Pelton turbine model when hydroelectric station works on grid

[vampslayer]

Hi!
When my Hydroelectric station is not connected to a GRID (islanded mode)
I control the frequency or the speed, and here are the equations for my turbine:
http://oi46.tinypic.com/3127ixe.jpg
Look the third equation.
I built that in Matlab Simulink, and my out put is turbine speed n[o/min]
http://oi48.tinypic.com/1zqc3o5.jpg

That's all good. But now I want to make simulation model when my hydroelectric station is connected to the GRID.
In that case I control the Power and my speed is constant, f=50Hz so n=const.
So I need equations for that case, and If I look my third equation
dw/dt=(1/J)*(Mv-Mtr-Mt)
dw/dt=0(because speed is const) -> Mt=Mtr-Mv
So my equation in this case is Mt=Mtr-Mv ?
Mt is load torque and I think it should be my output if I want to control the power
Mv- is Torque developed on turbine blades(Torque of water)
Mtr - friction torque

I just want to check am I right about this guys?

 

[jim hardy]

Your equation looks right to me, steady state output is equal to input minus friction losses.
Any torque imbalance goes into accelerating the machine, and you've locked ω at synchronous which is close to how they work in real world.

But I don't pretend to know a thing about Matlab.

Is your Matlab model stable?
Real generators have something called "Amortisseur Windings" to prevent small oscillations about synchronous speed, which are called "hunting".
They'd add to your torque equations a term proportional to ω-ω_n , making M_t a function of ω.

But that's a refinement for quite a bit later on when you unlock ω to make your modelling more rigorous.

Have fun with this - it looks real interesting. I hope you'll continue the thread.

It is refreshing to see your progress and increasing confidence.
I look forward to learning from you !

old jim

 

[vampslayer]

Your equation looks right to me, steady state output is equal to input minus friction losses.
Any torque imbalance goes into accelerating the machine, and you've locked ω at synchronous which is close to how they work in real world.

But I don't pretend to know a thing about Matlab.

Is your Matlab model stable?
Real generators have something called "Amortisseur Windings" to prevent small oscillations about synchronous speed, which are called "hunting".
They'd add to your torque equations a term proportional to ω-ω_n , making M_t a function of ω.

But that's a refinement for quite a bit later on when you unlock ω to make your modelling more rigorous.

Have fun with this - it looks real interesting. I hope you'll continue the thread.

It is refreshing to see your progress and increasing confidence.
I look forward to learning from you !

old jim

Ty Jim and I will continue threading, but I don't think you can learn a lot from me, I am just a student with no experience(just a lil bit in Matlab)
I'm not sure what do you mean about "Is your MATLAB model stable", but for beginning I will model my generator very simple Pgen=η*Mt*ω
and I will make feedback Pgen(that's when I control the power).

This is my turbine model when I control the power(based on the equations in the first post)
http://oi45.tinypic.com/fkqz2s.jpg

So let's see what should happen in a moment when I switch from OFF GRID mode to ON GRID mode and vice versa.
Right now, when I switch from OFF Grid to ON Grid Mode, the speed is const and Mt is rising until some value(I still don't have PID power governor, but I think here comes his functionality, power governor must keep this Mt or Pgen close to (Pref-reference value of Pgen)), but when I switch from ON Grid to OFF Grid, speed is rising and my PID speed governor must act.

 

[vampslayer]

And Jim, what do you think about how I simulate the affect of generator in OFF GRID mode.
I did it as a load torque, Mt=P/ω(Turbine see Generator through load torque)
P- power which turbine gives to generator., and that Power should be equal to the power of what load is connected to the generator.
http://oi46.tinypic.com/3127ixe.jpg
look third equation.

What do you think about this at all?

 

[jim hardy]

I did it as a load torque, Mt=P/ω(Turbine see Generator through load torque)
P- power which turbine gives to generator., and that Power should be equal to the power of what load is connected to the generator.
http://oi46.tinypic.com/3127ixe.jpg
look third equation.

What do you think about this at all?

I said earlier I think it makes sense - you summed torques and calculated acceleration.
And I think it's correct for both ONGRID and OFFGRID.
Ahh, physics from first principles is beautiful !

I'm not sure what do you mean about "Is your MATLAB model stable", but for beginning I will model my generator very simple Pgen=η*Mt*ω

I just meant is it hunting or oscillating...

So let's see what should happen in a moment when I switch from OFF GRID mode to ON GRID mode and vice versa.
Right now, when I switch from OFF Grid to ON Grid Mode, the speed is const and Mt is rising until some value(I still don't have PID power governor, but I think here comes his functionality, power governor must keep this Mt or Pgen close to (Pref-reference value of Pgen)), but when I switch from ON Grid to OFF Grid, speed is rising and my PID speed governor must act.

Not quite sure I followed all that. If power(Mt ?) is settling out at some value it sounds like you're close.
If it's continuing to rise , that doesn't sound right.

Regarding your PID controller -

Our governor was not Integral, just P with some D.
It sensed speed by a shaft mounted centrifugal pump.
Pump discharge pressure was opposed by a spring and through hydraulic amplifiers controlled the steam inlet valves to maintain constant speed.
Spring force was the speed setpoint.
The spring was adjusted remotely by a knob in control room..
To control power you just adjusted the speed setpoint by compressing or releasing the spring a little bit.

It had gain of about 33% valve travel(steamflow) per % speed error.
So at synchronous speed and no load, speed setpoint was 100% of synchronous.
At synchronous speed and 100% load, speed setpoint was 103% synchronous.
...3% speed error X gain of 33 = valves wide open, full power at synchronous speed.
If you lose load, turbine accelerates, governor closes valves at rate of 33% valve travel per % speed error, so turbine would settle to 103% of 1800 RPM = 1854 RPM with valves (almost completely)closed.
That type governor control scheme is often called "Droop Control", ie power droops when turbine speeds up.

PID governors do exist and they'll hold speed at exactly synchronous due to their integral term. They are useful when islanded (off grid) to keep customers' electric clocks correct. In fact our switchboard had two wall clocks - one from a 60hz outlet and one a battery powered quartz. Integration to keep clocks correct would be done manually... But that's just a boring anecdote...
Point was - your PID governor will probably have to have limits else it'll call for more power than your source (water pipes) can deliver. Start with a simple proportional one.

I think the proper term for PID governor is "Isochronous", as in 'same time' as with the clocks. I never worked on one of those.

See if this helps you with governors, if not ignore it.

http://www.control.com/thread/1026221805

I hope I'm not just muddying things for you.

Your model looks like a lot of fun. I'm afraid to look into Matlab - I might get hooked.

 

[vampslayer]

Not quite sure I followed all that. If power(Mt ?) is settling out at some value it sounds like you're close.
If it's continuing to rise , that doesn't sound right.

Actually this is what i get
Mt=Mv-Mtr=-Mtr
And why?, I think that's because Mv(torque of water is only acting when I use water to change speed)
When I switch to ON Grid mode(I still don't have power governor so its open feedback)
my speed feedback is not acting so Mv=0.

Of course when I add PID power governor Mv won't be 0

I will post my whole hydroelectric model Jim if you wish.
Matlab is good, and it should be very easy for someone like you, with experience

 

[jim hardy]

I envy you, for your math is still fresh !

(torque of water is only acting when I use water to change speed)

I don't understand... water acts on the wheel whenever it's admitted, even if the only opposing torque is friction - your Mtr.

So you need a minimum water flow just to overcome friction.
Any leftover torque goes into either load Mt, or acceleration dω/dt.

