Question

Water flowing at 70 kg/min is heated from 25°C to 65°C in a counter-flow double-pipe heat exchanger using hot oil. The oil enters at 110°C and exits at 65°C. If the overall heat transfer coefficient is 300 W/(m²·K), the heat exchanger area is ___________ m² (rounded off to 1 decimal place).

Hint:

For a counter-flow heat exchanger, use the log mean temperature difference (LMTD) to calculate the heat exchanger area.

Answer 1

The heat duty (Q) for the heat exchanger is given by the equation: \[ Q = m \cdot c \cdot \Delta T \] Where:
\( m = 70 \, {kg/min} = \frac{70}{60} \, {kg/s} = 1.167 \, {kg/s} \) \( c_{{water}} = 4.2 \, {kJ/(kg°C)} \) \( \Delta T_{{water}} = 65 - 25 = 40^\circ C \) Now, calculate the heat duty: \[ Q = 1.167 \cdot 4.2 \cdot 40 = 196.1 \, {W} \] For a counter-flow heat exchanger, the heat exchanger area \( A \) can be calculated using the equation: \[ Q = U \cdot A \cdot \Delta T_{{lm}} \] Where:
\( \Delta T_{{lm}} \) is the log mean temperature difference (LMTD). For a counter-flow heat exchanger:
\[ \Delta T_{{lm}} = \frac{(T_{{in, hot}} - T_{{out, cold}}) - (T_{{out, hot}} - T_{{in, cold}})}{\ln\left(\frac{T_{{in, hot}} - T_{{out, cold}}}{T_{{out, hot}} - T_{{in, cold}}}\right)} \] Substitute the given values: \[ \Delta T_{{lm}} = \frac{(110 - 65) - (65 - 25)}{\ln\left(\frac{110 - 65}{65 - 25}\right)} = \frac{45 - 40}{\ln\left(\frac{45}{40}\right)} = \frac{5}{\ln(1.125)} \approx 15.2^\circ C \] Now, solve for the area \( A \): \[ A = \frac{Q}{U \cdot \Delta T_{{lm}}} = \frac{196.1}{300 \cdot 15.2} = 15.2 \, {m²} \]