Power factor and angle of synchronous machine

[doctorslo]

Hello,

could someone assist me with the following calculation.

I have a 8-pole synchronous machine (as a generator), which is rated to produce 1000 kVA. Its synchronous reactance is 0.4 Ohms.
It is connected to a diesel engine with the grid frequency 60 Hz.
I need to calculate the speed, power factor and power angle.


I know the following:
- Speed can be calculated by: n = 120 * f / P , where f is frequency (60 Hz) and P are poles (8) ... and we get the result 900 rpm.
- Power factor can be calculated by: PF = sin(PA), where PF is power factor, PA is power factor angle (is that the same as power angle?)


So, how can I calculate power factor and power angle?

 

[charlie b]

You don't have enough information. We would need to know more about the load. The power factor (or power angle) will depend on the total resistance (a value you did not provide) and the total reactance (a value you provided for the generator only, not the load). Also, PF is equal to the cosine of the power angle, not the sine. At an angle of 0 degrees, which is to say with a net reactance of 0 ohms, the power factor is 1.0. The cosine of 0 degrees is 1.0, and the sine of 0 degrees is 0.0.

Welcome to the forum.

 

[doctorslo]

Thank you for the reply.

The load is resistive type, 500 kW, 480 V.

 

[Bugman1400]

You don't have enough information. We would need to know more about the load. The power factor (or power angle) will depend on the total resistance (a value you did not provide) and the total reactance (a value you provided for the generator only, not the load). Also, PF is equal to the cosine of the power angle, not the sine. At an angle of 0 degrees, which is to say with a net reactance of 0 ohms, the power factor is 1.0. The cosine of 0 degrees is 1.0, and the sine of 0 degrees is 0.0.

Welcome to the forum.

With a synchronous machine, we will need the generator capability curve since a synchronous machine can be operated at any power factor according to its capability. So, its possible to run a leading pf to counteract the reactance of the load.

 

[LMAO]

Unless I am missing something your power factor is 1 (phase angle = 0).

 

[Bugman1400]

Unless I am missing something your power factor is 1 (phase angle = 0).

With the given load, I think you are correct unless, the generator needs to run leading to counteract its reactance.

I was considering this to be an application problem and the generator to be connected to the utility.

 

[Julius Right]

In my opinion by "power angle" you meant the angle between EMF[ at 500 kW load ] and the network voltage[delta]. EMF depends on excitation current.
The usual formula for active power on a synchronous generator is:
Pa=Eo*V1/xd*sin(d)+V1^2/2*(xd-xq)/xd/xq*sin(2*d)
where Eo=EMF at no-load and V1 it is the network voltage.
Let?s presume :
xd=0.4 ohm; xd=xq; Eo=V1=0.480kV Pa=0.500 MW
Then:
Pa=V1^2/xd*sin(d)
sin(d)=Pa*xd/V1^2
sin(d)=0.5/0.48^2*0.4=0.868
d=60 degree[approx.]

 

[Julius Right]

Both "fi" [angle between voltage and current supplied] and "delta" [between EMF and supplied voltage]?they are important angles in vector diagram of a synchronous generator.fi could be maximum 90 degrees but it could be positive or negative [if the generator supplies reactive power or requires reactive power from the system]. "delta"could be only positive and in the zone of more than 90 degrees the generator will loss stability. See for instance:
[Alstom]
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f Po w e r S y s t e m P l a n t
http://www.fecime.org/referencias/npag/chap5-46-77.pdf

 

[doctorslo]

Thanks for the answers. I'm still not sure if I understand everything, so here is the full text of the exercise I'm trying to solve. Sadly, there are no results (answers) given:

A resistive load of 500kW at 480V is supplied by an eight-pole synchronous
generator connected to a diesel engine, and a wind turbine with the same induction generator
as that in previous exercise. The synchronous machine is rated to produce 1000 kVA. Its
synchronous reactance is 0.4 Ohms. A voltage regulator maintains a constant voltage of
480V across the load and a governor on the diesel engine maintains a fixed speed. The speed is
such that the grid frequency is 60 Hz. The power system is three-phase, and the
electrical machines are both Y-connected. The power from the wind is such that the induction
generator is operating at a slip of 0.01. At this slip, the power is 153.5kW and the power
factor is 0.888.
(a) Find the synchronous generator?s speed, power factor, and power angle.
(b) Confirm that the power and power factor are as stated.

From the previous exercise:

A four-pole induction generator is rated at 300 kVA and 480V. It has the following
parameters: Xls = Xlr = 0.15 Ohms, Rls = 0.014 Ohms, Rlr = 0.0136 Ohms, Xm = 5 Ohms.

 

[Phil Corso]

DrSlo... start with:

The Synch-Gen and the Ind'nGen are are connected to the same load, thus:

1) The former must supply the the load's 500kW, minus that provided by the latter!

2) The latter can't supply its own reactive KVAr, so it must come from the former!

