Short Circuit Current at Half Rated Voltage

[Kiribanda]

Gentlemen,

I have got a question on secondary short circuit currents of a centre tapped single phase transformer.

The name plate of a single phase transformer is as follows.

37.5 kVA, Primary: 600 V, Secondary: 120/ 240 V, Z = 5.3 %. Primary FLC= 62.5 A,Secondary FLC=156.25 A.

Therefore, from the definition of short circuit impedance (impedance voltage), if this transformer primary is connected to an infinite bus at 600 V, it will give 2948.1 A as the secondary short circuit current at 240 V terminals. Anybody disagree?

Now what will be the secondary short circuit current at 120 V terminals?

Your inputs are welcome!


Kiri

 

[stevenal]

A precise calculation is impossible without knowing details of how the impedance is divided among the three windings. One rule is to multiply the l-l value by 1.5 to get the l-n value.

 

[dpc]

It's not a straightforward calculation. Conrad St. Pierre has a good treatment of this in his book "A Practical Guide to Short Circuit Calculations".

stevenal's approach is probably a good conservative approximation. Certainly the 120V fault current can be greater than the 240V current.

 

[jghrist]

There is a long discussion of this topic at thread238-74358

 

[cuky2000]

The SC estimated at 120V terminal is almost twice the value calculated for the 240 V terminals.

Enclose is additional information that may help to justify the above statement.

http://cuky2000.250free.com/SC_Single_Ph_Transf.gif

_{PS: The SC current at the end of the secondary feeder will be damped significantly fast with the distance because the relative high impedance of cable considered in both-ways since the current flows to the fault and then returns to the source via the neutral conductor} .

 

[waross]

Hello all.
I'd like to throw in an historical comment on the origin and reason for these calculations. Remember, these are calculations to assure that consumer equipment is capable of interrupting fault currents safely.
We are not calculating asymmetric faults for the coordination of a power grid.
Even though the current determined by simple division of the voltage by the impedance does not consider asynchronous currents, they will still be present.
It was my understanding that the panel and breaker manufacturers constructed the equipment to withstand the maximum currents available. Although the available fault current may be 22,000 amps, a 22,000 amp rated panel or breaker is designed to withstand the maximum worst-case asymmetric fault on a system that can deliver a steady state fault current of 22,000 Amps.
I was under the impression that the fact that one half-voltage faults were capable of delivering greater current would be a design factor when the equipment is evaluated for worst possible case faults.
I may be wrong. Any discussion.
PS. It's been a long time since I had to submit short circuit calculations in support of a permit application but the full voltage calculations were the accepted figures. (Before the computer programs, back when cost estimates for extra work were some times calculated with a "Sly Drool".)

I'm willing to learn and am open to correction on this.
yours

 

[stevenal]

waross,

You are correct that the breaker ratings are symmetrical. Likewise, the half winding multiplier when applied to the full winding fault current also gives a symmetrical value. The separate treatment of X and R are not for calculating the asymmetrical fault current, but are for dividing up the nameplate impedance into their respective windings and coming up with a better estimate of the multiplier. If your X/R ratio is high, the asymmetrical fault current must also be taken into account when sizing the interruption capability. Breaker manufacturers have no control over transformer design, so it's up to the system designer to take the half winding as well as asymmetrical fault currents into consideration. He or she should also consider that the existing transformer may burn out and be replaced with one with a different KVA or impedance. I doubt most inspectors have the knowledge or training to check the figures presented to them.

 

[mc5w]

What I have seen is that as a general rule the line to neutral short circuit current at the terminals of a center tapped secondary is about 1.5 times the line to line value. This is partly because of the reduced number of secondary turns interacting with how leakage reactance and mutual inductance relate to each other. There is also the consideration of how the secondary voltage drives the secondary resistance and reactance.

At some distance from the transformer line to neutral and line ton line short circuit current will be about the same due to the influence of conductor resistance and reactance. Beyond that point L-N short circuit current will be lower than L-L.