Now - if you shut off all water, the generator will spin the turbine by drawing power from grid. In which case yes, "load" becomes negative and equal to friction losses.

I guess I was thinking too much of normal operation and envisioning the machine. You never stood next to it and marveled at how that huge steel shaft can twist... and I never worked Matlab.

I think in pictures. So let me change my mental image for a moment and look harder at your equations forgetting the machinery.
Hmmm ( scratches head icon)
So I think now you are right, if water admission Mv is zero, then power Mt is equal and opposite to friction Mtr.
The computer has no idea whether there's a source for that power, it just reports the number to satisfy the equation. . . Of course - how silly of me ! That there's no grid from which to draw power is immaterial to that equation . This is a calculating engine not a steam engine !

Keep on please - i'll get the hang of this yet !

Sorry for so many words, but I am a plodder.

Seems you're doing fine.

old jim

 

[vampslayer]

I guess I was thinking too much of normal operation and envisioning the machine. You never stood next to it and marveled at how that huge steel shaft can twist... and I never worked Matlab.

That's exactlly what I need, someone with real life experiences, and I hope one day I will work on a real stuffs and stand next to huge steel shaft and work on it.

Sorry for so many words, but I am a plodder.

No, no it's ok, you are explaining me a lot of things, and with this stuffs(power engineering) you must be ample. So here is my whole model. I still didn't find the parameters for governors but I will. I learned in college year ago something about that, but we would have the transfer function of process, and it was in "s-domain(Laplace transform)"
http://tinypic.com/view.php?pic=35d50rl&s=6

I want to ask you now a lil bit more about this Pref(you can see on image). Who determines the amount of Pref? We are connected to the grid, so grid maybe? If grid needs X amount of power, to compensate its own needs(grid needs), then my Pref must be equal to X

Is this right?

 

[SirAskalot]

In a true system, Pref is set by the power company. That is, the amount of power the generator owner has bid into the pool for that hour/day etc. This amount depends on the market price (merit order).

First of all, what is the purpose of your simulations? Is it transient stability, steady state analysis, is it a generic or specific system you have in mind?

What your value of Pref should be depends on the purpose of the analysis. But true for most analysis the system should be in steady state at time=-0. So the input should equal the output.

As for the turbine governor controller, in most power plants connected to a regular grid, as mentioned, only proportional controlled (Kp) is used. Multiple integral controllers controlling the same system(grid) would make the system unstable. (every controller, hundreds of them, would try to oppose each other when a error/deviation occurs). The controller measures speed/frequency and acts on the error/deviation between the setpoint and measurement. That is Kp is the slope of the "droop" in a "house diagram". See this link:
http://www.ece.uAlberta.ca/~knight/electrical_machines/synchronous/parallel/house.html

When your system islands this will happen:
- First stage; Load torque will change causing a acceleration in the rotation, torque angle will oscillate and in response the power "produced" by the generator. This may cause the system to become unstable and trip the protection. Time duration: seconds to minutes.
- second stage; Turbine governor starts to react to the speed change trying to counteract the dropping/increasing frequency. Speed will settle to steady state with a lower or higher frequency than 50/60 Hz. Due to the lack of integral action. Time: several minutes. (water flow in a large hydro power plant(kaplan, francis turbine) can't change so fast due to momentum of the water in the piping. With pelton turbine this action can be faster due to the ability to shift the nozzle away from the turbine)
-third stage; System operator will change the setpoint (manually) to restore the correct frequency. Time:several minutes after the fault.

 

[SirAskalot]

In your model:
The feedback/control of the output power to Pref is not a parameter that is usually controlled. The negative feedback should come from the speed/frequency of the system and subtract from the setpoint(50/60 Hz or xxxx rpm)

 

[jim hardy]

If grid needs X amount of power, to compensate its own needs(grid needs), then my Pref must be equal to X

Islanded (or "Off Grid") : You and your load ARE the grid. It's a local grid. So when you control frequency you are matching power generation to power consumption, if any. If none you're just supplying your internal losses.

"On Grid" : Much of your customer load is rotating motors which have inertia and a speed/power curve of their own. If total grid generation and use are mismatched speed will change just as with any other machine. Power mismatch will accelerate every rotating mass on the grid.
Normally one plant is not large enough to affect grid frequency, though. That's why we call it "Infinite Bus" and treat it math-wise as an immovable object..

------------------ some miscellaneous thoughts ----

as SirAskalot said, a central dispatch office tells you how much power he wants you to pump into the grid. That becomes your Pref. In my plant we set that Pref with the governor speed control knob that adjusts that spring I mentioned.

Here's a real world example of islanded:
My site had a fossil generator adjacent the nuclear plant. They shared a switchyard.
One foggy morning we lost all transmission lines out of the switchyard. There'd been a dry spell and insulators were loaded up with salt(the lines ran near the ocean), when the fog rolled in it moistened the salt and insulators started flashing over.

The nuke was shut down, I don't remember anymore if it tripped or we were in process of heating up... But its internal consumption was forty megawatts.

The operators in the fossil control room saw load decrease to forty megawatts and thought something was awry upstate, so they raised their governor to try and send more power upstate.
Well - megawatts stayed at just about forty but frequency went up to 61 hz.
So they tried again. At 62 hz with generation still ~ forty megawatts, they realized what had happened and held frequency there. I was in the nuke plant and noticed our reactor flow was up to nearly 103%. All the pumps had sped up, of course.

So we were a small grid unto ourselves, one plant generating and one consuming, where obviously the one and only generator WAS large enough to change grid frequency.

Boring anecdote I know, I relate it only for the purpose of helping you visualize this machinery.
When you can work it 'in your head' you will have more faith in your equations.
At least that's what I have to do. When I can resolve the math with a mental model I feel more confident.

The governor on a steam turbine is fast - the steam inlet valves can travel full stroke in ~ one tenth of a second.
I would think a hydro plant must have turbine bypass valves so a sudden load rejection and inlet valve closure doesn't cause a hydraulic ram effect that'd wreck the inlet pipes..

I hope you become interested in power systems, they're interesting.
I was just a curious plant instrument guy who looked into the power system through a very small window. I don't have power system expertise, just worked around some folks who did.

old jim

 

[dlgoff]

If I can jump into this very interesting and informative thread for a minute, ...The North American Electric Reliability Corporation (NERC) has a publication that fits into this thread that's "intended to explain the concepts and issues of balancing and frequency control." BALANCING AND FREQUENCY CONTROL to ensure the reliability of the bulk power system

