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In your example the "B1" and "B_stator" are not buses. They are three phase current and voltage metering blocks which are probably re-scaled to save some space.

Hi Slavag, Sorry it is getting late, the earlier post only shows the faulted bus currents and their RMS, I forget to attach the plots from the winding currents. http://files.engineering.com/getfile.aspx?folder=9b0ac6e7-2443-4357-a9fa-870238056ea8&file=Zig_Zag_Transformer_Currents_.pdf

Hi Slavag, The attached file shows the phase (winding) currents and the neutral current when there is a bolted fault on the phase A (24 kV supply).http://files.engineering.com/getfile.aspx?folder=5e2ef44c-86e7-4c28-a832-c84170f25df2&file=Zig_Zag_Transformer_Currents_.pdf

Your model is wrong: 1.) You should use Thyristor bridges instead of IGBT bridges when you using a pulse 12 pulse generator. (GenPulsos1) 2.) Use the ”MedicionR” block is only for AC power. 3.) Use a “mean value” block to smooth the DC Current and voltage before the DC power calculation.

Additionally the factor a (7)sqrt(10) = 1.39 is also used e.g. 25mm^2*1.39 = 35 mm^2 ; 50mm^2*1.39 =70mm^2; etc.

The geometric series factor for IEC cables are calculated as: (5)sqrt(10) = 1.585 where the numbers rounded off

ScottyUK, The following excerpt from IEC 947-4-1 "5.3.5.5 Starting and stopping characteristics of starters c) one direction of rotation, or two directions of rotation as in item b), but with the possibility of infrequent inching (jogging). For this service condition, direct-on-line starters...

Occasional inching and plugging could be the reason.

jraef I think it means that the motor is still rotating and the motor terminal voltage is approx 20% of the line voltage due to the trappad flux in the rotor circuit.

Gunnar, My background is also with ASEA. Sorry but I did not realize that you know DC converters.

Gunnar, 1. Using an insulated oscilloscope or high voltage differential probes check first that that the control (synchronizing) voltage is in phase with the bridge voltage. 2. Check the alpha and beta limit (the rectifier and inverter limit) of the control. The easiest way to do it not...

To calculate the X/R ratio from the peak and Rms current check the IEEE 242 -2001 chapter 2.4 (eq. 2-3)

Asymptote, The parallel resonance frequency of your system can be estimated by the following formula n = sqrt(Ssc/Qcap) Based on the given data I calculated the approximately 144 MVA short circuit power @ the 6.6 kV bus Therefore in your case n = sqrt(144/0.4) =sqrt(360) = 1439 Hz which is...

Use a PI regulator instead of PID, it will improve your speed control stability, the D part of the controller is not very suitable for speed control application. If you looking for a very accurate and tight speed control your current controller response time has to be 6-8 times faster than a...

The failures happened during start up or the system was running? If it happened switching do, you have a vacuum breaker or switch and three-phase cable feed from the switchgear to motors?

I agree with GordS, my experiance is that the duration of the inrush current will be effected by the ressitive components of the load.

How long is your 12 V control wire? If the main contacts of the relay is not welded it is possible the the relay winding may stay energize by the 12 volt control wire capacitancealso make sure the yor 12 volt relay coil is energized by dry contact.

Electricpete, To find out what is going on use a storage oscilloscope with three DC current probes, pre-trigger the scope on the relay tripping and record about 500ms before the relay trips. It will reveal the transient condition and the duration of the current transient.

Sorry for my English and let try again The peak (subtransient) value of the transient AC current can be calculated as follows: Ipeak = Irms*sqrt(2)*k Where the Irms = steady state rms current k=1.02+0.98^-3*R/X The R is the resistance of the circuit and X is the inductive reactance of the...

Electricpete, You were right of the duration of transient inrush current, when I made a quick simulation just using X and R of the circuit at pf = 0.1 the DC component was practically gone after 6 cycles. The subtransient peak current was 3343 A which is 1.73 times of the 1567 A, but the 1.73...