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Transformer voltage drop based on impedance

Transformer voltage drop based on impedance

Transformer voltage drop based on impedance

(OP)
Hello,
I would appreciate a sanity check.  I have a 1.5MVA transformer, 13.8kV to 480, 5.67% impedance.  I have a 600Hp motor (720 FLA) to start.  I am trying to make sure my voltage drop is less than 20%.  I have assumed 6 times FLA for starting.  From this information, I have calculated a voltage drop of 13%.  Does this seem reasonable?

thanks,
EE  

RE: Transformer voltage drop based on impedance

Yes, if you ignore cable impedance, which I would not.

Rafiq Bulsara
http://www.srengineersct.com

RE: Transformer voltage drop based on impedance

A rigorous solution would include the X:R ratio and the power factor.
However, the impedance is probably mostly reactance and the starting current is mostly reactive so calculations based on impedance and starting current alone may be surprisingly accurate. Not exact but you may introduce more error by ignoring the cable impedance.
When you include the cable impedance you should calculate two voltage drop values.
One at the transformer terminals to evaluate the effect on other loads and one at the motor terminals to evaluate the effect on motor starting.

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

RE: Transformer voltage drop based on impedance

(OP)
Thanks for the check.  I am not ignoring cable impedance.  The big motor is an existing load.  I only needed the transformer voltage drop because I am adding a new load to this transformer, which will consist of several starters and a PLC.  And I want to make sure the voltage drop caused by the big motor would not be enough to affect the controls.

RE: Transformer voltage drop based on impedance

I agree with rbulsara the voltage drop calculated is correct if you neglect the voltage drop on cable and if the transformer is free from other load. As usual I should take an initial load of 80% steady state on transformer[ including motor FLA].In this case the voltage drop could be 14 % [at least] at transformer terminals.

RE: Transformer voltage drop based on impedance

If this is existing by far the most accurate method will be to measure it.
"It's nice when actual conditions agree with the simulation."
A designer with infinite faith in his computer simulation.

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

RE: Transformer voltage drop based on impedance

You are right, waross. First of all, I think the rated motor LRA is 7.5 at 480 volt and 6*FLA at 480-20% drop [0.8*480= 384 V]. But also 13.8 kV could be up to 5% less that means 0.95*13.8=13.11 kV.
The frequency also could be 60+/-1Hz.The transformer copper losses may be 15-20 kW .Also, the transformer impedance has a tolerance of 7.5-10%. So the voltage drop could be at least 10% more in these conditions.
This is only a theoretical calculation, indeed.
 

RE: Transformer voltage drop based on impedance

The transformer impedance is only an accurate indication of short circuit current after it has reached steady state conditions.
Normally transformer voltage drop depends on the transformer regulation which is heavily influenced by the power factor of the load.
A rigorous solution of voltage drop requires knowledge of the load power factor and the transformer X:R ratio as well as the impedance.
However, in this instance, given the information available and the fact that the load (motor starting) is highly reactive, the simple calculation may be surprisingly accurate.
By dint of offsetting assumptions and conditions the original calculation is probably within acceptable limits of error.

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

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