transformer inrush

[don_resqcapt19]

I asked the supplier for the inrush currents for a 30 KVA transfomer so I could make sure the primay OCPD would not trip from the inrush current. They sent me the following sheet.

TYPICAL CALCULATED INRUSH CURRENT

MODEL NO. 9T83B3872G01

RATING 3PH 60HZ 30.0KVA 480 +2 -4(2.4%TAPS) 208Y/120


INPUT VOLTS 480.00 480.00
FREQUENCY - HZ 60.00 60.00

1. PEAK AMPERES MAXIMUM

CYCLE 1 460.66 460.66
CYCLE 2 308.84 308.84
CYCLE 3 222.40 222.40
CYCLE 4 168.15 168.15
CYCLE 5 131.74 131.74
CYCLE 6 106.08 106.08
CYCLE 7 87.29 87.29
CYCLE 8 73.11 73.11
CYCLE 9 62.14 62.14
CYCLE 10 53.47 53.47

2. AMPERES AVERAGE MAXIMUM

CYCLE 1 134.93 134.93
CYCLE 1 THRU 2 103.61 103.61
CYCLE 1 THRU 3 83.64 83.64
CYCLE 1 THRU 4 69.86 69.86
CYCLE 1 THRU 5 59.83 59.83
CYCLE 1 THRU 6 52.22 52.22
CYCLE 1 THRU 7 46.27 46.27
CYCLE 1 THRU 8 41.50 41.50
CYCLE 1 THRU 9 37.59 37.59
CYCLE 1 THRU 10 34.34 34.34

3. INTEGRAL OF I SQUARED

CYCLE 1 816.3 816.3
CYCLE 1 THRU 2 1111.3 1111.3
CYCLE 1 THRU 3 1240.0 1240.0
CYCLE 1 THRU 4 1303.6 1303.6
CYCLE 1 THRU 5 1338.1 1338.1
CYCLE 1 THRU 6 1358.1 1358.1
CYCLE 1 THRU 7 1370.4 1370.4
CYCLE 1 THRU 8 1378.2 1378.2
CYCLE 1 THRU 9 1383.5 1383.5
CYCLE 1 THRU 10 1387.1 1387.1
PREPARED BY: Prajwal DATE: MAY 30, 2002

What is the information in the last table, "integral of I squared" used for?

 

[gar]

121220-1402 EST

Don:

Almost certainly it is a measure of energy.

i²*R = instantaneous power

Integrate this over time, not stated that the integral is relative to time, but a good assumption. The units are not specified, nor waveform being integrated, and exactly what they really do in their integrating process.

The table shows increasing values as you would expect, since the value is the sum of all previous values with the addition of the latest cycle's contribution. To a large extent a thermal breaker or fuse over a short time period will have a tripping point in time related to the energy put into the thermal element. Over long time periods heat loss from the thermal element relative to average power input is the determining factor, but for short times it is primarily input energy to the thermal element.

I can not tell you how to relate the numbers in the table. These have to be correlated with the characteristics of the breaker or fuses.

.

 

[G._S._Ohm]

A 60 Hz cycle is 0.017 seconds so
IIRC 461^2 x 0.017 is an I^2T rating of 3613 amps-squared-seconds. The CB needs an I^2T rating of greater than this. Use peak with this short time because the breaker responds to peak values but is calibrated in RMS amps. Do the same for the other cycles and make sure that no cycle I^2T intersects the lower edge of the CB trip curve.

The last table may be the area under the amps vs. time sine wave as seen on a scope.

 

[G._S._Ohm]

If it helps, I plotted 1 vs. 2, 1 vs. 3 and 2 vs 3 and they are all linearly related, with 3 being inversely linearly related to 1 & 2.

Using Excel's SLOPE and INTERCEPT function you can get the equations relating these variables.

But I'd ask the maker for an explanation.

