Question 30.4: A tubular membrane with a diameter of 2 cm and a water perme......
A tubular membrane with a diameter of 2 cm and a water permeability of 250 L/m² -h-atm is being used for UF of cheese whey. The whey proteins have an average diffusivity of 4 × 10^{- 7} cm²/s, and the osmotic pressure in atmospheres is given by Jonsson’s equation^{23} :
π = 4.4 × 10^{- 3}c- 1.7 × 10^{- 6}c² + 7.9 × 10^{- 8}c³
where c is the protein concentration in grams per liter. (a) Calculate the effect of Δp on the flux for a clean membrane if the solution velocity is 1.5 m/s and the protein concentration is 10, 20, or 40 g/L. Assume the gel concentration is 400 g/L and the rejection is 100 percent. (b) If the membrane permeability is reduced fivefold by plugging, what is the effect on the permeate flux?
Diameter of the tubular membrane: 2 cm
Water permeability of the membrane: 250 L/m²-h-atm
Whey proteins diffusivity: 4 × 10^(-7) cm²/s
Osmotic pressure equation: π = 4.4 × 10^(-3)c – 1.7 × 10^(-6)c² + 7.9 × 10^(-8)c³
Solution velocity: 1.5 m/s
Protein concentrations: 10 g/L, 20 g/L, 40 g/L
Gel concentration: 400 g/L
Rejection: 100%
Membrane permeability reduction due to plugging: fivefold
Final Answer
(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)
\begin{aligned}N_{ Sh } & =0.0096 \times(30,000)^{0.913} \times(25,000)^{0.346}=3.9 \times 10^3 \\ k_c & =\frac{3.9 \times 10^3\left(4 \times 10^{-7}\right)}{2}=7.8 \times 10^{-4} cm / s\end{aligned}
\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.
