Question 31.6: A packed tower is used to reduce the ammonia (NH3) concentra......

The Flooding correlation will be used to estimate the tower diameter. To use the Flooding correlation, the mass flow rates of the gas and liquid in countercurrent flow must be determined. For gas absorption, the highest mass flow rates of both gas and liquid are located at the bottom of the tower. Therefore $A G_{1}^{\prime}{\mathrm{~and~}}A L_{1}^{\prime}$ must be determined. The inlet molar flow rate of gas is determined from the volumetric flow rate using the ideal gas law:

$A G_{1}=\dot{V}_{1}\frac{{ P}}{R T}=\left(0.20\frac{{\mathrm m}^{3}}{s}\right) \frac{(1.0\ atm)}{\left(0.08206{\frac{\mathrm{m^{3}}\cdot{\mathrm{atm}}}{\mathrm{kgmole}\cdot\mathrm{K}}}\right)(293\,\mathrm{K})} =8.32\times10^{-3}\,\mathrm{kgmole}/\mathrm{s}$

The average molecular weight of the inlet gas is determined by

$M_{w,G_{1}}=y_{A_{1}}M_{A}+(1-y_{A_{1}})M_{B}=(0.040)(17)+(1-0.040)(29)=28.5\,\mathrm{{kg/kgmole}}$

where component B refers to the carrier gas (air). The mass flow rate of entering gas is

$A G_{1}^{\prime}=A G_{1}\cdot M _{w,G_{1}} = (8.32\times10^{-3}\mathrm{~kgmole}/s)(28.5\mathrm{~kg/kgmole})=0.237\mathrm{~kg/s}$

The inlet molar flow rate of the pure water solvent $(x_{A_{2}}=0)~{\mathrm{is}}$

$A L_{2}=A L_{2}^{\prime}/M_{w,L_{2}}=(0.231\ {\mathrm{kg/s}})/(18\ {\mathrm{kg/kgmole}})=1.28\times10^{-2}\ {\mathrm{kgmole}}/{\mathrm{s}}$

The outlet gas molar flow rate is calculated by

$A G_{2}=\frac{A G_{S}}{1-y_{A_{2}}}=\frac{A G_{1}(1-y_{A_{1}})}{1-y_{A_{2}}}=\frac{(8.32\times10^{-3})(1-0.040)}{(1-0.003)}=8.01\times10^{-3}\,\mathrm{kgmole}/s$

From an overall balance on the terminal streams, the outlet liquid molar flow rate is

$A L_{1}=A L_{2}+(A G_{1}-A G_{2}) =1.28\times10^{-2}+(8.32\times10^{-3}-8.01\times10^{-3})=1.31\times10^{-2}\mathrm{kgmole}/{\mathrm{s}}$

A balance of solute A around the terminal streams is given by

$y_{A_{1}}A G_{1}+x_{A_{2}}A L_{2}=y_{A_{2}}A G_{2}+x_{A_{1}}A L_{1}$

Or, since $x_{A_{2}}=0,$

$x_{A_{1}}=\frac{y_{A_{1}}A G_{1}-y_{A_{2}}A G_{2}}{A L_{1}}= {\frac{(0.040)(8.32\times10^{-3})-(0.0030)(8.01\times10^{-3})}{1.31\times10^{-2}}}=0.024$

The average molecular weight of the outlet liquid stream is

$M_{w,L_{1}}=x_{A_{1}}M_{A}+(1-x_{A_{1}})M_{B}=(0.024)(17)+(1-0.024)(18)=18\;\mathrm{kg/kgmole}$

where component B refers to the solvent (water). Finally, the outlet mass flow rate of liquid is

$A L_{1}^{\prime}=A L_{1}\cdot M_{w,L_{1}}=\left(1.31\times10^{-2}\ \mathrm{kgmole}/s\right)\left(18\mathrm{~kg/kgmole}\right)=0.237\ \mathrm{kg/s}$

With the mass flow rates known, the x-axis on the Flooding correlation will now be determined. First, the ratio of the liquid to gas mass flow rate at the bottom of the tower is

${\frac{L^{\prime}}{G^{\prime}}}={\frac{A L^{\prime}}{A G^{\prime}}}={\frac{0.237\ {\mathrm{kg/s}}}{0.237\ {\mathrm{kg/s}}}}=1.00$

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

$\rho_{G}=\frac{P}{R T}M_{W,G_{1}} = \frac{(1.0\ atm)}{\left(0.08206{\frac{\mathrm{m^{3}}\cdot{\mathrm{atm}}}{\mathrm{kgmole}\cdot\mathrm{K}}}\right)(293\,\mathrm{K})}(28.5\,\mathrm{kg/kgmole})=1.19\,\mathrm{kg/m^{3}}$

Therefore, the x-axis on the Flooding correlation (Figure 31.23) is

$\frac{L^{\prime}}{G^{\prime}}\left(\frac{\rho_{G}}{\rho_{L}-\rho_{G}}\right)^{1/2}=1.00\left(\frac{1.19}{998.2-1.19}\right)^{1/2}=0.034$

which is a dimensionless quantity. At a pressure drop of 200 N/m² per meter of packing depth, the y-axis value of the Flooding correlation is 0.049 if the x-axis value is 0.034. Consequently,

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

This expression for the y-axis is rearranged to determine the required superficial molar velocity of gas, G′ :

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

From Table 31.2 $c_{f}=155$ for 1.0-inch Raschig rings. Therefore, the required superficial mass velocity of the gas is

$G^{\prime}=\sqrt{\frac{0.049\left(1.19\frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)\left(998.2-1.19\frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)(1.00)}{\left(155\right)\left(993\times10^{-6}\,\mathrm{kg/m}\cdot s\right)^{0.1}(1.0)}}=0.865\,\mathrm{kg/m^{2} \cdot s}$

The cross-sectional area of the tower is backed out from G′ by

$A={\frac{A G^{\prime}}{G^{\prime}}}={\frac{0.237\,{\mathrm{kg/s}}}{0.865\,{\mathrm{kg/m}}^{2}\cdot{\mathrm{s}}}}=0.274\,{\mathrm{m}}^{2}$

and so the tower diameter is

$D=\sqrt{\frac{4A}{\pi}\ =}\sqrt{\frac{4(0.274\,\mathrm{m}^{2})}{\pi}}=0.59\,\mathrm{m}$