Reactive power compensation in power system

[cnh1995]

I am trying to understand the solution provided to a numerical problem based on reactive power compensation.

Here's the problem:
Two buses A and B are connected by a transmission line of reactance j0.5 pu. Voltages at both the buses are 1 pu. Bus A is connected to a generator. If the complex power demand at bus A and bus B is 1 pu each, what is the rating of the capacitor bank connected at bus B. (There is one capacitor bank at bus 2).

Since there is no real power source at bus B, all the real power at bus B must be coming from bus A, through the transmission line. Fine.

In the solution provided, they say "since voltages at both the buses are 1 pu, real power at bus B is also 1 pu." I am having trouble understanding this. How can we assume this? The question says "complex" power at bus B is 1 pu. Why is real power at bus B 1 pu?
(Final answer given: rating of the capacitor bank= 0.268 MVAR).
Are they asking for the "minimum" rating?

Thanks a lot in advance!

 

[anorlunda]

There must be more in the problem statement than you posted.

With zero resistances, real power at B must be the same as at A. Where else could the power go?

 

[cnh1995]

Here's the original problem (Q 29).

 

[magoo]

For each bus, you can write:

S = V I * where I* equal I-conjugate

So, at bus 1 (or A), V = 1/0 or 1 pu at an angle of 0 degrees.

Since V = 1 and S = 1, I* must equal 1 + j0 for the load at that bus. So, I also = 0.

At bus 2, S is also equal to 1 pu.

The maximum power transfer equation for what is sent from bus 1 to bus 2 is:

P1-2 = V1 * V2 / X * sin(θ1-2) where V1 and V2 are just magnitudes

So, with 1 pu power demand at bus 2, you can solve this equation:

1 = 1 * 1 / 0.5 * sin(θ1-2)

or

sin(θ1-2) = 0.5

Inverse sine tells you θ1-2 represents an angle of 30 degrees

Therefore, bus 2 lags bus 1 by 30 degrees. Sending bus always leads the receiving bus.

So with S2 = 1, what is the reactive power requirement at bus 2?

We know that I12 = (V1 – V2)/jX

I12 = (1 – 1/-30)/j0.5

= (0.134 + j 0.5)/j0.5

= 1 - j0.2679

Since the real part (1 pu) is sent from bus 1, the reactive part is 0.268 pu, which the capacitor bank supplies.

 

[cnh1995]

For each bus, you can write:

S = V I * where I* equal I-conjugate

So, at bus 1 (or A), V = 1/0 or 1 pu at an angle of 0 degrees.

Since V = 1 and S = 1, I* must equal 1 + j0 for the load at that bus. So, I also = 0.

At bus 2, S is also equal to 1 pu.

The maximum power transfer equation for what is sent from bus 1 to bus 2 is:

P1-2 = V1 * V2 / X * sin(θ1-2) where V1 and V2 are just magnitudes

So, with 1 pu power demand at bus 2, you can solve this equation:

1 = 1 * 1 / 0.5 * sin(θ1-2)

or

sin(θ1-2) = 0.5

Inverse sine tells you θ1-2 represents an angle of 30 degrees

Therefore, bus 2 lags bus 1 by 30 degrees. Sending bus always leads the receiving bus.

So with S2 = 1, what is the reactive power requirement at bus 2?

We know that I12 = (V1 – V2)/jX

I12 = (1 – 1/-30)/j0.5

= (0.134 + j 0.5)/j0.5

= 1 - j0.2679

Since the real part (1 pu) is sent from bus 1, the reactive part is 0.268 pu, which the capacitor bank supplies.

Thanks @magoo for your answer. I understand the math here.
My question is, how do I know that the 1pu demand at bus 2 is the active power demand? How can I know that S_B=1+j0 pu and not 0.8+j0.6 pu (or something else)?

I understand that there is no generator at bus B, hence, its 1 pu power demand is fulfilled by bus A through the transmission line. From the given data, how should I determine that S_B=1+j0 (i.e. S_B has no reactive component)?

 

[magoo]

The rule of thumb I was taught was that:

- A difference in voltage angle between two busses results in MW or power transfer

- A difference in voltage magnitude between two busses results in MVAR or reactive power transfer

Since we don’t have a difference in voltage magnitude, there will be no reactive power transfer.

We know that P1-2 is 1 pu. Therefore, the reactive load is supplied by means of the capacitor bank at Bus 2.

 

[cnh1995]

A difference in voltage magnitude between two busses results in MVAR or reactive power transfer

Q=|V₁||V₂|cosδ/X_L-V₂²/X_L.

This means, even if voltages of the two ends are equal, there can still be some reactive power flowing, depending on δ.

 

[cnh1995]

What if δ=0° and the load at bus B is purely reactive? This makes P_AB=Q_AB=0. But since load on bus B is reactive, S_B=Q_B=1pu=0+j1.
In this case, entire reactive power demand on bus 2 is fulfilled by the capacitor bank, so its rating should be 1 pu.

Is it possible that they are asking "minimum" rating of the capacitor bank?
This makes sense, because if S_B=P_B= 1 pu, Q_B=0. So the reactive power supplied by the source at bus A is absorbed by the capacitor bank at B, and no reactive power flows to the load at bus B. This will be the minimum reactive power required to be supplied by the capacitor bank.

 

[magoo]

Yes. I agree with you.

I interpreted your problem as a transmission line, like 220 kV or 400 kV. Our transmission lines in the U.S. are typically interconnected so that in many situations you can control the voltage magnitudes at each end to be close to 1 pu and thereby minimize var flow.

If the line is radial in nature, then you would get a voltage drop in both magnitude and phase angle from A to B or 1-2.

I didn't get a chance to deal with your latest post which I just received and will look at it tomorrow.