# Insights AC Power Analysis – Part 2, Network Analysis - Comments

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1. Mar 30, 2016

### Staff: Mentor

2. Mar 30, 2016

### ProfuselyQuarky

Part 1 was amazing! Can't wait to dig into this one

3. Mar 30, 2016

### Greg Bernhardt

Very interesting topic I never gave much thought to. Looking forward to the third part.

4. Mar 30, 2016

### Staff: Mentor

We all take infrastructure for granted as long as it's working.

5. Mar 30, 2016

### Staff: Mentor

Jim Hardy suggested a mechanical analogy that did not make it into the article.

Imaging a cylindrical rotating shaft. How do we know if it is transmitting power from one end to the other?

Stop the rotation, and paint a dotted line along the top surface. Then rotate again with you riding on the shaft like a rider on a horse. The shaft appears stationary to you. If the shaft is transmitting power from my end to the far end, it will twist and I can see the twist because of the dotted line. If the far end ships power to my end, the shaft twists the other way. Twist is another way of saying that the angle at my end is different than the angle at the far end.

To increase the twist, i make the RPM at my end slightly faster than at the far end for a short duration. To do that, I must temporarily put in more power at my end, then return the power to the previous value.

In the steady state, when twist is not changing. I can state with certainty that the RPM at my end is identical to RPM at the far end, and I don't need a computer to figure that out.

It would be possible to send the same power down the shaft at different RPMs, so making the twist stop changing and making the RPMs return to normal, are two independent goals.

What happens if load at the far end increases? Both ends of the shaft will start slowing down. I'll have to put more power in at my end, until the angle stops changing and the RPM is back to normal. When I get there, I should see that the twist has increased, and that the extra amount of power I need to achieve that, exactly matches the increase in load power.

Which angle is zero? It doesn't matter. Twist is the same, no matter what angle you choose to be zero.

A shaft has only two ends, but the power grid has many nodes with an angle at each node, so the analogy is not perfect. Analogies never are. The analogy is also primarily DC, and it does nothing to help us understand VARs, unless we start adding springs, flywheels, and oscillators making a simple analogy complicated.

6. Mar 30, 2016

### donpacino

I like those two articles. They provides a good explanation of a concept that can be very confusing.

Have you thought about writing on piece on the effects of an increase in distributed loads such as windmills, solar panels, home battery banks (for off peak energy use) and plug in hybrid/electric vehicles on the grid? Particularly their effect on the operation and operation philosophy of grid control companies?

The reason I ask is I feel with your background you might be able to tackle that problem.

7. Mar 30, 2016

### Staff: Mentor

Thanks for the kind words.

I'll give that some thought. If I did write such an article, it will cover both engineering and economics. As you can already see perhaps, discussion of the grid is inherently multidisciplinary.

8. Mar 30, 2016

### jim hardy

Sure glad we have a genuine power system guy here . I've only looked over the shoulders of some.

Electrical, mechanical, control theory(downright right burly math - search on "Power System Stabilizer"), and a rapidly increasing share of computer science.

Last edited: Mar 31, 2016
9. Apr 1, 2016

### larsa

Could you please explain a bit more the proposition that flow of reactive power is proportional to the magnitude of voltage difference between nodes?

10. Apr 1, 2016

### cnh1995

Let me see if I understood it correctly.
Suppose a purely resistive load R is fed from an ac supply V through a transmission line with reactance X(negligible resistance). Transmission line reactance X is fixed. Now if load R is increased, more current is drawn from the supply. This causes a voltage drop across X and load voltage is reduced, increasing the difference Vsource-Vload. If the load is decreased, voltage across the load increases and Vsource-Vload decreases, reducing the reactive power flow.

11. Apr 1, 2016

### larsa

Thank you for your comment. But if resistance of load is responsible for reactive power flow, what is responsible for flow of active power?

12. Apr 1, 2016

### donpacino

imagine an AC voltage source with an inductor and resistor in series. The inductor will resist change in current.
You can analyze this in the time or frequency domain. Looking at it in the frequency domain, the inductor will impart a phase shift on that signal. The larger the inductor is in relation to the resistor, the closer that phase shift will move to 90 degrees. Keep in mind at 90 degrees, the power would be purely reactive. At 0, the power would be purely real (at zero it would just be a resistor).

