Hi,

Even if it a is an old thread, it in important to note the Kre's comment: The power factor is a very important value to know the real power taken by a load with windings inside. These windings have a characteristic named inductance that modifies the simple formula used by Kanoa50. To correct this formula we need to use a factor named Power Factor.

When we use a DC supply, in a passive circuit (no transistors or electronics), the current is stable in a very short time. With no voltage variations, no current variations.

When an alternating current is used, there is a continuos variation in the voltage applied (60 times per second) and of course in the current (Amps) circulating. In the case of a pure resistive load, the voltage variations are followed by proportional current variations at exactly the same time. Voltage and current are in phase.

When the load contains coils or windings (as in motors or transformers) and are fed with AC, these coils react sending a voltage in the opposite direction, reducing the effective voltage applied. This reaction takes a time so the current increases a little after the voltage raises. It is the current lag with respect to the voltage and this value in a trigonometric representation forms an angle whose cosine is named cos φ (cos Phi)or power factor.

This PF varies in a motor or transformer according to the load applied. The cosine function varies from 0 to 1, but in the common electrical loads it is normal to found values between 0.4 to 0.95 (+or-) and is the factor that reduces the Volts x Amps product. At less load, less power factor.

Volts x Amps = Apparent power in VA or KVA

Volts x Amps x cos φ = real power Watts or KW.

In this case, assuming a PF = 0.8

Apparent power: 240 V x 12 Amp = 2880 VA or 2.8 KVA

Real power: 240 V x 12 Amp x 0.8 = 2304 Watts or 2.304 KW

But...This is the real power absorbed by the motor, but this motor is not perfect, it dissipates heat, electromagnetic waves, mechanical vibrations, noise etc etc. All this energy waste comes from our incoming 2304 Watts, so the mechanical output in the form of a rotational force applied to the pump's shaft can't be the same as the input power. The relationship between the output power and the input power is the efficiency and it will never reach the unit (100%).

In this case assuming an 80% efficiency :

Input power x Eff. = Power in the shaft

2304 Watts x 0.8 = 1843 Watts

1843 Watts / 746 W/Hp= 2.47 HP

This is only an approximation, assuming PF and Eff. This values change ,as Kre explains, with the flow of water (GPM) and the pressure (PSI). Every pump has its own set of curves that determine most of the variables found on it.

Regards