Power System expertise 6: Generation
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- čas přidán 13. 04. 2020
- To use the background simulator yourself go to www.ecsp.ch. How does a power generator work in a power system? Demonstration of the generation torque control, generation magnetic field control and the phase angle control between the power system and the synchronous generator. Watch more videos: / @georgschett801
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Great video, best in the series so far. Thanks.
Dude! Thanks for this. So I'm a novice at this but it seems like you are saying the power flow from the generator is a function of the phase angle and generator voltage, although this appears nominal at the grid. Is that right? Other basics to get me up to speed would be appreciated
The phase angle controls the real power flow, the voltage controls the reactive power flow.
@@georgschett801 got it. I'm a little thrown in your simulation like at 835 time of your video. You are showing voltage of two sources doing the phase shift as blue, but power as red. Why is power twice the frequency of the voltage, I would think the kw would be the same frequency....what am I missing? Is this not the kw seen at the infinite gird or some other power?
@@nonyafletcher601 Power = U*I, thus always positive if both are positive or both are negative at the same time. It is explained in textbooks in details, you may draw a curve with excel. It has nothing to do with two sources. One source and an inductive and/or ohmic impedance is enough to get it right.
Hello Georg, thanks for very good video. I wonder, for stiff grid Ug =Un and at Delta = 90 (maximum power output), it seems to me that the load current has do be capacitive to make the phasor diagram complete? Thinking this scenario: Ug= Un+ jIaXs. Ia= armature current/load current and Xs synchronous reactance, referencing Un at zero degrees and Ug leading Un by 90 degrees, Ia must then lead Un by approximately 45 degrees so voltage drop jIaXs completes the phasor diagram. But then again, this seems strange as per phase output power from generator can also be written as Pout= IaUn x Cos(phi), where maximum power transfer is when cos(phi) is 1, that is when Ia is in phase with Un. What I guess i'm wondering is, how can this be true: (Referencing the video) (UgUn/2Xd)*sin(d) = IaUn*Cos(phi)? At maximum power output, Delta = 90, but can Phi be equal to 0 then? Cheers Magnus
Hello Magnus, I am not sure if I get your question right. There might be a confusion between impedance (load) angle and the angle between sending and receiving voltage source. You can use my free version of the simulator www.ecsp.,ch in order to play with the 2 sources and change the angle between two sources. So you can observe the impact of the angles and of the source voltages on the power exchange between the 2 sources. All the best, George
Thanks, I thought that the angle between Ug and Un was the load angle, is this not the case?@@georgschett801
I have assumed this: Load angle is equal to torque angle of synchronous generator pluss angle due to line impedance
No Magnus, it is the angle between the two sources.
@@MagneManet
The total source angle between the 2 sources is equal to the torque angle of G1 + the angle between one side of the line and the other side of the line plus the torque angle of G2. @@MagneManet
Which modulation software you use?
it is my own software, the link is on all the slides in the video: www.ecsp.ch. The version I used is a beta, will be uploaded soon.
It is an own online browser based software: www.ecsp.ch
Why not a induction generator?(asynchronous)
There are async. generators in the power grid, but mostly these are smaller units for example in some wind turbines. In order to build a functioning grid you need synchronous generators, they are black start capable. In order for an async. generator to work, you need it to be connected to an existing functioning grid. Adding smaller and cheep async. generators to a grid is a good way to increase the power supply at lower costs.
@@georgschett801 Thank you for the info, we always hear about induction type machines, that's why I asked!