Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations SSS148 on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Short Circuit Calc Trouble 1

Status
Not open for further replies.

kingtutley

Electrical
May 11, 2007
49
I have beat this horse before, but this time it comes from a different source. I hav attached a page from a manual produced by the Army Corp of Engineers for their training. I have a real problem understanding where they are getting the 18.271A for the fault at LC BUS. Nothing I do gets that number or even close.

As I understand it (and this is ALL the information given about this simplistic system), the fault just from the 1MVA transformer would be (1M/(480*sqrt(3)))/.0475 = 25.3kA.

So what am I doing wrong, ,or what did they do wrong, or what am I missing?

Understand, there is NO other information given about the system. This is NOT a homework problem, and I KNOW this is probably very simple.

I ask here because I would rather look like an idiot in front of a bunch of strangers than ask some one in my office and have someone I KNOW think I'm an idiot.

I hope some one here will please answer this, as I really don't understand this.
 
Replies continue below

Recommended for you

You are including only the impedance of the 13.8 kV-480 V transformer. You need to include the impedance of the 34.5 kV source (calculate from the 500 MVA available) and the 34.5 - 13.8 kV transformer.
 
If you intended 18.271ka, then is the 1MVA rating of the transformer the top rating, and not the base rating?
If you intended 18.271a, then what about the ground resistor that the US Army likes so well?
 
Cranky108 >> As I stated, this page represents ALL the information given about this system. There is no other information given. I have no other information to give you. This is ALL the information I have. There is no mention of any kind of using a grounding resistor. I have no further information.

jghrist >> You lost me..... If I include the other transformer (ie, take the Thevinine equivalent of the system from the fault point) I will get the following for the total series/parallel impedance: 4.75+8||28=8.76%

Then the SC fault will be, what, (1M/(480*sqrt(3)))/.0876? = 13.73kA. Admittedly closer, but still almost 25% error. I would think the Army is a bit more meticulous than that.
 
A 1 MVA or 1000 kVA transformer is a standard padmount rating, so it's impedance would be at 1 MVA.

Select your MVA base for the calculation. Hint: 1 MVA is pretty convenient.

Next look at the impedances in order:

Source:
500 MVA ==> represents an impedance of 0.2 % on 1 MVA base

3750 kVA transformer
8% Z at 3750 ==> 2.133% at 1 MVA (1/3.75*8%)

1000 kVA transformer
4.75% Z at 1 MVA

Total impedance (Z)
7.083% (= 0.2 + 2.133 + 4.75)

Short circuit current (3-phase)
Ibase = 1000/(0.48 * 1.732) = 1202.8 A

Ishort-circuit = Ibase/Z = 1202.8/0.07083
= 16,982 A

This represents the three-phase fault current. The single-line-to-ground fault current could be higher, but we don't have enough information in the problem statement. You'd need to have the zero sequence impedances in order to solve this.

This result is in the ballpark of what the Army says it should be, but not as precise as we would like it.

There's no combination of series and parallel impedances in this situation. It's all series impedances.
 
I think that the page is either being taken out of context or there is a typo, or both. How about posting the text from the preceding and following pages?

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
You also need to assume, since it does not say on the diagram, that you are neglecting the cable impedance and feeder length. You have a motor contribution to the fault as well limited my cable impedance and motor impedance. Try your fault calculation for the first transformer for the fault shown in between the two transformers...you'll see that the calculation comes close (I got 1961 amps) they show 1934amps so they may be accounting for the cable impedance which we don't see. Anyway, use 500MVA as your base and calculate this in per unit. You'll have only a positive sequence contribution and you'll basically have the source impedance (1.0pu) in series with Z1 (transformer one on a 500MVA base) in series with Z2 (transformer two on a 500MVA base) all in parallel with Zmotor at the fault point. Add up the three, take the parallel combination of the three and the motor %Z and that equals your fault impedance. Now take your base MVA, 500 and divide it by the fault impedance...this equals your total fault MVA. Now take the total fault MVA divided by the fault voltage (in this case 480 volts) multiplied by sqrt 3 since this is three phase. This gives you your fault current at that point. If you want to look at different points on the circuit, you just have to rearrange the impedances depending on the fault point. Clear as mud?
 
I just noticed that there's a generator or motor shown below the 480 V bus which I ignored in my previous response.

200 kVA with 28% impedance gives 859.2 A of additional fault current to the 480 V bus.

Therefore, the total fault current is now

16982 + 859 = 17,841 A

which is still less than the 18,271 A shown on the diagram.

Close enough for government work!
 