A phase shift translates to a time delay in the time domain. with no time delay the sin waves will be the same (pure real power. With some time delay the power will lag slightly, because one signal lags, the difference between the 2 voltages will increase, proportionally to the reactive power (the closer to 90 degrees you get, the greater the instantaneous difference, and the greater the reactive power).

note: i personally think that power is driven by the voltage and current, and the difference between voltages. But thats how I think of it. Voltage is MUCH easier to measure than current.

13. Apr 1, 2016

### cnh1995

The difference between angles of Vsource and Vload i.e. δ determines the active power. At δ=90°, you get maximum active power. But that reduces the terminal voltage much below the permissible limit and satbility is compromised. So, δ is kept 15-20°.

14. Apr 1, 2016

### larsa

What you wrote is helpful. If i understand your point, the voltage difference is due to a time difference between generator voltage and load voltage. But this phase/time difference is the cause of the real power as well according to the elegant analysis of the article which i don't understand

15. Apr 1, 2016

### donpacino

ahhhh. so I answered the wrong question... my apologies.. I'll try to get you an answer

edit. to be honest. I'm having trouble coming up with a simple explanation. Einstein said if you can't explain it simply, you don't understand it well enough. I guess I need to hit the books.

Last edited: Apr 1, 2016
16. Apr 1, 2016

### cnh1995

Source voltage is 10V. You can see increase in load angle (by increasing the load) increases the active power (from bottom to top) but reduces the load voltage.

17. Apr 1, 2016

### Staff: Mentor

You can see a tutorial here, http://www.pjm.com/~/media/training...son5-power-flow-on-ac-transmission-lines.ashx

and a derivation of the equations here, http://home.eng.iastate.edu/~jdm/ee553/DCPowerFlowEquations.pdf

The article gives $P=\frac{V_A\cdot V_B}{X}\sin(\theta)$.

The equation for Q is $Q=\frac{V_A\cdot (V_A-V_B)}{X}\cos(\theta)$
but where $V_A$ and $\cos(\theta)$ are both close to one,
the approximate equation is $Q=\frac{(V_A-V_B)}{X}$

Note that both of these equations use only $V_A$, $V_B$, and $X$. We don't need to know the load resistance. That is why I try to discourage thinking about a load R at all. Focus on $V_A$, $V_B$, and $X$

I could assume that the load is some computer controlled device that manipulates stuff to draw the P and Q that it wants. If the load P and Q do not vary with $V_B$ then the load is not an impedance or a resistance. The load is just a load and how the load varies with voltage or frequency we don't know, but we don't need to know to calculate P and Q as above. So once again, I urge forget the load R.

18. Apr 8, 2016

### larsa

What you wrote is very helpful. I want to ask you something. Suppose that we have a generator that supplies electric power to a factory. When the workers of the factory start a motor, in order for a magnetic field to be created at the stator of the motor the stator of the generator is demagnetized. So the field current of the generator is increased in order to compensate. Is this correct?

19. Apr 8, 2016

### Staff: Mentor

• It is rare for one generator to supply one factory. Think of the whole grid providing the supply.
• The generator field is part of the voltage regulator loop. It will probably increase because extra VARs will be needed.
• The generators main response to the motor will be to increase its angle.
• Power generated does not come from the field. It is theoretically possible (but impractical) to use permanent magnets for the field. So you should stop thinking more power more field.

20. Apr 8, 2016

### larsa

Thank you for your reply. Small grids with only one or two generators running in parallel are very rare but are the rule in merchant and military navy. At these small grids the proposition that the flow of reactive power is proportional to voltage difference is still valid?

21. Apr 8, 2016

### Staff: Mentor

Yes still valid.

22. Apr 10, 2017

### ElectricRay

I hope this thread is not to old to add some comments here. I work for some years now as a commissioning engineer of synchronous generators and your document I should have read many years ago it is very interesting. But as others the two wonderfull formulas to calculate P and Q confuses me a bit. As well I like to mention that it is not rare units supply to one factory. But in this whole discussion there's not mentioned any difference between Island mode and Grid mode. Island mode you see a lot e.g. ships, countries with unreliable grid where factories disconnect when things get bad and there are more situation.

As far I understand i need more field current when load is increased (P or Qind). The reason you need to excite more when P is increased is because the terminal voltage tends to drop or sufficient reactive power needs to be exported.
Connected to the grid and assuming it is a "infinite strong grid" which means that my 1 generator can not influence the voltage of the grid result in a different scenario. If connected to such a grid and I increase my field current I will start to export more reactive power (leading current) the opposite happens when I decrease the field current.
For that reason I dont understand the two formulas. Maybe I look to it from a completely different and therefore wrong point of view. But I was taught that a big grid like we have in europe e.g. you cant change the voltage. But here we express the amount of Q in relation to voltage difference and reactance?