Here are a few visuals/pages.https://docs.google.com/viewer?pid=bl&srcid=ADGEESiI-TizqsV0cRjlLirKA_z0t7Sg1hsteOk29JX1Wu0LfXIVlhJpJXOHR3xtC-FcAieRSp3uPaWhZbrgB2wXGtf-CIHmzaHG9olSaaRKDfbNbS1tk97WA-HtdGK_7DVaZBw_hV9j&q=cache%3A_-4EPn3Ba8kJ%3Awww.nerc.com%2Fdocs%2Foc%2Frs%2FNERC%2520Balancing%2520and%2520Frequency%2520Control%2520040520111.pdf%20&docid=b11d8c36a8eae7b4e95faf80431fd2a9&a=bi&pagenumber=7&w=891

https://docs.google.com/viewer?pid=bl&srcid=ADGEESiI-TizqsV0cRjlLirKA_z0t7Sg1hsteOk29JX1Wu0LfXIVlhJpJXOHR3xtC-FcAieRSp3uPaWhZbrgB2wXGtf-CIHmzaHG9olSaaRKDfbNbS1tk97WA-HtdGK_7DVaZBw_hV9j&q=cache%3A_-4EPn3Ba8kJ%3Awww.nerc.com%2Fdocs%2Foc%2Frs%2FNERC%2520Balancing%2520and%2520Frequency%2520Control%2520040520111.pdf%20&docid=b11d8c36a8eae7b4e95faf80431fd2a9&a=bi&pagenumber=8&w=891

https://docs.google.com/viewer?pid=bl&srcid=ADGEESiI-TizqsV0cRjlLirKA_z0t7Sg1hsteOk29JX1Wu0LfXIVlhJpJXOHR3xtC-FcAieRSp3uPaWhZbrgB2wXGtf-CIHmzaHG9olSaaRKDfbNbS1tk97WA-HtdGK_7DVaZBw_hV9j&q=cache%3A_-4EPn3Ba8kJ%3Awww.nerc.com%2Fdocs%2Foc%2Frs%2FNERC%2520Balancing%2520and%2520Frequency%2520Control%2520040520111.pdf%20&docid=b11d8c36a8eae7b4e95faf80431fd2a9&a=bi&pagenumber=9&w=891

https://docs.google.com/viewer?pid=bl&srcid=ADGEESiI-TizqsV0cRjlLirKA_z0t7Sg1hsteOk29JX1Wu0LfXIVlhJpJXOHR3xtC-FcAieRSp3uPaWhZbrgB2wXGtf-CIHmzaHG9olSaaRKDfbNbS1tk97WA-HtdGK_7DVaZBw_hV9j&q=cache%3A_-4EPn3Ba8kJ%3Awww.nerc.com%2Fdocs%2Foc%2Frs%2FNERC%2520Balancing%2520and%2520Frequency%2520Control%2520040520111.pdf%20&docid=b11d8c36a8eae7b4e95faf80431fd2a9&a=bi&pagenumber=15&w=891You can get the pdf here: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&sqi=2&ved=0CC8QFjAA&url=http%3A%2F%2Fwww.nerc.com%2Fdocs%2Foc%2Frs%2FNERC%2520Balancing%2520and%2520Frequency%2520Control%2520040520111.pdf&ei=vwVeUeuRGKbz0gHHnIGwBQ&usg=AFQjCNFNovutwDN31cNKZd1hFc-Hk5H_sQ&bvm=bv.44770516,d.aWc

Okay. I'm done. Thanks.

 

[jim hardy]

Okay. I'm done. Thanks.

You seem to have a knack for finding just right graphics for any concept. They added immensely to every discussion you've joined.

So I really hope you're NOT done.

Thanks, old jim

 

[vampslayer]

Thank you all on your posts, I've read them all, and they are helping me to understand this matter.
Now I will say a bit bit closer what I am trying to do. I want to simulate situation when my hydroelectric power station is connected to the grid, and for some unknown reason link between GRID and my Hydroelectric power plant breaks suddenly. This can be caused by I don't know, some power system lines breaks or what so ever. At that situation my hydroelectric power plant works in islanded mode and my hydroelectric power plant must supply the consumers, it must compensate the GRID as much as possible. In that scenario the Load would increase a lot, and that's where I want to act. Speed would drop a lot and that's exactly what's going on in my MATLAB simulation.
When I change Load for example from 150kW to the 450kW my speed of course suddenly drops, from 1000 r/min to the 900 r/min and then governor acts and in a few seconds speed is back to 1000 r/min again. Ok, so what do I want? This drop is huge, 1000->900, that's 10% which means frequency drops from 50Hz to the 45Hz. I would like to somehow limit my speed drop to the let's say 4% or so. Do you guys have any Idea how this stuffs works and what should I do, and even if you don't I would like to read other opinions.

One more important stuff. While hydroelectric power plant is on GRID , the turbine needle is opened 100% and jet shifter is opened about 50-60%.Why? Well because jet shifter is faster than
turbine needle, and we use him to react fast. So when load increases from 150kW to 450kW jet shifter opens to the 80% or so.

jet shifter is pelton turbine deflector

 

[jim hardy]

Do you guys have any Idea how this stuffs works and what's should I do, and even if you don't I would like to read other opinions.

In my utility there were relays scattered about the grid that disconnect feeders, knocking out whole neighborhoods, at various stages of underfrequency. It was called "Load Shedding" , an attempt to quickly match load to available generation.

I saw this mentioned in a paper dlgoff linked. It gave typical mw/hz load shedding.

Our steam turbine would allow 2 minutes below 58 hz, then it disconnected itself to protect against blade resonance. So system had that long to get itself back in shape, ie load matched to available generation.

At 56.1 hz our reactor shut down to protect itself against low flow and that tripped the turbine.

Once upon a time we had a system disturbance that left a big metropolitan area way short of generation. One of our turbines failed to disconnect on underfrequency signal. When it tried to supply the whole city by itself, its shaft snapped.

Hope that's some help.

I would like to somehow limit my speed drop to the let's say 4% or so.

You're getting the hang of this !

You'll have to match load to available generation. Add some load shedding relays.
The dynamics will be fascinating - you have an inertia term so you'll see a rate of frequency decay proportional to generation shortfall, you can shed load in chunks according to that rate.


surely somebody with genuine power system experience can give you proper terminology and real world algorithms.

old jim

 

[vampslayer]

In my utility there were relays scattered about the grid that disconnect feeders, knocking out whole neighborhoods, at various stages of underfrequency. It was called "Load Shedding" , an attempt to quickly match load to available generation.

So one of the solutions of this problem is cutting off feeders/consumers or like you call it "Load Shedding"? In that case, if you look my example(Load jumps from 150kW to 450kW), you want me to shed load with load shedding relays to prevent large frequency fall. Price of this is that some areas stay out of electricity.

Our steam turbine would allow 2 minutes below 58 hz, then it disconnected itself to protect against blade resonance. So system had that long to get itself back in shape, ie load matched to available generation.

Can you explain me what is the blade resonance and what it causes. What bad can happen if frequency is below 58Hz for over 2 minutes. I know that the frequency should be 50/60Hz to keep balance between resources and customer demand but I thought that condition must be "true" just to avoid additional costs. Why would I produce more than customers need, that's waste of money. Is there any other reason for this beside the money?

At 56.1 hz our reactor shut down to protect itself against low flow and that tripped the turbine.

Once upon a time we had a system disturbance that left a big metropolitan area way short of generation. One of our turbines failed to disconnect on underfrequency signal. When it tried to supply the whole city by itself, its shaft snapped.

That's what I need, I need to prevent frequency to fall that low, but to be honest I don't know what bad that can cause. If turbine don't disconnect at low frequency it's possible that its shaft would snap?

 

[dlgoff]

Something to consider.

Pumped-storage hydroelectricity

Maybe.

 

[jim hardy]

http://www.basler.com/downloads/loadshed.pdf

http://store.gedigitalenergy.com/FAQ/Documents/SFF/GET-6449.pdf section 3 is background info.


http://www.midwestreliability.org/01_about_mro/overview/mro_manual/MRO_UFLS_Program_06-03-10.pdf

a lot has happened after I retired ; http://www.publicpower.org/Media/magazine/ArticleDetail.cfm?ItemNumber=19303


......

Can you explain me what is the blade resonance and what it causes. What bad can happen if frequency is below 58Hz for over 2 minutes.

The turbine blades are very large and under tremendous centrigugal force.
So the metal in them sees tremendous tension at speed.
Designers do not want the turbine operated at a frequency that can excite vibration at one of a blade's natural frequencies. Fatigue can set in quickly.
Our turbine had that limit on underfrequency operation, 2 minutes at 58hz or lower.
If a blade fatigues and breaks off, it makes quite a mess of other blades as it works its way out of the turbine. It is not uncommon for them to enter the condenser at considerable speed and shear some tubes, then you have both a turbine and a condenser to repair.


