Complex Imaginary Variation 2 problem 75

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ElecPwrPEOct11

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Hi, I'm hoping this is a quick one to explain.

This problem has a 6.6kV utility service with source impedance of 0.94%. It is connected to a transformer serving a building at 480V. The transformer has 4.25% impedance and rated current of 1,720A. It asks for Isc at the transformer's secondary.

The problem solution simply takes:

Isc = FLA / Zeq

Zeq = 0.0094 + 0.0425= 0.0519 pu

FLA = given = 1720A

Isc = 33,141A

This is a fairly simple pu problem but I don't understand why you don't need to convert voltage bases for the utility source impedance. My solution was to find a new Zsource and then use FLA/ Zeq.

It seems like the utility voltage should matter in this sort of problem, but the solution doesn't include it at all. Can anyone explain why? Thanks for any last minute help!

 
I guess the simplest answer is if you think of the source as a Thevenin equivalent you are only given two factors, either voltage and Thevenin resistance or voltage and current. In this case you are given the utility voltage and resistance and hence you can figure out the third variable.

Another explanation is if you had two utility sources with the same "strength" but different voltages at each point. The %Z would be different to compensate for the different voltages such that the strength would be the same.

No idea if this helps but thought I'd try.

 
^ The Thevenin analogy is a good one, thanks. I guess the take-away is that the % impedance already has effect of the magnitude of the voltage factored into it, so you don't need to convert it (again) to use with the transformer impedance. Ex- a 100MV source would have a lower source impedance so you wouldn't need to convert voltages for this problem.

Now I need to go back and figure out the problems where converting bases was the right thing to do...

Maybe this would help me, what is the definition of a source's % impedance?

 
Guess I'll try to explain my simple view of it. Let's say you have a 120 V ideal source (yes, ideal like in your circuits textbook) and in series with this source you have a 0.3Ω resistor (call this your Thevenin equivalent). This gives a short circuit current of 400A and if you leave that on very long, the smoke comes out. Now let's say this source is rated for a full load current of 5A.

So at rated current flow (1 pu if you like) you would have a voltage drop of 1.5V across this internal resistance. 1.5/120 is 1.25% which is the %Z. For best regulation, i.e. voltage stability with varying load, you want a small %Z as the voltage that is dropped directly subtracts from the supply which results in varying voltage at the customer. Unfortunately, to reduce the short circuit current capability and be able to use lower priced panels/breakers you want a high %Z so those are the competing factors.

 
Hi, I'm hoping this is a quick one to explain.

This problem has a 6.6kV utility service with source impedance of 0.94%. It is connected to a transformer serving a building at 480V. The transformer has 4.25% impedance and rated current of 1,720A. It asks for Isc at the transformer's secondary.

The problem solution simply takes:

Isc = FLA / Zeq

Zeq = 0.0094 + 0.0425= 0.0519 pu

FLA = given = 1720A

Isc = 33,141A

This is a fairly simple pu problem but I don't understand why you don't need to convert voltage bases for the utility source impedance. My solution was to find a new Zsource and then use FLA/ Zeq.

It seems like the utility voltage should matter in this sort of problem, but the solution doesn't include it at all. Can anyone explain why? Thanks for any last minute help!
------------------------------------------

What does mean transformer impedance 4.5%? It means, when you will apply 4.5% of primary voltage (4.5% * 6.6 kV) to the primary side of transformer with secondary shorted, transformer secondary will have rated current (here 1720 A). So %Z is not representing high side or low side of transformer and you don't need to convert for either voltage level.

(or vice versa also true, aka...you can apply 4.5% of secondary voltage to secondary side with primary shorted to obtain the rated current).

 
Last edited by a moderator:
Hi, I'm hoping this is a quick one to explain.

This problem has a 6.6kV utility service with source impedance of 0.94%. It is connected to a transformer serving a building at 480V. The transformer has 4.25% impedance and rated current of 1,720A. It asks for Isc at the transformer's secondary.

The problem solution simply takes:

Isc = FLA / Zeq

Zeq = 0.0094 + 0.0425= 0.0519 pu

FLA = given = 1720A

Isc = 33,141A

This is a fairly simple pu problem but I don't understand why you don't need to convert voltage bases for the utility source impedance. My solution was to find a new Zsource and then use FLA/ Zeq.

It seems like the utility voltage should matter in this sort of problem, but the solution doesn't include it at all. Can anyone explain why? Thanks for any last minute help!
------------------------------------

In PU system, source be 1.0 pu

So I (pu)= 1.0/(.0094+0.0425)=19.27 pu

I (SC)= 19.27*1720=33140 A

Simply calculate the PU current and multiply by rated current.

(Please note that I mentioned in earlier post why you don't need to convert transformer impedance).

Thanks,

 
Guess I'll try to explain my simple view of it. Let's say you have a 120 V ideal source (yes, ideal like in your circuits textbook) and in series with this source you have a 0.3Ω resistor (call this your Thevenin equivalent). This gives a short circuit current of 400A and if you leave that on very long, the smoke comes out. Now let's say this source is rated for a full load current of 5A.

So at rated current flow (1 pu if you like) you would have a voltage drop of 1.5V across this internal resistance. 1.5/120 is 1.25% which is the %Z. For best regulation, i.e. voltage stability with varying load, you want a small %Z as the voltage that is dropped directly subtracts from the supply which results in varying voltage at the customer. Unfortunately, to reduce the short circuit current capability and be able to use lower priced panels/breakers you want a high %Z so those are the competing factors.
Thanks all to the responses. DK- your description really helped me understand this on a conceptual level.

 
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