# Automatic Load Frequency Control in power system

1. Nov 15, 2016

### cnh1995

I am studying ALFC and I have a few questions about it.

Consider the flyball governor in this system. It senses the changes in speed of the turbine. If load is increased, the balls will slow down and will move inwards, moving point B upwards. Now, this will open some additional valves to allow more fuel. What is the role of speed changer in this? Basically, what is a speed changer? How is reference power set in the speed changer? Please explain.
Also, is it correct that the primary ALFC loop (containing governor) is used to prevent sudden changes in frequency and there remains some steady state error in frequency even after opening or closing the additional valves?

Last edited: Nov 15, 2016
2. Nov 15, 2016

### Staff: Mentor

I'm glad to hear that you're studying it. First, let's clean up the nomenclature. Speed governors work inside one power plant. The central control for all the power plants in the region, used to be called LFC, later ALFC, and most modern AGC (for Automatic Generator Control).

Second, governors work with proportional control. That means that frequency error is proportional to power. The reciprocal of constant of proportionality is called droop. The classical value for droop was 5%. That means a 5% change in speed gives 100% change in power, or a gain of 20:1 per unit power / per unit speed.

I'm sure that there are some mechanical flyball governors still in service, but most plants do it electronically or digitally today. The "speed changer" sets the speed reference point (which is proportional to power reference point). Many plants call it the speed/load changer to emphasize that it does both functions at once. During startup, before synchronization with the grid, it changes speed. After synchronization, the same adjustment it is used to change load (i.e. generation).

That picture is of James Watt's original governor. The turnbuckle on the vertical member to the right could be used as the speed changer. If you adjust the turnbuckle, you change the amount of power without changing speed. An adjustment nut on the top of the flyball assembly could do the same thing.

Take away the word "sudden" from your description; frequency is incapable of sudden changes.

The governor acts to change power in response to speed changes. A proportional controller does not force the speed error to return to zero, so yes an error remains in steady state. The key thing to remember is that speed governors act locally, and swiftly. They do not need information other than the locally measured frequency. AGC does need central measurements and remote communications, and it controls all generators in the area, not just one. ACG is necessarily slower than governor action.

See the PF Insights articles.

https://www.physicsforums.com/insights/what-happens-when-you-flip-the-light-switch/
https://www.physicsforums.com/insights/ac-power-analysis-part-1-basics/
https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/
https://www.physicsforums.com/insights/ac-power-analysis-part-3-cyber-resilience/

The central grid control does two main things. It determines the optimum allocation of total load to the various generators and sends that signal to the "speed changers" in each plant. We call them power base points in the central control. In the old days, we call this function ED for economic dispatch.

Central control also biases the power base points in response to grid frequency and interchange power errors. That is what we call AGC. AGC forces the grid frequency and interchange errors to zero in the steady state. Once in a while, all the control centers in the entire interconnection will change the frequency set point slightly for a short period of time to correct the time on electric clocks.

All of this was laid out in the 1959 book Economic control of interconnected systems by Leon Kirchmayer. Kirchmayer was my first boss when I graduated from college.

3. Nov 15, 2016

### cnh1995

Thanks for the detailed explanation! It is really helpful.
And that is how you "set the reference power" in speed changer?
So the speed changer responds to both the governor and the central grid control? I guess that has something to do with the 'raise' and 'lower' commands. Is it done by some device, like a motor?

4. Nov 16, 2016

### Staff: Mentor

Yes exactly, at least in those othe fashioned systems. Today it its mostly electronic or digital.

5. Nov 16, 2016

### cnh1995

This governor is in our syllabus. So, when the turnbuckle is adjusted for setting reference power, is there some additional controlling force that is applied to the governor flyball mechanism to maintain its speed?
Also, I read that the reference power set is higher than the actual power, which increases the proportional error and more fuel can be allowed on the turbine. Is this reference power setting always constant for a machine?
I read this expression from block diagram:
Input to the governor block was written as
ΔPg=ΔPref-Δf(s)/R. I can understand that the term Δf(s) is a direct feedback from the electrical grid but where does this ΔPref come from? Why change the reference power?

6. Nov 16, 2016

### Staff: Mentor

You're missing an important point. Before synchronizaiton, the governor controls speed. After synchronization, the generator is locked to grid frequency, and a singe generator can do almost nothing to change the frequency of the entire grid. Therefore, when synchronized, that same speed control changes power, not speed. You should review the step-by-step chronology in the first PF insights article.

Yes, if the set point is the same as the feednback, the output would be zero, meaning no steam or no fuel, zero power. Therefore, in the steady state an error remains. The error times the gain is proportional to the steady state power.

Pref is the speed/load changer we were talking about. It is adjusted by the operator and also by the grid (i.e. the raise/lower pulses you mentioned.)

I think you are getting confused by the different names used for these things. Historically these things have picked up many names over the years. We just have to live with that. Learn to focus more oh the topology of the feedback loops rather than the labels of the components. For each loop, what gives you a steady state?