One must also be sure nothing is connected to the generator that could excite the natural torsional frequency of the shaft, around 7hz for ours. Generator's voltage regulator should be kept relatively slow.
I would think numbers for a waterwheel are different than a steam turbine.
You don't need to model that for anything you've asked - just be aware such things exist.
Search on "sub synchronous resonance" when you have some spare time.

Power systems are an interesting machine.

 

[jim hardy]

I know that the frequency should be 50/60Hz to keep balance between resources and customer demand but I thought that condition must be "true" just to avoid additional costs.

That is the standard frequency. As machines got larger their design became more refined and less tolerant of off-frequency operation.

Why would I produce more than customers need, that's waste of money. Is there any other reason for this beside the money?

my reference to 58 hz was for our 60 hz system.
At 58 hz we are making less not more than customers need..

Maybe I'm confusing you with these extras.
Sorry, i Just wanted to make it interesting, if instead I'm causing you pain i'll stop.

I guess I'm enthusiastic to see somebody else interested in power systems.

Keep us posted about your simulation.
You can refine it as much as you want to, one step at a time.
Unlike a real one your simulated generator won't be hurt by what would amount to mechanical abuse.
So have fun with it!

old jim

 

[dlgoff]

http://www.basler.com/downloads/loadshed.pdf

http://store.gedigitalenergy.com/FAQ/Documents/SFF/GET-6449.pdf section 3 is background info.http://www.midwestreliability.org/01_about_mro/overview/mro_manual/MRO_UFLS_Program_06-03-10.pdf

a lot has happened after I retired ; http://www.publicpower.org/Media/magazine/ArticleDetail.cfm?ItemNumber=19303

Loadshed, SCADA, retired ...

Gets me to thinking what I worked on for the power utility.

... combined the SCADA, and AGC functions, with the forecast and scheduling
functions.

with a little Economic Dispatch thrown in

... a model of the power system, using load flow techniques, then up-date
the model with real time data so that it closely resembled the actual system in real time.

using a custom designed Harris Controls system.

I think you'll enjoy this Jim.

"scadahistory.com/resources/SCADA+History.docx"

 

[SirAskalot]

...
When I change Load for example from 150kW to the 450kW my speed of course suddenly drops, from 1000 r/min to the 900 r/min and then governor acts and in a few seconds speed is back to 1000 r/min again. Ok, so what do I want? This drop is huge, 1000->900, that's 10% which means frequency drops from 50Hz to the 45Hz. I would like to somehow limit my speed drop to the let's say 4% or so. Do you guys have any Idea how this stuffs works and what should I do, and even if you don't I would like to read other opinions.

Here is where the settings of the turbine governor comes in. Since its (in real cases) only proportional controller, the gain of the controller must be adjusted.
4% decrease in speed equal a frequency of 48 Hz, within the same range a 450-150 kW=300 kW load has increased. That is the generator must increase its generation by 300kW per 2 Hz drop in frequency, or 150kW/Hz. This is the gain of your controller.

One more important stuff. While hydroelectric power plant is on GRID , the turbine needle is opened 100% and jet shifter is opened about 50-60%.Why? Well because jet shifter is faster than
turbine needle, and we use him to react fast. So when load increases from 150kW to 450kW jet shifter opens to the 80% or so.

jet shifter is pelton turbine deflector

This is a poor operation principle, why? Some of the water just passes the turbine without transferring any energy. That is, you throw money out the window. To my knowledge the only(main) purpose of the jet shifter is to deflect the water away from the turbine when the loading of the generator DECREASES, this is why its fast reacting, in order to prevent damage to the generator and turbine due to over speed. When loading increases the needle is the reacting part in the plant.

You must remember one thing. Most plants are built and operated in a way where multiple plants work together creating a strong and stiff grid where large load changes has little impact on each generator.

 

[jim hardy]

Loadshed, SCADA, retired ...

Gets me to thinking what I worked on for the power utility.I think you'll enjoy this Jim.

"scadahistory.com/resources/SCADA+History.docx"

You must've had fun, don !

Good link, and it helped me out with something... thanks.

When I started we dispatched by radio. They were just installing basic Leeds & Northrup systems that transmitted the "Raise-Lower" beeps, what a marvel ! We still used pneumatic instruments and recorders were vacuum tube...

In late 1970's our system folks procured a huge Control Data Corp system. I only knew of it peripherally - some friends worked on it and I once got a tour.

That's a whole 'nother world to me. I envy you your understanding of the power system.I hope vampslayer becomes an industry guru.
I once met Chester Rackzowski one of Westinghouse's great men.
We indeed stand on the shoulders of giants.

... seen secondhand lions ? ...old jim

 

[vampslayer]

That is the standard frequency. As machines got larger their design became more refined and less tolerant of off-frequency operation.my reference to 58 hz was for our 60 hz system.
At 58 hz we are making less not more than customers need..

Maybe I'm confusing you with these extras.
Sorry, i Just wanted to make it interesting, if instead I'm causing you pain i'll stop.

I guess I'm enthusiastic to see somebody else interested in power systems.

I know reference in USA is 60Hz in Europe 50Hz and in Japan it's both 50 and 60.
My sentence wasn't regard on that. I said that to illustrate something but nvm I don't know how to explain it. Point is that I have thought that we keep frequency equal to 50Hz or 60Hz in USA just because of money. I didn't know that low frequency and probably high frequency can cause mechanical damage on turbines. I know that I need to prevent huge frequency/speed drop but I'm not the best sure WHY. So reasons are money and mechanical damages which can cause low frequency because devices are designed for nominal frequency?

And Jim, don't even think about stopping. My goals as a future electrical engineer are to learn as much possible about power systems. I didn't open thread to focus just on hydroelectric turbine.I want to know how things works, I want to hear for new terms like for example Load shedding. One more important thing is my English power engineering vocabulary which I want to expand, you probably noticed English is not my native.
To summary: I want to be able to create a picture in my head how all those things works

And Jim, you know what? I envy you because you have already has that picture in your head and even more you have real life experience working on real turbines.

 

[vampslayer]

Here is where the settings of the turbine governor comes in. Since its (in real cases) only proportional controller, the gain of the controller must be adjusted.
4% decrease in speed equal a frequency of 48 Hz, within the same range a 450-150 kW=300 kW load has increased. That is the generator must increase its generation by 300kW per 2 Hz drop in frequency, or 150kW/Hz. This is the gain of your controller.
This is a poor operation principle, why? Some of the water just passes the turbine without transferring any energy. That is, you throw money out the window. To my knowledge the only(main) purpose of the jet shifter is to deflect the water away from the turbine when the loading of the generator DECREASES, this is why its fast reacting, in order to prevent damage to the generator and turbine due to over speed. When loading increases the needle is the reacting part in the plant.

You must remember one thing. Most plants are built and operated in a way where multiple plants work together creating a strong and stiff grid where large load changes has little impact on each generator.

Yep I agree, if we keep jet shifter opened to the 50% half of the water power goes away. But How do you think to react on huge loading increases when the needle is slow.?

I was thinking about this: Can we somehow measure the current? When load increase my governor waits for speed change and then governor reacts but electrical changes(current) are a lot faster than mechanical(speed). So in a moment when Load increase from 150kW to the 450kW
the current should raise too? right? and I was thinking to make pre-regulation, so my jet shifter can react even faster.

