Peak Wattage of AC generator

[elogesh]

Dear all,

I am doing one project on electromagnetic analysis of AC Generator of two wheeler.
I have found out Open Circuit Voltage and Short Circuit Current of the same.
I need to find out the Peak Wattage of this AC Generator.
Please help me out.

Any suggestion on this is welcome.

Thanks

Loganathan

 

[waross]

Peak wattage depends on the power of the prime mover. (And a little inertia).
When the load out (watts) exceeds the power in (Power of the engine) the generator slows down.
The other limit is the heating effect of the KVA. (More correctly the amps (A) of the KVA)
Running in excess of the KVA rating will result in overheating.
Attempted running in excess of the power capability of the prime mover will result in the prime mover slowing down.
On a small generator Kw often equals KVA. On a larger generator the KVA = Kw over PF (power factor). A typical power factor is 80%. 80 Kw / .8 = 100KVA
yours

 

[elogesh]

Thank you very much for instant reply.

But what I need to know to calculate the peak wattage of

my AC Generator?

Please let me know this.

Loganathan

 

[waross]

It depends on the power of the motor.
We loaded up a 600KW Diesel genset and as soon as the KW (Wattage) went over 600KW the motor started to slow down.
Smaller sets often have more reserve power, but;
What size is your set? What type of motor do you have? What is the horsepower of the motor?
Horsepower times 746 will give you an answer that will feel good, but may be meaningless.
BTW, why do you need to know "Peak wattage"?
yours

 

[VEBill]

"...Open Circuit Voltage and Short Circuit....need to find out the Peak Wattage."

Peak Wattage is NOT peak voltage times peak current. Open circuit voltage is zero amps thus zero watts. Short circuit current is (roughly) zero volts and thus (roughly) zero watts.

The best way to determine the max wattage is to look it up in the documentation.

 

[LionelHutz]

To get maximum power transfer you use a load impedance that is equal to the source impedance. The source impedance is the open circuit voltage divided by the short circuit current.

 

[VEBill]

"To get maximum power transfer you use a load impedance that is equal to the source impedance."

In the real world, with a typical electrical generator, that might last a few seconds. The source impedance is often very much lower than the lowest allowed load.

 

[VEBill]

By lowest allowed load, I mean lowest 'resistance' (highest load).

Many generators would be current limited.

 

[elogesh]

AC Generator I have is AC generator of two wheeler.

This Genset is mechanically coupled to vehicle engine.When engine starts,it gives power directly to the generator.
So from 500 rpm to 10000 rpm when engine accelerates,AC generator produces AC voltage varying according to the speed.
So the bHp of two wheeler will be around 7.5.
I have very little knowledge why my customer wants to know peak wattage.Excuse my ignorence.
The size of my generator is 112 mm dia outside diameter.
It is permanent magnet AC generator with rotor sitting outside and stator inside.
regards,
Logesh

 

[Skogsgurra]

..in a bike..

Gunnar Englund

http://www.gke.org

 

[elogesh]

yes it is AC generator of a bike which is called as MAGNETO

 

[waross]

I think that most of us thought that by two wheeler you meant a trailer mounted diesel generator. No problem. The quick answer is maximum voltage (probably at full speed) times rated current.
For a more exact figure, you must calculate the current at different frequencies and voltages and find the highest. Voltage and frequency will both increase with speed and the load impedance may change with the frequency. The highest current may not be at full speed but probably is.
yours