Question 30.4: Trichloroethylene (TCE), a common industrial solvent, is oft...

Relevant physical property data are provided below. The process is very dilute so that the bulk gas has the properties of air and the bulk liquid has the properties of water. The equilibrium solubility of TCE in water is described by Henry’s law of the form

p_{A}=H \cdot x_{A}

where H is 550 atm at 293 \mathrm{~K}. The binary gas phase diffusivity of TCE in air is 8.08 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s} at 1 \mathrm{~atm} and 293 \mathrm{~K}, as determined by the Fuller-Shettler-Giddings correlation. The binary liquid-phase diffusivity of TCE in water at 293 \mathrm{~K} is 8.9 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}, as determined by the Hayduk-Laudie correlation.

With this physical property information in hand, our strategy is to estimate the gas film coefficient k_{G}, the liquid film coefficient k_{L}, and then the overall mass-transfer coefficient K_{L}. First, the bulk velocity of the gas is

v_{\infty}=\frac{4 Q_{g}}{\pi D^{2}}=\frac{4 \cdot\left(2.0 \times 10^{-3}  \frac{\mathrm{m}^{3}}{\mathrm{~s}}\right)}{\pi(0.04 \mathrm{~m})^{2}}=1.59  \frac{\mathrm{m}}{\mathrm{s}}

The Reynolds number for air flow through the inside of the wetted wall column is

R e=\frac{\rho_{\text {air }} v_{\infty} D}{\mu_{\text {air }}}=\frac{\left(1.19  \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)\left(1.59  \frac{\mathrm{m}}{\mathrm{s}}\right)(0.04 \mathrm{~m})}{1.84 \times 10^{-5}  \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}}=4113

and the Schmidt number for TCE in air is

\mathrm{Sc}=\frac{\mu_{\mathrm{air}}}{\rho_{\mathrm{air}} D_{\mathrm{TCE}-\mathrm{air}}}=\frac{1.84 \times 10^{-5}  \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}}{\left(1.19  \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)\left(8.08 \times 10^{-6}  \frac{\mathrm{m}^{2}}{\mathrm{~s}}\right)}=1.91

where the properties of air are found from Appendix I. As the gas flow is not laminar (\operatorname{Re}>2000), equation (30-17)

\frac{k_{c}D}{D_{A B}}\frac{P_{B,L B}}{P}=0.023\,\mathrm{Re}^{0.83}\mathrm{Sc}^{0.44}       (30-17)

is appropriate for estimation of k_{c}. Therefore

\begin{gathered} k_{c}=\frac{D_{A B}}{D} 0.023 \operatorname{Re}^{0.83} \mathrm{Sc}^{0.44}=\left(\frac{8.08 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}}{0.04 \mathrm{~m}}\right) \\ (0.023)(4113)^{0.83}(1.91)^{0.44}=6.17 \times 10^{-3} \mathrm{~m} / \mathrm{s} \end{gathered}

for a dilute solution where p_{B, l m} / P is essentially 1 . The conversion to k_{G} is

k_{G}=\frac{k_{c}}{R T}=\frac{6.17 \times 10^{-3} \frac{\mathrm{m}}{\mathrm{s}}}{\left(0.08206 \frac{\mathrm{m}^{3} \cdot \mathrm{atm}}{\mathrm{kg} \mathrm{mol} \cdot \mathrm{K}}\right)(293 \mathrm{~K})}=2.57 \times 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}}

The liquid-film coefficient is now estimated. The Reynolds number for the falling liquid film is

\operatorname{Re}_{L}=\frac{4 \rho_{L} \dot{V}_{L}}{\pi D \mu_{L}}=\frac{4 \cdot\left(998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)\left(5 \times 10^{-5} \frac{\mathrm{m}^{3}}{\mathrm{~s}}\right)}{\pi \cdot(0.04 \mathrm{~m})\left(9.93 \times 10^{4} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}\right)}=1600

and the Schmidt number is

\mathrm{Sc}=\frac{\mu_{L}}{\rho_{L} D_{\mathrm{TCE}-\mathrm{H}_{2} \mathrm{O}}}=\frac{\left(9.93 \times 10^{-4} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}\right)}{\left(998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)\left(8.90 \times 10^{-10} \frac{\mathrm{m}^{2}}{\mathrm{~s}}\right)}=1118

where the properties of liquid water at 293 \mathrm{~K} are found in Appendix I. Equation (30-20)

{\frac{k_{L Z}}{D_{A B}}}=0.433(8{\bf c})^{1/2}\left({\frac{\rho_{L}^{2}g z^{3}}{\mu_{L}^{2}}}\right)^{1/6}({\bf R e}_{L})^{0.4}     (30-20)

is appropriate for estimation of k_{L} :

\begin{gathered} k_{L}=\frac{D_{A B}}{z} 0.433(\mathrm{Sc})^{1 / 2}\left(\frac{\rho_{L}^{2} g z^{3}}{\mu_{L}^{2}}\right)^{1 / 6}\left(\mathrm{Re}_{L}\right)^{0.4} \\ =\frac{8.9 \times 10^{-10} \frac{\mathrm{m}^{2}}{\mathrm{~s}}}{2 \mathrm{~m}} 0.433 \cdot(1118)^{1 / 2} \\ \left(\frac{\left(998.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right)^{2}\left(\frac{9.8 \mathrm{~m}}{\mathrm{~s}^{2}}\right)(2 \mathrm{~m})^{3}}{\left(9.93 \times 10^{-4} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}\right)^{2}}\right)^{1 / 6}(1600)^{0.4}=2.55 \times 10^{-5} \mathrm{~m} / \mathrm{s} \end{gathered}

Since the process is dilute, Henry’s law constant in units consistent with k_{L} and k_{G} is

H=550 \mathrm{~atm} \frac{M_{\mathrm{H}_{2} \mathrm{O}}}{\rho_{L, \mathrm{H}_{2} \mathrm{O}}}=(550 \mathrm{~atm})\left(\frac{18 \mathrm{~kg}}{\mathrm{~kg} \mathrm{~mol}}\right)\left(\frac{1}{993.2 \mathrm{~kg} / \mathrm{m}^{3}}\right)=9.968 \frac{\mathrm{atm} \cdot \mathrm{m}^{3}}{\mathrm{~kg} \mathrm{~mol}}

The overall liquid mass-transfer coefficient, k_{L}, is

\frac{1}{K_{L}}=\frac{1}{k_{L}}+\frac{1}{H k_{G}}=\frac{1}{2.55 \times 10^{-5} \frac{\mathrm{m}}{\mathrm{s}}}+\frac{1}{\left(9.96 \frac{\mathrm{atm} \cdot \mathrm{m}^{3}}{\mathrm{~kg} \mathrm{~mol}}\right)\left(2.57 \times 10^{-4} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^{2} \cdot \mathrm{s} \cdot \mathrm{atm}}\right)}

or k_{L}=2.52 \times 10^{-5} \mathrm{~m} / \mathrm{s}. As K_{L} \cong k_{L}, the process is liquid film mass-transfer controlling, which is characteristic of interphase mass-transfer processes involving a large value for Henry’s law constant.