 

[dlgoff]

I envy you your understanding of the power system.

And I envy yours.

I hope vampslayer becomes an industry guru.
I once met Chester Rackzowski one of Westinghouse's great men.
We indeed stand on the shoulders of giants.

... seen secondhand lions ? ...

He's on his way.

Secondhand Lions? Yes. My daughter made me watch it; Thanks Sweetie.

 

[SirAskalot]

Yep I agree, if we keep jet shifter opened to the 50% half of the water power goes away. But How do you think to react on huge loading increases when the needle is slow.?

I was thinking about this: Can we somehow measure the current? When load increase my governor waits for speed change and then governor reacts but electrical changes(current) are a lot faster than mechanical(speed). So in a moment when Load increase from 150kW to the 450kW
the current should raise too? right? and I was thinking to make pre-regulation, so my jet shifter can react even faster.

In many cases where parts of the grid are disconnected from each other and few generators only supply the load other factors like transient stability and voltage stability play a role in the health of the network. Not only steady-state frequency like we discuss. So the aim of system operators are not only to hold the correct frequency but ensure the stability of the system with regards to other parameters. In your system i would not be surprised if the system collapses before the low frequency would be a concern. This is a more advanced topic, but still interesting to have some basic knowledge about.

I have done a project on such an topic and may share some of the results we found. The network is taken from the book "P. Kundur - Power system stability and control". The analysis is done in SimPow (http://simpow.com/)
Here is a overview of the system:

The network comprises of two areas (left and right), with two generators each (AC source symbol), loads (black triangle) and capacitor banks. The two areas are connected with two parallel tie-lines securing the power flow from one area to the other.
At t=1 sec a three phase fault appears on the bus in the middle (50ms duration). Here is a plot of the voltage on the actual bus:

As noted the voltage oscillates and settles to a lower value after a few seconds.
In the next plot you see the active power generation of each generator during and after the fault:

Notice the large oscillations. But the system remains intact and stabilizes to a new steady state.

The last plot may give you insight into your solution regarding current measurement. After a disturbance the current fluctuate, but would you like the governor to react to any of these transients? What about the time delay in the system?

 

[SirAskalot]

In the same case (short-circuit) the speed of the 4 generators are shown below:

As noted the generators in each area 180 degrees phase shift in rotor speed. Hence they swing "against" each other and power is sent back and fourth between the areas causing instability.
The next plot is regarding voltage stability, here a short circuit has occurred and one of the two tie-lines is disconnected.

Here the voltage collapses after 5 sec. In other real world scenarios there has taken considerably longer time for the voltage to collapse, giving the impression that the system was stable after the fault but in reality it was not.

For sure this topic is very interesting, but the problem one is faced with are very complex systems, with many factors working together. A network consists of several hundreds-thousands generators and loads all connected and contributing to the stability. Where I live generators in the north of the country swings against generators in the south during a disturbance, that is 2000 km of overhead power lines in between them.

Regarding PLC, this is more of a engineering profession when projecting and installing power plants and such. Not in the sense of power system analysis. If you ever are going to visit a power plant, the basic knowledge of PLC operation will come in handy, but every manufacturer has its own software so specializing in one type might be to early. PLC these days are more like computers, with all its flexibility, than those old ones with only timers, AND, OR gates etc.

 

[jim hardy]

That simulation program looks like an awful lot of fun. The charts you made are very interesting.

Does load in that program have rotating inertia?

I can't quite make out the printing on your one line diagram so can't tell which generator is which.
As one would expect the generators all accelerate during the fault , if I read the charts right?

Are the two that accelerated most (red and black) nearest the fault, or way out at the ends?

just curious, not challenging the simulation.

old jim

 

[SirAskalot]

Generators are in the following order from left to right: G1, G2, G4, G3. See the attached PDF for an overview in larger format.

As for the distance to the fault I think all generators on each side has equal distance. That is G1 and G3 have equal distance to the center bus (fault bus). G2 and G4 the same distance. I can't say for sure as I don't have the data in front of me right now.

As for the initial state one can see in the attachment that power flows from left to right due to the larger load in the region on the right. All generators has equal initial production.

The generators models are 4. or 5. order, that is, 4 or 5 differential equations for each generator. So yes inertia is an important parameter. Other parameters are the transients reactances in d and q axis. , time constants etc.

 

[jim hardy]

As for the initial state one can see in the attachment that power flows from left to right due to the larger load in the region on the right. All generators has equal initial production.

Thanks for the enlargement.

The generators on the side with excess generation sped up more than those on the side short of generation.
That makes sense.

Perhaps the most interesting feature of your one line drawing is it shows angles. One sees the phase shift along every line.
Those power lines look absolutely static when we see them crossing country on their towers.
But really they are analogous to mechanical drive shafts - power transmitted is equal to phase difference between ends divided by stiffness, in this case 'electrical stiffness' (admittance, 1/Z).

So in reality the power system is a dynamic thing, with inertias connected by flexible links and plenty of energy input.
That's a spring-mass system subject to Newton's laws and harmonic motion. Just like we studied in freshman physics and sophomore dynamics.
It might as well be living and breathing. It just looks static.

That is shown in your speed charts where red and black's speed criss-crossed blue and green's speed. The system was in a torsional oscillation like a rotary pendulum. In your mind's eye grab both ends of that one line diagram and give it a twist. When you let go it'll rebound in a (hopefully) damped torsional oscillation.

I have stood next to a generator and shined a stroboscope synched to machine terminal volts on the shaft. You can see it hunting about its nominal power angle, and see it find a new one when excitation is changed.
Back when we had grid stability troubles we would sometimes see our units hunting in unison against the north part of the state. You've described that happening where you live.
If you still have that, try switching a voltage regulator anyplace in the system to manual . That'd damp our divergent oscillation.

For me awareness of this gives meaning to the phrase "Smart Grid" .

There's just too much to learn in one lifetime. You guys armed with today's computers will really have a good time.

Dig in Vamp - the industry needs people who like this stuff.

 

[vampslayer]

I have got a problem. I want to expand my turbine model into multiple-jet turbine model.
The only difference between single-jet and multiple-jet is in the Q(m^3/s - water flow)
So far I calculated Q on this way:
Q=k1*k2*Q_max
Q_max-max water flow
k1-needle openness factor (from 0% till 100%)
k2-deflector openness factor(from 0% till 100%)
In islanded mode servo motor regulated k2 and k1 was const. and in GRID mode k1 should be controlled and k2 set to 100%

My problem is how to expand this into multiple-jet, let's say the number of jet's is "n".

I also don't understand how many needles and deflectors turbine has with multiple jets.
For example let's say turbine has 4 jets, but how many needles? 1 or 4 too, and how many deflectors?

 

[SirAskalot]

Source

A needle is located inside each jet/nozzle, also there is a deflector on each jet. So the number of needles and deflectors is equal to number of jets.

As for the water flow, I would treat all the jets as one single unit, hence multiply your formula with n (no. of jets), if Q_max is max flow per jet.

Each jet is usually not controlled individually (if ever?), so k1 and k2 would be variables controlling the total water flow.

 

[vampslayer]

A needle is located inside each jet/nozzle, also there is a deflector on each jet. So the number of needles and deflectors is equal to number of jets.

As for the water flow, I would treat all the jets as one single unit, hence multiply your formula with n (no. of jets), if Q_max is max flow per jet.

Each jet is usually not controlled individually (if ever?), so k1 and k2 would be variables controlling the total water flow.

