the KWH is unit for a energy.for calculating the power we use this formula p=root3 vl il cos(angle between voltagecurrent).
I think the consumption should be KWH=1.73 x V x I, thus the losses should be included.. your meter reads Apparent power not Real power..
Your calculations are almost correct. The startup current can be neglected, since it should only last less than a few seconds. You should use a power factor, which is dependent on the motor specifications and its load level. Typical values would be about 0.75 - 0.85 for this small motor. You don't mention if the voltage is measured line-to-neutral, but this is a reasonable assumption in the U.S. Your calculation of daily energy consumption sounds about right for a small motor (about 2.5 hp) running for only about a half hour per day.
Power = voltage x current In a motor you have to factor in the power factor (1 or less) and the effects of the 3-phase wiring. In 3-phase you have the current divided across the phases. Take the total current and multiply by the square root of 3 (about 1.7), you will also need the power factor (if you don't have it, use 1). Use the formula (KW = 1.73 * volts * amperes * power factor/1000). KWH is the power in KWs over time.
