- Thread starter jcbeck
- Start date

- Location
- SE USA as far as you can go

Sounds like a homework problem!

To know how much power can be transmitted, you need to figure out the maximum current capability of the line, but to know that you have to know what the conductor size is. Then subtract your losses, and you should get your answer.

Yes, correct - it's part of my course. The problem is I haven't been given the conductor size! It's really left me scratching my head as to how to solve it!Sounds like a homework problem!

To know how much power can be transmitted, you need to figure out the maximum current capability of the line, but to know that you have to know what the conductor size is. Then subtract your losses, and you should get your answer.

- Location
- SE USA as far as you can go

And I know the book I had for this type of problem had a conductor chart in the back...

with different conductors you can look up listing R, X, and C values. I wonder, if you converted the r and x to ohms per mile, and there was such a chart, what you might find?

- Location
- Boca Raton, Fl, USA

JCBeck...

3-phase or single-phase?

Regards, Phil Corso

3-phase or single-phase?

Regards, Phil Corso

Given the total reactance and total resistance per phase should be enough information to make conductor size nonessential...Yes, correct - it's part of my course. The problem is I haven't been given the conductor size! It's really left me scratching my head as to how to solve it!

...though a total resistance of 0.2 ohms per phase over 30 miles seems to indicate a very, very large conductor, even if paralleled multiple times over... :huh:

30 miles = 158.4kft

0.2ohms ? 158.4kft = 0.00126ohms/kft

That's less resistance than ten (10) paralleled 1,000kcmil copper conductors hmy:

Additionally, to determine effective Z, you also need to know the power factor.

There are definitely missing parameters to arrive at any determination... :happyyes:

This ideal power PSIL=VL-L^2/Zc. If we consider x=6.3 ohm as characteristic impedance of the line then Zc~6.3 ohm.PSIL=33^2/6.3=172.86 MW per all three phases or 172.86/3=57.62 MW per phase.

I don't know whether or not it was ignored but it makes little difference to the calculation.Julius, thanks for that. My initial calculation was to include the resistance of 0.2 Ohms per phase with the reactance - why is the resistance ignored?

X 6.3

R 0.2

Z 6.303173804

Vl 33 kV

Vph 19.05255888

19052.55888 V

3022.692928 A

57590035.01 VA

57.59003501 MVA

Note that the answer is MVA, not MW.

How did you determine the effective Z???I don't know whether or not it was ignored but it makes little difference to the calculation.

X 6.3

R 0.2

Z 6.303173804

Vl 33 kV

Vph 19.05255888

19052.55888 V

3022.692928 A

57590035.01 VA

57.59003501 MVA

Note that the answer is MVA, not MW.

Sqrt(X[SUP]2[/SUP]+R[SUP]2[/SUP])How did you determine the effective Z???

I've seen this before when power factor is unknown. I don't understand. Can you explain, briefly?Sqrt(X[SUP]2[/SUP]+R[SUP]2[/SUP])

- Location
- California

I'll take a crack at it. Resistance and Reactance are vectors situated at 90 degrees from one another -- resistance on the real axis and reactance on the imaginary axis. Their vector sum is impedance.I've seen this before when power factor is unknown. I don't understand. Can you explain, briefly?

You can solve for impedance using the Pythagorean Theorem (Z[SUP]2[/SUP] = R[SUP]2[/SUP] + X[SUP]2[/SUP]), which is what Besoeker did. Or you can use the Power Factor, which is really just the cosine of the angle between the impedance vector and the real axis.

Perhaps my mad MS Paint skills will help:

I think it's a very poorly framed question.I've seen this before when power factor is unknown. I don't understand. Can you explain, briefly?

Working back from the result, the (unstated) assumption appears to be a short circuit at the end of the 30 miles so we are looking only at the impedance of the conductors.

I understand it as presented. At some past point in time, I knew this. But what I don't understand is how this might relate to a realisticI'll take a crack at it. Resistance and Reactance are vectors situated at 90 degrees from one another -- resistance on the real axis and reactance on the imaginary axis. Their vector sum is impedance.

You can solve for impedance using the Pythagorean Theorem (Z[SUP]2[/SUP] = R[SUP]2[/SUP] + X[SUP]2[/SUP]), which is what Besoeker did. Or you can use the Power Factor, which is really just the cosine of the angle between the impedance vector and the real axis.

Perhaps my mad MS Paint skills will help:

View attachment 9331

According to my calculations, to arrive at an

- Location
- Ann Arbor, Michigan

The original post question lacks adequate definitions.

The title is "Power Loss Calculation".

The second sentence is

"How much power can be transmitted over a 30 mile long 33kV line which has a total reactance of 6.3 Ohms per phase and total resistance of 0.2 Ohms per phase?"

The first implies a measurement of power loss. Insufficient information.

The second implies maximum power transfer from the source to the load. This occurs when the source resistance equals the load resistance. Assuming the source reactance is inductive, then for maximum power transfer to the load, the load needs to be 0.2 ohms in series with a capacitive reactance equal to the source inductive reactance. This amounts to an extremely large amount of power. Clearly not the question for the answer given.

The original post needs to be rewritten with a correct statement of the question. It appears that the correct question may be something like --- what is the short circuit input volt-amperes from one line to neutral? This has nothing to do with maximum power transfer over a 30 mile transmission line.

Maximum power transfer will occur somewhere between a short circuit load and no load.

Without the capacitor in the load the maximum power transfer will occur around a load resistance of 6.3 ohms. I do not want to spend the time to do an exact calculation. Maximum power per leg is about 28.6 MW. Under these load conditions the load voltage is about 19*0.707 = 13.4 kV.

Check my math.

.

ItAccording to my calculations, to arrive at aneffective Zof 6.303173804 ohms using R=0.2 and X = 6.3 would require a power factor of approximately 0.03. That seems quite unrealistic to me...???

I didn't say impossible, just unrealistic. So just what large area and spacing are you implying? How does that relate to using sqrt(X?+R?) for Z[SUB]e[/SUB]?Itmightbe plausible for an overhead transmission line where the conductors are spaces some distance from each other. You than have a single turn loop with a large area and an inductance of about 20mH seems feasible.