Ok, that sounds good :P
Following that Water flow is Q=n*k1*k2*Q_max_per_single_jet
In this case I don't need to care about how many nozzles/jets should currently work(1,2, or even total number n) to compensate current load demands, all I have to care about is value of k1 or k2, depending in which mode I am.

Ok, I have got one more thing to ask. When I am controlling the needle(GRID mode) should I use same servo-motor output which I use in islanded mode during the deflector control.
Same servo-motor moves the needle and deflector too? I guess in reality that's not the case, but I don't see the problem in a case of simulation.

 

[SirAskalot]

The two valves (needle and deflector) got different actuators (motors), and different control scheme. The problem you face is related to time constants. The control circuit for the needle has a "fictive" time constant for how fast the needle can operate. That is, if a step response is applied (decreased power output) the needle moves much slower than physically possible. This time constant is independent of grid or island mode.

The deflector has a different time constant.

When I come to think about it, (I basically only have a text-book knowledge about the system), I would guess the deflector operates in states; on or off. Not anything in between. The reason for my statement is that once you interfere with the water-stream you will shift the direction of the stream and the impact point on the turbine-wheel. This in connection with the precision of the deflector would give a system with poor control of power output. So k2 would either be 1 or 0, this again would reduce the system degrees of freedom (control variables). You may research the topic further.

 

[vampslayer]

Just let you know, my simulation is going good. I'm really not sure should I ask this in a new topic but I will let the administrators to decide.

I ask for a opinion, what would happen in a moment when my hydroelectric station switchs to the islanded mode from the on grid mode(due to some failure in the grid). What it would be with the speed. I know speed will increase if the load is lower in islanded mode than in "on grid" mode. But is it just like that simple?

 

[jim hardy]

In a steam plant -

The turbine of course begins to accelerate so its governor drives the steam inlet valves closed.
Other valves open to bypass steam around the turbine straight to the condenser.
That's because the the boiler cannot be immediately driven from full power to near zero power.
As fuel to boiler cuts back to match power demand, the bypass valves modulate shut.

Your hydro plant probably has analogous equipment.
The water in the inlet pipes cannot be immediately decelerated, so something must allow it to go around the turbine instead.

That's the basic concept. Others here are obviously better versed in hydro plant equipment and terminology than I am.

 

[vampslayer]

Your hydro plant probably has analogous equipment.
The water in the inlet pipes cannot be immediately decelerated, so something must allow it to go around the turbine instead.

Yeah, that would be deflector in hydro plant

 

[vampslayer]

I found that terms isochronous and droop speed control are closely related to this topic. I've just read some free ebooks. Would be good if someone of you would explain main differences isochronous and droop mode.

Also one more questions. What would be huge load change? For example if my nominal turbine Power is Pn=1734kW, what it would be huge load disturbance? How many percentage of nominal power?
And if my the load changes for 30%, how much I should let the frequency to be changed?
What are some normal borders. I'm asking for relation between load change and frequency change.

 

[Localise]

Microgrid with dump load

Hello everybody,

with interest I read the threads on this topic so far. I really enjoyed having found this! I am not sure whether it's ok to post into this thread, as the last post is already two months old. But I will let the admins decide.

Me, too, I want to simulate a little grid in Matlab/Simulink. I am studying Renewable Energies and this might become my thesis. I am at the very, very, very beginning at the moment and my questions are very silly compared to the stuff that has already been discussed here... I am thankful for every hint, though!

The grid consists of two small hydro turbines (combined output is about 3 kW), two permanent magnet synchronous generators, a PI controller, a dump load (that controls the speed) and a consumer.

The following data will be given: the turbine characteristics, the consumer characteristics and the torque.

The input signal will be rpm, the output signal will be the speed deviation that goes to the dump load and the disturbance is the consumer. I think so anyway, does that make sense?

How will the two parallel turbines behave with the PI controller? I do not know how they would react to a sudden load variation in reality – so I do not know what to bear in mind while doing the simulation.

And than I am not sure which equations I will need to describe and simulate this system.

So far only these came to my mind:

(1) P = ρgHQη

(2) \frac{dω}{dt} = \frac{1}{J} (M_t + M_l)

η = efficiency of turbine and generator
M_t= turbine torque
M_l= load torque
J=rotating mass

I think my problem is that I do not really know how to start systematically. Would it be better to start “drawing” the system in Matlab/Simulink?

Thank you very much for taking the time to read all of this and thank you in advance for any hints you might think of!

Lise

 

[SirAskalot]

I will try to share some of my experiences when it comes to matlab/simulink, as I have scratched my head to many times trying to get it to work, and I hope you don't have to.

First you need to have a opinion and goal on what your model is to describe. What is the output variables you are interested in. In power system analysis one can distinguish between stationary and transient behavior. The first can be described by algebraic equations, the latter with differential equations.

As far as modelling goes, one would "always" try to simplify the model as much as possible without the result being invalid. Thus one can try to make a statement and a assumption regarding the system, and model accordingly. However one should always verify that the assumptions are correct in some way. Either by incorporating the more complex phenomena and notice no change in the output variable in the two models. Or based on earlier research.

You should always acquire the equations you need to describe the system before you start modeling. Most advanced power system analysis books describe these in dept. Trying to make a model in matlab/simuling without equations, I would say is a bad approach.

Simulink incorporates several toolboxes, one popular in power system is the "power system toolbox". It operates with currents and conductors like in pspice etc. and not the usual transfer functions in ordinary simulink. Making it easier to start with. I am not a big fan of this because I have seen many students "drawing" huge systems only by dragging and dropping components. When its time to run the simulation it either displays many warnings or errors or shows unexpected results. They then spend most of their time trying to resolve these problems. And this takes a lot of time since one now needs to figure out how each block/component work.

I believe the problem lies in the how easy it is to build a large network, and one neglects the mathematical equations and misses the true understanding of how the system works.

Just think back to your course in ordinary differential equations. A RL or circuit is governed by a simple 1.st order ODE. Taking the laplace transform and modelling in simulink is simple once one has mastered it. On the other hand, drawing the electrical circuit in pspice or "sim power systems" is also easy, but one only gets the results and not the physical/mathematical behavior of a inductor.

Doing all the calculations and laplace transforms in a large network, like a 3 phase inverter is hard and time consuming. But here is where the simplifications comes into play. In the simplest form a converter is only a controlled voltage or current source. Why not use such a simplification instead of modelling all the switches?

Anyhow I could ramble on for hours about this, but to make it short:

Start small, verify the behavior of the model, and expand it as you go, always having full control on how each "block" behaves and what the expected output should be.

In power systems, start with the mechanical system. Depending on the simplicity, the flow in the turbine could be constant at first. Then you got Newtons 2nd law of rotation. (1 order ODE). Output variable is speed.

Speed is linear related to frequency. Amplitude of the voltage is controlled by the excitation system controlled by something (proportional controller with a setpoint maybe?).

Then you got your load. Is it constant or dependent on voltage or frequency? Constant may be the easiest to start with ( you can change it manually to see the response)

The load (power) and speed is proportional to the (counter) electromagnetic torque(M_t) produced by the generator. Which is the input to your 1. order ODE you started with.

Run the simulation, using proper initial values if you like ( initial speed of rotation). Change the load and observe the speed. If it behaves as you expect continue to expand the model. Expectation comes from experience, and in this simple example you probably know how Newton laws works with acceleration etc.

Next step would be to implement a turbine governor (input power) to make the system stable. (Start with a simple proportional controller.)

Just ask if something is unclear. A photo (print screen) of the network is always helpful.

PS: There are many software's dedicated to power system analysis like; Simpow, DIgSILENT PowerFactory etc. If your university has a license you might want to look into these.

