Week 3 - Exercise#

Problem Statement#

Compute the total membrane area necessary to produce 10 l/h purified water (0.03 wt% NaCl) through a single stage Reverse Osmosis separation of seawater (water 4 wt% NaCl). The feed pressure is 100 bar and the composition in the retentate stream is 7 wt%.
The permeate stream is at ambient pressure (1 bar), all streams are at 25\(^o\) C, and the osmotic pressure in bar is captured by the semiempirical expression:

\[ \pi=\frac{1.12}{14.5}T\sum_{i=1}^Nc_i \]

Where T is the temperature in K, \(N\) is the total number of ionic species in solution, and ci is the concentration expressed in \([mol/l]\).

The asymmetric membrane used for the separation has a permeance of \(10^{-5}\,[m h^{-1} bar^{-1}]\).

Once you have solved the problem consider the following questions:

  • Can you evaluate the salt rejection in this process, and salt passage?

  • How would you modify the operation of the system to increase the productivity of the process?

Solution trace#

Data:

NaCl weigth percentage:
\(w_{NaCl,P}=0.03\)
\(w_{NaCl,F}=4\)
\(w_{NaCl,R}=7\)

Water density: \(\rho_{H_2O}\)=1\(\times\)10\(^3\) [g/l]

Working expression:

\[ c_{NaCl,i}=\frac{w_{NaCl,i}}{100}\times\frac{\rho_{Solution,i}}{MM_{NaCl}} \]

Where: \(i\) indicates the stream we are referring to, \(\rho_{Solution,i}\) is the mass density of the NaCl solution in each stream, and \(MM_{NaCl}\) is the molar mass of NaCl.
The mass density of the solution \(\rho_{Solution}\) can be computed known the NaCl weight percentage and the density of water as:

\[ \rho_{solution}=\frac{\rho_{H_2O}}{1-\frac{w_{NaCl,i}}{100}} \]

Computing the Osmotic pressure difference across the membrane
The expression provided in the exercise assignment allows to compute the osmotic pressure associated with an assigned concentration of NaCl in a solution of assigned NaCl concentration. Note that the concentration appearing in such expression is the total ionic concentration, which can be computed from the NaCl concentration knowing the unit formula of NaCl.

\[ c_{TOT,i}=c_{Na^+,i}+c_{Cl^-,i}=2c_{NaCl,i} \]

Note that in this case it is typically reasonable to assume the concentration on the feed/retentate side of the membrane to be the average between concentrations in the feed and retentate.

\[ c_{TOT,F/R}=2 \left({\frac{c_{NaCl,F}+c_{NaCl,R}}{2}}\right) \]

Alternatively one can consider the feed/retentate side to be perfectly mixed and its composition constant and equal to the retentate one.

The difference in the osmotic pressure can thus be computed as:

\[ \Delta{\pi}=\frac{1.12}{14.5}T\left(c_{TOT,F/R}-c_{TOT,P}\right) \]

Computing the membrane area
The flow rate through the membrane is computed by the expression:

\[ F_{p,wat}=J\times{A}=A\,P_{wat}\,\left(\Delta{P}-\Delta{\pi}\right) \]

hence:

\[ A=\frac{F_{p,wat}}{P_{wat}\,\left(\Delta{P}-\Delta{\pi}\right)} \]

Salt rejection
Given the concentrations computed earlier salt rejection can be computed from its definition:

\[ R_s=1-\frac{c_{NaCl,P}}{c_{NaCl,R}} \]

Salt passage is given by:

\[ P_s=\frac{c_{NaCl,P}}{c_{NaCl,R}} \]
import numpy as np
import matplotlib.pyplot as plt 
from matplotlib.pyplot import cm

#
PM=1E-5; #[m/h/bar]
MMNaCl=58.5; #[g/mol]

#Feed Retentate Permeate
wtp=np.array([4, 7, 0.03])/100; #mass fractions
water_density=1E3; #[g/l]

c=wtp*water_density/(MMNaCl*(1-wtp));  #[mol/l]

C=1.12/14.5;
T=298;
pi_P=2*C*T*c[2];
pi_R=2*C*T*np.mean(c[0:2]);
Dpi=pi_R-pi_P; # [bar]

DP=100-1; # [bar]
Fpwat=10E-3; # [m3/h]

A=Fpwat/PM/(DP-Dpi)

SP=c[2]/c[1]
SR=(1-SP)

print("\nPermeate concentration: ", f"{c[2]:.4}", " [mol/l]")
print("Feed concentration: ", f"{c[0]:.4}", " [mol/l]")
print("Retentate concentration: ", f"{c[1]:.4}", " [mol/l]")

print("\nOsmotic pressure: ", f"{Dpi:.4}", " [bar]")

print("\nMembrane area: ", f"{A:.4}", " [m^2]")
print("Salt rejection: ", f"{SR:.4}", "[-]")
print("Salt passage: ", f"{SP:.2}", "[-]")

Permeate concentration:  0.00513  [mol/l]
Feed concentration:  0.7123  [mol/l]
Retentate concentration:  1.287  [mol/l]

Osmotic pressure:  45.77  [bar]

Membrane area:  18.79  [m^2]
Salt rejection:  0.996 [-]
Salt passage:  0.004 [-]

Contributions#

Debugged by Nikita Gusev, 29 Jan 2021