Synchronous generator under load

[cnh1995]

A synchronous generator is connected to the grid through a reactane of j0.65 pu. The grid voltage is 1 pu and the generator is feeding 0.7+j0.6 pu power to the grid. After some calculations, the generated emf is found to be 1.46 ∠18.12° pu. If the active power is increased such that new complex power is 0.8+j0.6 pu (reactive power unchanged), the generated emf comes out to be 1.48∠20.51°.
I can understand the fact that the power angle δ increased so as to feed the extra real power. What I don't understand is why the generated emf increased. According to the equation E=V_t+ I*X, the value of E should increase since the current increased. But since the reactive component of the armature current is unchanged, why should the excitation be increased? If the armature current is lagging, its reactive component has a demagnetizing effect on the air gap and hence, terminal voltage drops. To bring the terminal voltage back to 1 pu, excitation is increased. In this problem, there is no change
in the reactive component of armature current. So why is the excitation increased when the reactive power is constant?
Thanks a lot in advance!

 

[jim hardy]

Try the phasor approach described in this June thread
https://www.physicsforums.com/threads/voltage-and-reactive-power-relationship.876346/#post-5505312

i think your question lends itself to graphical solution
which reduces it to a geometry problem.

but i didn't try it with your numbers.

 

[cnh1995]

Try the phasor approach described in this June thread
https://www.physicsforums.com/threads/voltage-and-reactive-power-relationship.876346/#post-5505312

i think your question lends itself to graphical solution
which reduces it to a geometry problem.

but i didn't try it with your numbers.

Thanks. I understand now that for reactive power to be constant, Ecosδ should be
constant and for active power to be constant, Esinδ should be constant.

 

[cnh1995]

Hi @jim hardy..
I have a small question about this diagram..

Here, before raising the excitation, the VARs were leading and to bring them back to zero, excitation was increased. Right?
Before that, the rotor was given extra mechanical torque and hence, it pulled ahead. My question is, how does the rotor rotate at the previous speed? It accelerates while pulling ahead and hence, shouldn't its speed be more? Real part of the load current stabilizes the rotor and it moves with a constant speed after pulling ahead but shouldn't that speed be more than the no load speed?

 

[jim hardy]

It accelerates while pulling ahead and hence, shouldn't its speed be more? Real part of the load current stabilizes the rotor and it moves with a constant speed after pulling ahead but shouldn't that speed be more than the no load speed?

I'm on a trip but will take a minute to answer.
How can it ?
You haven't thought that question through.
The speeds are unequal only while the rotor is changing its power angle.
It accelerates, gets ahead and decelerates. Probably it'll hunt a little as it finds new equilibrium power angle.

thought experiment......

What if you and i are both running side by side around a one mile circular track at 15mph, doing repeated 4 minute miles?(Unlikely for me at my age )
If you sprint a little faster to get a few feet ahead and then slow back down to resume 15mph,
we will both continue to do 4 minute miles with constant distance between us
and both at 15mph.
You do not need to run faster to stay that few feet ahead, nor can i run slower and stay same distance behind.

Watch children on a merry-go-round.

old jimkeep it simple.

 

[cnh1995]

It accelerates, gets ahead and decelerates. Probably it'll hunt a little as it finds new equilibrium power angle.

Yes. That's intuitive. Also, I forgot that the differential equation governing relation between power angle and time is of second order, which explains the oscillations of the rotor.
Thanks! Enjoy your trip!