Question

An adiabatic pump of efficiency 40% is used to increase the water pressure from 200 kPa to 600 kPa. The flow rate of water is 600 L/min. The specific heat of water is 4.2 kJ/(kg°C). Assuming water is incompressible with a density of 1000 kg/m³, the maximum temperature rise of water across the pump is \_\_\_\_°C (rounded off to 3 decimal places).

Hint:

To calculate the temperature rise across an adiabatic pump, use the first law of thermodynamics and account for the efficiency of the pump.

Answer 1

The work done on the water in an adiabatic pump can be calculated using the following relation: \[ W = \dot{m} \cdot \left( \frac{P_2 - P_1}{\rho} \right) \] Where:
\( \dot{m} = \rho \cdot Q = 1000 \cdot \frac{600}{60} \, {kg/s} = 10 \, {kg/s} \),
\( P_2 - P_1 = 600 \, {kPa} - 200 \, {kPa} = 400 \, {kPa} = 400 \times 10^3 \, {Pa} \),
\( \rho = 1000 \, {kg/m³} \).
The real work done on the water is: \[ W_{{real}} = \frac{W_{{ideal}}}{\eta} = \frac{10 \cdot 400 \times 10^3}{0.4} = 10^7 \, {J/s} \] The temperature rise \( \Delta T \) is given by: \[ \Delta T = \frac{W_{{real}}}{\dot{m} \cdot c} \] Substitute the values: \[ \Delta T = \frac{10^7}{10 \cdot 4.2 \cdot 10^3} = 38.5 \, {°C} \]