Transmission Line Voltage drop/rise equation 1

[shattenjagger]

Hello All,

I am having trouble rectifying an equation for voltage drop....

IEEE Red book states that voltage drop on a line can be calculated as
Vdrop = IRcos(phi) + IXsin(phi)

where R, X are the line impedance and phi represents the load angle. I is a magnitude.

I am using the short TLine model where

Vsending = Vreceiving + I(R+jX)

and it seems that I(R+jX) should be approximately equal to the Vdrop as suggested by the RedBook. However, I seem to notice quite a bit of difference between the two, especially as line reactance goes up.

Mathematically it makes sense that they are not equal as
I(R+jX) = IRcos(phi) + IXsin(phi) + j(IXcos(phi) - IRsin(phi)) where the vector I = Icos(phi) +jIsin(phi)
and it seems that IEEE is just taking the real portion of the result for their Voltage drop.

My Question is, when does the IEEE version stop being applicable and you are forced to use the Short Line Model? Or should you ever use the IEEE model? Or have I missed something and am merely having a "doh!" moment?

 

[magoo2]

[and it seems that I(R+jX) should be approximately equal to the Vdrop as suggested by the Red Book]

This is the same as saying tha voltage drop equals I*Z. This is a bad assumption.

I don't know if the attachment helps. I have a diagram that goes with it but I can't locate it. I believe it is also in the Red Book.

 

[robertplant22]

Resistance and reactance per feet or meter are usually given in tables at a specific power factor or a power factor of 1.

If your power factor is not 1; you would then need to use Vdrop = IRcos(phi) + IXsin(phi) to account for the difference in power factor.

phi = arcos(pf).

To answer your question, the IEEE version stops being applicable whenever the power factor at hand is not 1. Some tables have already done the calculation for you and tell you what the effective Z, or R and X are for a given power factor. See table 9 of NEC for an example of this; this table can be found on CH9 at the end of the code book.

 

[jghrist]

Vsending = Vreceiving + I(R+jX)

and it seems that I(R+jX) should be approximately equal to the Vdrop as suggested by the RedBook. However, I seem to notice quite a bit of difference between the two, especially as line reactance goes up.

Voltage drop is the difference in the magnitude of Vsending and Vreceiving. I(R+jX) is the vector difference.

 

[magoo2]

If your power factor is not unity, the approximate voltage drop equation is still fairly accurate.

 

[magoo2]

There was a 1982 IEEE paper on the subject of exact versus approximate voltage drop equations. Author was Claude Brice of Texas A&M. As I recall, the error was within a few percent over a wide range of power factors.

 

[shattenjagger]

Thanks for the quick responses, you guys got me a nudge further down the path.....

I kept working on the problem this afternoon and came up with the following......

My new assumption is this (or my new understanding of what IEEE assumption of Vdrop approximation is)....
IEEE is calling voltage drop something similar to Voltage Regulation....
(|Vs| - |Vr|) / |Vr|
and it seems to work pretty decently as long as you have a lagging PF and your X/R ratio is small.

I ran thru some examples and it seems to fall apart rather quickly once your PF is leading and the X/R > 2

The attached sheet is what I have been using to help me understand... (if you can decipher my terrible formatting)

I am really just trying to make the IEEE model fit and recognize when their model is no longer sufficient.

 

[shattenjagger]

Thanks for the tip on the paper, got a flight this weekend and I found it in the IEEE archives.

 

[stevenal]

Red Book Fig 3-11 shows it all pretty well. Jghrist has it right, both the estimated and actual Vdrop are on the real axis while IZ is not. Scroll down to the end of the section to get the actual Vdrop formula, also a real number.

Also note the Error on the diagram. This is the difference between the magnitude of Es and the real part of Es. The error gets larger as the phi approaches zero, or as X increases.

 

[shattenjagger]

Yeah it does show the error, but what it doesn't show, even in their exact solution, (and the source of my confusion)is the couple percent of error you start to see as your lines get longer (more inductive).

The main problem may be that I am trying to use the equation outside of it's application as my lines are getting to be more than a mile long.

Thanks to everyone for their help.

 

[stevenal]

Quite a spreadsheet. I confess I haven't got through it all, but it looks like you are considering |I*Z| to be an approximation of Vdrop? The IEEE formula and diagram do not support this. The arc on the diagram is not IZ rotated to the real axis, it is Es.

I believe the diagram does show the error increasing with X. If X is extended, Es will be longer and higher making the difference between its magnitude and real part greater.

Why is your pf leading? Both phasor diagrams show Ir lagging Vr or Er. If you want to look at leading pf, I would suggest redrawing the diagrams.

 

[shattenjagger]

The diagrams are actually straight out of IEEE, just there as a reminder to myself as I was working. My PF is leading because I am a generator considering what is happening further down the line.

I am really just trying to relate the Short Tline model (the top part of the sheet) with the IEEE equations (the bottom part of the sheet) for purposes of equivalency. It threw me for a loop when IEEE's exact equation did not match up with what I was seeing using the short Tline model.

Here is an updated version that has the exact IEEE calc included (as well as Magoo2's equation from the document he supplied). You can edit the highlighted cells to change the line variables. I put in a UG and an OH sample line for fun too.

 

[stevenal]

The pf will be the same at the generator. If the loads are lagging, the generator is supplying the lagging kvars and the diagram will look like the one in the book.

After playing some with your last spreadsheet, I believe the "exact" IEEE formula is not all that exact. I agree with your |Vs|-|Vr| column.

 

[shattenjagger]

Thanks a bunch, needed that peer review to make sure i wasn't losing it!

 

[stevenal]

Keep in mind also:

acos(pf) does not always = theta where pf =cos(theta).

Negative theta yields a positive acos. For some reason my results worsen when I substituted a signed angle, though.

The second term under the radical in your "exact" formula should be squared. Again the correction makes it further off.

Using Magoo2's formula for Er and then using the result in the next formula below makes no sense. Er (Vr) is a given in your question.

 

[shattenjagger]

Yeah, that is why I had the user enter the current and voltage angles, so the sign could be used.as.needed. I will check that again.

Thanks for the tip on the calculation errors, I was bound to miss something with all that activity.

I had magoo's equation using the calculated voltage from the tline model and then back calculating, figured that it should work in reverse any way. The magnitude of that calculation should definitely be correct.

 

[stevenal]

Magoo2 had Es as the reference angle, while your method (like IEEE) is to use Er. Cannot mix the two.

Your latest spreadsheet has ABCD values that no longer short line. Now all bets are off, since IEEE only applies to short lines. Your Pi model will have an Is that differs from Ir.

I take back what I said earlier about the angles. IEEE's definition of phi indicates that acos(pf) is correct, and this is also consistent with the diagram.

 

[shattenjagger]

The sheet is capable of calculating the medium line model, but you can set the suceptance equal to zero, which is what I was doing.

You can mix in Magoo's equation, because it references the angle of the sending end that mine calculates and performs the calculation in reverse. His result is very close to what IEEE is getting in all the samples that I have run. Now that I do not need it for a sanity check anymore, I will probably delete it as the IEEE form makes more sense to me.

Thanks again for the input.

 

[SixOn]

Electrical Engineers also have a red book!
Where can I find the soft copy?

 

[shattenjagger]

If your organization has a subscription to IEEE standards you can access it online. Otherwise you can purchase it as part of the color books package.