All-righty then. The only thing missing is the kinetic energy of getting that flow going. So you get a V^2 * rho / 2 factor that'll raise the input power slightly. I understand where you're trying to get, and practically speaking your diagram "A" matches your previous statement about a 100% efficient pump/motor.
I think more to the point is the actual motor power. Motor nameplates merely state the maximum power the motor will consume. This does not mean that they consume that amount of power irrespective of flow and/or head.
That's the problem with skulemate's data on page 1. He assumes that his motor is drawing 180 watts with ~46 watts showing up as thermal energy within the water. In the graph I posted, the operating point (as indicated by the red right-angle) calls for ~1.45 hp. At other flows/heads, it would list a different number. We use a 1.5 hp motor because it can handle all combinations of flow/head that we use. This does not mean that the motor draws 1.5 hp.
|