from the Power system department...
Real power flow
 Posted by Anonymous on 17 March, 2007 - 12:27 am
It is usually said that the real power flows from the end where the magnitude of voltage phase angle is greater (delta 1 > delta 2). The reactive power flows when magnitude at sending end is higher than the receiving end voltage magnitue i.e |Vs| > |Vr|. Help required in this regard. Any equations?

 Posted by Phil Corso, PE on 24 March, 2007 - 1:10 am
Anonymous, I'm not exactly sure what your question is, but I'll try to answer it. Let me know if my answer satisfies your query!

I have interpreted your question as, "Given a power-cable (or transmission-line) carrying current, and its end voltages, as well as the anglular displacement between them, are known, then does real power flow from the higher-voltage end to the lower-voltage end?"

The simple answer is... not necessarily! What is required is knowledge of the nature of the load. For example, if the load current is lagging (inductive load), then the direction of power "flow" is from the higher-voltage end toward the lower-voltage end. The resultant voltage-drop is positive. But, if load current is leading (capacitive), then the resultant voltage-drop COULD be negative. Hence, power "flow" is now from the lower-voltage end towards the higher-voltage end. The equation follows but I warn you that it may be difficult to understand without the use of the dreaded V-word, vectors, or the more modern P-word, phasors!

First, some terminology. Let,

o E1 = voltage at one end, ph-ph.
o E2 = voltage at the other, ph-ph.
o dE = the difference between E1 & E2.
o A = angular displacement between E1 & E2.

Then, Cos(A) = (|E2|-|DE|)/|E1|.

I hope this helps. If a more detailed explanation is required, then a vector or phasor diagram is a necessity. Let me know!

Phil Corso, PE (cepsicon@aol.com)

 Posted by Phil Corso, PE on 24 March, 2007 - 7:21 pm
Anonymous, remove the absolute lines from the DE value in the equation I provided earlier.

Also, Let me know if you need the equation for DE!
Phil Corso, PE (cepsicon@aol.com)

 Posted by Anonymous on 13 April, 2007 - 12:57 am
Mr. Corso,

I read from one of the technical papers:
"Active- and reactive-power flows in a power system are relatively independent. For a transmission line connecting two sources, active-power flow depends mainly on the angle between the voltages at the line terminals. Active power flows from the leading-voltage line end to the lagging-voltage line end. On the other hand, reactive-power transfer depends mainly on the voltage magnitudes. Reactive power flows from
the line terminal having higher voltage magnitude to the line terminal with lower voltage magnitude."

Is there a mathematical proof of what is said above? By which equations can we prove this? In most of the power system books, we get only max. power transfer or power transfer equations, which do not prove the above said. I shall be grateful if you could kindly help in this regard.

 Posted by Phil Corso, PE on 14 April, 2007 - 2:01 am
Anonymous, the best way for me to clarify my interpretation of your source's two postulates is by example. Two power stations, connected via a single tie-line, have the following parameters:

Station A.
o Generator provides 18MVA @ 0.55 pf, lagging.
o Local load is 6MW and 10MVAr.
o Bus voltage, Va, is 1.0 @ pwr angle, alpha.

Station B.
o Generator provides 10.4MVA @ 0.96 pf, lagging.
o Local load is 14MW and 8MVAr.
o Bus Voltage, Vb is 1.0 @ pwr angle = 0-deg.

Tie-line C.
o The tie-line must carry 4MW and 5MVAr to Sta B.
o Line impedance, Z, is 0.05 pu.

Real Power Flow.
Bus Va voltage power angle is calculated from the fact that power input to bus A equals power input to Bus B. Hence, Pa = Pb = (|Va||Vb|/Z)Sin(alpha). Then, alpha=30 deg. Thus, your source's first postulate is correct, i.e., power flow is from Bus A to Bus B, and Va does have the larger power angle!

