Question

A crude oil with a flowrate of 1000 kg/hr is to be cooled using water in a double-pipe counter-flow heat exchanger from a temperature of 80\(^\circ\)C to 40\(^\circ\)C. The water enters the exchanger at 20\(^\circ\)C and leaves at 40\(^\circ\)C. The specific heat capacities of the oil and the water at constant pressure are 2 kJ kg\(^{-1}\) K\(^{-1}\) and 4.2 kJ kg\(^{-1}\) K\(^{-1}\), respectively. The overall heat transfer coefficient is 0.25 kW m\(^{-2}\) K\(^{-1}\). Neglecting the heat loss and using the log mean temperature difference (LMTD) method, the minimum heat exchanger area (m\(^2\)) required for the operation is ____________ (rounded off to two decimal places).

Hint:

For counter-flow heat exchangers, always calculate LMTD using both terminal temperature differences before applying the area formula.

Answer 1

Correct Answer: 3

Heat removed from oil: \[ Q = \dot{m}_o \, C_{p,o} \, (T_{hi} - T_{ho}) \] \[ Q = 1000 \times 2 \times (80 - 40) = 80{,}000\ \text{kJ/hr} \] Convert to kW: \[ Q = \frac{80{,}000}{3600} = 22.22\ \text{kW} \] Temperature differences (counter-flow): \[ \Delta T_1 = (80 - 40) = 40^\circ C, \quad \Delta T_2 = (40 - 20) = 20^\circ C \] LMTD: \[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{40 - 20}{\ln(40/20)} = \frac{20}{\ln 2} = 28.85^\circ C \] Heat-exchanger area: \[ A = \frac{Q}{U \times \text{LMTD}} = \frac{22.22}{0.25 \times 28.85} \approx 3.08\ \text{m}^2 \] Thus, the required heat-exchanger area lies between 3.00 and 3.15 m\(^2\). Final Answer: 3.00–3.15 m\(^2\)