kW Consumption Calculations

Your method is correct. You need to check the actual amperage and voltage and calculate again. Verify the operation hours and days running. The motor may not be running full load at all times.

Ancillary equipment often gets overlooked, in a large plant just the lighting can account for a considerable amount. Just one building I worked in had 30KW of lights burning day and night, that was just one of over 30 buildings.

I don't know if this will help but when we provided Solar power to the pumping stations, we were told to buy very big and very expensive, DC to AC Inverters, to change the Solar power DC supply to 240v AC, to run the AC motors. These Inverters cost 3 times more than DC motors would cost. So to save money, we bought DC motors instead and ran them directly from the DC Solar Power supply.

The Interesting part, is that the DC motors now pumped 300% MORE water, than the AC motors of the same wattage. So we were able to SAVE 70% of the power needed.

This would indicate that you should save Electricity cost, if you change to DC motors. BUT if you only have AC power supply available, then you would have to convert this to DC, which may have high losses in the conversion. I would guess that you would still save energy with DC motors, even after conversion losses. (OR you could buy some Solar Power systems, to save the conversion losses and have NO electricity bill at all)

The formula is correct.But as suggested by forum members, u have to take the actual voltage and PF and current and calculate.Other option is to take a calibrated watt hour meter and compare the readings of both meters.

make your connections and start your motor.when in normal operation,measure your voltage input,operating current and with the pf,calculate the KW=√3VIcosφ.

For the energy(kwh) consumption,multiply by hours run.

Actually speaking in √3VIcosΦ formula all were variables except 1.732. Hence the thereotical calculation you do with right formula by calculating the current and voltage at a instant may give you a margin of actual figure only.

Are you taking the motor NAMEPLATE amperes or are you MEASURING the amperes? The same for voltage, are you ASSUMING the voltage or are you measuring it? And even if you measured, does it change during the day / week / month? Most likely all of those things change constantly. If you go out and measure on one day at one moment, you are only taking a snapshot of consumption, not a full picture.

Think of it this way: if you stood on the side of a road with a radar gun and measured speed of a single passing car at noon on a weekend, does that tell you anything about the average speed on that road? What if you had taken a single radar reading at commute time on a week day? Is that any more indicative of the average speed?

If you really want to determine consumption, you must determine an avaerage consumption periodic cycle length and measure that entire cycle. Anything else is just a snapshot and should not be used to determine anything important.

By the way, your calculation is also missing the motor efficiency, and that too will vary by loading, voltage etc.