Is anybody willing to clarify this a bit more?

Thanks a lot

23. Apr 10, 2017

### Staff: Mentor

@ElectricRay ,
I shouldn't be surprised by your confusion. The viewpoint of the grid operators and the power plant operators are very different, and there is not much cross-training.

Connected to the Euro grid, you feel like a drop of water in an ocean. But suppose that all generators in Europe were ordered to raise their terminal voltages by the same amount at the same time. If all generators act in unison, then the behavior of the continent is more like that of a ship at sea but at a larger scale. More important and more realistic, suppose you are ordered to raise voltage and another plant on the other side of the continent is ordered to lower voltage at the same time. What would the effect of that be? Why would the grid operators order that? Those are questions that can't be answered solely from the viewpoint of a single power plant.

Consider the power grid to be like a copper window screen. Each place where two wires cross and connect, is a node (call it a bus if you like). Each strand of copper between nodes is a transmission line. The distinctive feature of the grid is that there are very many paths connecting every node to every other node (even if there are numerous broken strands).

The job of the power plant operator (or the load manager) is to focus on how many MW and MVAR are injected or withdrawn from the grid at their local node. The job of the grid analyst (explained in the Insights article), is to calculate the voltages at all nodes and the flows of MW and MVAR through all the wire strands, regardless of which nodes are generators and which are loads.

The formulas in the Insights article calculate MVAR flow as a function of the voltage difference between nodes. How do you see that in your power plant? You don't. You probably don't even have a measurement of the voltage at the "other end" of the transmission line you connect to. In fact, you probably connect to several lines so that there are several "other ends".

But from a logical sense, what happens if you raise the terminal voltage at your plant by 1%? It is reasonable that the voltage at the "other end" of the transmission line also raises. However it will probably raise less than 1%, therefore the voltage difference between the two ends of that line will increase. It is those differences used in the formulas (angle difference for MW and voltage difference for MVAR), something that you can't measure locally at your plant.

How does the grid analyst account for the generator at the power plant node? He doesn't. It makes no difference to the analyst if there is a synchronous generator with field windings, or batteries, or solar panels, or whatever. It is just a node. The formulas are written so that any amount of MW and MVAR (plus or minus) may be injected or withdrawn at any node.

So the article was written purely from the point of view of grid analysis, deliberately ignoring the details of the power plants. So no wonder that it sounds alien to you. Of course real life grid controls take account of the specific transient capabilities of each plant, but that is "advanced control" the article's ambition was to only explain "basic control."

My target audience was neither power plant operators nor grid operators, but rather people who know about electric circuits, and who get puzzled when they try to apply Ohm's law, KCL and KVL to understand how the national power grid works. I'm also a bit of a history fan, so it delights me to say that if I had written that article in 1888 rather than 2016, not a single word would have to change.

Perhaps I should write a part 4 to the AC Power Analysis series to elaborate on the questions you ask. It could be fun and useful for cross training. "Ying and Yang: Grid operations for power plant operators, and power plant operations for grid operators."

Last edited: Apr 10, 2017
24. Apr 10, 2017

### ElectricRay

@anorlunda ,

Wow what a very nice explanation! I hope you were not offended maybe I wrote my message a bit harsh but that was surely not the intention. I just love my job and always want to learn more. Funny thing I noticed here that i even didnt know that there is a clear difference between a transmission line or a bus (which i see the most). In my work I dont think further as the differential protection of a step-up strafo and synchronization to the grid. We pump in the net what is allowed or requested.
This is a completely different point of view as for someone who doing grid analysis. This is very clear now to me, which must be an awesome job by the way ;)

But this all makes me curious how the gird analyser (if so) communicates) with the power plants? I would think there must be some communication somewhere. As loads fluctuates during the day and seasons like rush hour, factories starting in the morning, summer & winter differences.

Im also very happy to know that it was Steinmetz who "invented" the way we can calculate easy with loads. Im using this on a daily base without knowing where it came from. So thanks for that as well.

I can't look in your agenda but a follow up on your articles would be very cool. But will create most probably more interesting questions en will surely be a pleasure to read.

25. Apr 10, 2017

### Staff: Mentor

Thank you for the questions. As you can tell, I like these subjects.