 

[G._S._Ohm]

Apparently the integration is used to find the harmonic contribution to the inrush current.
http://xa.yimg.com/kq/groups/233494...rush+current+in+a+single-phasetransformer.pdf

 

[don_resqcapt19]

I guess my real question is if a breaker with a 600 amp trip point that is vertical on the time trip curve to 0.2 seconds will withstand this inrush without tripping?

 

[G._S._Ohm]

I guess my real question is if a breaker with a 600 amp trip point that is vertical on the time trip curve to 0.2 seconds will withstand this inrush without tripping?

This kind of question has always nagged me. It concerns the energy dumped over time into the breaker, which will heat it.
10 cycles is 0.17 seconds, so if the RMS amps of each cycle stays less than 600A RMS then I'd say, He!! no, it won't trip.

List #1 probably represents an exponentially damped sine wave and so the energy contained under that curve can be calculated for 1 to 10 cycles by integration. With your breaker I see no need.

I think if each cycle was at 600A RMS and we went 10 cycles, it might be close.

 

[G._S._Ohm]

This is still bugging me.

Using this curve as a reference and assuming 10-50 temp and a 1A breaker.


http://www.geindustrial.com/publibrary/checkout/GES-6119D?TNR=Time Current Curves|GES-6119D|generic


At time T= 0 I put in a current pulse of 10A and it lasts 0.3 sec.
Log scales have no zero so this pulse first registers at 0.01 seconds and 10A on the graph.
This pulse heats up the thermal part of the breaker but the pulse is soon gone and so the breaker cools off again.
Q: All agreed this will not trip the CB?


At time T= 0 I put in a current pulse of 1A and it lasts 1000 sec.
This pulse heats up the thermal part of the breaker but the pulse is eventually gone and so the breaker cools off again.
Q: All agreed this will not trip the CB?


Now I have an inrush current vs time curve for a huge machine or transformer that gives me 10A @ 0.3 sec., 7A @ 0.8 sec., 4A @ 3 sec., etc.
Q: Since the breaker is 'preheated' will it eventually trip even though no part of this decaying current is above the trip curve lower boundary?

 

[jim dungar]

Since the breaker is 'preheated'

Breaker curves are drawn assuming all of the internal parts of the breaker are normal operating temperature. The internal parts of smaller ampere ratings (like your curve) have such a low thermal mass, there is effectively no difference between cold and hot operating times. For breakers that have not reached their stable operating temperature, you might be able to use the adjusment factors for non-standard ambient ambients, but it is rarely ever worth the timeand effort.

 

[hurk27]

Don is this for utility or the NEC side?

I ask because 30kva @ 480 3ph is 36 amps, and @250% is 90 amp breaker max per table 450.3(B) if secondary protection is used, if not then your limited to 125%, but in your last post you said a 600 amp breaker?:?

 

[hurk27]

If the vertical line in the trip curve is at x12 as it is in the GE diagram that will give you a 40 amp breaker to meet the 460.66 amp peek in one cycle.

460.66/12=38.38 amps
@ 10x you will be at a 50 amp breaker with a trip curve of 24 cycles well out of the inrush range.

36 amps @ 125% is 45 amps next size up is a 50 which will put your inrush handleing at 500 amps which is above the one cycle rating the manufacture gave you.

 

[__dan]

I guess my real question is if a breaker with a 600 amp trip point that is vertical on the time trip curve to 0.2 seconds will withstand this inrush without tripping?

Inrush in a transformer is a fast transient event in the first cycle or two, closing into an unloaded transformer at a particular part of the cycle, and the transformer is very close to the source busbar. If the breaker and transformer have a 100+ ft or so of feeder, if the closing is at a good part of the cycle, or if the transformer has a load on the secondary, all those factors reduce inrush.

Peak current will be in the first quarter cycle, or first 4.16 milliseconds, when the primary is not generating counter EMF and the winding is closer to a short than a load. This is shown in table 3 where the first two cycles have a high intergral I squared but that number does not go up a lot for the immediately following cycles. In this range for this problem, you would be concerned with the instantaneous magnetic trip setting and not a short time delay thermal trip in the 1 to 10 second range.

