A pump transfers shaft work to a liquid, usually to raise its pressure.

For steady, approximately adiabatic flow with negligible kinetic and potential-energy changes,

$$w_{in}=h_2-h_1.$$

If the liquid is nearly incompressible and its temperature change is small,

$$h_2-h_1\approx v(p_2-p_1),$$

so

$$w_{in}\approx v\Delta p.$$

For water with $v\approx0.001\ \mathrm{m^3/kg}$ pumped through a pressure rise of $2\ \mathrm{MPa}$,

$$w_{in}\approx0.001(2000)=2\ \mathrm{kJ/kg}.$$

At $10\ \mathrm{kg/s}$, the ideal shaft-power input is about $20\ \mathrm{kW}$.

Real pump losses increase the required shaft work above this simple reference estimate. The steady-flow energy equation remains the bookkeeping framework for both ideal and real operation.