Asychnronous Motor-Generator 2

[jimmynora]

We have an asynchronous motor (slipring rotor) to run our short-circuit Generator which produces electricity for our testing procedures. Power factor = 0.9

During a capacitive current test we need only a capacitive current. During operation the output from our generator consists of both active and reactive power (am i right?), and the capacitive current will come from the reactive power side.

So is it possible to save the active power part of the generators output if it is not needed ? How could we do that ?

 

[Skogsgurra]

Not easy to understand your reasoning.

If your load consumes only reactive power then that's it. Where do you think the active power goes?

Is it the 0.9 power factor you are tninking about? That is only part of the specification - not anything that always applies. If load power factor is close to zero (as capacitive load is) then the generator will see (and deliver) just that load. No active power - except some losses.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[rbulsara]

Asynchronous generator (induction machine) does not "supply" any reactive power, it in fact "absorbs" reactive power (hence inductive) from the utility (or other large source) to which it is connected to.

Your last two paragraphs do not make much sense, but I gather you are asking if this machine can be used as a condenser (like capacitors), then the answer is no. Synchronous machine can do that as its excitation can be controlled.


Rafiq Bulsara

http://www.srengineersct.com

 

[jimmynora]

So the generator is delivering only reactive power during the capacitive test procedure ?

Does the load always dictate what kind of power mix is being output ?

 

[rbulsara]

What is a capacitive test for an induction machine?

Rafiq Bulsara

http://www.srengineersct.com

 

[jimmynora]

We are running a capacitive current test (making/breaking) on a circuit breaker. The electricity provided for our tests is always through our generator which is being power by an asynchronous motor.

 

[rbulsara]

I am sorry, I misread a bit.

The asycn motor (the prime mover) does not supply reactive power, ever. Only active power.

The power factor of a synchronous generator in stand alone mode is solely dependent on nature of the load.

Rafiq Bulsara

http://www.srengineersct.com

 

[rbulsara]

So if your load is only capacitive (connected to the generator), no active power is supplied by the prime mover, except for the losses.



Rafiq Bulsara

http://www.srengineersct.com

 

[Skogsgurra]

It is not probable that your asynchronous machine will be able to deliver any reactive power (or rather make your capacitors deliver reactive power) if there isn't a 'governing' voltage from the grid.

The question about deliver or consume reactive power is dependent on what one has defined as source and sink. The convention is that inductances consume reactive power and that capacitors produce reactive power. But the definition is a lot more arbitrary than it is for active power.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[waross]

The phase angle of the current through a short circuited breaker will depend on the X:R ratio of the total circuit. That is both the generator and the connections to the breaker under test. If you want a capacitive current you must add capacitors to the circuit.
Active power and reactive power is a way to quantify the effects of a phase shift between the voltage and current.
Your active power will be I[sup]2[/sup]R. If R is taken as the R of the load and connections to the generator terminals, If the R of the generator is added, this will be the total active power including internal generator losses.
The Reactive power is I[sup]2[/sup]X. Normally, X is taken as the load reactance, but under short circuit conditions the generator X becomes significant.
If you use series capacitors to get a reactive current, your generator will supply active power based on the test current and the total resistance, internal generator resistance and external test connection resistance.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rasevskii]

for jimmynora:

Your test generator is a synchronous machine driven by an induction motor (aka asynchronous motor) a slipring machine. There is no electrical connection between the test generator and the supply to your induction motor.

The generator is absorbing reactive VARs from the capacative test load, and supplying some few KW of active power to the test setup (losses). It is irrelevant that the generator nameplate reads 0.9 PF. The PF in this case is nearly zero and is dictated by the test setup only.

No you cannot "save the unused active power" because there is not any to speak of.

Get back to the basics of electrical machines, especially if your work in the test room requires this knowledge.

Such test setups were common in electrical manufacturers factories in the old days. The knowledge seems to have departed unfortunately.

rasevskii

 

[LionelHutz]

I agree with the others - It sounds like you have a squirrel-cage induction motor coupled to a generator.

Does the load always dictate what kind of power mix is being output ?

Seriously, did you never take any circuits courses? Your load always dicates what power is provided by the source. A source won't just have it' rated power "go away to nowhere". Any power produced by the source has to go to a load.

During operation the output from our generator consists of both active and reactive power (am i right?),

No, the generator will not just "output" a nameplate power. The generator will supply whatever current and power factor the load requires from it. Yes, you can get both an active and reactive power output if the load requires both.

The generator power factor rating is used to know how much reactive power you can also get from that genset. For example, a 1MW machine rated 0.9 power factor means it can also provide around 484kVAR while providing 1MW. The machine may also have a curve showing the kVAR for different MW loads.

 

[Samiyoussif]

The PF in the name plate is not a concern, you should pay attention to the VAR you absorb not to exceed the limit of the generator that will damage the diodes or thyristors, and since it is a capacitive load you may not feel the gen is overloaded