7. Nov 16, 2016

### cnh1995

Thanks.
Here's what I think have understood so far.

Say a generator rated at 100MVA is delivering its rated power at 50Hz frequency and governor droop is 5% (which means no load frequency will be 105%).The speed changer setting is such that ithe machine will deliver 100MW at 50Hz. Now, if the load in the grid increases, this generator will slow down. The governor will sense the reduction in speed, which is say 1%. So the linkage mechanism attached to the governor will operate the speed changer (turnbuckle) and the power input will be increased by 20%.

If the load in the grid is constant and the generator wants to take up more load from the grid, then the speed changer will be operated by a motor to open additional valves. This is how alternator takes up load after synchronization with the grid.
Is this correct?

I read the first PF insights article you mentioned. It says that after about 15 seconds, central generation control unit senses the changes in speed and issues new set points to the individual generators. Does it include setting of Pref? When the speed changer receives this change in Pref, does it immidiately set the new power reference? Doesn't that affect its current operation? I mean, if the generator is supplying 100MW when the Pref signal is received from the central control, and the new Pref is 120MW, how does the speed changer respond?
I know I am asking too much but I can't get my head around many things and I am having trouble correlating various blocks of the system.

Last edited: Nov 16, 2016
8. Nov 16, 2016

### Staff: Mentor

You are making it too difficult than need be. A two-step approach is easier to understand.
1. One generator connected to an infinite grid. The grid frequency is always 50 hertz, you can't change it. For the generator to run at 100% power, the speed reference must be set at 105% (or 52.5 hertz). Then the 5% error between set point and speed, times the gain of 20 gives 100% power. To reduce power to 50%, reduce the speed ref to 102.5%. Load changes on the infinite grid are always 0% of grid load, so they frequency never changes, and the frequency feedback to the generator never changes.

2. All grid generators have the same droop and the same Pref. Now the grid is not infinite, but rather has capacity equal to the sum of all generator ratings by definition. Now a load change causes frequency to start dropping. Every generator responds with a 20% change in power per 1% change in frequency, until we come to a new steady state where power generated matches load. If the load increase was 10% of grid capacity, then in the steady state, frequency would sag 1%/20 or 0.05%. Then the grid AGC control reacts. It increases the Pref of every generator by 10%, and frequency returns to 50 hertz.
In #7 it sounded like you were mixing up the cases with one generator versus all generators, making it harder. Keep those two cases separate in your head, and it should be easier.

Note that the above text could be expressed in terms of Pref rather than speed ref. They are the same thing except for the numerical value of 20.

Do you see why power engineers are fond of "per unit" expressions. One generator with a capacity of 1 MW and another with a capacity of 1000 MW, both could run at 50% power and both could respond with a 10% change up/down. In real units the two are a factor of 1000 different, but in per unit they are the same. If total load increases by 10% then after AGC acts, every generator responds by increasing its generation by 10% regardless of the units rated capacity.

9. Nov 16, 2016

### cnh1995

Thanks a ton! It is much clearer to me now. I will do some more reading and post if anything is unclear.

10. Nov 16, 2016

### cnh1995

Since the new frequency is less than 50Hz, the error between reference speed and actual speed will be more than 5% and this amounts to a power more than 100%. Is this why the speed changer does not act during this period and only the governor works? To make the speed changer work, the droop line should be moved up and hence, it will work only after Pref is increased, which is done by the AGC a few seconds later.
Is this correct?

11. Nov 16, 2016

### Staff: Mentor

It may try to make more than 100% but it won't succeed.

To make your example work with load changes in either direction, start at initial point $0\lt X \lt 100$%

12. Jan 17, 2017

### cnh1995

Before the grid AGC reacts, the grid is operating at a reduced frequency (0.05% sag) and the power output of each generator is now 10% more (than its previous output before increase in the grid load). So, before AGC reacts, each generator is delivering more power at reduced frequency. Is that right?

Now, when the AGC reacts, speed reference of each generator is increased and frequency is brought back to 50Hz. But what happens to the power now? Does it increase too? Because, to bring back the frequency to 50Hz, speed will have to be increased, which means more mechanical torque should be applied. This is increasing both torque and speed. So, when the frequency is brought back to 50Hz, wouldn't the power output increase too?

Or does AGC increase the speed reference by keeping the power output constant? If yes, how does it do that?

What happens to the power output during this change in speed changer setting?

One more question, is terminal voltage assumed to be constant throughout this process?

Last edited: Jan 17, 2017
13. Jan 17, 2017

### dlgoff

#### Attached Files:

• ###### NERC Balancing and Frequency Control.pdf
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14. Jan 17, 2017

### Staff: Mentor

Yes, all voltages are presumed constant during this hypothetical lesson. We also ignore power losses.

For your other questions, I think a numerical example may be easier to understand than words.

