# Synchronous Generator in Isolated Mode

1. Jan 4, 2012

### I_am_learning

I tried simulating a simple synchronous generator driving a resistive load in isolated mode (not connected to the grid/infinite bus), using Matlab Simulink, SimPowerSystems.

First I used, Input mechanical power of 1 pu, and output resistive Load of 1 Pu. And Excitation Field Voltage of 4 pu. This resulted on output terminal Voltage of 1 pu, Output Electrical Power of 1 pu, and,
and Rotor Speed of around 0.5 pu.

I then Increased the excitation voltage to around 6 pu and everything remaining same, the rotor speed drooped to 0.3 pu.

All this seems logical, in that, when I am over exciting, this result in higher Pole fields. In order to maintain same terminal voltage, the rotor speed must decrease.
Same terminal voltage needs to be maintained, because, since the load I used was resistive load, only by maintaining fixed terminal voltage, can the electrical output power equal the fixed input mechanical power.
As we know, the two powers (input mechanical power and output electrical power) must always equal, their difference creates an accelerating torque that restores the rotor to correct speed.

I hope everything I said is correct?

I will ask my question once I get the confirmation that I was correct.
Thanks.

2. Jan 4, 2012

### I_am_learning

Ok, I will ask my question right off.
Is there any chance, the RPM of synchronous Generator to keep on increasing and increasing, when its connected to a fixed resistive load and the excitation voltage is fixed.
Ok, at the begining, the applied mechanical power is greater than the ouput electrical power, but when its speed keeps increasing, so would its generated voltage. I think, at one point, the voltage would be large enough that the power consumed by the resistive load would be big enough to counter balance the applied mechanical power.

But this don't seem to happen in my simulation. When I set the excitation voltage at around 1 pu, The speed just keeps increasing on and on but somehow, the terminal voltage remains fixed. Although I have fixed excitation voltage, I guess the terminal voltage should keep increasing with speed??

This seems so simple, but yet Matlab simulation is against me. Just want to hear your views?

3. Jan 4, 2012

### jim hardy

sounds to me like matlab may be getting confused. i don't know a thing aboout it.
maybe it has a voltage regulator that you need to turn off?

in a real utility generator the driving machine will be equipped with a speed controller to hold it very near synchronous speed. sounds like matlab is not.

that aside,

go back to your e=n*d(phi)/dt,, 'n' being number of turns in the generator winding and phi = flux (~excitation).

at fixed excitation, you'll have pretty constant flux so terminal volts will be almost linear with rotational speed.

at fixed speed , since e=n*d(phi)/dt, at any given speed the voltage is almost linear with excitation. up to a point .

in a real generator the amount of iron it's built with is sufficient to make rated terminal volts (at rated speed) and not much more.

Generators have a limit called "Volts per hertz". that's because -
when you over-excite an isolated machine as you describe, ie add excitation beyond rated, you make more magnetic flux than the iron can handle and the iron heats up. If pushed hard, this will literally melt the iron in a pretty short time. Terminal voltage will no longer increase with excitation but will become an increasingly distorted pseudo-sinewave , a peaky looking mess with lots of third harmonics.
In a big utility generator it will literally blow a hole in the end-bell and molten iron will run out.

i do not know if that is where matlab put the excess energy in your first question. is matlab that detailed? does it show stator temperature?

anyhow that's a case where input and output power will not equal, you are losing a lot inside the machine melting its core.

happens in real world a time or two per year, someplace. if Matlab reproduces it , a kudo to those programmers!

4. Jan 4, 2012

### I_am_learning

Thanks jim.
First off all, the matlab software I am using isn't that detailed as to simulate Iron heating and melting, though, it claims to simulate the electrical things resulting from core saturation.
That aside.
You seem to have lot of practical knowledge, so its my privilege to meet you.
Suppose, A generator is running at isolated mode and is at steady state with 100% excitation, 100% rated terminal voltage and 100% input and output power. (Internal losses asides for now).
What happens when we over-excite slightly, say 120% now?
My guess is, the generator's speed (frequency) should slightly decrease.
Similiarly, if we slightly under-excite, I guess, the generators speed should go up and stabilize somewhere. Clearly, it shouldn't go up and up and up as my simulation showed.
Can you through some light in this aspect?

