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# Max Power Transfer Question - Updated NCEES Power #531

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I am working on problem 531 and I personally do not agree with the NCEES answer.  Help!

The question is asking for the Var losses in a transmission line during the transfer.

I used the equations I found to calculate the active power transmitted, done to simply verify that the system did in fact 5,000 MW from System A to System B, using the equation ONE below.  Next I calculated the var losses using equation TWO below.

Based on the equation to find vars transmitted from A to B I calculate

Yet NCEES says the answer is (D) 10,000 MVARS.

So what am I missing?  It does not seem like we find the vars transmitted from System A to System B as well as from System B to System A which would double.  I say this because the problem states 5,000 MW transmitted from A to B.

Any help is greatly appreciated.

FYI-> I just found a similar equation to the one I used on the NERC website and I would hope they get it correct.

Edited by jnspark
Added a note about finding a NERC equation similar to the one I used.

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I found the easiest way to do this problem is to solve for the current and then use S = I x I* x Z to find the real and imaginary components of the power.

So in this case the current is the difference between the voltages divided by the impedance, or I = (Va - Vb) / X.

Substitute actual variables for I = (500<90 kV- 500 kV) /  50j  = 10 + 10j kA.

Then the total power is S = I x I* x Z, or S = (10 + 10j)(10 - 10j)(50j)(1000)(1000)= 10000j x 10^6  or 10000 MVAR.  All power is imagary, and answer D.

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I talked to a co-worker, who took the April test, and he explained to me the "j" that is dropped in the NCEES equation makes it purely reactive and therefore the I^2 * R equation becomes the reactive power.  All that confusion for the lack of the "j" in the denominator.

Thanks BirdGrave, but per chance did you forget the root 3 in the Current equation since these are line-to-line?

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16 hours ago, jnspark said:

Based on the equation to find vars transmitted from A to B I calculate

Hi @jnspark, the equation for reactive power that you're trying to use (quoted above) calculates the reactive sending power.  NCEES is asking for the reactive power dissipated (lost) in the line reactance when the sending active power is at it's maximum. Slight but important difference.

"When the sending power is at its maximum" tells you what angle to use for your complex voltages. NCEES was even (surprisingly) kind enough to tip you in the right direction of what angle to use for delta if you are unfamiliar with max power conditions (sin(90º) = 1).

To calculate the power lost in the line reactance just treat it like a two terminal device. If you know the voltage across a two terminal device and the impedance of the two terminal device, then you can calculate the complex power consumed by the impedance (or supplied by it if the device is a source):

S = V^2/Z
(O+jQ) = V^2/(0+jX)
Q = (VA-VB)^2/X

Where VA and VB are the line-to-line complex voltages with VA leading VB by 90º, X is the per phase line reactance, and Q is the three phase reactive power consumed by the per phase line reactance.

Alternatively, you could plug in the line-to-neutral complex voltages VA_1ø and VB_1ø with the same angular displacement of VA_1ø leading VB_1ø by 90 degrees to solve for the per phase reactive power loss, then multiply by three to calculate the total three phase reactive power loss since it is a three phase transmission system:

Q_1ø = (VA_1ø-VB_1ø)^2/X
Q_3ø =3·Q_1ø
Q_3ø = 3·(VA_1ø-VB_1ø)^2/X

Or you could solve by calculating the current first using Ohm's law then solve for reactive power like others have pointed out in this thread similar to the NCEES solution. All roads lead to Rome.

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I have a question about this problem. Theoretical power was given as 5000 MW how would you normally determine that?

Would that be :

P = (3|V_s|*|V_r|*sin lambda) / X

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On 9/26/2018 at 9:50 AM, jnspark said:

I talked to a co-worker, who took the April test, and he explained to me the "j" that is dropped in the NCEES equation makes it purely reactive and therefore the I^2 * R equation becomes the reactive power.  All that confusion for the lack of the "j" in the denominator.

Thanks BirdGrave, but per chance did you forget the root 3 in the Current equation since these are line-to-line?

It listed the problem as being line current so I just neglected to add the sqrt(3) as I knew it would be multiplied and then divided out again.  Sorry for the confusion.

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