To my surprise, when the pump kicks on around 20 psi, my amp draw is close to 15 amps. As the pressure builds, the amp draw decreases to 12.2 amps. The more the pressure, the less the amps. That seems odd to me, I would have thought just the opposite.
Yep this is a non-positive displacement pump, specifically a centrifugal pump. There is no sealed displacement area in this type of device the water is just slung around the volute area, and as pressure builds there is increased slippage back around the pump rotor and volume output drops off.
If you valve off either the input or the output the motor current will drop.
The same thing happens with centrifugal fan or blower blocking off the air flow will lower the motor current.
I have 2 55 mic 440 Volt capacitors. By connecting a capacitor to the end of the white cord, I could plug the capacitor into my 3 ft extension cord and also have the well pump plugged into the same cord. This puts the 55 mic in parallel with the pump and the total current decreased by 1.1 amps. Putting the second 55 mic cap in parallel with the first, my total current went from 12.2 amps to 9.99 amps. I was impressed with my results. Even had a little excitement too, especially after unplugging the capacitors and shorting them. I confirmed the big ban theory.
The lagging current in the induction motor is being PF corrected by the leading current flow in the capacitor, over correction with too large of capacitor will begin to create a leading PF and as the PF shifts from lagging to leading the current will begin to increase as the PF passes through unity.
Separate the leads going to the capacitor and the motor and with your setup use your ammeter to monitor the current flows into and out of the capacitor, the line, and the current flows into and out of the motor.
So, are you not able to power a universal motor?
The torque developing current flows are not induced in the series universal motor but are a result of current flow in the series field and armature circuit therefore the motor current is nearly in phase with the line voltage. So there is no out of phase current to phase correct.
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