{min}}=\frac{A G_{S}(Y_{A_{1}}-Y_{A_{2}})}{X_{A_{1},\mathrm{min}}-X_{A_{2}}}=\frac{4.85\times10^{-4}(0.0365-0.0131)}{(0.0453-0)}=2.51\times10^{-4}\,\mathrm{kgmole}/s$

Finally, the ratio of the operating solvent flow rate to the minimum solvent flow rate is

$\frac{A L_{S}}{A L_{S,\mathrm{min}}}=\frac{5.26\times10^{-4}\,\mathrm{kgmole}/s}{2.51\times10^{-4}\,\mathrm{kgmole}/s}=2.09$

Thus, the operating solvent flow rate is 109% above the minimum solvent flow rate.