Question

In a counter current heat exchanger, the hot fluid enters at 175 °F and exits at 100 °F. The cold fluid enters at 75 °F and exits at 85 °F. For the calculation of heat transfer rate, consider the tube surface area (per unit length) to be 0.26 ft²/ft and a tube length of 40 ft. The overall heat transfer coefficient of the exchanger is 100 BTU/hr-ft². The minimum number of tubes required in the exchanger for a heat duty of \( 15 \times 10^5 \) BTU/hr is \(\underline{\hspace{2cm}}\) (round off to nearest integer).

Hint:

To calculate the number of tubes, use the heat duty equation and the log mean temperature difference (LMTD) method to find the required surface area and then divide by the surface area per tube.

Answer 1

Correct Answer: 21

The heat duty equation for a heat exchanger is: \[ Q = U \cdot A \cdot \Delta T_m, \] where:
- \( Q \) is the heat duty (given as \( 15 \times 10^5 \) BTU/hr),
- \( U \) is the overall heat transfer coefficient (100 BTU/hr-ft²),
- \( A \) is the surface area of the tubes,
- \( \Delta T_m \) is the log mean temperature difference.
The log mean temperature difference \( \Delta T_m \) is calculated as: \[ \Delta T_m = \frac{(T_1 - T_2) - (T_3 - T_4)}{\ln \left( \frac{T_1 - T_2}{T_3 - T_4} \right)}, \] where:
- \( {T_1 = 175 \, }^\circ \text{F} \) (hot fluid inlet),
- \( {T_2 = 100 \, }^\circ \text{F} \) (hot fluid outlet),
- \( {T_3 = 75 \, }^\circ \text{F} \) (cold fluid inlet),
- \( {T_4 = 85 \, }^\circ \text{F} \) (cold fluid outlet).
Substitute the values: \[ \Delta T_m = \frac{(175 - 100) - (75 - 85)}{\ln \left( \frac{175 - 100}{75 - 85} \right)} = \frac{75 + 10}{\ln \left( \frac{75}{-10} \right)} {\approx 70.6 \,}^\circ \text{F}. \] Now, calculate the heat transfer surface area required: \[ Q = U \cdot A \cdot \Delta T_m \Rightarrow A = \frac{Q}{U \cdot \Delta T_m} = \frac{15 \times 10^5}{100 \cdot 70.6} = 212.6 \, \text{ft}^2. \] The surface area per tube is given as 0.26 ft²/ft, and the tube length is 40 ft. The surface area per tube is: \[ A_{\text{tube}} = 0.26 \times 40 = 10.4 \, \text{ft}^2. \] Finally, the minimum number of tubes required is: \[ \frac{212.6}{10.4} \approx 20.5 \Rightarrow \text{round to nearest integer: } 21 \, \text{tubes}. \] Thus, the minimum number of tubes required is 21.