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Confusion around base kV and PU voltage

asked 2019-10-30 08:08:13 -0500

kewasiuk gravatar image

updated 2019-10-31 11:37:27 -0500

Edit:

To further illustrate my point, consider a sequence of buses. 1, 2a, 2b, 3. All lines impedances are negligible. Now, I have my generator set at 1 pu controlling bus 1 on a 1 kV base. So, for my study, I could confidently say that bus 1, 2a, 2b and 3 are all at 1 pu, or 1 kV. But what if the base of bus 2b was 2 kV? The PSS/E solution tells me that it is 1 pu on a 2 kV base, so I would record it as 2 kV. Clearly this is incorrect as the generator is regulating the voltage at 1 kV. Fortunately, this network is simple and I can figure this out. But what if I had a much larger network, and connected two buses of different base kVs. The only way I would be able to tell you the actual kV (and true base kV for that matter) is to trace the lines back to a voltage regulated bus, and take note of the transformer tap ratios along the way. To me, this seems like a needless step as PSS/E should incorporate this already with the base kV. As far as I can tell, base kV is only used for equipment ratings, but other than that it is misleading.

O-|--|--|--|->

..1..2a.2b.3..

Original:

I encountered what I thought would be obvious and intuitive, but has left me confused. I made a simple 3 bus radial system to illustrate my point: a swing generator (1), connected to an intermediate bus (2), serving a load (3). Everything is set to 1: Vswing = 1 pu (controlling bus 1), load is 1 MW, all buses are at a base KV of 1 kV. Branch impedances are negligible.

O-|--|--|->

..1..2..3..

So everything solves without error, and leaves all the voltages at 1 pu. Now, if I change the middle bus's (2) base kV, to let's say 2 kV, and then solve, then shouldn't it's new PU voltage be 0.5 pu? Since the swing is controlling the volage of bus 1 to 1 PU on a 1 kV base, that is 1 kV, and bus 2's voltage should also be the same (negligible impedance between). The way I see it, the base kV for a type 2 bus is just an interface for us users to see the pu voltage. When PSS/E is solving the loadflow in the background, I imagine it uses whatever base it so desires, and then converts it to a PU number on the base we chose.

If I change the base to an arbitrary number, the PU voltage stays the same--shouldn't this change? How do I know what the actual base that PSS/E is using to derive the PU values.

I'm using 33.10.0

Here's the raw data if anyone is interested.

0, 100.00, 33, 0, 1, 60.00 / PSS(R)E-33.10 WED ... (more)

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answered 2019-10-31 12:59:08 -0500

ademuth93 gravatar image

A couple comments that probably don't help you at all:

  • It's not good modeling practice to connect buses of different nominal voltages without a transformer
  • As a couple of others have said, this is the nature of the pu system. All calculations are done based on the per unit values, which will all be 1 pu in your case as expected. What you are seeing is expected behavior, and you will need to add a couple 2:1 transformers if you want to see 0.5 pu voltage on your 2 kV bus.

RAW data with transformers to create the 0.5 pu value:

