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?

 

[jraef]

It depends on how accurate you want to be.

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

The problem I see is your ignoring the power factor. When starting, the power factor is very low and the current very high, with the pf increasing and current decreasing as the motor accelerates. This means that your equations for kW are dynamic, not static. So if you have a pf of 0.2 at initial start, your kW at that point is 4.16 x 2496 x .2 x 1.73 = 208kW, but that is a moving target because 10 seconds later, if the pf is .8 and you have dropped to FLA, your kW is 4.16 x 312 x .8 x 1.73 = 1798kW.

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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[jraef]

Dag-nabit, I knew to check my math but forgot to check the first calc. I knew it wasn't right, just forgot to go back and find the error before clicking "Submit".

Should be 3597kW at initial start!


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[rockman7892]

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?

 

[jraef]

Compared to "thousands of dollars", you are absolutely right! The differences in accuracy would be in fractions of a dollar.

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.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[waross]

For a rough and ready number, use the average of the starting kW and the running kW.
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

 

[Laplacian]

You are complicating it a bit more than required.

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.

 

[Laplacian]

Divide my previous calcs by 60

 

[edison123]

Why don't you just use a power analyzer and measure the actual start-up power for each motor ? That kinda "practical" proof would be easier for the bean-counters to "understand".

 

[Laplacian]

17,963 * 10s * 1min/60s * 1hr/60min = 49.9 KVAh

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.

 

[dpc]

I think you're in the ballpark. Running motors is more expensive than starting them. And sometimes NOT starting one can be very expensive..

If someone thinks big money can be saved in their power bill by reducing the number of motor starts, they are usually wrong.

 

[itsmoked]

Or worse, they think they can save money by frequently starting a very large motor.

Too frequently...

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

The only way motor starting could be costly is if the motor is pushed to it's thermal limit during the start meaning the stator windings or rotor bars are basically damaged during each start. So, doing a lot of starts vs leaving it running could mean a premature failure of the motor costing a lot of money. I know this is an indirect cost but it's a cost due to starting the motor all the same.

I guess another way could be if the motor or load experiences a large vibration as it accelerates causing vibrational damage.

 

[cswilson]

This reminds me of the canard I used to hear frequently in offices that the act of turning on a fluorescent bulb consumed as much power as keeping the bulb on for the whole day. I would have loved to see the fuses and wiring to support this, just as I would love to see them for a motor that consumes "thousands of dollars of electrical energy" in a 10-second start.

(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

 

[rockman7892]

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?

 

[dpc]

1) Maybe in the old days, but newer digital meters are sampling the current and voltage multiple times per cycle, so they don't miss much.

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.

 

[itsmoked]

1) There is a cost associated with starting a motor as it is 'harsh service' in a motor's life. Generally the smaller the motor gets the more negligible starting is to it.

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

 

[waross]

A 10 second start time may cost as much as running the motor loaded for 15 or 20 seconds.
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

 

[Jmadamek]

Are you sure the demand is averaged over a time span? I thought the old style meters just had a KW dial with one fixed and one floating pin for demand. So the fixed pin would push the floating pin to whatever max occurred over the time period. Then the meter reader would reset the floater for the next billing period.

 

[pablo51]

You have remember that your demand charge is set by the highest use over (usually) a 15 minute sliding window -- you add up all of the energy used over that period to set the demand... that charge will then apply for the month or sometimes longer.... starting one motor may not cost much, it does adds to your overall use in that given window; but starting several large motors within such a window can significantly add to your demand -- this charge is independent of the energy charge... you should treat them separate