The speed governor is a simple proportional control. $P_{gen} = P_{ref}-20*(f_{error})$
Consider both power and frequency scaled in per unit.

1. Initial state: $P_{ref}= 0.9, f_{error}=0)$ therefore $P_{gen} = .9-0=.9$ Since it it steady, $P_{gen} = P_{load}$
2. Load increases , frequency sags by .005 per unit. $P_{gen} = .9-20*(-.005)=1$ If frequency stops decreasing, then $P_{gen} = P_{load}$ must be true again.
3. AGG increases $P_{ref} to 1$, $P_{gen} = 1-20*(-.005))=1.1$ That is bigger than the load so frequency starts increasing again.
4. Final state: $P_{ref}= 1, f_{error}=0)$ therefore $P_{gen} = 1-0 = 1$ Since it it steady, $P_{gen} = P_{load}$

The beauty of per unit, is that the numbers stay the same no matter how we scale the size of the system. We could be talking about a 0.1 MW change on a 1 MW system or a 100 MW change on a 1 GW system. With multiple generators of different sizes, each generator responds the same in it's own per unit base.

This whole example could be repeated with some import/export of power to other areas, but the principles remain unchanged.

The document @dlgoff recommended should be helpful.

If you want to understand these things, I recommend focusing more on math than words, and focusing more on the time rate of change of quantities than on absolute values. In other words, some things must balance to keep things steady. If they are not balanced, what changes in which direction, and what are the equations to establish a new balance?

15. Jan 18, 2017

### cnh1995

But I am having trouble understanding this from the machine point of view. (I am sure I am missing something very basic, but can't find what it is..)

The 0.05% sag will give 49.975 Hz frequency.
Now,
1)Pref1=Pg1=0.9 pu at 50Hz.
So, Torque T1=0.9 pu at frequency f1=1pu.

2) Load is increased by 10%, which means new load is 1 pu.
So, we "finally" want Pref3=Pg3=1 pu at
frequency f3=1 pu. (50 Hz)
This means, torque T3=1 pu.

3)When governor reacts, Pref2=Pref1=0.9 pu, Pg2=1 pu but f2=0.9995 pu. This means torque T2=1.00050 pu.

4) When AGC reacts, finally, we have
Pref3= Pgen3= 1 pu, f3=1 pu and torque T3= 1pu.

But we can see that torque T2 is more than torque T3.
Does this mean that when AGC increases Pref, mechanical input is first increased and then decreased?

16. Jan 18, 2017

### Staff: Mentor

I congratulate you on achieving so much understanding already without using math. But if I answer your question in the manner posed, it will just lead to another question and another. You are wasting my time and your time because your approach is so much more difficult than using mathematics.

My recommendation is to temporarily stop studying AGC and control systems, and spend your efforts to first learn basic calculus, and especially differential equations. Once you have those, then come back to power system dynamics and I think that you will find that is it much simpler than you can imagine today, and you will be able to calculate transient curves like the ones below, step-by-step.

17. Jan 19, 2017

### jim hardy

Doesn't this suggest that ?

Or did i fail to read it thoroughly (again)? Seem's i'm always proving the adage "Haste and Waste are Birds of a Feather."

Anyhow
my governor was analog hydraulic, a variation on the theme of flyball.
It measured speed as oil pressure via a centrifugal pump literally inside the turbine shaft
and Pref was the force exerted by a spring . An electric motor and leadscrew compressed or relaxed the spring to set the Pref force.
Oil pressure acting on area of a piston made a force proportional to speed , opposing the Pref spring. Imbalance between those forces admitted oil to a hydraulic amplifier that controlled the steam valves.

So yes, whenever speed error changes the valves will move.

Thinking about your question, does it take some extra power to accelerate the inertia of machinery?
Use your everyday experience - you mash your car's gas pedal to get up to cruising speed then back off.

As anorlunda said, thinking about acceleration speed and inertia soon leads to differential equations .

old jim

Last edited: Jan 19, 2017
18. Jan 19, 2017

### dlgoff

As far a maintaining frequency, the generator governor is considered "Primary Control". (ie, in the time frame of seconds) But for maintaining frequency for a control area (where ACE is being calculated there), that's where AGC comes into play. The setpoint on the generators (when set to automatic mode) is determined where both proportional and integral control is used. This is considered "Secondary Control". (ie, in the time frame of minutes)

From the NERC Balancing and Frequency Control document I posted:

19. Jan 19, 2017

### cnh1995

Yes.
Yes. I was not sure but later I understood it.
During initial normal operation, the torque is T1.
During reduced frequency operation, torque T2 is more than T1 since frequency is reduced but power is the same.

After Pref is increased, the initial torque T3 is more than T2, which suggests opening of additional valves. This will accelerate the machine and the governor flyballs will move downward until the frequency error is zero. Downward movement of the flyball collar will close the valves and the new steady state torque T4 will be such that T1<T4<T2<T3.