As you said, greatly over-exciting can lead to catastrophic problems. Does, greatly under-exciting have any such problems?

5. Jan 4, 2012

### jim hardy

in these thought experiments it's important to agree on ALL conditions

i assume you're assuming constant input power

so torque will go up if speed decreases

power = torque X rpm X some constant for units

""Suppose, A generator is running at isolated mode and is at steady state with 100% excitation, 100% rated terminal voltage and 100% input and output power. (Internal losses asides for now).
What happens when we over-excite slightly, say 120% now?""

assuming no magnetic saturation effects, and resistive load -
voltage will rise
load should demand more kw as result
so machine will slow back down to some speed where input and output powers balance
probably something like 110% voltage and 1/110% speed

try it with simple algebra
volts and torque both linear f(rpm) and see if you can't write simple equations

we had similar situation in our power plant one foggy morning when lost all transmission lines
he kept voltage constant but when attempt to increase power output only raised frequency to 63 hz, we all knew we were islanded.

but our pumps really pumped great!

6. Jan 4, 2012

### I_am_learning

Hi, Jim hardy
sorry, but I think at the final steady state condition, the voltage must be 100%, because, only then the output power will be 100% again. (to balance the constant (100%) input power).
But since the excitation has increased to 120%, the speed must now be, 100/120 % (So as to produce the same 100% voltage)
Thanks.

7. Jan 4, 2012

### commelion

i have an idea of what is happening for you but there is a question i need answered

you seem to be trying to run a synchronous motor as a generator ? if this is correct then your motor should not speed up at all, i see this sometimes in wind turbines (induction motors i know but think of the princible being the same) your generator parameters are incorrect because your generator is motoring. in effect the generator has wants to act like a motor because your settings may be wrong. i would first try an increase the imput mechanical power and decrease your excitation current try 50 % increase/decrease respectively, adjust your speed to your required frequency also.

let me know how this goes, also matlab gives problems like this in motor simulation all the time. for example its possible to get a motor giving minus torque in matlab !!!, dont worry to much about the matlab end of things, it looks as if your ratio's need to be tweeked.

let me know how she blows !!

8. Jan 4, 2012

### I_am_learning

Hi commelion,
Sorry, but I am not trying to run synchronous motor as generator. I am using a genuine synchronous generator.
I am using a preset generator model, so the parameters must have been set correctly. To keep things clear, the only thing bugging me is that, the generator terminal voltage some-how don't increase with the machine RPM (the field current remaining same)

I have attached my simulation model, if you would like to view.
(the excitation block is currently un-used).
Note that after about 6 seconds, the field current stabalizes and remains constant, the generator RPM keeps on increasing, but some-how the terminal voltage remains constant , its supposed to increase with RPM.)

File size:
10.5 KB
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9. Jan 4, 2012

### commelion

yo

let me know if this looks close to being what you are after ?

you mention your dissapointment at Vt not increasing with speed, it does at the start just remember that the voltage across any resistive load will depend on the current x resistance,
assuming you are using correct power ratings for load the voltage should spike and than stay constant. id say your scope results look pretty good, its doing what it says on the tin !!

i got rid of a sqrt term on the voltage calculator as is would not run on my vershion,plus i used auto scale function to put all scopes in to perspective.

i dont see anything else wrong with this at first glance to be honest, but let me know and i will investigate futher.

regards

#### Attached Files:

• ###### Syn gen output.doc
File size:
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10. Jan 4, 2012

### jim hardy

i think you're right. my bad, i was in a hurry (grandkids calling) and i didnt think it through.

okay - you've got that part nailed. GOOD!

we could write some simple simultaneous equations

power proportional to volts^2
volts proportional to speed X excitation

power1 = k X (speed1 X excitation1)^2
power2 = k X (speed2 X excitation2)^2

your condition that power1 = power2
lets us divide those first 2 equations and get

1= [(speed1 X excitation1) / (speed2 X excitation2)]^2

since excitation2 = excitation1 X 1.2 , what is speed2/speed1?

neglecting internal losses and magnetic nonlinearity of iron , of course

please check my thinking, as you did before,
and you dont have to be be so polite when calling out my mistakes....

old jim

11. Jan 4, 2012

### I_am_learning

Hi,
I have highlighted my concern below,

Although, I agree that the voltage across resistive load is proportional to current through it, But in our case above, when the speed is increasing, what might be preventing the terminal voltage (and load current) to increase ?