Reactive Power Flow.
Now determine the reactive power for each bus.
Qa=((|Va|^2)/Z)-(|Va||Vb|/Z)Cos(alpha). Then, tie-line reactive power, Qc, equals 2xQa, or about 5.4MVAr.

Note that both busses have equal voltage magnitudes. Thus your source's second postulate is incorrect!

If the above is not clear enough, let me know thru control.com, or directly via my e-mail address.

Regards, Phil Corso, PE (cepsicon@aol.com)

 Posted by krishnan on 16 April, 2007 - 9:15 pm
Phil, I agree with your first postulate, but I had some doubts about the eg you have provided is a simple power system where real and reactive power of STN b is met by STN A and tie feeder.

But if we see the values for STN b
Generator provides 10.4MVA @ 0.96 pf, lagging.

> o Local load is 14MW and 8MVAr.
> o Bus Voltage, Vb is 1.0 @ pwr angle = 0-deg

Is it possible to have 1pu value for voltage at bus b as the KVA requirement is much more, which should eventually lead to reduction in bus voltage at B (theoritically)?

If that's the case the reactive power transfer seems to follow from high voltage mag (A bus and tie feeder assuming tie is at 1 angle0) to low magnitude (B Bus).

 Posted by Phil Corso, PE on 16 April, 2007 - 11:17 pm
Krisnan, to recap:

1) Gen B is supplying only 10 of the 14MW load connected to Sta B. Therefore, Gen A is supplying the balance, 4MW, thru the tie-line.
2) Gen A is providing 6-MW for Sta A load, and 4MW thru the tie-line. Thus, its total output is 10-MW.
3) Total generation is 20MW, 18MVAr @ 0.56 pf.
4) Total load is 20MW, 18MVar @ 0.98 pf.
5) Yes, Sta A bus-voltage is 1.0pu @ 30.0deg.
6) Yes, Sta B bus-voltage is 1.0pu @ 0.030.

I just completed a study that demonstrates that Postulates "1" and "2" can't always be true. The configuration follows:
o Sta 1 generation is connected to the HV-grid thru a step-up transformer. No local load.
o Sta 2 generation is connected to the HV-grid thru a step-up transformer. No local load.
o Sta 3 has a syn-motor connected to HV-grid thru a step-down transformer.
o Sta 4 has no local load, but used only as tie.

The study shows that bus-voltage angles do not support Postulate "1!" And I mean neither the LV-busses, nor the HV-busses! Furthermore, the bus-voltage magnitudes differ by less than 1%, thus negating Postulate "2!"

In closing, one shouldn't rely on bus-voltage angles and magntude to determine the direction of real and reactive power flow.

I can e-mail a copy, if you are interested!

Regards, Phil Corso, PE (Cepsicon@aol.com)

 Posted by krishnan on 18 April, 2007 - 12:26 am
I would be glad if you could send me the copy.

krishmenon02 @ gmail. com

 Posted by Phil Corso, PE on 17 April, 2007 - 11:50 pm
Krishnan, I believe you have confused bus-voltages with generator-emfs. The latter, obviously, are not known.

Please note that pf value(s) for generation and load are incorrect. The 20MW and 18MVAr values should have been 26.9 MVA @ 41.4Deg, yielding a 0.74pf for each category!

Regards, Phil Corso, PE (Cepsicon@aol.com)

 Posted by petros on 10 October, 2012 - 6:03 am
If a generator is supplying a load (three phase system) and the load has a leading power factor, can we determine the direction of the active and reactive power?

 Posted by Phil Corso on 10 October, 2012 - 1:04 pm
Petros... the answer is an emphatic YES!

However, please clarify your question; are you referring to the generator only or the "Station Bus" to which it is connected!

Regards,
Phil Corso

 Posted by CSA on 10 October, 2012 - 2:54 pm
A synchronous generator supplying a load, by definition, is "pushing" active power out of the machine to supply the load requirements.

From a generator perspective, a leading power factor means--and I'm gonna take a hit for using this terminology, but, here goes--reactive "power" is "flowing into" the generator.

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