Motor Starting Current Calcualtion Double check

[zimou888]

System 11KV 50KA 3Sec; 60MVA 132/11KV transformer; Exsiting 11KV load 7.9MW, Motor 11KV,2950MW,2990rmn, LFA=6.2In, power factor 20%, locked torque =0.47TN, whether and how long the motor could be started?

 

[7anoter4]

Hi zimou888
First of all I guess the motor rated power is 2950 KW instead of 2950 MW.
Secondly there are many other data you need to calculate the motor starting time.
You need:
Load and motor torque curve [see the attached sketch].Also you need the GD^2[or WK^2] of the motor and of the load.
You need also uk% of the transformer, the length and the cross section of the 11kv supply cable in order to calculate the voltage drop.
In order to calculate all this I appreciated the motor dimensions and all other data and
I think that an approximate starting time will be 3 seconds. Also the voltage drop at the transformer terminal and at the motor terminal will be approximate 4-5% at the start.
Best Regards

 

[zimou888]

Thanks you very much.

According to the calcualtion of EDSA, starting time is about 3.47s, voltage drop 6%. The motor is 60kgm2, the load is 6kgm2.

Aanother question regarding the starting time, In my mind, it is related with starting torque, the bigger the quicker. But I met a problem about a small pump, 11kV system,132/11kV transformer 20MVA, SC=20KA,1S for 11kV, 11KV load 5MW, Motor 11KV,585KW,2975rmn, LFA=6In, power factor 15%, locked torque =0.6TN, motor is 16.25kgm2, load is 5kgm2. I found it takes 15s to start based on EDSA result, I was confused the longer starting time although it will not impact on the voltage drop. So the GD2 has the critical parameter during starting, is it right? I also suspect the EDSA result.