 

[Localise]

Thank you so much for your quick reply! Your explanations clarified a lot for me. Or at least I think so ;) I have been mulling over a few things and here indeed are some more questions...

Am these assumptions correct:

1) In a stationary system there are no oscillations after a sudden change in the circuit (e.g. a load step), but a steady state is assumed directly after the change. As opposed to a transient system.
2) Every system could be described as transient or stationary, assuming a stationary state is just a simplification.
→ Is there an advantage in simulating a transient system? Or what indicates that I should assume a transient state?

I had a go at drawing a block diagram of what I want to draw in Simulink: http://tinypic.com/r/533hn4/5

The corresponding equations would be:

1)ΔM = M_t + M_l
2)ΔM \frac{1}{J} = n_{eff}
3)Δn = n_{ref} + n_{eff}

and about the part around the PI controller I am not yet sure. Does this make any sense somehow?

And then I have two questions on Matlab/Simulink itself:
Is it correct that it would be possible to do the simulation without Simulink?
Is it simpler to do it with Simulink?

Thank you so much again for any hints.

 

[jim hardy]

and about the part around the PI controller I am not yet sure. Does this make any sense somehow?

in my steam plant that would have been just a P controller, with gain perhaps 25.
..


The dynamic system will be much more informative
but probably difficult to keep stable - the computer is perfectly happy to extend your linear equations well past the physical limits of the machinery.
For example it sees nothing wrong with a valve's being 10000% open and it may oscillate your system between huge negative and positive numbers if the math says it'd happen.
So one must address limitations on travel and rate of travel (slew rate) of the parts he is modelling. And order of execution becomes important - program must flow in agreement with cause-effect else you have a paradox- something gets calculated from a value that follows not precedes it..

As suggested - start simple and build. Add plenty of 'meters' for debugging.

Welcome to simulation !

old jim

 

[dlgoff]

To add a little to old jim's "dynamic system" and "simulation welcome", here's a little online PID Control animation. Kind of fun, albeit it's not a Pelton turbine.

 

[Localise]

Thank you for your warm welcome!

So I will go with a stationary system...
The PID controller animation was very nice to see, thank you for that dlgoff.

I tried a more detailed diagram of what I want to simulate: http://tinypic.com/r/2e4esqt/5

Is this what you described, Sir Askalot? Did I forget anything?

For the transfer to Simulink: To “draw” this in Simulink I have to find the functions for the PI controller and the rest to press it into the form shown in the following diagram, correct? :
http://tinypic.com/r/nwymg6/5

I am sorry, if these are stupid questions. I have been fiddling around with this so long and even after a break I feel that I am unable to think straight about Simulink at the moment... Thank you all for your patience. It always takes me a while to get used to stuff that is very new to me.

 

[Localise]

I have been fumbling around for a bit more with the equations and that is how far I have come:

1.\ P_a(t)=P_T-P_L
2.\ P_T(t)=M_a(t)ω(t)=M_a(t)2πf(t)
3.\ M_a(t)=\frac{P_a(t)}{ω(t)}
4.\ M_a(t)=J_T\frac{dω}{dt}=J_T2π\frac{df}{dt}
5.\ \frac{df}{dt}=\frac{1}{2πJ_T}\frac{P_a(t)}{2πf_0}=\frac{1}{T_A}\frac{df_0}{P_0}P_a(t)
6.\ T_A=\frac{J_Tω_0^2}{P_0}
7.\ ΔP\frac{1}{sT_a}=Δf
8.\ ΔP_L(t)=V_LΔf(t)
9.\ ΔP_T(t)=V_PΔf(t)
10.\ s=\frac{1}{V_P}\frac{P_N}{f_N}

These are the equations that lead to the following block diagrams in Simulink:

The inertia of the grid itself: http://tinypic.com/r/1zmh0r5/5

The grid's self-regulation: http://tinypic.com/r/2d10fn4/5

Implemented P-controller: http://tinypic.com/r/9zowoh/5

Implemented PI-controller: http://tinypic.com/r/28aqeeh/5

All values are example values, so I can check whether it actually is possible to run the simulation.

So far I think this makes some sense to me. I just do not really know how to work in the data that are given. It's just the turbine characteristics (P-n-curve). For now I am supposed to assume constant speed. Where does that come in? How do I get the transfer function for the turbine?

For the moment I “modelled” the turbine as a lag element, as I have seen this in examples. Is a turbine always a lag element? Or approximated as a lag element?

Is it the Laplace transform of the second equation?

Am I totally wrong here? Thank you for any hints.

 

[SirAskalot]

Sorry for the late answer, but during the vacation I forgot this thread.

I re-read your first post, and you mentioned a "dump load". Can you clarify what you mean by that? I have yet to encounter such a "object" in a normal power system. But it may be possible to incorporate such a object with the pros and cons that follows.

1) In a stationary system there are no oscillations after a sudden change in the circuit (e.g. a load step), but a steady state is assumed directly after the change. As opposed to a transient system.
2) Every system could be described as transient or stationary, assuming a stationary state is just a simplification.
→ Is there an advantage in simulating a transient system? Or what indicates that I should assume a transient state?

In a stationary system you normally use the (simplified) algebraic equations. Eg. a http://www.ece.uAlberta.ca/~knight/electrical_machines/synchronous/parallel/house.html or load flow analysis
And as such, the time variable is neglected, as opposed to transient system analysis where one uses differential equations where time is a important variable.

There are pros and cons to using both methods. As an example, a experienced engineer could get the result and give an answer to a problem using stationary (or simple transient analysis) analysis, because of his experience on what is the governing factors. Where as a novice (as myself) would have to build experience from scratch, starting simple and increasing complexity, backtracking and noting the difference (or lack of difference) in the results. So on the next project one use this experience and uses the appropriate model, with regard to complexity. Saving project time.

In your case, even though its a thesis, you should gain experience with power systems and modelling. Not only produce "cutting edge research" which could have been obtained with a simpler model. You should either way justify the validity of your model and results.

Is it correct that it would be possible to do the simulation without Simulink?
Is it simpler to do it with Simulink?

Stationary analysis could be done with pen, paper and a slide rule. Transient simulation could be done with pen, paper and a plotting tool for single ordinary differential equation (ODE). For system of differential equations such as advanced transient analysis, numerical ODE solvers are a "must". Remember that Simulink is only a graphical interface to the ODE solvers in Matlab. You could always solve the same problem by using the ODE solver in Matlab by writing the system of equations in a function/script file.

Generally Simulink is gaining popularity because of its simple interface and reduced project time and expenses.

I tried a more detailed diagram of what I want to simulate: http://tinypic.com/r/2e4esqt/5

Doing stationary analysis, using eq. (2) in your initial post: dw/dt=0. In other words, in stationary state (steady state) nothing is changing ( no change in speed/frequency ). Solving eq. (2) (or similar equation) with dw/dt=0 renders inertia (J) to be omitted. Hence inertia has no effect on the system, with the right assumptions. Otherwise it looks ok. I did not have in mind the dump load object. And one might want to omit it to start with, in order to build the system on known facts and equations.

A normal system controls the frequency / speed with a proportional (P) controller which adjusts the input power (P_turbine).
In normalized values (p.u):
Δn * K_p = P_turbine
Δn = n_ref - n_shaft

where K_p is the gain (multiplier) in the controller. This is the equations in the "house diagram".

I have been fumbling around for a bit more with the equations and that is how far I have come:

Now you have started doing transient analysis.

