Question 31.5: An existing packed tower with a packing height of 12.0 m was......

To use the design equations for a packed tower, it is first necessary to characterize the compositions and flow rates of all terminal streams. From the initial specifications of the process, the following flow rates and compositions are known: G_{1}=0.0136\,\mathrm{kgmole}/\mathrm{m^{2}}\cdot\mathrm{s},\;\;L_{2}=0.0272\,\mathrm{kgmole}/\mathrm{m^{2}}\cdot\mathrm{s},\;\;y_{A_{1}}=0.020,\;\;y_{A_{2}}=0.050,\ x_{A_{2}}=0. To characterize the remaining unknown streams, we first estimate the carrier gas and solvent flow rates, which are inert components:

G_{S}=G_{1}(1-y_{A_{1}})=(0.0136\,\mathrm{kgmole}/\mathrm{m^{2}}\,\cdot\,\mathrm{s})(1-0.020)=0.0133\,\mathrm{kgmole}/\mathrm{m^{2}}\cdot\mathrm{s}

L_{S}=L_{2}(1-x_{A_{2}})=\left(0.0272\,\mathrm{kgmole}/\mathrm{m}^{2}\cdot{ s}\right)(1-0.0)=0.0272\,\mathrm{kgmole}/\mathrm{m}^{2}\cdot{s}

Consequently,

G_{2}=G_{S}/(1-y_{A_{2}})=(0.0133\,\mathrm{kgmole}/\mathrm{m^{2}}\cdot{\mathrm{s}})/(1-0.0050)=0.0134\,\mathrm{kgmole}/\mathrm{m^{2}}\cdot{\mathrm{s}}

With G_{2} known, an overall material balance around the terminal streams is used to determine L_{1}:

G_{1}+L_{2}=G_{2}+L_{1}

L_{1}=G_{1}-G_{2}+L_{2}=0.0136-0.0134+0.0272=0.0274\,\mathrm{kgmole}/\mathrm{m}^{2}\cdot s

Finally, the outlet mole fraction of solute A in the liquid (x_{A_{1}}) is determined by

L_{S}{=}L_{1}(1-x_{A_{1}})

or

x_{A_{1}}=1-L_{S}/L_{1}=1-(0.0272)/(0.0274)=0.0073

Since the equilibrium line is linear, the pinch point at the minimum solvent flow rate will occur at the gas feed, so that

x_{A_{1},\mathrm{min}}=y_{A_{1}}/1.75=0.020/1.75=0.0114

Therefore, since x_{A_{1},\mathrm{min}}\gt x_{A_{1}}, we are assured that operating solvent flow rate is above the minimum solvent flow rate.

All mole fraction compositions are at or below 2.0 mole%, which indicates a dilute system. Furthermore, the equilibrium line is linear. Based on these two considerations, the appropriate design equation for the countercurrent gas absorption tower is equation (31-34). In order to estimate the log-mean mole fraction driving force \left(y_{A}-y_{A}^{*}\right)_{l m}, values for y_{A}^{*} at the top and bottom ends of the tower are needed. Therefore,

z=\frac{G}{K_{G}a\cdot P}\frac{\left(y_{A_{1}}-y_{A_{2}}\right)}{\left(y_{A}-y_{A}^{*}\right)_{l m}}          (31-34)

y_{A_{1}}^{*}=1.75x_{A_{1}}=1.75(0.0073)=0.0130

y_{A_{2}}^{*}=1.75x_{A_{2}}=1.75(0.0)=0.0

Consequently, from equation (31-35), the overall log-mean mole fraction driving force for mass-transfer of solute A based on the gas phase is

\left(y_{A}-y_{A}^{*}\right)_{l m}={\frac{\left(y_{A}-y_{A}^{*}\right)_{1}-\left(y_{A}-y_{A}^{*}\right)_{2}}{\ln[(y_{A}-y_{A}^{*})_{1}/(y_{A}-y_{A}^{*})_{2}]}}={\frac{\left(0.020-0.0130\right)-\left(0.020-0\right)}{\ln[(0.020-0.0130)/(0.020-0.0)}}=0.0123

Since the gas flow rate does not change significantly from the bottom to the top of the tower, an average value is used

G={\frac{G_{1}+G_{2}}{2}}={\frac{0.0136+0.0134}{2}}=0.0135\,\mathrm{kgmole}/\mathrm{m^{2}}\cdot\mathrm{s}

Finally, K_{G}a is backed out from by a rearranged form of equation (31-34):

K_{G}a=\frac{G}{z\cdot P}\frac{\left(y_{A_{1}}-y_{A_{2}}\right)}{\left(y_{A}-y_{A}^{*}\right)_{l m}}=\frac{\left(0.0135\mathrm{~kgmole}/m^{2}\cdot s\right)}{\left(12.0\ \mathrm{m}) (1.2\ \mathrm{atm} \right)}\frac{\left(0.020-0.0050\right)}{\left(0.0123\right)}=1.14\times10^{-3}\frac{\mathrm{kgmole}}{{\mathrm{~m^{3}\cdot s\cdot a t m}}}

As an aside, for dilute systems, the packed-tower height is commonly described as the product

z=H_{O G}N_{O G}            (31-39)

where N_{O G}   is the number of transfer units, given by

N_{O G}={\frac{\left(y_{A_{1}}-y_{A_{2}}\right)}{\left(y_{A}-y_{A}^{*}\right)_{l m}}}={\frac{\left(0.020-0.0050\right)}{\left(0.0123\right)}}=1.22          (31-40)

and H_{O G} is the packing height of the transfer unit, given by

H _{O G}={\frac{G}{K_{G}a\cdot P}}={\frac{(0.0135\ \mathrm{kgmole}/{ m}^{2}{ s})}{(1.14\times10^{-3}\ \mathrm{kgmole}/{ m}^{3}\cdot{ s}\cdot\mathrm{atm})(1.2\cdot\mathrm{atm})}}=9.86\,{ m}        (31-41)