Question 29.3: In an experimental study of the absorption of ammonia by wat...
In an experimental study of the absorption of ammonia by water in a wetted-wall column, the overall mass-transfer coefficient, K_{\mathrm{G}} was found to be 2.74 \times 10^{-9} \mathrm{~kg} \mathrm{~mol} / \mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}. At one point in thecolumn, the gas phase contained 8 mol ammonia and the liquid-phase concentration was 0.064 kg mol ammonia/m³ of solution. The tower operated at 293 K and 1.013 \times 10^5 \mathrm{~Pa}. At that temperature, the Henry’s law constant is 1.358 \times 10^3 \mathrm{~Pa} /\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^3\right). If 85 % of the total resistance to mass transfer is encountered in the gas phase, determine the individual film mass-transfer coefficients and the interfacial compositions.
The total resistance in both phases, according to equation (29-12), is
\frac{\text { resistance in the gas phase }}{\text { total resistance in both phases }}=\frac{\Delta p_{A, \text { gas film }}}{\Delta p_{A, \text { total }}}=\frac{1 / k_G}{1 / K_G} (29-12)
\frac{1}{K_G}=\frac{1}{2.74 \times 10^{-9} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}}}=3.65 \times 10^8 \frac{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}}{\mathrm{kg} \mathrm{mol}}
As the resistance in the gas phase, 1/K_{\mathrm{G}}, is 85 % of the total resistance, we may evaluate the individual gas-phase coefficient by
\frac{1}{k_G}=0.85\left(3.65 \times 10^8 \frac{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}}{\mathrm{kg} \mathrm{mol}}\right)=3.10 \times 10^8 \frac{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}}{\mathrm{kg} \mathrm{mol}}and
k_G=\frac{1}{3.10 \times 10^8}=3.226 \times 10^{-9} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}} .The liquid-phase coefficient, k_{\mathrm{L}} is evaluated using equation (29-16)
\frac{1}{K_G}=\frac{1}{k_G}+\frac{H}{k_L} (29-16)
3.65 \times 10^8=3.10 \times 10^8+\frac{1.358 \times 10^3 \mathrm{~Pa} /\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^3\right)}{k_L}
k_L=2.47 \times 10^{-5} \mathrm{~kg} \mathrm{~mol} / \mathrm{m}^2 \cdot \mathrm{s} \cdot\left(\mathrm{kg} \mathrm{mol} / \mathrm{m}^3\right)
At the stated point in the column
p_{A, G}=y_A P=(0.08)\left(1.013 \times 10^5 \mathrm{~Pa}\right)=8.104 \times 10^3 \mathrm{~Pa}c_{A, L}=0.064 \mathrm{~kg} \mathrm{~mol} / \mathrm{m}^3
Upon introducing Henry’s lawconstant, we find the partial pressure, p_A^*, in equilibrium with the bulk liquid concentration
p_A^*=H c_{A, L}=\left(1.358 \times 10^3 \frac{\mathrm{Pa}}{\mathrm{kg} \mathrm{mol} / \mathrm{m}^3}\right)\left(0.064 \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^3}\right)=87.1 \mathrm{~Pa}The mass flux, as expressed by equation (29-10), becomes
N_A=K_G\left(p_{A, G}-p_A^*\right) (29-10)
=\left(2.74 \times 10^{-9} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}}\right)\left(8.104 \times 10^3 \mathrm{~Pa}-87.1 \mathrm{~Pa}\right)
=2.20 \times 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}
The interfacial composition can be determined using equation (29-6)
N_{A, z}=k_G\left(p_{A, G}-p_{A, i}\right) (29-6)
N_A=k_G\left(p_{A, G}-p_{A, i}\right)
2.20 \times 10^{-5} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}=\left(3.226 \times 10^{-9} \frac{\mathrm{kg} \mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s} \cdot \mathrm{Pa}}\right)\left(8.104 \times 10^3 \mathrm{~Pa}-p_{A, i}\right)
p_{A, i}=1284 \mathrm{~Pa}
and using Henry’s law
p_{A, i}=H c_{A, i}(1284 \mathrm{~Pa})=\left(1.358 \times 10^3 \frac{\mathrm{Pa}}{\mathrm{kg} \mathrm{mol} / \mathrm{m}^3}\right) c_{A, i}
c_{A, i}=0.946 \mathrm{~kg} \mathrm{~mol} / \mathrm{m}^3