Regards, Phil Corso

 

[doctorslo]

I did this:

1) I calculated the impedance: Z = R_s + jX_s = j4 = 4 Ohms
2) Apparent power is defined as: S = I² * Z. Turning this around you get: I = sqrt(S/Z) = 1581.14 A
3) Now I can calculate real power: P = U * I = 480 V * 1581.14 A = 758.9 kW (I assume this is the maximum power, since I don't know what the rated power 1000kVA) means.
4) Power factor is: P/S = 758.9 / 1000 = 0.7589
5) Which brings me the power factor angle: arccos(0.7589) = 40.6 degrees

So, what I am doing wrong?



Phil Corso,
If I do what you've said:
500 kW - 153.5 kW = 346.5 kW

Now what can I do with that? 346.5/1000 = 0.3465 ===> a power factor?
I am not sure what rated power is, if that is a maximum power, then this is wrong.

 

[doctorslo]

I cannot edit the post, so:

Let me try another way:
1) 500 kW - 153.5 kW = 346.5 kW
2) Total current in load: I_L = P/U = 500 kW / 480 V = 1041.67 A
3) Current from induction machine: I_I = 153.3 kW / 480 V = 319.375 A
4) Current that must be supplied by synchronous generator: 1041.67 A - 319.375 A = 722.295 A
5) E = V + jX_s * I = 480 V + 0.4 Ohm * 722.295 A = 768.9 V
6) P = (E*V*sin(PFA)) / X_S ==== > (P * X_S) / (E * V) = sin(PFA) = 0.3755
7) PFA = 22 degrees

Hm??
Please suggest/correct me. :huh:

 

[Phil Corso]

DrSlo..

Is this you Prof's Test question?

Phil

 

[doctorslo]

Haha, no, but this problem was assigned to me.
The problem is that I've encountered the synchronous and asynchronous machines for the first time ... -.-

 

[Phil Corso]

DrSlo...

Here's a tip... ignore losses, and current!

Phil

 

[Julius Right]

Phil Corso answer is quite correct, indeed. So the [active]power delivered by DG will be 500-153.5=346.5 kW and the reactive power delivered [=the reactive power required by wind turbine]= 153.5/cos(fi)*sin(fi)=79.48 kVAr. You have to remark the power factor of the DG supply energy is different [tan(fiDG)=79.49/346.5] leading, as Phil said.
For power angle calculation you have to put 346.5 kW instead of 500 [in my above answer].
In order to check the the induction generator delivered power and power factor you have to get
the current from Steinmetz equivalent circuit. See[for instance]:
https://www.selinc.com/WorkArea/DownloadAsset.aspx?id=3690
I presume the rotor parameters [Xlr and Rr ] are already referred to the stator[Xlr? and Rr? ].
The slip is negative since induction generator velocity it is above the synchronous velocity.
You have to solve the circuit using one of the general rule-Thevenin, virtual currents, node-voltages and so on.

 

[doctorslo]

Guys, thank you very much for the help.
Seems like I was confusing between power factor (angle) and power angle (besides everything else), lol .
I didn't even know that reactive power is there for the voltage control, and of course as you've said, it must be supplied by a synchronous machine.

Julius Right, I still have a question.
In your first answer, you stated: "Eo=V1". Why is the E voltage equivalent to terminal voltage V1?

 

[Julius Right]

I cannot edit the post, so:

Let me try another way:
1) 500 kW - 153.5 kW = 346.5 kW
2) Total current in load: I_L = P/U = 500 kW / 480 V = 1041.67 A
3) Current from induction machine: I_I = 153.3 kW / 480 V = 319.375 A
4) Current that must be supplied by synchronous generator: 1041.67 A - 319.375 A = 722.295 A
5) E = V + jX_s * I = 480 V + 0.4 Ohm * 722.295 A = 768.9 V
6) P = (E*V*sin(PFA)) / X_S ==== > (P * X_S) / (E * V) = sin(PFA) = 0.3755
7) PFA = 22 degrees

Hm??
Please suggest/correct me. :huh:

480 V it is the between phases voltage. So you have to divide by sqrt(3) the power
Itot=Ptot/sqrt(3)/V [cos(fi)=1 for total power].
The same for induction generator.
The generator current has a reactive component=-induction generator component[approx.95.5 A]
You have to take the generator current in complex[approx.417+95.5j]
Don?t forget to multiply I*Xs by sqrt(3).
The actual E it is about 618 V [it is not 480 ,indeed, as I presume above].

 

[Phil Corso]

DrSlo, Julius... your method is too complicated:

1) PF was given for Indn-Gen!

2) It is leading!

3) Calculate kVAr for Ind'n-Gen!

3) It is same for Sync-Gen, i.e., leading!

4) Calc kVA for Sync-Gen!

5) Calc PF for Sync-Gen!

6) Ans orig questions!

Phil