Short Circuits Calcs w/ Transformer Taps 1

[AFerguson]

Given:

10 MVA, 46 kV - 12.47 kV Transformer
Z = 10.4%
Primary Tap -2.5% (45 kV)
Secondary Tap +5% (13.09 kV)
3P Secondary Short Circuit Current = 4290 A
1P Secondary Short Circuit Current = 4620 A

I am trying to determine the utility (primary) short circuit contributions but seem to be missing something in my equations. Most appreciated if someone could outline the method to solve the above problem (either using SCC or MVA method).

 

[musicguy1800]

This is a three-phase transformer?

 

[AFerguson]

Yes, this is a three-phase transformer.

 

[jbartos]

Suggestion: Normally, the system impedance Zsys and System Voltage, Vsys, behind the Zsys are used to account for the contribution from the Utility Source. Contact the Utility for the above data.
Reference: IEEE Standard 141-1993 Red Book

 

[AFerguson]

The utility supplied the transformer information and secondary short circuit currents. Normally the secondary SCC would be acceptable, but the model I am creating requires that I develop a primary utility contribution using the given data.

 

[jbartos]

Suggestion: Contact the Utility for the transformer primary side parameters, most likely a transmission line parameters or switchyard parameters.

 

[musicguy1800]

I have a theory, though it may not be correct. I am a brand-new engineer (guys with experience, please help me out if I'm wrong!!). Couldn't one argue that the short-circuit MVA is going to be the same for a fault on the primary side as on the secondary side? If so, then:

MVA,sc-primary = MVA,sc-secondary

VpIp = VsIs , where s is secondary and p is primary.
--------------------------------------------
3-phase fault:

(45 kV)Ip = (13.09 kV)(4290 A)

Ip = 1248 A

3P Primary Short Circuit Current = 1248 A
--------------------------------------------
1-phase fault:

(45 kv)Ip = (13.09 kV)(4620 A)

Ip = 1344 A

1P Primary Short Circuit Current = 1344 A
--------------------------------------------

What do you all think (bracing myself for criticism)?

 

[skiier]

Aferguson:

Sorry I don't understand the question as posed and I am a little confused by some of the replys.
Am I missing something? Why has no one asked about the available short circuit capacity ahead of your t/f? Or in all the jibberish has someone inadvertently made a clear statement that I missed?
Go to

http://www.bussmann.com/library/docs/EPR_Booklet.pdf

and read. Your answers are there.
Lots of generic short circuits calcs there.

 

[jschwei314]

Mr. AF, I would work backwards, say the 4290 amps is what the utility is supplying on the 12.47kV side. What is the fault on the 46kV side?

Start with an infinite bus calculation, 10000kva / 46kv x sqr(3) x .104 equals about 1250 amps.

This writer likes to use the point to point method for accuracy.

f= SCA(12.47) x V(12.47) x %Z / 100,000 x KVA(transformer)

M= 1/ (1+f)

SCA(46kv) = V(12.47)/V(46) x M x SCA( 12.47)

or about 590 amps at the primary of the transformer.

note slide rule accuracy.

 

[jghrist]

If you consider the utility source as 12.47 kV and do not model a transformer, are the short-circuit currents acceptable inputs for your program? These are adequate to determine the source impedance if you assume that the source impedance is inductive. You may need the X/R ratio for breaker rating calculations, however.

 

[skiier]

Aferguson:
When you talk about a primary fault are u talking about all the feeders and conductors "in front" of the t/f faulting to gnd? If so, would you not have to know the available fault current or available short circuit "MVA" from the utility source. Also, would you not have to consider loads feeding back from your equipment thru the t/f to the point of fault?
If the utility tells you that 4290A is available as a fault current at the secondary side of your t/f why would they not tell you what is available at the primary? And don't assume that you can use the t/f turns ratio here to find the primary fault current! The 4290A assumes the t/f is in the cct when the fault occurs. If you fault the primary of the t/f, say at the primary t/f taps,we no longer have a t/f "choke" offering 10% impedance in the cct and common sense would tell you that will get alot more than 4290A at the fault. The fault at the primary is now only limted by the conductors feeding the t/f. A huge difference in impedance to a fault.
You need to know the availble MVA that the utility can give you right at the point of connection to their system. Then you can work back thru your system to properly protect it.
You have almost all the information you need to answer your own question. Get the availble MVA from the utilty.

