Understanding Asynchronous Generators - Page 1
Forum section: electronics / general

author: quiteblock
author_rank: Newbie
timestamp: April 17, 2024, 11:40:55 am
content: Hey. I'm a Mechanical Engineer and I think I'm pretty familiar with traditional synchronous generators (I work at a plant).You supply the rotor windings with DC, the excitation results in a magnetic field.Rotate the rotor and you now have a rotating magnetic field.Slap on some windings around that field and now the rotating magnetic field from the rotor induces an alternating current in the stator windings. Simple enough.If you want to connect it to the grid you need to have your AC be in phase with the grid so that your rotor becomes magnetically locked with the grids rotation, supply it with torque and you help the grid's current move along. Makes sense.Recently I've been getting involved with a wind farm and I can't wrap my head around the asynchronous generators here. I don't get how the generator rotor can spin at a different RPM/frequency than the grid. It says the generators nominal speed is 1500rpm (4 poles = 50Hz) and it can still generate power at +- 30% of that speed.What I have so far:Instead of supplying the rotor with DC you supply it with AC.Slip = variation between the rotor frequency and stator frequency.But is it the frequency of the shaft rotating that matters? Or is it the combined rotation of the magnetic field from the excitation AC of the rotor + the rpm of the rotor?Ex: if the AC being supplied to the rotor for excitation is 12 Hz. And the rotor is physically rotating at 40 Hz, the effective rotation of the magnetic field will be (12+40) 52 Hz right? So is the slip that 52 Hz vs the stator grid frequency of 50 Hz => 52/50 = 4% slip?And to actually generate power the rotor magnetic field rotation frequency has to be greater than tye stator so it effectively pulls it behind it as it rotates ahead of it? Right? There has to be positive slip for it to generate.So for example if the generator shaft was rotating at 30Hz and you wanted to generate power, you would have to feed the rotor 22hz of AC to result in a total magnetic field rotation of 52Hz which has positive slip vs the 50Hz of the grid and would effectively pull it along and generate power right?Other example, if the shaft was already rotating at 52hz. Would you supply it with 0hz AC? I.e. DC? to get a positive slip of 4% and generate power.Last example, if the shaft was rotating at 60hz (faster than the grid), would you supply it with -8hz AC (opposite direction) to reduce the effective rotation of the magnetic field to 52Hz and maintain that 4% positive slip to generate power?When I say you, I don't mean manually control the frequency, as I understand there's a frequency control system which automatically does this obviously. I think?And lastly lastly.The manual states that the generator can switch between star and delta connections. At lower speed it uses a star connection and at higher speed it uses a delta connection. Supposedly the star connection increases the speed range allowing generation at even lower RPMs, but why?Star connection has a lower output voltage right? Why does the variation in voltage allow lower RPM for generation?Actually last question.For terminologies, induction/asynchronous/wound rotor are all synonymous right. What about DFIG, are all induction generators DFIG?

author: ejeffrey
author_rank: Super Contributor
timestamp: April 17, 2024, 03:23:29 pm
content: What you need to generate power is a non-zero angle between the stator and rotor magnetic fields. Slip (measured in Hz) is specifically for induction motors where the rotor field is induced from the stator. The rotor field is constantly dieing down due to resistive losses and being recreated with an in phase component aligned with the stator field at the moment. This causes the rotor field to spin slightly faster than the rotor itself. That is the slip.Synchronous motors, whether they have permanent magnets or an externally applied excitation should not have slip in normal operation.For a given winding voltage, wye configuration will have higher phase to phase voltage than delta. Since you need to hit a specific operating voltage (the grid voltage), changing from wye to delta allows operating at higher winding voltage at higher power.

author: David Hess
author_rank: Super Contributor
timestamp: April 17, 2024, 04:51:18 pm
content: I think you are referring to a doubly fed electric machine:https://en.wikipedia.org/wiki/Doubly_fed_electric_machine#Brushless_wound-rotor_doubly_fed_electric_machineThe excitation is variable frequency AC so that the excitation by itself causes the magnetic field to rotate, and that combined with the physical rotation produces the output frequency. So the AC excitation frequency is made to track the rotor RPM to produce a constant 60 Hz output, allowing the rotor RPM to vary, and the phase is also neccessarily controlled so the output may provide reactive power as well.

author: woodchips
author_rank: Frequent Contributor
timestamp: April 18, 2024, 08:41:03 am
content: Search for induction generator, will then find lots of info.An induction generator is simply an induction motor that is being driven by an external power source faster than its speed as a motor, so faster than 1500rpm for a 50Hz 4 pole motor.BUT, an induction generator gets its excitation from the power lines it is connected to, or capacitors, what must be understood is that this excitation power is leading power factor. An induction generator will not supply power to a lagging load, typically an induction motor. As such its use becomes somewhat less.