Counter-flow heat exchanger with equal heat capacities: A hot stream enters at 100°C and leaves at 60°C. A cold stream enters at 40°C. If C_hot = C_cold, what is the mean temperature difference (LMTD) between the two fluids?

Introduction / Context:
The logarithmic mean temperature difference (LMTD) is a key parameter for sizing heat exchangers when the overall heat transfer coefficient and duty are known. Special simplifications occur when the capacity rates of the two fluids are equal.

Given Data / Assumptions:

• Counter-flow heat exchanger.
• Hot fluid: T_h,in = 100°C, T_h,out = 60°C.
• Cold fluid: T_c,in = 40°C.
• Equal capacity rates: C_h = C_c ⇒ ΔT changes linearly and the cold outlet temperature rise equals the hot temperature drop.

Concept / Approach:
With equal capacity rates, energy balance dictates that the cold stream temperature rise equals the hot stream temperature drop (40°C here). Therefore T_c,out = 40 + 40 = 80°C. Once the terminal temperatures are known, compute LMTD for counter-flow using the two end temperature differences.

Step-by-Step Solution:

Verification / Alternative check:
If ΔT_1 = ΔT_2, the logarithmic mean reduces to that common difference directly; thus LMTD = 20°C without further calculation.

Why Other Options Are Wrong:

• 40°C or higher values do not reflect the computed terminal differences.
• 66.7°C is unrelated to these endpoints and overestimates the driving force.

Common Pitfalls:
Using parallel-flow formula or assuming the cold outlet is unknown; with equal capacities it is immediately 80°C, making the end differences equal.

Final Answer:
20°C