Magoo2, you are great!!!
less from result, xmmm, 2-3%, I think armu use more digits than you :).
Regards.
slava
 
magoo2 >> Yes, I was about to point that out, but in your calc above, you converted everything to the 1MVA common base. Wont that have an effect on your 859A motor contribution?

Assuming the motor had a va base of 200, then the pu impedance on 1mva is 140% or 1.4pu right? That makes the contribution 172A, right?

Personally I really hate the pu system. I believe it causes more problems than it solves. Just my opinion to which I am entitled.

 
kingtutley,
The 859 A shouldn't change depending on the base that you use. I'll verify that below.

200 kVA with 28% impedance gives 859 A at 480 V. ( i.e., 200/(.48 * 1.732)* 1/0.28 )

If you did the calc with the 1 MVA base, you'd simply use the base current that goes along with 1 MVA and the corresponding impedance that also goes along with the 1 MVA base.

8% at 200 kVA is equivalent to 140% at 1 MVA. Fault current should be the same, or 1000/(0.48 * 1.732) * 1/1.4. We increased the impedance, but we also increased the MVA so the result is the same.

I think the per unit system is a much easier way to deal with systems that have different voltage levels. In this example, we ignored cable/conductor impedance. Had we included them in, you'd have a corresponding base impedance for each voltage level. Then you'd have to take the conductor impedance and divide it by the appropriate base impedance. Once you get used to it, you don't want to go back to another way.

Thanks, slavag. It's always good to hear positive feedback.

 
All,
I think, the reason why the figures didn't come close to the figure "18,271A" is because magoo2 treated the utility short circuit available MVA like a transformer instead of as a "current source". 500MVA SC availability tells us that the system can, at that point of interconnection, deliver that short-circuit MVA.
Disregard the 0.2% and add the pu impedances of the 2 trafo = 2.1333+ 4.75 = 6.8833; The short circuit from upstream = 1202.8/(0.068833) = 17,474Amps
Add the motor contribution of = 200kva/489*(sqrt(3)/0.28 = 859amps: total SC = 17,474 + 859 = 18,333 amps; very close to 18,271amps!
 
The 200 is probably horsepower, not kVA, but this won't make much difference because a 460 volt induction motor has a full-load current of 240A per NEC Table 430.250, or 191 kVA.
 
burnt2x,

I don't agree with you.

You said:

I think, the reason why the figures didn't come close to the figure "18,271A" is because magoo2 treated the utility short circuit available MVA like a transformer instead of as a "current source". 500MVA SC availability tells us that the system can, at that point of interconnection, deliver that short-circuit MVA.

Your recommendation: Disregard the 0.2%.

I represented the source impedance in per unit as 1/500 or 0.2% on 1 MVA base. This is an impedance in series with a voltage source, so it is in effect a current source.

Since the 500 MVA source is given, you can't simply ignore it as you suggest. Sounds like you're trying to rationalize how to get closer to the 18,271 A value - which is admirable, but you can't simply ignore the source impedance value which is given.

Playing the devil's advocate, how can you legitimately get closer to the 18,271 A value?

Starting with the 18,271 A value, subtract the generator contribution (859 A). This leaves 17412 A from the system.

To get 17414 A requires an impedance value of 6.913% rather than the 7.083% that I had previously calculated. The difference is 0.17%.

Could it be something as simple as a typo in one of the impedances? For example, if I changed the impedance for the 1 MVA transformer from 4.75% to 4.57%. I would get a result that virtually matches the 18271 A value.
 
The army used 861 A as the symmetrical motor contribution. This is given on page 4-8 of the original document. There are a couple of obvious typos on this page. More typos are possible.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I guess the truth is now clear as Bill noticed! More typos in that doc!
 
The original document states on p. 2-2:
"Standard tolerance on impedance is plus or minus 7.5 percent for two-winding transformers. The minus tolerance should be used for short circuit studies."

When the impedances of the transformers is reduced by 7.5% (impedance multiplied by 0.925), I get the short circuit current to be 18300A without the contribution of the motor.

I think that contribution of the the motor can be neglected, because this is an example in coordination, and the current transformers do not "see" the current from the motor.
 
ijl, I agree with the foresight that when trying to coordinate protections, the motor contributions can't be "seen" by the CT due the fact that residual flux on the motor collapses quickly and appears only for a few cycles!
Back to the OP, it should not be a cause for alarm if figures are a bit off as long as these figures are within the ballpark, so to speak.
magoo2, thanks for pointing out the things I missed, It's good to have people reminding others.(Utility Zpu = 1X(100/SCA); SCA being short-circuit MVA available)
 
hey burnt2x, it's always good to have more than one set of eyes looking at a problem. We all seem to see things differently and then we all benefit from the other's perspective. So, thank you too.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor