Answer
$P_{f}=6.757kW$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$
The potential power needed is:
$P_{p}=1000\frac{kg}{m^3}*(0.03\frac{m^3}{s})*9.81\frac{m}{s^2}*45m=13.243kW$
As the pump provides $20kW$, the mechanical power lost because of friction effects is:
$P_{f}=20kW-13.243kW=6.757kW$
