Suppose I would use Bernoulli equation to compare point 1 before the pump and point 2 after the pump:
$$ \frac{P_{1}}{\rho g} + \frac{v_{1}^2}{2g} + z_{1} + \Delta h_{p}= \frac{P_{2}}{\rho g} + \frac{v_{2}^2}{2g} + z_{2} + \Delta h_{f} $$
By the continuity equation, $$ v_{1} = v_{2} $$
Moreover, neglecting the pressure drop across the pump, $$ P_{1} = P_{2} $$
So, the previous equation becomes:
$$ \Delta h_{p}= \Delta h_{f} $$
If this analysis is correct, does it show that, when a pump is not needed (no energy input would be needed to overcome any fittings or friction in the pipe), the head that it produces is the same as the possible friction that it produces? I'm imagining this situation as if the pump stops the fluid and accelerates it again.
Thanks!