Question Bank • QB_DFE_UOP_008 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Unit Operations in Dairy and Food Engineering

Subtopic: Blanching

Question Statement:

A reverse osmosis (RO) membrane system desalts cheese whey at $25.0^\circ\text{C} = 298.15\text{ K}$. The feed osmotic pressure is $\Pi_f = 1.20\text{ MPa}$ and the permeate osmotic pressure is $\Pi_p = 0.05\text{ MPa}$. The hydrostatic pressure applied across the membrane is $\Delta P = 4.50\text{ MPa}$. If the membrane water permeability coefficient is $A_w = 4.0 \times 10^{-6}\text{ kg/(m}^2\cdot\text{s}\cdot\text{kPa)}$, the pure water permeate flux $J_w = A_w [\Delta P - \Delta \Pi]$ in $\text{kg/(m}^2\cdot\text{h})$ is ________ (round off to one decimal place).

Answer Key:

48.2

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Net Driving Force Calculation** ``` Hydrostatic ΔP = 4500 kPa =======> (Driving Force) Osmotic ΔΠ = 1150 kPa <======= (Opposing Back-Pressure) Net Driving Pressure ΔP - ΔΠ = 3350 kPa ``` 1. Net osmotic pressure across membrane: $\Delta \Pi = \Pi_f - \Pi_p = 1.20 - 0.05 = 1.15\text{ MPa} = 1150\text{ kPa}$ 2. Applied hydrostatic pressure: $\Delta P = 4.50\text{ MPa} = 4500\text{ kPa}$ 3. Net driving pressure difference: $\Delta P - \Delta \Pi = 4500 - 1150 = 3350\text{ kPa}$ 4. Permeate water flux: $J_w = A_w (\Delta P - \Delta \Pi) = (4.0 \times 10^{-6}\text{ kg/(m}^2\cdot\text{s}\cdot\text{kPa)}) \times 3350\text{ kPa}$ $J_w = 0.0134\text{ kg/(m}^2\cdot\text{s)}$ 5. Convert to $\text{kg/(m}^2\cdot\text{h})$: $J_w = 0.0134\text{ kg/(m}^2\cdot\text{s)} \times 3600\text{ s/h} = 48.24\text{ kg/(m}^2\cdot\text{h}) \approx 48.2\text{ kg/(m}^2\cdot\text{h})$ **Method 2: Direct Formula Multiplication** $J_w = 4.0 \times 10^{-6} \times 3350 \times 3600 = 48.24\text{ kg/(m}^2\cdot\text{h})$
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