Short Circuit Calc Trouble
Short Circuit Calc Trouble
(OP)
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.
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.






RE: Short Circuit Calc Trouble
RE: Short Circuit Calc Trouble
If you intended 18.271a, then what about the ground resistor that the US Army likes so well?
RE: Short Circuit Calc Trouble
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.
RE: Short Circuit Calc Trouble
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.
RE: Short Circuit Calc Trouble
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Short Circuit Calc Trouble
RE: Short Circuit Calc Trouble
http://
This page is in chapter 4.
RE: Short Circuit Calc Trouble
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!
RE: Short Circuit Calc Trouble
less from result, xmmm, 2-3%, I think armu use more digits than you
Regards.
slava
RE: Short Circuit Calc Trouble
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.
RE: Short Circuit Calc Trouble
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.
RE: Short Circuit Calc Trouble
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!
RE: Short Circuit Calc Trouble
RE: Short Circuit Calc Trouble
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.
RE: Short Circuit Calc Trouble
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Short Circuit Calc Trouble
RE: Short Circuit Calc Trouble
"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.
RE: Short Circuit Calc Trouble
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)
RE: Short Circuit Calc Trouble
RE: Short Circuit Calc Trouble