Question 30.4: A tubular membrane with a diameter of 2 cm and a water perme......

(a) Assume the bulk solutions have the same density and viscosity as water:

$D=2 ~cm \quad \bar{V}=150 ~cm / s \quad \rho=1 ~g / cm ^3 \quad \mu=0.01 ~g / cm – s$

$N_{ Re }=2 \times 150 \times 1 / 0.01=30,000 \quad N_{ Se }=\frac{0.01}{1 \times 4 \times 10^{-7}}=25,000$

From Eq. (21.55),

$N_{ Sh }=0.0096 N_{ Re }^{0.913} N_{ S e}^{0.346}$    (21.55)

$\text { For } c_1=10~ g / L \text {, pick } v=10^{-3}$ cm/s or 36 L/m²-h. From Eq. (30.53)

$v=k_c \ln \frac{c_s}{c_1}$    (30.53)

$\begin{aligned}\ln \frac{c_s}{c_1} & =\frac{v}{k_c}=\frac{10^{-3}}{7.8 \times 10^{-4}}=1.282 \\ c_s & =3.60 c_1=36 ~g / L \end{aligned}$

At $c_s, \pi=4.4 \times 10^{-3} \times 36-1.7 \times 10^{-6}(36)^2+7.9 \times 10^{-8}(36)^3=0.16 atm$. For complete rejection of protein $\Delta \pi=\pi=0.16:$

$Q_{ m }=\frac{250~ L }{ m ^2- h – atm } \times \frac{1}{36,000}=6.94 \times 10^{-3}~ cm / s – atm$

From Eq. (30.50)

$v=Q_{ m }(\Delta p-\Delta \pi) \frac{\mu_{ H _2 O }}{\mu}$    (30.50)

$\begin{aligned}\Delta p-\Delta \pi & =\frac{10^{-3}}{6.94 \times 10^{-3}}=0.144~ atm \\\Delta p & =0.144+0.16=0.304 ~atm\end{aligned}$

Note that over half of the driving force is needed to overcome the osmotic pressure difference caused by concentration polarization.

The maximum flux is obtained from Eq. (30.53) with $c_s=400$:

$v_{\max }=7.8 \times 10^{-4} \ln \frac{400}{10}=2.88 \times 10^{-3} ~cm / s =104~ L / m ^2- h$

At this point,

$\Delta p-\Delta \pi=\frac{2.88 \times 10^{-3}}{6.94 \times 10^{-3}}=0.41~ atm$

For c = 400, π = 6.54 atm,

Δp= 6.54 + 0.41 = 6.95 atm

Similar calculations are made for other values of v from $10^{- 4}$ to $v_{max}$, and the flux is plotted against the pressure drop in Fig. 30.29. The predicted flux is constant for Δp > 6.95, and a gel layer of increasing thickness is presumed to form as the pressure increases, but in practice, the flux might decrease slightly because of compression of the gel layer.
The curves for the three concentrations are similar in shape, but the flux is zero until the pressure difference exceeds the osmotic pressure of the solution, and the intercept is more noticeable for the higher concentrations.

$\text { (b) If } Q_m=250 / 5=50~ L / m ^2 \text {-h-atm, } v_{\max }$ is not changed, but a greater Δp is needed for any value of v. For example, at $v=10^{-3} ~cm / s \text { and } c=40~ g / L \text {, }$

$Q_m=50 \times \frac{1}{36,000}=1.39 \times 10^{-3}~ cm / s – atm$

$\Delta p-\Delta \pi=\frac{10^{-3}}{1.39 \times 10^{-3}}=0.719~ atm$

$\begin{aligned}\frac{c_s}{c_1} & =3.6 \quad \text { as in part }(a) \\ c_s & =3.6 \times 40=144 \\\pi & =0.834=\Delta \pi \\\Delta p & =0.719+0.834=1.55~ atm\end{aligned}$

The dashed line in Fig. 30.29 shows that the largest effect of the lower membrane permeability is a 30 percent reduction in flux at low pressure drops.