Question Bank • QB_EXP_DFE_HEX_001 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat Exchangers: LMTD & NTU-Effectiveness Methods

Subtopic: Counterflow heat exchanger LMTD calculation

Question Statement:

In a counterflow milk chiller shown in the temperature profile diagram, hot milk is cooled from $85^\circ\text{C}$ to $45^\circ\text{C}$ by chilled water entering at $20^\circ\text{C}$ and leaving at $42^\circ\text{C}$. The Logarithmic Mean Temperature Difference ($LMTD$) is ________ $^\circ\text{C}$ (round off to two decimal places).

Answer Key:

33.19

Step-by-Step Solution & Explanation

$\Delta T_1 = T_{h,in} - T_{c,out} = 85 - 42 = 43^\circ\text{C}$. $\Delta T_2 = T_{h,out} - T_{c,in} = 45 - 20 = 25^\circ\text{C}$. $LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{43 - 25}{\ln(43/25)} = \frac{18}{\ln(1.72)} = \frac{18}{0.5423} \approx 33.19^\circ\text{C}$.
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