In a counter-current concentric tube heat exchanger, hot water and cold milk have identical heat capacity rates ($C_h = C_c = C_{min} = 2500\text{ W/K}$, so capacity ratio $C_r = 1.0$). The overall heat transfer coefficient is $U = 500\text{ W/(m}^2\cdot\text{K)}$ and the heat transfer area is $A = 7.5\text{ m}^2$. Using the effectiveness-NTU relation for counter-flow with $C_r = 1.0$ ($\epsilon = \frac{NTU}{1 + NTU}$), the thermal effectiveness of the heat exchanger is ________ (answer in decimal).
Answer Key:
0.6
Step-by-Step Solution & Explanation
1. Number of Transfer Units ($NTU$):
$NTU = \frac{U A}{C_{min}} = \frac{500\text{ W/(m}^2\cdot\text{K)} \times 7.5\text{ m}^2}{2500\text{ W/K}} = \frac{3750}{2500} = 1.50$
2. Effectiveness for counter-flow when $C_r = 1.0$:
$\epsilon = \frac{NTU}{1 + NTU} = \frac{1.50}{1 + 1.50} = \frac{1.50}{2.50} = 0.60$