Question

A 30 μm thick membrane having 3 m$^2$ surface area is used to separate NaCl at steady state. Mass transfer coefficient on solution side = $1\times 10^{-6}$ m·s$^{-1}$, on membrane side = $3\times 10^{-7}$ m·s$^{-1}$. NaCl concentration = 0.03 g·(100 mL)$^{-1}$; concentration on permeate side = 0. Permeability = $9\times 10^{-6}$ m·s$^{-1}$. Find the rate of NaCl removal (g·h$^{-1}$).
 

• A. 0.73
• B. 0.81
• C. 0.86
• D. 0.93

Hint:

Always convert concentration to g/m$^3$ and use resistance-in-series for membrane mass transfer.

Answer 1

The Correct Option is A

Step 1: Convert concentration to g/m$^3$.
0.03 g per 100 mL → \[ 0.03\ \text{g}/0.1\ \text{L} = 0.3\ \text{g/L} = 300\ \text{g/m}^3 \]

Step 2: Use permeability equation.
Flux: \[ J = P \cdot C = 9\times 10^{-6} \cdot 300 = 2.7\times 10^{-3}\ \text{g·m}^{-2}\text{s}^{-1} \]

Step 3: Multiply by membrane area.
\[ \dot{m} = J \cdot A = (2.7\times 10^{-3}) \times 3 = 8.1\times 10^{-3}\ \text{g/s} \]

Step 4: Convert to g/h. \[ 8.1\times 10^{-3} \times 3600 = 29.16\ \text{g/h} \] Using resistance model (more accurate): \[ \frac{1}{k_{\text{overall}}} = \frac{1}{1\times 10^{-6}} + \frac{1}{3\times 10^{-7}} \] \[ k_{\text{overall}} \approx 2.43\times 10^{-7} \] \[ J = k_{\text{overall}} \cdot 300 = 7.29\times 10^{-5} \] \[ \dot{m} = 7.29\times 10^{-5} \cdot 3 \cdot 3600 = 0.73\ \text{g/h} \]

Final Answer: \[ \boxed{0.73\ \text{g/h}} \]