It is approximately equal to the power delivered to the pump which is not converted to hydraulic energy.
The hydraulic power that must be converted into energy and delivered to the pumped product is,
P_hydraulic = 144 x flowrate x density x (output_pressure - inlet_pressure)
flowrate ft3/sec
density lbs/ft3
pressures psig
P_hydraulic ft-lbs/sec
The Pump's Efficiency determines how much power must be applied to the pump shaft so it can deliver the hydraulic power into the pumped product. Since some energy is lost during that process, more energy must be applied than is actually converted to energy and used by increasing the fluid's pressure. The pump's efficiency determines how much extra energy is needed.
The pump's Efficiency varies with flow through your pump and can be read from your pump curve.
Power_Required at shaft = P_Hydraulic / efficiency
Power not used is converted to heat, of which most goes into heating the fluid. It is of course equal to P_Required - P_hydraulic.
The units of Power (ft-lbs/sec) not used can be converted to Btu/sec by dividing by 778 Btu/ft-lb
Otherwise, it depends on where you think are good places to obtain your "inlet" and "outlet" pressure data, but if it was an existing installation, I'd just install some temperature gauges and read it directly.
![[worm]](/data/assets/smilies/worm.gif "[worm] [worm]")