Motor current calcs

[mcguirepm]

I have been handed a project that is outside my comfort zone.

A 3-phase 2hp motor that drives a conveyor stops and starts frequently to control product flow. Each time the motor starts the amp draw spikes to approximately 6 x FLA for say 1/2 second. It then levels out to around 1/2 FLA for the rest of the cycle. If the cycle is 1 minute long then I have 60 stop-starts per hour.

What I need to find out is how much does that spike cost in terms of 'demand charges' from our utility. I can calculate the normal energy usage, but I am clueless on how to calculate the kWh for these momentary spikes.

Any help would be appreciated.

 

[dpc]

This is generally far less of an issue than commonly believed. The utility demand charge is based on the maximum 15 min or 30 min average power demand of the preceding month (or possibly 12 months). This is computed in the meter using a rolling window (normally).

Even when the motor is starting every minute, this peak current is only a second or so.

In addition, the demand charge is generally based on peak kW, not current. A starting motor has high current but a very low power factor - maybe 0.2, until it gets up to speed.

And this is a 2 hp motor, so it's all much ado about almost nothing anyway. Even if it was at locked rotor current for the full 15 minutes, the demand would be maybe 3 or 4 kW or so.



David Castor

http://www.cvoes.com

 

[electricpete]

Good points by David. It’s hard to get excited about the energy used during starting.

What I would be more interested in is the wear and tear on the motor. One reference put out by NEMA:

http://www.joliettech.com/allowable_starts-intervals.htm

2hp motor has max starts per hour of 11.5 for 2-pole, 23 for 4-pole, 26 for 6-pole based on examination of column A. And possibly less based on examination of inertia using column B. And rest time between starts per column C.

I realize 60 per hour may have been a “talking number”, but still something to watch out for.


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[DRWeig]

McGuirePM,

Your utility meter will average demand over some time period, usually 30 minutes but sometimes 15. You won't see the entire starting power of the motor on your demand bill. If it's really only 0.5 seconds at 6X (a lot of which is reactive VA not watts that you pay for) per minute, then per minute you'll have 0.5 seconds at 6x plus 59.5 seconds at 1X or an average of 1.04X for the whole minute. Taking start-up power factor into account, it'll be much less.

You could meter your motor for an hour with a kWh meter, divide by 60 minutes, and get a good approximation of the demand.

Maybe someone else has a good example...

Good on ya,

Goober Dave

 

[LionelHutz]

A 2hp motor might have about 10 to 12kVA inrush. You indicate this is one motor of a whole process. I'm betting your demand changes are much higher than 12kVA per month meaning this motor makes very little difference.

At any rate, Dave has described it above. The motor will be say 12kVA inrush for 0.5 seconds and around 0.9kVA (1/2 the motor rated 1.8kVA) for 59.5 seconds. On average this gives around 1.09kVA that will be contributing to the demand charges.

If you only pay for demand kW then running at 1/2 current means you are at less than 1/2 motor load. So, your motor will be contributing less than 746W to the demand charges.

Others have pointed out that you only pay for the demand kW but this is not true. Often, the demand for an industrial customer is based on the kVA when they have a poor power factor.

 

[DRWeig]

Lionel's right, I often don't think about industrial customers having KVA billing -- but it is available from most utilities.

You should check your rate schedule before coming to any conclusions.

Double good on ya,

Goober Dave

 

[waross]

For one, 2HP motor, the cost of your time and a reasonable rate for the responders will have possibly been equal to or greater than the starting demand charges for your motor for a number of years into the future.

Bill
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"Why not the best?"
Jimmy Carter