Calculation of Equivelant Impedance

[7JLAman4]

I'm attempting to calculate Zeq at the buss level. I have three generator impedances (Z1, Z2, Z3) connected in parallel and the cable impedances (Z12, Z23) connecting each of the generator busses. How do I calculate Zeq at the buss? Are they all in parallel or something else?

Z1 nodes 0-1
Z2 nodes 0-2
Z3 nodes 0-3
Z12 nodes 1-2
Z23 nodes 2-3

 

[dpc]

Sorry, we don't help with homework.

 

[7JLAman4]

Sorry if you might have misunderstood. This is not for homework. Just looking to do something by long-hand rather than always using EasyPower, EDSA, or SKM. This is not my usual set of calculations. Looking to better understand what I may not have been taught properly before.

 

[7JLAman4]

Typically I assume that the buss ties are negligible and therefore Z1, Z2 and Z3 are easily paralleled for Zeq. I was just looking to understand how to compute if I were to assume buss tie impedance.

 

[dpc]

Just draw out an impedance diagram and solve it using normal network reduction methods. If you put in a bus tie impedance, you will have to combine the parallel branches on each side, and then solve the remaining network, depending on which side of the tie you are interested in.

 

[7anoter4]

If I well understood your scheme –three generators connected to three bus bars itself interconnected by means of two tie breakers shunted by cables –then [developing what dpc said]:
1) If both the ties are closed then Zeq=Z1||Z2|Z3 [if the tie impedance is zero].
2) If both the ties are open then if a solid three-phase shortcircuit will occur on the middle[no.2] bus bar then Zeq=(Z1+Z12)||Z2||(Z3+Z23)
3) If both the ties are open then if a solid three-phase shortcircuit will occur on the
No.1 bus bar then Zeq= (Z1)||{[Z2||(Z3+Z23)]+Z12}

 

[7JLAman4]

anoter- Understand Case1. That is what I currently assume. Re: Case2 and Case3 I'm confused. If the bus ties are open, wouldn't the Zeq at any of the busses only be Z1, or Z2, or Z3? or did you mean to state that the bus ties were closed for Case2 and Case3?

 

[7anoter4]

As I said "if I well understood your scheme".What I understood it is as follows:

 

[7JLAman4]

Sorry. Z12 and Z23 are bus tie cable impedances. There is a breaker on either side of the bus tie cable.

 

[rbulsara]

7janoter:

For the starters here is a hint: No current would flow through Z12 and Z23 when the respective tie breaker is open. How do you suspect the source of Z2 (or Z3) will contribute to the fault on bus 1, when the Tie 12 is open?

You need to start with a correct and complete impedance diagram. What you have drawn is representation of a one-line diagram. There are no voltage sources and no "common" (reference) bus is shown.

With all due respect, you need to refer to some text books and standards such as IEEE Violet Book™, IEEE Std 551™-2006— Recommended Practice for Calculating Short-Circuit Currents in Industrial and Commercial Power Systems (used to be part of the Buff Book) and/or the Red Book to learn how to prepare impedance diagrams and how to perform short circuit calculations.

Before that a refresher in electrical circuit network analysis (mesh or loop analysis) would help as to how to solve currents and voltages in a mesh.









Rafiq Bulsara

http://www.srengineersct.com

 

[7anoter4]

My dear rbulsara,
Thank you for your valuable remarks. I am very sorry for my European representation of impedance diagram but I still prefer the IEC 60909 for short-circuit calculation.
If the diagram was as simply as seems to me now to be I don't see what the problem was.
By the way, the dashed line provided with 2 arrows on the ends represent, in my world, a cable with 2 potheads connected DIRECTLY to the busbar and the tie breaker connect two busbars at the end of each busbar. If the connection between busbars is the cable connected to the breaker it is a very simple case without any problem. From the O.P. with my poor English I understood wrong. I am sory for the inconvenient.

 

[rbulsara]

7anoter4:
IEEE or IEC, your diagram is still wrong and incomplete. Electricity does not flow differently in Europe, certainly not through an open breaker!.

Your own post:

Sorry. Z12 and Z23 are bus tie cable impedances. There is a breaker on either side of the bus tie cable.

which is correct and hence the two diagrams on the right make no sense. The cables are in series with the breaker, you are showing them in parallel.

You do have a problem!


Rafiq Bulsara

http://www.srengineersct.com

 

[7anoter4]

A few remarks to your last post, rbulsara:
1) Your quotation is from 7JLAman4 post of 28 Aug 09 10:41.
2) The impedance diagrams [both] are according to schematic diagram from left hand.
3) I agree –and I apologize again-I didn't understand well as I thought the cables are in parallel with the breakers-and not in series as usual.
I quote from my own post of 28 Aug 09 0:34:
"If I well understood your scheme –three generators connected to three bus bars itself interconnected by means of two tie breakers SHUNTED by cables –then [developing what dpc said]:"
"Shunted" does mean "parallel", does it?

 

[rbulsara]

Shunted=Paralleled, yes.

If the cables shunt the breaker, the breaker is out of the circuit. The circuit is always closed via the cables. Opening or closing the breaker will have no effect.



Rafiq Bulsara

http://www.srengineersct.com