 

[reactive]

We will assume trfr at nominal tap (don't want to get into adjusting trfr impedances). This method will get you very close (it assumes the source and trfr are purely inductive):
For the 3-phase fault :
MVAscct = 4290 x 12.47 x sqrt(3)/1000 = 92.66 MVA
MVAsec = kV^2/Xtotal
Xtotal = kV^2/MVAsec (Xtotal = Xsupply + Xtx)
Xsupply = kV^2/MVAsec - Xtx
Xsupply = kV^2/MVAsec - kV^2/MVAtx x Zp.u.
Xsupply = 46^2/92.66 - 46^2/10 x 0.104 = 0,8298 ohms
MVAsupply = kV^2/Xsupply = 46^2/0,8298 = 2550 MVA or 32 kA.

For the single phase earthing arrangements and vector groups need to be known.

 

[jbartos]

Suggestion to musicguy1800 (Electrical) Jan 20, 2004 marked ///\\I have a theory, though it may not be correct. I am a brand-new engineer (guys with experience, please help me out if I'm wrong!!). Couldn't one argue that the short-circuit MVA is going to be the same for a fault on the primary side as on the secondary side?
///Yes and no. Yes, if you consider the MVAsc,sec for a fault on the transformer secondary. No, if you consider two different fault locations, one on the transformer primary and one on the transformer secondary. The primary MVAsc,prim will be bigger than the secondary MVAsc,sec since the transformer impedance will reduce MVAsc,prim, if it is intact by the short. If the short circuit burns the transformer and the transformer impedance becomes zero, then MVAsc,prim=MVAsc,sec.\\ If so, then:
MVA,sc-primary = MVA,sc-secondary

 

[skiier]

Hey guys multiply all the numbers you like together with the data given. You know what? It doesn't mean a thing! What you are calculating is not the primary short cct current should the fault occur on the primary of this t/f. The real important missing data here is the short cct capacity of the utility "in front of " the t/f. 4290A x anything here doesn't mean a thing if the fault occurs on the primary side of the t/f. Why is it that the obvious is so easily overlooked? one needs to work back with the numbers given to calculate the available short cct MVA form the utility. With this number you can calculate the fault in front of the t/f by determining the impedance of the conductors feeding the fault - from the utility connection point to the point of fault. The t/f may or may not be in the cct. Does anyone agree?

 

[davrom]

Sorry for this late post, but I have had a long time reading and studying the info and formulae given by various members and trying to match them with the shc theory that I know (which I have learned at school). I come from a diferent school which sometimes (manytimes) has different approaches to the various subjects. As I didn't know what MVA method means I have preferred to stay expectative for a while; in addition, some posts confused me for a while.

After reactive posted (mathematics - the perfect universal language) I realized that we all are talking about the same Mr. Jones but wearing different hats...

The jbartos's reply to musicguy1800 (MVAsccsec vs. MVAsccprim) confused me a little, as it has been the first time when I have heard of these two MVAs, but at a carefully look I realized that this is consistent with the reactive's method/formulae.

The difference between the MVA method and my shc theory is that: the MVA method considers the TX impedance as being "blind" to voltage level and the MVA as being "sensitive" to them while in my theory is reverse.

I think that the two MVAs are not real but are artificially created in order to sustain the MVA method/theory. I would be interested to know how these MVAs are introduced/defined in the MVA shc method/theory; maybe someone would clarify me.

My shc theory introduces the TX impedance (which is "sensitive" to voltage level) very simple: the TX theory shows that this machine cannot be studied in the absence of an unitary electrical diagram to link the parameters of one side to the parameters to the other side, as the sides are galvanical separated circuits; for this purpose, the parameters of one side are corrected/refered/"brought" to the voltage level of the other side by the mean of TX ratio ku=U1/U2 in order to make both sides compatible. The shc theory which I know does not make exception from this, so the impedances and currents are corrected by:
(1) Zp'=Zp*(Us/Up)^2
(2) Ip'=Ip*(Up/Us)
where:
- Zp' resp. Ip' are the primary side data corrected to the secondary side
- Zp resp. Ip are the real values of data on the primary side
- Us is the sec voltage, Up is the primary voltage
A short checking with the TX theory will confirm these above (I hope there will not be dofferent approaches!!!).
The formulae given correct the primary data to the secondary side; if you want to make reverse, just change the indeces 1 and 2 between them for each item of equations.
A note to musicguy1800: go get a bier yourself!

