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/m\(^3\), 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. The work done on the fluid is related to the pressure difference and flow rate.

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:
\( \dot{m} \) is the mass flow rate, \( P_1 \) and \( P_2 \) are the initial and final pressures, \( \rho \) is the density of water. The mass flow rate \( \dot{m} \) is related to the volumetric flow rate by: \[ \dot{m} = \rho \cdot Q \] Where \( Q \) is the volumetric flow rate. Given:
\( \dot{m} = 1000 \cdot 600 \, {L/min} = 600 \, {kg/min} = 10 \, {kg/s} \)
The specific heat of water \( c = 4.2 \, {kJ/(kg°C)} \)
The efficiency of the pump is 40%, so the work done on the water is: \[ W_{{real}} = \frac{W_{{ideal}}}{{Efficiency}} = \frac{10 \cdot (600 - 200)}{0.4} = 10000 \, {J/s} \] The temperature rise \( \Delta T \) is calculated from the first law of thermodynamics: \[ \Delta T = \frac{W_{{real}}}{m \cdot c} \] Substitute the values: \[ \Delta T = \frac{10000}{10 \cdot 4.2} = 0.139 \, {°C} \] Therefore, the maximum temperature rise is 0.139°C.