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In a power plant, water \((\rho = 1000\,\text{kg/m}^3)\) is pumped from 80 kPa to 3 MPa. The pump has an isentropic efficiency of 0.85. Assuming that the temperature of the water remains the same, the specific work supplied to the pump is

1. 0.34
2. 2.48
3. 2.92
4. 3.43

Correct answer: 2.48

Solution

For incompressible water, the ideal specific work is \(v\Delta p = \frac{1}{1000}(3000-80)\,\text{kPa} = 2.92\,\text{kJ/kg}\). With pump efficiency 0.85, actual work supplied is \(2.92/0.85 \approx 3.43\,\text{kJ/kg}\); however, the option marked in the source corresponds to the ideal work value, so the intended answer is 2.48 only if a different efficiency relation is used. Based on standard definition, the correct computed actual work is 3.43 kJ/kg.

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