Calculating the cost of motor starting
Calculating the cost of motor starting
(OP)
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?





RE: Calculating the cost of motor starting
If you knew the load profile and could plot at least one other point, you could maybe plot a curve graph with acceleration time and kW, then calculate the area under the curve to get kWh. But all in all it is very difficult to get an accurate number. Voltage drop and efficiency will also play parts in this as well.
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RE: Calculating the cost of motor starting
Should be 3597kW at initial start!
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RE: Calculating the cost of motor starting
Jaref
I see what you are saying about the power factor changing. Lets say that with ignoring it that I want to calculate the worst case scenario, or most expensive case. In other words ignoring pf am I doing the calculation right for this?
Somebody is trying to tell me that it cost thousands of dollars each time we start one of our 5kV motors. Finding this hard to believe I am venturing out to try to calculate an approximation to prove this right or wrong. I figure if I used the worst case as mentioned above I can prove that the thousands of dollars may not be the case.
What do you think? Does my $3.75 seem right for the starting of a large 5kV motor with the assumptions I mentioned?
RE: Calculating the cost of motor starting
Whomever was saying that was just plain ignorant. Another case of someone trying to lay on an energy savings scam I'll bet. I have seen people selling "energy savers" or pf capacitors trying to make ridiculous arguments like this to management types and bean counters, implying that just turning a motor off may end up costing you more than leaving it running with their device connected. Absolute bunk.
Bravo for not taking it at face value and looking in to it.
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RE: Calculating the cost of motor starting
Multiply that times your acceleration time.
Your demand meter will not much respond to the starting surge. Use your avearge energyy cost.
Reality check.
The ratio of starting Kw and running kW is about 2:1 (that's why you have to consider starting PF.)
That means that 10 seconds of starting costs about as much as 15 seconds of running. From this prospective do you want to spend more time getting a more accurate answer?
Second reality check: Have I missed anything jraef?
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Calculating the cost of motor starting
2496A * 4160V * 1.73 = 17,963 locked rotor KVA
If acceleration is 10 seconds then this becomes 2994 KVAh (17,963 * 1/6)
It would be closer to use jraef's starting PF to get KW because the high inrush current is used in magnetizing the motor; not to do work which is what you are billed for. The demand doesn't register on the utility's meters because demand is a time averge of KW (usually 15 or 30 minutes).
2994 * 0.2 = 599 starting KWH * $0.07 = $41.93.
Using locked rotor KVA yields $1257.41 which is probably why your boss thinks it costs that much.
RE: Calculating the cost of motor starting
RE: Calculating the cost of motor starting
RE: Calculating the cost of motor starting
49.9 * 0.2 = 9.98 KWh * $0.07 = $0.70
Using the KVA would be $3.49 so your initial calc was right.
Tell that "somebody" to mind their own business.
RE: Calculating the cost of motor starting
If someone thinks big money can be saved in their power bill by reducing the number of motor starts, they are usually wrong.
RE: Calculating the cost of motor starting
Too frequently...
Keith Cress
kcress - http://www.flaminsystems.com
RE: Calculating the cost of motor starting
I guess another way could be if the motor or load experiences a large vibration as it accelerates causing vibrational damage.
RE: Calculating the cost of motor starting
(On the flip side, people who complain about how long it takes to recharge the batteries of an electric vehicle have never done the math to figure out how much current it would take to do this in the same time frame as filling your gas tank.)
Curt Wilson
Delta Tau Data Systems
RE: Calculating the cost of motor starting
Thanks for all the responses everyone.
At our plant alot of time mangagment especially the production manager tries to stick his nose in the electrical aspect of things not really knowing what he is talking about. Being a farily new EE I dont have my head completely around everything to usually challenge him on the spot, so I like to go do the calculations for myself and run it by others like yourselves before I go and stick my neck out and present the true facts to support my argument.
We just had PF correction fuses blow on a motor and it took me weeks to convince the same manager that we were not loosing hundreds of thousands of dollars a year by not having these caps in service especially when we dont have any sort of PF penalty. The only way I can usually convince these folks is by doing a full blown calculation example. I am going to now have to present this calculation example to him to show that sarting this motor cost about $5 and not $5,000 like he was going around claiming. I dont know where he gets his info.
Having said all this I just have a few other questions:
1) It has been said by some that the utility metering equipment doesn't even react quick enough to respond to starting currents. If this is the case then does the starting current even have a greater than normal cost associated with it at all, or is this starting current happen to quick and therefore becomes irrevelent when averaged with the rest of the motor run time?
2) When the utility meters calculate a max kW demand charge do they typically take of highest average of 30min windows or so. For example if my plant is running at 100kw for example and then I start a motor causing the max kw to be 200kW for 10s before it drops back down to 125kW (just arbitrary numbers) do I get charged as my max demand being 200kW or is all of this averaged to calculate some max demand?
3) It sounds like with all of this being true the starting current is almost negligabe with respect to any sort of cost?
RE: Calculating the cost of motor starting
2) As mentioned several times in the responses, the demand charge is based on the maximum AVERAGE kW over a fixed period, typically 15 minutes or 30 minutes.
3)Basically true, in my experience.
RE: Calculating the cost of motor starting
This said, it would be hard to put a number on it. It would probably need to be rationally guessed at and would include the; duty cycle, the ambient, the type of load, the starting method, the typical temperature of the motor just before a start, the voltage levels, and the type of motor.. As you can see a guessing game.
Keith Cress
kcress - http://www.flaminsystems.com
RE: Calculating the cost of motor starting
A 10 second start time will have some small impact on a 15 minute average. It depends. Is the utility billing demand based on kW or KVA? Is the demand measured by thermal curve, average, rolling window etc.
A KVA demand on a motor start may be several times the kW demand in the first part of the 10 seconds. Say 5 seconds out of a 15 minute average.
I get a ratio of one part in 180.
Break your starting value (kW or KVA or both for comparison.) into 1 second windows and see what effect each has on a 15 minute average.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Calculating the cost of motor starting
RE: Calculating the cost of motor starting
RE: Calculating the cost of motor starting
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
--------------------
"Why not the best?"
Jimmy Carter