12. Jan 4, 2012

### I_am_learning

Clearly, speed2 = speed1 / 1.2.
Thanks.

13. Jan 4, 2012

### jim hardy

i thank you, my friend.

good job you did there on your thought experiment. a real valuable learning tool.

now - maybe you can fix matlab!

14. Jan 4, 2012

### Staff: Mentor

What happens to the frequency of the generated waveform when RPM soars? Yes, I know frequency should rise in step with RPM, but what does MATLAB show?

Last edited: Jan 4, 2012
15. Jan 4, 2012

### I_am_learning

I couldn't find any method for direct frequency measurement. But I could trace the instantaneous terminal voltage. Calculating from the time period of sinusoidal voltage waveform, Yes, the frequency was in step with RPM.

16. Jan 5, 2012

I have not looked at your model, but from the scope plots it looks like your running in "phasor mode" ( in your powergui) ?

To my knowledge the Phasor simulation mode solves the algebraic equations describing the system, not the differential equations. That is, your 'w' (omega, fundamental frequency) is constant. Then (i think) the equations for your machine will not give "correct" answer you are looking for. The terminal voltage (or more correct, the emf) at constant excitation will be governed by E=k*jw*phi where w is constant at 50 or 60 Hz set in your powergui.

I don't know the scope of your simulation, but it seems like you would like the machine to operate at variable frequency, like in reality. Then you probably want to run in "continuous simulation mode". Remember to convert the measuring signals from instantaneous to RMS.

17. Jan 5, 2012

### Jobrag

In the real world an synchronus generator will have two controls a speed control that keeps the machine at constant speed (by controling the amount of fuel entering the prime mover) and a voltage control that keeps the line terminals at a constant voltage (by altering the excitation current). The two controls are independent of each other.

18. Jan 5, 2012

### I_am_learning

Hi,
The powergui block is set at continuous mode. And, as I previously told, from the instantaneous Voltage waveform, it can be actually confirmed that the frequency is increasing with the RPM. So, no, w doesn't seem to be constant at 50 Hz.
Perhaps, you can spot something, please have a look at the model.
Thanks.

19. Jan 5, 2012

### commelion

hi again iaml

I agree that the voltage is doubling however, have you considered that this is a step response.that said matlab will put this step on to anything you want however, what you will see is the step response in time. Can you let me know what declimatiion your simulation is running at, as it appears you are running over 80 seconds 8 seconds 0.8 seconds ect ?

I quickly ran your simulation over one second and applied your constant value of 0.5 ? (base unit ?) your voltage stabilises after 0.5 seconds which is a direct result of shaft speed and attached load. I then ran the simulation for 20 seconds, if you look at the stator current as it produces its rotating field this tends to reduce the field from the excitation winding. This will reduce the excitation voltage by an amount Ear. Ear is most easily modelled as the voltage drop across an inductive reactance Xar. let me know the answers for above and i will take another look. be carefull in your defined model selection, this is a synchronous machine. as you know these machines can operate as motors or generators depending. your model and graphs attached are running at your own values so it might be a under/over driven problem. On quick opinion I cant see much wrong with this generator set up, I have seen far worse simulations !!!

attach the graph below

Red = field current
Green = rotor rpm
Blue = terminal voltage

#### Attached Files:

• ###### synchronous generator.fig.doc
File size:
37 KB
Views:
87
20. Jan 5, 2012

### jim hardy

aha

Xar indeed cause "armature reaction" and is basis of synchronous impedance.

for a purely resistive load it sholdn't be much of an effect ( for reasonable amount of Xar ) since the vectors are at 90deg.

does Matlab let you set such parameters?

if so, what is your Xar?

we had a 894mva machine with 171% which the old-timers considered high.

hope you dont mind me peeking over your shoulder.
this simulation you guys use today is amazing to me,
maybe i'll learn something.

old jim