Question Bank • QB_DFE_HMT_004 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Working principles of heat exchangers

Question Statement:

In a dairy milk pasteurizer operating at steady state, raw cold milk enters the regenerator at $4.0^\circ\text{C}$ and is pre-heated to $64.0^\circ\text{C}$ by hot pasteurized milk entering at $74.0^\circ\text{C}$. Both streams flow at identical mass flow rates and have equal specific heat capacities. The percentage heat regeneration (heat recovery efficiency $\eta_{\text{regen}} = \frac{T_{\text{preheated}} - T_{\text{cold,in}}}{T_{\text{pasteurized}} - T_{\text{cold,in}}} \times 100$) is ________ % (round off to two decimal places).

Answer Key:

85.71

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Direct Regeneration Formula** $\eta_{\text{regen}} = \frac{T_{\text{preheated}} - T_{\text{raw}}}{T_{\text{pasteurized}} - T_{\text{raw}}} \times 100\%$ $\eta_{\text{regen}} = \frac{64.0 - 4.0}{74.0 - 4.0} \times 100\% = \frac{60.0}{70.0} \times 100\% = \frac{6}{7} \times 100\% = 85.714\% \approx 85.71\%$ **Method 2: Heat Duty Ratio** Actual heat transferred: $q_{\text{actual}} = \dot{m} C_p (64.0 - 4.0) = 60 \dot{m} C_p$. Maximum thermodynamically possible heat recovery: $q_{\text{max}} = \dot{m} C_p (74.0 - 4.0) = 70 \dot{m} C_p$. $\eta = \frac{q_{\text{actual}}}{q_{\text{max}}} = \frac{60}{70} = 85.71\%$
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