Calculating the cost of motor starting

[rockman7892]

I wanted to calculate what kind of cost was associated with starting some of our larger motors here in the plant. The best way to do this was to first equate a kWH rating for the motor starting and then work this number into my electrical rates.

To get a kWH rating for the motor starting, I calculate a kilo-watt second kWS value for the motor starting and then convert this to a kWH value. The following is an example I did.

2300hp motor at 4.16kV

FLA=312A
LRA= Dont have exact value but will estimate at 8x FLA =2496A
Acceleration time (Acc_time) = 10s

kWS used during start = (V)(LRC)(Acc_time)(1.73)

Note: I perform this calculation example ignoring PF for
the time being and assume all current is real.

Therefore kWS = (4.16)(2496)(10)(1.73) = 179632 kWs

Convert kWs to kWH = (179632kWs/1)(1m/60s)(1h/60m)=49.89kWH

We do not get charged a PF penalty however we do have a kW demand charge. Averaging out the kWH rate and kW demand charge we get charged about .075cents/kWH.

Therefore: 49.89kWH * .075 c/kWH = $3.75

Does $3.75 sound right for starting this large motor or am I missing something here?

The only thing I am not factoring in, is how this start effects the demand charge for the rest of the monthly kwH however ignoring this for a second and looking strictly at cost of motor starting is this calculation method correct?

 

[waross]

Hello Jmadamek;
The old meters had a thermal element that took about 15 minutes to stabilize with a step change in load. You are correct in that the thermal element acted mechanically on the indicating pointer. The indicating pointer stayed at the highest reading that occurred during the month. The pointer indication was noted and then was reset and resealed at the same time that the KWHr reading was taken.
The thermal characteristics were such as to give a "front end loading to the reading. I a load persisted for 15 minutes the meter would register substantially the total demand. If a load persisted for 3 minutes out of 15 minutes, the meter would still register 67% of the increase of the demand. That is, if a 10 Kw load was added, in three minutes the demand registration would show an increase of 6.7 KW. (Or KVA if the demand is based on KVA).
Hi pablo51;
The demand is set by the 15 minute window that has the highest average reading. The window may be fixed or rolling and the average may be linear or weighted.
If the motors are pumps or some similar load, you may be able to make a difference to the demand charge by starting 5 or 10 in rapid succession. With a load such as a conveyor that starts unloaded, starting more motors in rapid succession while the running motors are unloaded may present less demand than the group of motors will when they are running loaded. And there are a lot of "It depends".
KVA or kW demand, power factor, power factor correction, the ratio of the power to drive an unloaded belt versus the power to drive a loaded belt.
Don't forget. Demand is only a factor if the plant is already at the previous maximum demand.
Based on this, it is possible that only the last motor of a rapidly started group will impact the demand, and possibly not then depending on the disposition of other machinery in the plant.
Demand only costs you when it exceeds the previous peak for the month.

Bill
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"Why not the best?"
Jimmy Carter