P.U. impedance for a circuit

[mull982]

As another example consider the equation for calculating a reactor to install on the supply side of a vfd application. The % impedance of the reactor is based off of the VFD's base impedance using the V^2/VA method.

So with this equation lets say we wanted to install a 5% reactor which is common for this application. This doesn't necessarily mean that the reactor is 5% of the total circuit impedance since this 5% value is only based on the base impedance. Is this correct?

We would have to know the actual impedance of the drive to determine the what % of the circuit the reactor represented and to determine the voltage drop across it in steady state. Is this correct?

Aside from determining the base voltage drom V^2/VA I have also seen base imp = V/A. Can this method also be used for determining the correct base voltage.

Thanks for all the help with this.

 

[jim dungar]

Has this been a theoretical discussion, or are you trying to confirm the sizing requirements of a reactor to feed a VFD?

Choosing an input reactor of 3% or 5% for feeding a VFD has little to do with the actual input impedance of the drive. I have never, personally, calculated the actual Xl of a drive reactor, I have always let the reactor manufacturer make the recommendations.

 

[mull982]

Has this been a theoretical discussion, or are you trying to confirm the sizing requirements of a reactor to feed a VFD?

Choosing an input reactor of 3% or 5% for feeding a VFD has little to do with the actual input impedance of the drive. I have never, personally, calculated the actual Xl of a drive reactor, I have always let the reactor manufacturer make the recommendations.

This has simply been a thoeretical discussion about % impedance and I was using the common application reactor example as an example to help better understand the topic.

Another example is a technote I recently read for sizing a rector on the primary of a TR set for an precipitator. The tech note says that the reactor should be sized at 50% of the circuit impedance with the circuit impedance being determined by the transformers rated V/I.

So in this case as well the article is saying to use the transformers base impedance for calculating reactor. The transformer is a single phase 480V primary 100kVA unit with a 5.7% impedance. So with this I come up with the following.

Base impedance = 480V/208A = 2.3ohms
Acutal transformer impedance= 2.3ohms * .057 = .113ohms
Actual Reactor impedance = 2.3ohms * .50 = 1.15ohms

So based of this it would appear that the majority of the impedance in this circuit is due to the reactor which is about 10 time larger than the impedance of the transformer. I this case the majority of the voltage would drop across the reactor only about 9% of the voltage will drop across the transformer. Does this seem right?