Help needed calculating the voltage drop on a 3-phase line reactor 2

[bdn2004]

Given: L = 13 mH which equates to 0.047 ohms / phase.
FLA = 170 Amps
Line Voltage = 480V
3% reactor

V = I x Z .... 0.047ohms X 170amps = 8 Volts.
The literature states this will cause a 3% voltage drop across the reactor. 277V x .03 = 8 Volts, which works out with that calculation.

And,
480V x 0.03 = 14V
8 X 1.732 = 14V

It appears the voltage will be 480V - 14V = 466V. Am I correct on this?

 

[waross]

Probably not.
The voltage drop across the reactor may be 8 Volts or 14 Volts but you can't just subtract. The voltage drop in most reactors is at almost 90 degrees to the applied voltage.
A rigorous solution requires knowledge of both the resistance and the inductive reactance of the reactor, of the source transformer and of the load.
For resistive loads you can use the approximation √ (480V)² - (14V)² = 479.796 Volts
As the power factor of the circuit drops the circuit voltage will drop.
To visualize this, get in a small plane and fly 480 Miles straight east, now fly 14 miles straight south. How far in a straight line back to your starting point?
If you put this reactor in series with a transformer with a rated full load current of 170 Amps you can add the percent impedance of the reactor to the % impedance of the transformer to estimate the Available Short Circuit Current.
Again a rigorous solution requires knowledge of the X/R ratios of both the transformer and of the reactor.

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

 

[wroggent]

Assuming you meant 0.13mH and that 0.047 Ohms is the correct value, 466V would be worst case.