The inertia of the grid itself: http://tinypic.com/r/1zmh0r5/5

Where is the inertia (or equivalent)? w=ΔT * (1/J) * (1/s) <-- integrator
You may also set the initial value of the integrator to 2*pi*f_0.

Implemented P-controller: http://tinypic.com/r/9zowoh/5

What is V_L? Remember what ΔM (ΔP) is. Its the turbine power _minus_ the load power. So your summation sign must change.

For the moment I “modelled” the turbine as a lag element, as I have seen this in examples. Is a turbine always a lag element? Or approximated as a lag element?

Yes, its mostly is, at least for hydro power. The physical explanation is that the flow in the pipes cannot change instantly. And closing the turbine valve has a built in time constant due to safety reasons.

Is it the Laplace transform of the second equation?

No, to model it correctly one needs to take into account the physical dimensions of the water pipe, flow equations etc.

For now I am supposed to assume constant speed. Where does that come in?

What do you mean? With the model you now have the speed does change.

As a last advice/question. Have you run the simulation ? Is the starting values correct? Are the steady state values correct? Speed frequency etc.? Compare with the steady state calculations.
As a example:
Intial: P_turbine = 1kW, P_load=1 kW. f_0=50 Hz. Kp= 1 kW/Hz.
Step change in load: P_load = 2 kW
Results: f = 49 Hz

 

[Localise]

Thank you so much for this massive answer, SirAskalot. I hope I can clarify a few things, that I had not described properly:

I re-read your first post, and you mentioned a "dump load". Can you clarify what you mean by that?

A dump load is a ballast that absorbs the excess output of the generator. So there is always a fixed total load on the generator at all times. So not the flow through the turbine is controlled, but the load on the generator. If the demand falls to zero then the ballast or dump load must be capable of absorbing the full output power of the turbine. The turbine can therefore run continuously at full flow and there is no longer a need for a flow-regulating mechanism.

Remember that Simulink is only a graphical interface to the ODE solvers in Matlab.

By inertia I meant to describe the response time of the system. I am not sure about the correct English term here, sorry. Does “response time” help? I mean the fact that there is a time delay between a load change and a response in frequency.

You may also set the initial value of the integrator to 2*pi*f_0.

How did you see that I had not done that? Engineer's magic?!

What is V_L?

In contrast to the grid-”inertia”-diagram, I wanted to take into account the frequency dependency of the load (V'L) in the grid-self-regulation-diagram. With frequency dependency I mean the frequency response resulting from inductive and capacitive components in the load impedance caused, for example, by lines, transformers and motors, that draw less power when the frequency decreases. The frequency decreases and a permanent deviation
builds up. If a higher amplification factor of the self-regulating effect is chosen, this deviation is smaller. A smaller amplification factor is results in an almost linear decrease in frequency.

Does this make sense? Do I need to take this into account?

Is it the Laplace transform of the second equation?

No, to model it correctly one needs to take into account the physical dimensions of the water pipe, flow equations etc.

I do not have these dimensions and I won't get them. But I thought I do not need those as the system is “governed” by a dump load instead.

For now I am supposed to assume constant speed. Where does that come in?

What do you mean? With the model you now have the speed does change.

Ok, it took me a bit to get my head round that. Matlab is quite massive and something very new to me.

Now you have started doing transient analysis.

I now realized that I had not fully understood the difference between transient and stationary. Your explanations helped a lot. So indeed, I am working on a transient system.

The inertia of the grid itself: http://tinypic.com/r/1zmh0r5/5

Where is the inertia (or equivalent)? w=ΔT * (1/J) * (1/s) <-- integrator

By inertia I meant to describe the response time of the system. I am not sure about the correct English term here, sorry. Does “response time” help? I mean the fact that there is a time delay between a load change and a response in frequency.

You may also set the initial value of the integrator to 2*pi*f_0.

How did you see that I had not done that? Engineer's magic?!

What is V_L?

In contrast to the grid-”inertia”-diagram, I wanted to take into account the frequency dependency of the load (V'L) in the grid-self-regulation-diagram. With frequency dependency I mean the frequency response resulting from inductive and capacitive components in the load impedance caused, for example, by lines, transformers and motors, that draw less power when the frequency decreases. The frequency decreases and a permanent deviation
builds up. If a higher amplification factor of the self-regulating effect is chosen, this deviation is smaller. A smaller amplification factor is results in an almost linear decrease in frequency.

Does this make sense? Do I need to take this into account?

Is it the Laplace transform of the second equation?

No, to model it correctly one needs to take into account the physical dimensions of the water pipe, flow equations etc.

I do not have these dimensions and I won't get them. But I thought I do not need those as the system is “governed” by a dump load instead.

For now I am supposed to assume constant speed. Where does that come in?

What do you mean? With the model you now have the speed does change.

Is this explained by what I said regarding the dump load?

As a last advice/question. Have you run the simulation ? Is the starting values correct? Are the steady state values correct? Speed frequency etc.? Compare with the steady state calculations.

As a example:
Intial: P_turbine = 1kW, P_load=1 kW. f_0=50 Hz. Kp= 1 kW/Hz.
Step change in load: P_load = 2 kW
Results: f = 49 Hz

This again raises the question about the transfer function of the turbine and the input of the turbine characteristics (P-n-curve). Where do I enter P_turbine? And P_dump_load…

 

[SirAskalot]

How did you see that I had not done that? Engineer's magic?!

I didn´t, just a guess based on own experiences.

Does this make sense? Do I need to take this into account?

Ok, it makes sense. But the extent depends on how large the grid is and what load sinks there are. Have you done some research and found a proper value?

I do not have these dimensions and I won't get them. But I thought I do not need those as the system is “governed” by a dump load instead.

Ok, I have misunderstood your system. I was talking about the whole system. I think you may talk about the "efficiency" of the turbine. If you want the generator and turbine to produce a constant amount of power the governor and turbine models may not be necessary.

This again raises the question about the transfer function of the turbine and the input of the turbine characteristics (P-n-curve). Where do I enter P_turbine? And P_dump_load…

By P-n curve you mean power vs. speed on the turbine? Do you have this data?
If its a linear equation you can use a gain or other math expression where speed is the input and turbine power is the output. If its data points, you can use a "look up table". n is the speed and is related to frequency and the pole pair number of the generator. ( n / w / f is the output of the integrator).

So to clearify:
- Do you want to regulate the input to the turbine/generator?
- How is the dump load controlled? Only w.r.t. frequency to start with?
- Does the dump load have some time constants? (What type of element is it? Switched resistors?)

As a last advice, you should use the equations known and implement them into a diagram (like the one you have made) making sure to name all the signal ('wires', outputs, etc.). This will help a lot in the understanding.

 

[Localise]

By P-n curve you mean power vs. speed on the turbine? Do you have this data?

Yes, I will get this data. And as far as I know it is not data points but a curve.

So to clearify:
- Do you want to regulate the input to the turbine/generator?

No, the input to the turbine will be constant. I want to regulate the load on the generator, or rather as the input to the turbine is constant, the load on the turbine/generator has to be constant: If there is less demand by the consumer, the dump load “provides the additional demand”.

- Does the dump load have some time constants? (What type of element is it? Switched resistors?)

Most probably ceramic heating resistances will be used. I am sorry, for this lack of information. Also, I do not yet have any info about time constants or any other parameters of the dump load.

As a last advice, you should use the equations known and implement them into a diagram (like the one you have made) making sure to name all the signal ('wires', outputs, etc.). This will help a lot in the understanding.

I will do this and hopefully have more information asap. You have already helped me a lot in understanding what I want to do here, SirAskalot.