Peak current in table one at 460 amps probably corresponds better to the instantaneous trip rating of the breaker but this would have to come from the manufacturer to confirm this view. You are saying the breaker trip curve is ~ 10 times the handle rating, horizontal for the first 200 milliseconds (12 cycles) at 600 amps.

Intergal I is the area under the I curve and will give you an average per unit time or total current over the total time period, which would be for thermal effects. I squared would be used to make the negative part of the cycle positive and sum to the aggregate rather than cancel as half the cycle is negative.

If the transformer is right next to the switchgear and there is nothing to limit inrush, closing the breaker into an unloaded transformer can trip on instantaneous magnetic trip. For this (nuisance tripping) code may allow a larger breaker.

 

[don_resqcapt19]

The installation is for a transformer on the load side of a UPS maintenance/bypass switch. The breaker supplied by the UPS manufacturer will have a 40 amp rating plug. Here is the breaker curve. I am looking to see if I need to go up the the 50 or 60 amp plug.

 

[__dan]

The installation is for a transformer on the load side of a UPS maintenance/bypass switch. The breaker supplied by the UPS manufacturer will have a 40 amp rating plug. Here is the breaker curve. I am looking to see if I need to go up the the 50 or 60 amp plug.

In the 25 millisecond range, ~ 1.5 cycles, the note 2 trip rating is the same for the 40, 50, and 60 amp plugs at 8727 amps. It looks like a pretty good breaker with an electronic trip. As long as the first 25 milliseconds does not exceed the current limiting trip point > 8000 amps, the note 2 says the breaker will hold closed.

If it's critical load that is always on, load on the secondary reduces inrush below the theoretical maximum. If the UPS is remote from the transformer, the length of the feeder also reduces inrush. Available fault (inrush) current at the load side of the UPS will be limited at the output to somewhere ~ 200% of the UPS rating for ~ 2 seconds, the UPS downstream fault clearing rating. On bypass, available fault current will be much higher.

I see the 30 kVA, 480 v, primary line amps at 36 amps. 125% is 45 amps making the 50 amp trip plug applicable. If the tranny is close to the load in conditioned space, the 50 could be small. The concern in IT space would be carrying the load when a redundant, B, source trips and the load swaps to the A cords. The bigger breaker is a plus when that happens, it's the weak link.

 

[don_resqcapt19]

Dan,
Thanks for the information and help.

The UPS is rated at 20 kva so the normal load is not really an issue, just the inrush. The power source is from a MCC about 50 wire feet from the transformer. The MCC feeder is parallel 350 kcmil ~140' long. The feeder breaker is in switch gear that is connected via bus to 4160 to 480 volt transformer. I don't remember the KVA of that transformer.
If the UPS sees a short circuit on the load side, it will switch to bypass because it can't source enough current to quickly open the OCPD.

 

[__dan]

Dan,
Thanks for the information and help.

The UPS is rated at 20 kva so the normal load is not really an issue, just the inrush. The power source is from a MCC about 50 wire feet from the transformer. The MCC feeder is parallel 350 kcmil ~140' long. The feeder breaker is in switch gear that is connected via bus to 4160 to 480 volt transformer. I don't remember the KVA of that transformer.
If the UPS sees a short circuit on the load side, it will switch to bypass because it can't source enough current to quickly open the OCPD.

It looks like a question for the UPS manufacturer. When the UPS output sees a transient high current fault, will it attempt to clear the fault, have a fault clearing mode, where it tries, say, 200% output for 2 sec, or will it fail and switch to bypass. Something good to know before it happens unplanned with the load on it. Or if it does happen, why.

If the tranny were connected right at the service with 10 ft of wire, inrush could be a problem on bypass. With the tranny remote from the service, the feeder impedances add up to limit inrush to the point where nuisance breaker tripping should not be a problem.

 

[don_resqcapt19]

Thanks again Dan,
Have a Merry Christmas

 

[__dan]

Thanks again Dan,
Have a Merry Christmas

Your welcome Don.

Have a good Christmas.