The shc current is calculated by:
(3) Ik=U/(1.73*Zk)

Aferguson, your question is "How can I determine the primary/utility contribution to the secondary side of my TX if the shc current on seconadry is 4290A?". The answer is included in your question: the 4290 amps IS the primary side contribution to the secondary side. Your actual questions should be:
- "How can I calculate the shc current on primary side if I know that its value is 4290A on secondary side?"
- "Can I determine the utility impedance and short-circuit power (MVA) if I know the TX impedance and the shc current on secondary?"
From equatons (1) and (2) you get the answers.

A note: you can calculate the shc impedance in two ways:
- calculating the TX impedance for primary voltage (by equation 4 for U=U1) and correcting it to secondary side by formula (1), or
- you can calculate the TX impedance directly refered to secondary voltage (the impedance is "sensitive" to voltage level) by:
(4) Zts=(Z%/100)*(Us^2/St)
The results are the same.

Numerical application: Aferguson's case:
- Shc current on secondary side (4290A):
(5) Iks=Us/(1.73*Zks)
- But the secondary side shc impedance Zks is actualy the primary side impedance corrected to secondary side:
(6) Zks=Zkp'=Zkp*(Us/Up)^2
- Replacing (6) in (5) yields:
(7) Iks=Up^2/(1.73*Zkp*Us)
- Extracting Zkp from (7) yields:
(8) Zkp=Up^2/(1.73*Us*Iks)=46000^2/(1.73*12470*4290)=22.863 ohms
you will recognize the MVAsccsec below the fraction line.
- But the primary side impedance Zkp is the sum of TX impedance calculated for primary level (Up) Ztp and the utility/supply impedance Zu;
(9) Zkp=Zu+Ztp
(10) Ztp=(Z%/100)*(Up^2/St)=(10.4/100)*(46^2/10)=22.0064 ohms
- The utility impedance yields from (9):
(11) Zu=Zkp-Ztp=22.863-22.0064=0.8565 ohms
- The utitily short-circuit power available on primary side at the point of installation of TX is:
(12) Su=Up^2/Zu=46000^2/0.8565=2470 MVA
- The primary side shc current:
(13) Ikp=Iks*(Us/Up)=4290*(12.47/46)=1162,963A
or
(14) Ikp=Up/(1.73*Zkp)=46000/(1,73*22.863)=1162.99A
Q.e.d.

The value given by (13) and (14) is the shc current on primary side of TX and is the utility contribution on the primary side at the point of installation of TX; to see the utility contribution on secondary side you have to correct this current to secondary side, which yields 4290A.

A last note: the reactive's sentence "MVAsupply=32 kA" may make some people be tempted to consider this as the utility contribution to the short-circuit, but IS NOT. It is the utility short-circuit power available at the point of installation of your TX expressed in kA.

Hope I haven't confused you more!

Regards,
David/davrom

 

[jbartos]

Suggestion to the previous posting: There are devices, e.g. current limiting fuses, which reduce short circuit current level. The reduction of the short circuit current downstream of the fuses implies reduction of the short circuit MVAsc, e.g. hypothetically, if the current limiting fuse reduced current to a zero, there would be zero MVAsc since the short circuit current is zero.
The Utility often apply current limiting fuses in their services. Therefore, a transformer in the consumer part of service may see a relatively low MVAsc contribution from the Utility. It may be well within the transformer short circuit MVAsc rating.
A reference for MVAsc:
Alvin H. Knable "Electrical Power System Engineering," Problems and Solutions," McGraw-Hill Book Company, 1967,
page 230, Fig. 8-5 Example solved by short-circuit-kVA method

Obviously, the short-circuit-kVA method is very similar to the short-circuit-MVA method

 

[jghrist]

According to AFerguson's Jan 20 post, The utility supplied the transformer information and secondary short circuit currents. The utility owns the transformer, so the utility connection is on the secondary (12.47 kV) side of the transformer. There is no need to know the fault current for a fault on the primary side because the utility is responsible for this protection.

 

[tulum]

Skiier,

With regards to your last post. "Hey guys multiply all the numbers you like together with the data given. You know what? It doesn't mean a thing!"

I do not agree. I think reactive's post is a viable calculation method in this situation. The secondary fault levels were given by the Utility... this means that they "more than likely" used the Primary side data to get to that point. Hence, the Prim. SC can be found through reverse deduction methods.

 

[musicguy1800]

davrom, what did you mean by this:

A note to musicguy1800: go get a bier yourself!

??

 

[reactive]

Hey guys

All AF asked for was the fault level on the HV. I have no idea what he wants it for.