Question Bank • QB_EXP_DFE_HEX_034 MCQ • 1 Mark

Section 7: Dairy and Food Engineering

Heat Exchangers: LMTD & NTU-Effectiveness Methods

Subtopic: Counterflow heat exchanger LMTD calculation

Question Statement:

In the temperature profile diagram, if milk fouling increases the total fouling factor by $R_f = 0.0003\text{ m}^2\cdot\text{K/W}$ in a heat exchanger with clean $U = 2000\text{ W/(m}^2\cdot\text{K)}$, the fouled $U_f$ will be:

Options:

  • (A) 1250 W/(m^2 K)
  • (B) 1500 W/(m^2 K)
  • (C) 1000 W/(m^2 K)
  • (D) 1750 W/(m^2 K)

Answer Key:

A

Step-by-Step Solution & Explanation

$\frac{1}{U_f} = \frac{1}{U} + R_f = \frac{1}{2000} + 0.0003 = 0.0005 + 0.0003 = 0.0008\text{ m}^2\text{K/W}$. $U_f = \frac{1}{0.0008} = 1250\text{ W/(m}^2\cdot\text{K)}$.
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