Question Bank • QB_DFE_UOP_012 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Unit Operations in Dairy and Food Engineering

Subtopic: Pasteurization

Question Statement:

In a continuous High-Temperature Short-Time (HTST) milk pasteurizer operating at a throughput of $4500\text{ L/h}$, the stainless steel holding tube has an inside diameter of $D = 50.0\text{ mm} = 0.050\text{ m}$. To guarantee a minimum legal holding time of $15.0\text{ seconds}$ for the fastest flowing particle (with fastest-particle velocity ratio $\eta_v = \frac{v_{\text{avg}}}{v_{\text{max}}} = 0.83$), the required length of the holding tube in $\text{m}$ is ________ (round off to two decimal places).

Answer Key:

11.51

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Velocity Profile Kinematics** 1. Volumetric flow rate ($Q$): $Q = \frac{4500\text{ L/h}}{3600\text{ s/h}} = 1.25\text{ L/s} = 0.00125\text{ m}^3/\text{s}$ 2. Cross-sectional area of holding tube ($A$): $A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.050)^2 = 1.963495 \times 10^{-3}\text{ m}^2$ 3. Average flow velocity ($v_{\text{avg}}$): $v_{\text{avg}} = \frac{Q}{A} = \frac{0.00125}{1.963495 \times 10^{-3}} = 0.63662\text{ m/s}$ 4. Maximum (fastest particle) velocity ($v_{\text{max}}$): $v_{\text{max}} = \frac{v_{\text{avg}}}{\eta_v} = \frac{0.63662}{0.83} = 0.76701\text{ m/s}$ 5. Tube length ($L$) for minimum holding time $t_{\text{min}} = 15.0\text{ s}$: $L = v_{\text{max}} \times t_{\text{min}} = 0.76701\text{ m/s} \times 15.0\text{ s} = 11.505\text{ m} \approx 11.51\text{ m}$ **Method 2: Effective Volume Approach** $V_{\text{effective}} = \frac{Q \times t_{\text{min}}}{\eta_v} = \frac{0.00125 \times 15.0}{0.83} = 0.02259\text{ m}^3$ $L = \frac{V_{\text{effective}}}{A} = \frac{0.02259}{0.0019635} = 11.51\text{ m}$
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