Counter-current heat exchanger LMTD: Hot water (0.01 m^3/min) cools from 80 °C to 50 °C in the tube side of a counter-current shell-and-tube exchanger. Cold oil (0.05 m^3/min, density 800 kg/m^3, cp = 2 kJ/kg·K) enters at 20 °C. What is the approximate log-mean temperature difference (LMTD) in °C?

Introduction / Context:
The log-mean temperature difference (LMTD) provides the effective driving force for heat exchange in counter-current and co-current exchangers. Given inlet and outlet temperatures, LMTD captures end-point differences into one representative value for duty calculations.

Given Data / Assumptions:

• Hot water: 0.01 m^3/min ≈ 10 kg/min, cp ≈ 4.18 kJ/kg·K.
• Hot side: T_h,in = 80 °C, T_h,out = 50 °C.
• Cold oil: 0.05 m^3/min × 800 = 40 kg/min, cp = 2 kJ/kg·K.
• Cold side: T_c,in = 20 °C, counter-current layout.

Concept / Approach:
First compute the heat duty from the hot side, then find the cold outlet temperature from an energy balance. With both end temperature differences known, evaluate LMTD = (ΔT1 − ΔT2) / ln(ΔT1/ΔT2).

Step-by-Step Solution:
Hot capacity rate: C_h = 10 * 4.18 = 41.8 kJ/min·K.Heat duty: Q = C_h * (80 − 50) = 41.8 * 30 = 1254 kJ/min.Cold capacity rate: C_c = 40 * 2 = 80 kJ/min·K.Cold outlet: T_c,out = 20 + Q/C_c = 20 + 1254/80 ≈ 35.7 °C.End differences (counter-current): ΔT1 = T_h,in − T_c,out ≈ 80 − 35.7 = 44.3 °C; ΔT2 = T_h,out − T_c,in = 50 − 20 = 30 °C.LMTD = (44.3 − 30) / ln(44.3/30) ≈ 14.3 / ln(1.48) ≈ 14.3 / 0.39 ≈ 36.7 °C ≈ 37 °C.

Verification / Alternative check:
Using more precise cp for water alters Q slightly but does not change the LMTD choice from ~37 °C among the options.

Why Other Options Are Wrong:
32 and 45 °C deviate significantly from the computed LMTD; 50 °C would require equal-end differences around 50 °C, inconsistent with the data.

Common Pitfalls:

• Mixing co-current and counter-current end temperatures when forming ΔT1 and ΔT2.
• Forgetting to compute T_c,out via energy balance before LMTD.

Final Answer:
37