# Generator Synchronizing Current

1. Sep 2, 2011

### I_am_learning

I was reading about synchronizing of two AC generator. As an aid to grasp the concept, I tried to analyze what will happen if two generators of same frequency and voltage but slightly different phases were paralled.

I am slightly surprised to find out that active power flows in there even if there appears no resistive Loads. And that the direction of active power flow is governed by which generator lag in phase.
Also, I asked the question, why should the Generator 1 Provide power (real) to generator 2 simply because its phase is leading? Is this a natural duty for leading to power the Lagging?
It seemed quite asymmetric to me.

Of course maths have the answer. So, after trailing back calculation, I convinced myself that the asymmetry comes from the fact that in between the generators lies an inductive reactance as opposed to capacitive reactance.

I am looking for practical cases of what the active power and reactive power does in there? Does it help to make the phase equal? How?

Thanks.

2. Sep 2, 2011

### uart

Be careful about that second statement, it's not true (despite your calculations being correct).

What your calculations show is that G2 is supplying -17.36 Watts and also supplying +1.52 VAR (so absorbing real power and supplying VAR).

The above is correct. Further, it is correct to assume that the power absorbed by G2 is being supplied by G1, so G1 is therefore supplying 17.36 Watts.

It is however incorrect to assume that the reactive power being supplied by G2 is being absorbed by G1. If you do the calculations at G1 you will find that in fact both G1 and G2 are supplying reactive power.

Q. Where is the reactive power going?
A. To supply the reactive power absorbed by the inductor. (BTW. Inductors absorb reactive power and capacitors generate it).

3. Sep 2, 2011

### b.shahvir

But as per the figure, the generators are operating on NO-LOAD. Also, one can throw some light on the problem with purely resistive load across the generators.

4. Sep 2, 2011

### I_am_learning

I did calculation as you suggested. You are correct. Reactive Power is negative across G1. So, the inductor seems to consume reactive power from G1's side too.
But I am confused here. Is the inductor consuming 2*1.519 Var or just 1.519 Var. Are we finding same power from two sides or there are two powers? Seems simple enough but still confusing.
Thanks.
And for the OP, what do the active power do? Does it help bring G2 in phase with G1 since G2 will be in motoring mode (OR actually in Less-Generating mode?). Some insightful lecture is expected. :)

5. Sep 2, 2011

### uart

Yep it is 2 times 1.519 var being consumed by the inductor.

6. Sep 2, 2011

### uart

Yes, it's fairly easy to show that if $\theta$ is the phase difference between the two generators (and V1,V2 the rms voltages) that the real power flow is,

$$P = \frac{V_1 V_2 \sin(\theta)}{X_L}$$

And that the direction of power flow is from the generator with the relatively leading angle to the one with the relatively lagging angle.

In terms of the machine dynamics, this direction of power flow tends to try to slow the leading generator (causing it's angle to retard) and to speed up the lagging generator (causing it's angle to advance).

Last edited: Sep 2, 2011
7. Sep 2, 2011

### I_am_learning

So, in our example, the Two generators are likely to get stabilized at voltage of 10<-5 (10 angle 5 degree) ,
the midway between them, if we keep the mechanical driving power to them almost constant throughout.
I can understand that it wont be a smooth transition but may be associated with some overshoots and transients.
Sound fair enough. Thanks.

I have another question regarding synchronous generator, slightly off topic from this thread title.
Suppose I have a synchronous generator and I have connected an external load (say a resistive load of 10 KW). I now drive the generator with constant Torque of say, 1 KNm.
So as to balance the input power and output power, the generator rotational speed will be
P(m) = P(e). (Assume 100% efficiency)
so, T*w = 10 KW. So, w = 10 KW / 1K = 10 rad/sec.
(w is rotational speed (rad/sec), T is the driving torque, P(e) is electrical output power, P(m) is mechanical input power)

Now, if I increase the external electrical load to 11 KW, (suppose the generator is sufficiently rated), the from the power balance rule, w should increase to 11 rad/sec.

But my intuition and literatures tell that increase in electrical load is associated with decrease in speed and hence the frequency! Whats wrong here?

8. Sep 2, 2011

### uart

No, in that case (constant torque prime mover) the power output (of the prime mover) can't increase unless the speed increases first. What actually would happen is that as soon as you increased the load the extra energy would have to be extracted from the machines rotational KE, so it would slow, produce even less power, so slow some more and actually stall. I think you'll find it was an unstable (or marginally stable) operating point to begin with.

Remember that with real generators the prime movers usually have feedback control (eg control of a valve to increase or decrease the amount of steam to a turbine) to stabilize the speed.

9. Sep 2, 2011

### I_am_learning

Ah! I seem to remember it. An increased external electrical load will create increased electrical torque. Since, the mechanical torque is kept constant, the rotor will now deaccelerate and as you told, it will move towards Halting.
(however in practice, if the 11 KW load is just a simple resistive load, as soon as the voltage sufficiently drops by sufficient dropping of speed, the power delivered will also drop (to 10 KW) and new equilibrium will be established)

Ok. I don't want that to happen. I need to supply 11 KW at the same speed. So, my best bet is to increase the torque (by letting more steam or water) in there. Sounds fair enough.

But still there is one confusion. What determines at what speed generators will rotate in the beginning. Say, as I told, If 1 KN torque is being applied by prime mover and generator is being started from stand-still to power up a resistive load of 10 KW.
Can you tell me what will be the final steady state speed?
T*w = P(e), don't seem to work, as previously demonstrated (i.e. in the case 11 KW load was switched, the generator speed can't be determined using this equation)

10. Sep 2, 2011

### jim hardy

"" I tried to analyze what will happen if two generators of same frequency and voltage but slightly different phases were paralled.
""

they will very quickly come into phase.
As you have calculated they will exchange real mechanical power , one driving and one driven, forcing whatever machines are driving them to run at same speed.

