# Three-phase salient-pole synchronous generator

Discussion in 'Homework Help' started by Jess_88, Oct 1, 2012.

1. ### Jess_88 Thread Starter Member

Apr 29, 2011
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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

2. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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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

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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

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4. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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an example from an old textbook.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

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6. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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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.

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7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

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8. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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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.

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9. ### t_n_k AAC Fanatic!

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

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10. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
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

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

12. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Here is my phasor diagram.
Does it look ok?

I'm getting pretty confused with my trig

13. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
I noticed something in my notes... but its doesn't exactly make sense to me.
Is this true? (looking at Xq)

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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This is how I saw it ....

• ###### Phasor Diagram.pdf
File size:
26.8 KB
Views:
36
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15. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
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.

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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 $$$E_{a}-I_{d}X_{d}$$$.

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17. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Sorry, I think I'm a little slow...
Did you calculate Id and Iq as well to do this?

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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