Question 30.5: Ultrafiltration tests with a 1.5-cm tubular membrane at NRe ......
Ultrafiltration tests with a 1.5-cm tubular membrane at N_{Re} = 25,000 gave a permeate flux of 40 L/m²-h and 75 percent rejection for a 5 percent polymer solution. The polymer has an average molecular weight of 30,000, and the estimated diffusivity is 5 × 10^{- 7} cm²/s. (a) Neglecting the effect of molecular diffusion in the pores, predict the fraction rejected for a flux of 20 L/m²-h, and predict the maximum rejection. (b) Estimate the fraction rejected for the low-molecular-weight fraction of the polymer with M ≅ 10,000. (c) If the selective layer thickness is 0.2 μm, does molecular diffusion have a significant effect on the rejection for case (a)?
(a) Base case:
\begin{aligned}v & =40 \times 2.78 \times 10^{-5}=1.112 \times 10^{-3} ~cm / s \\ N_{ Se } & =\frac{0.01}{5 \times 10^{-7}}=20,000\end{aligned}
From Eq. (21.55)
N_{ Sh }=0.0096 N_{ Re }^{0.913} N_{ S e}^{0.346} (21.55)
\begin{aligned}N_{ Sh } & =0.0096(25,000)^{0.913}(20,000)^{0.346}=3060 \\ k_c & =\frac{3060\left(5 \times 10^{-7}\right)}{1.5}=1.02 \times 10^{-3}~ cm / s \end{aligned}
\frac{1-R}{R}=\frac{0.25}{0.75}=\frac{K}{1-K} \exp \frac{1.112 \times 10^{-3}}{1.02 \times 10^{-3}}
\begin{aligned}\frac{K}{1-K} & =0.112 \\ K & =\frac{0.112}{1.112}=0.101\end{aligned}
If the flux is reduced to 20~ L / m ^2- h \text { or } 0.556 \times 10^{-3}~ cm / s
\begin{aligned} \frac{1-R}{R} & =\frac{0.101}{0.899} \exp \frac{0.556}{1.02}=0.194 \\ R & =\frac{1}{1.194}=0.84\end{aligned}
As the flux approaches zero, R approaches 1 – K:
R_{\max }=1-0.101=0.90
(b) Use Fig. 30.24 for a rough estimate. Locate the point R_1 = 0.75 and M_1 = 30,000 on the graph, and draw a line similar to that for PM 30. At M_ 2 = 10,000, R_2 \cong 0.35. For an independent calculation, predict K and k_{ c } \text {. If } K_1=0.101=(1-λ_1)^2,
\begin{aligned} & \lambda_1=0.682=\frac{D_1}{D_{\text {pore }}} \\ & D_2 \cong D_1\left(\frac{10,000}{30,000}\right)^{1 / 3}=0.694 D_1 \\ & \lambda_2=0.682(0.694)=0.473 \\ & K_2=(1-0.473)^2=0.278\end{aligned}
The diffusivity of large molecules varies with -\frac{1}{3} \text { power of the size, and } k_c \text { varies with }D_v^{0-65} \text { or } M^{-0.22}:
k c_2=k c_1 \times 3^{0.22}=1.02 \times 10^{-3} \times 1.27=1.29 \times 10^{-3}~ cm / s
\text { At } v=1.112 \times 10^{-3} ~cm / s \text {, }
\begin{aligned} \frac{1-R_2}{R_2} & =\frac{0.278}{1-0.278} \exp \frac{1.112}{1.29}=0.912 \\ R_2 & =\frac{1}{1.912}=0.52\end{aligned}
This is appreciably higher than the estimate of 0.35, but in any case, a sharp separation is not possible for molecules differing only threefold in molecular weight.
(c) For M=30,000 \text { and } D_v=5 \times 10^{-7}~ cm ^2 / s, estimate
\begin{aligned}D_{\text {pore}} & =1 \times 10^{-7} ~cm ^2 / s \quad \varepsilon=0.5 \quad \tau=2 \\ D_e & =2.5 \times 10^{-8}~ cm ^2 / s \\ L & =0.2 \mu m =2 \times 10^{-5} ~cm \\ \frac{v L}{D_e} & =\frac{\left(5.56 \times 10^{-4}\right)\left(2 \times 10^{-5}\right)}{2.5 \times 10^{-8}}=0.445 \end{aligned}
From Eq. (30.63) with K = 0.101,
\frac{c_2}{c_s}=\frac{K \exp \left(v L / D_e\right)}{K-1+\exp \left(v L / D_e\right)} (30.63)
\frac{c_2}{c_s}=\frac{0.101 \exp 0.445}{0.101-1+\exp 0.445}=0.24
Diffusion in the membrane makes the permeate concentrations about twice as high as it would be if c_2=K c_s=0.101 c_s. This indicates that the partition coefficient is lower than that estimated in part (a).