0,   100.00, 33, 0, 1, 60.00     / PSS(R)E-33.11   THU, OCT 31 2019  13:06


     1,'            ',   1.0000,3,   1,   1,   1,1.00000,   0.0000,1.10000,0.90000,1.10000,0.90000
     2,'            ',   2.0000,1,   1,   1,   1,0.50000,  -0.0001,1.10000,0.90000,1.10000,0.90000
     3,'            ',   1.0000,1,   1,   1,   1,1.00000,  -0.0001,1.10000,0.90000,1.10000,0.90000
0 / END OF BUS DATA, BEGIN LOAD DATA
     3,'1 ',1,   1,   1,     1.000,     0.000,     0.000,     0.000,     0.000,     0.000,   1,1,0
0 / END OF LOAD DATA, BEGIN FIXED SHUNT DATA
0 / END OF FIXED SHUNT DATA, BEGIN GENERATOR DATA
     1,'1 ',     1.000,     0.000,  9999.000, -9999.000,1.00000,     0,   100.000, 0.00000E+0, 1.00000E+0, 0.00000E+0, 0.00000E+0,1.00000,1,  100.0,  9999.000, -9999.000,   1,1.0000
0 / END OF GENERATOR DATA, BEGIN BRANCH DATA
0 / END OF BRANCH DATA, BEGIN TRANSFORMER DATA
     2,     1,     0,'1 ',1,1,1, 0.00000E+0, 0.00000E+0,2,'            ',1,   1,1.0000,   0,1.0000,   0,1.0000,   0,1.0000,'            '
 0.00000E+0, 1.00000E-4,   100.00
0.50000,   0.000,   0.000,     0.00,     0.00,     0.00, 0,      0, 1.10000, 0.90000, 1.10000, 0.90000,  33, 0, 0.00000, 0.00000,  0.000
1.00000,   0.000
     2,     3,     0,'1 ',1,1,1, 0.00000E+0, 0.00000E+0,2,'            ',1,   1,1.0000,   0,1.0000,   0,1.0000,   0,1.0000,'            '
 0.00000E+0, 1.00000E-4,   100.00
0.50000,   0.000,   0.000,     0.00,     0.00,     0.00, 0,      0, 1.10000, 0.90000, 1.10000, 0.90000,  33, 0, 0.00000, 0.00000,  0.000
1.00000,   0.000
0 / END OF TRANSFORMER DATA, BEGIN AREA DATA
0 / END OF AREA DATA, BEGIN TWO-TERMINAL DC DATA
0 / END OF TWO-TERMINAL DC DATA, BEGIN VSC DC LINE DATA
0 / END OF VSC DC LINE DATA, BEGIN IMPEDANCE CORRECTION DATA
0 / END OF IMPEDANCE CORRECTION DATA, BEGIN MULTI-TERMINAL DC DATA
0 / END OF MULTI-TERMINAL DC DATA, BEGIN MULTI-SECTION LINE DATA
0 / END OF MULTI-SECTION LINE DATA, BEGIN ZONE DATA
0 / END OF ZONE DATA, BEGIN INTER-AREA TRANSFER DATA
0 / END OF INTER-AREA TRANSFER ...
(more)
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Comments

To your first point: For my particular study, I am looking at interconnecting our system to a neighbor with a different base voltage, so this is unavoidable. I definitely would not deliberately connect to different bases.

kewasiuk gravatar imagekewasiuk ( 2019-10-31 15:00:37 -0500 )edit

It's not that I really want to see the voltage as 0.5 pu--it's what I expected to see. In my case, I am stuck with a 110 kV bus connected to a 124 kV bus (by a 0 impedance line) both at 1.05 pu. I do not know what to report the voltage as! 1.05 on a 110 kV base, of 124 kV. It can only be one.

kewasiuk gravatar imagekewasiuk ( 2019-10-31 15:02:33 -0500 )edit
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answered 2019-10-30 12:00:38 -0500

perolofl gravatar image

PSSE is always using pu in the network calculations. Voltage in kV is only for user interface and it is of course Vpu*Vbase. All buses should have a pu voltage around 1 pu, regardless of the base kV for the bus.

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Comments

I understand that, but when I call `psspy.busdat(ibus, 'KV')`, wouldn't it only make sense to return the actual bus kV voltage? I can set bus 2 to an arbitrary base, and it still shows 1 PU. How do I know what the actual kV is? How do I know what the base used in the network calculation is?

kewasiuk gravatar imagekewasiuk ( 2019-10-30 14:53:35 -0500 )edit

Base kV is not used in the network calculation. Everything is in pu!

perolofl gravatar imageperolofl ( 2019-10-30 16:31:13 -0500 )edit

I understand that Base kV is not used in the solution.... But it is definitely used when I call for the bus voltage in kV, and it's giving me incorrect results. If I have a base kV of 1 kV and a bus at 1 pu, PSSE should tell me the bus kV is 1 kV. If I change the base to 2 kV, it should tell me 0.5

kewasiuk gravatar imagekewasiuk ( 2019-10-30 21:22:39 -0500 )edit

You don’t understand at all! The voltage will be 2 kV! You have completely misunderstood the whole concept of the pu system.

perolofl gravatar imageperolofl ( 2019-10-30 22:15:20 -0500 )edit

I think what @perolofl is trying to say is that if you change the base kV, the whole PU system moves along with it and thus it is gonna be 2 kV and 1 pu, not 0.5. This is actually how the PU system works.

drsgao gravatar imagedrsgao ( 2019-10-31 04:54:39 -0500 )edit

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Asked: 2019-10-30 08:08:13 -0500

Seen: 114 times

Last updated: Oct 31