# Three-phase salient-pole synchronous generator

#### Jess_88

Joined Apr 29, 2011
174
Hey guys

I'm having some trouble with this question.

Three-phase salient-pole synchronous generator has the following constants in per unit on the machine kVA rating as a base (all resistances are neglected):

Xd = 1, Xd' = 0.35, Xd''=0.25, Xq = 0.6
Ea = 1.8<15°, Vterminal = 1<0°

- Compute power angle (δ), generator current magnitude (in pu) and phase angle (Ia<θ).

Now, the power angle is easy (just 15°) but I have no idea what equations I can use to fined the generator current. I'v been reading every text book I can fined and nothing really seems to fit. There just doesn't seem to be much information on this sort of thing.

What has given me the most promising results is:
E = Vt + Ia(Xd + Xq)
Ia = (E - Vt) / (Xq + Xd)
... but I have been told this is wrong

I'm guessing I have to do a phasor diagram and use some trig... but I cant see how it could be done with the given information.

Any help would be great,
thanks guys

#### Kermit2

Joined Feb 5, 2010
4,162
The generator supplies only the current demanded by the motor in this scenario.

Synchronous motors draw/create current based on the power angle(load). No load - no lag angle - no current

#### t_n_k

Joined Mar 6, 2009
5,455
What has given me the most promising results is:
E = Vt + Ia(Xd + Xq)
Ia = (E - Vt) / (Xq + Xd)
... but I have been told this is wrong

I'm guessing I have to do a phasor diagram and use some trig... but I cant see how it could be done with the given information.
You are correct in assuming you must look at the synchronous generator phasor diagram. This is pretty standard - as you would presumably know. Ignoring any resistive terms [as the question states] makes the solution simpler. One would not include the transient or sub-transient reactances in the [steady state] analysis. The answer will "come out" using simple geometrical relationships for the direct and quadrature terms.

#### Kermit2

Joined Feb 5, 2010
4,162
an example from an old textbook.

#### t_n_k

Joined Mar 6, 2009
5,455
Hello Kermit2,

If you re-check the OP's post you'll note there's no mention of a motor in this question.

I believe in this case one simply assumes the generator is being driven [by something] at synchronous speed and the generator output [perhaps to an infinite bus] meets certain criteria based on the power angle [already known] and other stated parameters such as terminal voltage & machine reactances.

#### Kermit2

Joined Feb 5, 2010
4,162
I believe the vector summations would be the same in either case, but you are right, and the mention of the motor/gen example was probably not the best approach.

#### t_n_k

Joined Mar 6, 2009
5,455
I believe the vector summations would be the same in either case, but you are right, and the mention of the motor/gen example was probably not the best approach.
I have to disagree somewhat - the vector summations wont be the same as those shown in your attachment. With a salient pole synchronous machine one should consider both the direct and quadrature components of the machine reactances. This makes for a more [but not overly so] complicated vector diagram.

#### Kermit2

Joined Feb 5, 2010
4,162
I was suggesting an approach to the problem, and did not intend to answer the problem for him, just a shove in the right direction. I would be hard pressed to do any of these maths now anyway since my education is almost as old as the text book I used.

I will defer to your superior knowledge of the subject.

#### t_n_k

Joined Mar 6, 2009
5,455
You overestimate my knowledge on the subject and probably undervalue your own.

#### Jess_88

Joined Apr 29, 2011
174
Thanks for the input guys

So im pretty sure I have the theta angle worked out from my phasor diagram
x=tan^(-1)(0.6/1)
theta = x - 75 = -44
which looks good.

Now I'm pretty sure I wouldn't use a phasor diagram for the current magnitude (at lest I think...). I would need a formula incorporating the voltage drop across Xd and Xq and Vt. I'm just unsure which formula to use

#### t_n_k

Joined Mar 6, 2009
5,455
Thanks for the input guys

So im pretty sure I have the theta angle worked out from my phasor diagram
x=tan^(-1)(0.6/1)
theta = x - 75 = -44
which looks good.

Now I'm pretty sure I wouldn't use a phasor diagram for the current magnitude (at lest I think...). I would need a formula incorporating the voltage drop across Xd and Xq and Vt. I'm just unsure which formula to use
Unless you draw the phasor diagram you won't be able to solve the problem. Your answer for θ of -44° isn't what I calculate.

#### Jess_88

Joined Apr 29, 2011
174
Here is my phasor diagram.
Does it look ok?

I'm getting pretty confused with my trig

#### Jess_88

Joined Apr 29, 2011
174
I noticed something in my notes... but its doesn't exactly make sense to me.
Is this true? (looking at Xq)

#### t_n_k

Joined Mar 6, 2009
5,455
This is how I saw it ....

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#### Jess_88

Joined Apr 29, 2011
174
Thanks t_n_k,

Could you explain a how you got your 75 angle? I think there may be a relationship I'm not aware of.
I was looking at it from my phasor as, 90 degree difference from Iq to Id. So subtracting 15 from 90 to give -75 from the x axis.

#### t_n_k

Joined Mar 6, 2009
5,455
Yes - since IqXq is orthogonal with Ea it must be true with $$\delta$$ = 15° that the opposite angle is 75°. Just to be clear, this applies to the right-angled triangle formed by the sides $$V_{T} \ , \ I_{q}X_{q}$$ and $$\(E_{a}-I_{d}X_{d}$$\).

#### Jess_88

Joined Apr 29, 2011
174
Sorry, I think I'm a little slow...
Did you calculate Id and Iq as well to do this?

#### t_n_k

Joined Mar 6, 2009
5,455
Sorry, I think I'm a little slow...
Did you calculate Id and Iq as well to do this?
Not to find the 75° - which by the way is essentially superfluous information. It was meant to indicate the right angle triangle lying along the VT and Ea phasors.

The relevant relationships for finding Id & Iq can be extracted from the formulas shown at the top of page on the phasor diagram of post #14