Question Bank • QB_DFE_EXP_022 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Unit Operations in Dairy and Food Engineering

Subtopic: Pasteurization

Question Statement:

A High Temperature Short Time (HTST) pasteurizer processes milk at a flow rate of $3600\text{ L/h}$ ($0.001\text{ m}^3\text{/s}$). The holding tube has an internal diameter of $0.050\text{ m}$. To ensure the mandatory minimum holding time of $15.0\text{ seconds}$ for the fastest moving fluid particle (using laminar flow ratio $V_{avg} / V_{max} = 0.50$ or plug flow margin), for turbulent flow where $V_{max} \approx 1.20 V_{avg}$, if $t_{min} = \frac{L}{V_{max}} = 15.0\text{ s}$, with average velocity $V_{avg} = \frac{Q}{A} = \frac{0.001}{\frac{\pi}{4}(0.05)^2} = \frac{0.001}{0.0019635} = 0.5093\text{ m/s}$, taking $V_{max} = 1.20 \times 0.5093 = 0.611\text{ m/s}$, the minimum length of the holding tube $L$ in meters is ________ (round off to 2 decimal places).

Answer Key:

9.17

Step-by-Step Solution & Explanation

1. Tube cross-sectional area: $A = \frac{\pi}{4} (0.050)^2 \approx 1.9635 \times 10^{-3}\text{ m}^2$ 2. Average velocity: $V_{avg} = \frac{0.001\text{ m}^3\text{/s}}{1.9635 \times 10^{-3}\text{ m}^2} \approx 0.5093\text{ m/s}$ 3. Max velocity under turbulent flow ($V_{max} = 1.20 V_{avg}$): $V_{max} = 1.20 \times 0.5093 = 0.61115\text{ m/s}$ 4. Holding tube length $L$ for $t = 15\text{ s}$: $L = V_{max} \times t = 0.61115 \times 15.0 \approx 9.167\text{ m} \approx 9.17\text{ m}$
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