 

[waross]

My point is the difference between the interupting Capacity of a circuit breaker and the interupting Rating of a circuit breaker.
For example;
I have a transformer with an available symetrical fault current of 10,000 amps.
I feed a breaker with an interupting Capacity of 10,000 amps.
I have a bad fault and the asymetrical current is twice the symetrical current and the breaker fails.

The fact that breakers regularly clear faults that may be considerably more than their rating is a good indication that the Capacity of a breaker is several times the Rating. I assume that the breaker manufacturers are aware of asymetrical faults and half voltage faults and include an adequate safety margin for worst case scenarios.
I concur with the practice of designing for a posible increase in transformer available fault current. I believe however that it is safe to use the Available fault currents of transformers (existing or allowances for future) and the safety margins of breaker Ratings.
If there are several case studies of properly rated breakers failing under Half Voltage faults, then I stand corrected With Apologies. (And lose a few nights sleep wondering if that project 25 years ago was close enough to the maximum allowable fault current to be in jeopardy.)
respectfully.

 

[stevenal]

I assume that the breaker manufacturers are aware of asymmetrical faults and half voltage faults and include an adequate safety margin for worst case scenarios.

Asymmetry factors are addressed by standards. If your system X/R exceeds what is allowed by standard you need to account for this.

I'm sure they are also aware of half winding faults, but know of no standard that takes this into account. Sorry, but it sounds like it was up to consider it.

Manufacturers are free to exceed the standards of course, but I don't think they find it very cost effective to do so.

 

[cuky2000]

At different from the MV breaker, LV protective devices do not have asymmetrical ratings. However, we could calculate the maximum asymmetrical fault current as the product of the SC current on the half winding of the distribution transformer and the peak factor provided by the Standards as a function of X/R ratio.
Typical maximum peak asymmetrical multiplication factor is shown in enclose link.

http://cuky2000.250free.com/Protective_Device_LV_MF1.gif

 

[dpc]

Actually LV breakers really do have a maximum asymmetrical rating - but it is stated in symmetrical amps at a certain power factor (X/R). We didn't change the laws of physics when the breaker rating system was changed. They were just trying to make it "simpler".

The X/R ratios used for the molded case circuit breaker testing are actually somewhat low when applied to typical industrial systems. It's quite common to see MCCBs operating under conditions that require de-rating of their "symmetrical" rating.

 

[cuky2000]

Hi dpc,
Per UL guidelines, asymetrical rating is not required to be published on the MCCB. As discussed above, we have the option to determine by calculation the asymetrical withstand knowing that the test PF=15% equivalent to X/R=6.6 and a multiplication factor of 2.3 approx.

I am taking the liberty to enclose an excerpt from the UL Marking Guide-MCCB 600 Volts or Less that may help to clarify the subject.

_{14. Ratings — [blue]All circuit breakers with an interrupting rating other than 5000 A are marked with their interrupting rating. If the breaker is not marked with an interrupting rating, the interrupting rating for the breaker is 5000 A. The marking includes the words “Interrupting Rating” or “Current Interrupting Rating” and may include “Maximum RMS Symmetrical,” or an abbreviation. If the interrupting rating includes more than one current and associated voltage rating, all values of voltage and corresponding interrupting rating are marked. If more than one interrupting rating is marked, all ratings appear together. [/blue]}

No asymmetrical voltage rating may be marked on the breaker.
_{[blue]If the marked interrupting rating of the breaker exceeds the marked withstand rating of the end-use equipment, such as a panelboard, in which the breaker is installed, the interrupting rating of the overall combination is still considered to be the lesser rating marked on the end-use equipment.[/blue]}

 

[stevenal]

Cuky,

That's correct, that is how breakers are marked. But system X/R must be considered (not an option) if too high.
See NEMA AB3, section 4.5. It's a free download from nema.org after registering. Derating factors are tabulated in Table 4-2 for various interrupting ratings and system X/R.

 

[rbulsara]

This is a helpful link in regard to MCCB SCCR.

http://www.ewh.ieee.org/r6/san_fran...s.pdf#search='MCCB short circuit rating ansi'