If they are close enough together you can shine a strobelight on the shafts and watch them lock into step.
That opens up all kinds of dynamic possibilities - you have rotating inertias connected by the somewhat elastic magnetic fields inside the generators. Harmonic motion becomes a possibility.....

In a power grid you have dozens of machines all connected in parallel and spread across large geographical area. Imagine the balancing act to keep them all sharing power while trying to keep the generation as close as practical to where it is consumed.

you're touching on a fascinating area of electrical engineering called power system analysis. I hope you find it interesting.

--------------

"""" Now, if I increase the external electrical load to 11 KW, (suppose the generator is sufficiently rated), the from the power balance rule, w should increase to 11 rad/sec.

But my intuition and literatures tell that increase in electrical load is associated with decrease in speed and hence the frequency! Whats wrong here? ""

our minds can believe in things that are physically impossible.
You said your generator is driven by constant torque.
Then you increased the load on it.
Any engine i know of slows down when you put more load on it.
Your intuition was right. Unless you increase the power input to your generator to match load, it'll slow down to a stop .
You could increase either the torque or the speed. Your math showed that incresing speed would work.

That's the trouble with simulation - computers have no common sense..
_____________

"""And for the OP, what do the active power do? Does it help bring G2 in phase with G1 since G2 will be in motoring mode (OR actually in Less-Generating mode?). Some insightful lecture is expected. :) ""

op? i dont know what that is.

To grasp synchronous machines it is useful to think of the rotor as if it were a simple bar magnet with a N and S pole. It spins inside the stator.

Current flowing in the stator creates a magnetc field that can have two effects on the rotor-

1. It can exert torque on the rotor. It can either aid or oppose rotor's rotation.

If it aids you have a motor and real electrical power is flowing into machine;
If it opposes you have a generator and elecrtical power is flowing out of the machine.

2. It can enhance or diminish the strength of the rotor field.
If it enhances rotor's magnetic field, we say by convention that reactive power is flowing into the machine, and call that "leading power factor".
If it diminishes the rotor's field , we say by convention that reactive power is flowing out of the machine and call that lagging power factor.

Google on "direct and quadrature currents" to see if you can find a tutorial on
synchronous machines.

old jim

11. Sep 3, 2011

### I_am_learning

Thanks jim.
OP just meant original post , the first post. I was just going back to the question I was asking in the first post.

But I am still looking for answer to this.
Given a synchronous generator. I have connected a fixed electrical load at its output. I drive it with constant Mechanical Torque.
How can I be able to find at what speed (frequency) would the machine rotate?

I am trying to understand how an increase in Load would result in decrease in frequency. Don't that imply that if frequency is decreased, we can generate more power, so as to balance the increased load. If yes, how?

12. Sep 3, 2011

### jim hardy

""" But I am still looking for answer to this.
Given a synchronous generator. I have connected a fixed electrical load at its output. I drive it with constant Mechanical Torque.
How can I be able to find at what speed (frequency) would the machine rotate?

""""""""""'

"a qustion well phrased is half answered"...

Don't complicate things with unrelated ideas.
It's a physics question not an electrical one...

If i have a black box with something inside which accepts torque in on one side and puts out electricity on the other;
since power in = power out

the first approximation is :

(pardon the english units here - i'm an old guy, work in horsepower and revolutions per minute)
A horsepower is 550 ft-lbs per second or 33,000 ft-lbs per minute
it's also 746 watts

Soooo,,,
horsepower in = 2 X pi X torque X rpm / 33,000

set those equal and solve for rpm as function of load

rpm = (kw/0.746) X 33,000 / ( 2 X pi X torque)

you could make some adjustments for efficiency , but i think you were asking something more basic.

i hope i understood the question.

old jim

13. Sep 4, 2011

### jim hardy

""I am trying to understand how an increase in Load would result in decrease in frequency.""

The generator does not produce any power it only transforms it from mechanical to electrical.
Some engine is driving that generator. That engine is the source of power not the generator.
Any engine will slow down when you load it harder, and the generator has no choice but go along with it.

""Don't that imply that if frequency is decreased, we can generate more power, so as to balance the increased load. If yes, how? ""

Well sure, we can do it like this:
sense the decrease in speed and apply more throttle to the engine to speed it back up.
That is done by an attachment to the engine called a governor.

Look at your lawnmower. Inside the cover adjacent the flywheel is a little vane connected by a wire to the throttle.
That vane is pushed against by the air from the fan on the flywheel so it produces a force proportional to the square of the engine speed.
The vane tries to pull the throttle closed, and a spring tries to pull throttle open.
The vane and spring engage in a constant tug-of-war. When you hit tall grass and the engine slows down, air from the fan pushes less forcefully against the vane so the spring wins the tug-of-war and pulls the throttle open. You hear the mower engine work harder in tall grass, and that's why.
When you get out of the tall grass the engine speeds up,
so air pushes harder against vane,
so vane overcomes spring and pulls throttle closed slowing engine back down.

Perhaps you are giving the synchronous generator too much credit in your thinking above.
The generator does not set the speed, the governor on its driving engine does.
Industrial governors are a little more sophisticated than your lawnmower but most operate on the principle of tug-of-war between centrifugal force and a spring.

The generator converts engine power into electricity just as the mower blade converts engine power into shredded grass.
Frequency is speed which is set by the governor, not the generator or blade.

Good Questions, I M L...
This is the age of technology, so enjoy it. Take things apart and figure out how they work.
I was never smart enough to remember a lot of equations but found if i could figure out how something worked the formulas came naturally enough from basic physics.

old jim

Last edited: Sep 4, 2011