 # Kw/Day Calculations

3 replies to this topic

### #1 Mohammad.tahseen

Posted 29 March 2006 - 12:52 PM

Dear Sirs,
I have one question and I hope that you can help me! In My site I have 4 KV Motor with 650HP and 0.92 Power Factor, and I want to calculate the Kw/Day for the motor but I dont have the formula to calculate KW/Day. And another question that is the power rating which is written on the name plate (650HP) is it the same (650 HP/Hr)? And if not, what is the formula to find the HP/Hr from the rated power on the name plate?

I Hope that you will answers my doubts.

Many Thanks,

### #2 marke

marke
• Gender:Male
• Location:Christchurch, New Zealand

Posted 29 March 2006 - 06:38 PM

The rated power of the motor (KW or HP) is the rated output power of that motor. When the motor is operating at full load, it should be producing that nameplate power out the motor shaft.
The electrical input power will be higher because it will also include the losses in the motor.
If we assume that the motor has an efficiency of say 92%, then the input power will be the output power divided by the efficiency.

In order to calculate the KW/Hr used by the motor, you need to know the power consumed by the motor and this is dependant on the shaft power that the motor is producing. In otherwords, it is very load dependant. There is no magic formula that will give you this information unless you know what the output power is.

If we assume that the motor operates continuously at rated shaft power, then the KWHrs used would be 650 x 0.746 KW / efficiency x number of hours run. Note this assumes continuous operation at full load. This is very rarely the case!! At half load, the KWHrs will be almost halved.

Best regards,

### #3 Mohammad.tahseen

Posted 30 March 2006 - 06:15 AM

Thank you very much for your replay, you cleared my doubts, but when I asked about the rated power is it the same Kw/Hr what I mean, if the motor runs at the full load for one hour, is the Kw/Hr reading will be the same (name plate rated power?, and from what I understood from your replay that the name plate power is the mechanical power and to find the input electrical power we have to divide by the efficiency.
So if I have continuous reading of the current as a feed back to PLC system and already my voltage is fixed to around 4.0KV and I Know the power factor, Can I take one calculation each one minute (or less) for the power (sqr3*I*V*P.F) for one hour and find the average and consider it as Kw/hr then multiply this with working hours for one day to find Kw/Day??

Many Thanks and Best Regards

### #4 marke

marke
• Gender:Male
• Location:Christchurch, New Zealand

Posted 30 March 2006 - 10:11 AM 