CTs knee-point voltage for 6.6 kV motor feeder 1

[Schwatzernov]

Dear All,

I have one question regarding to the CT sizing for phase overcurrent relay. The question is that the CTs used for motor feeder normally is dimensioned based on continuous current of motor. For example motor 320kW, 6.6 kV, the 5P20 class,50/5 A CT 15VA is selected while the max. three-phase fault at point just below VCB for its feeder is 30kA.
This means that the fault current is 30,000A/50 A = 600 times of its rated primary current. Does this means that if fault occur, the CTs will be always saturated ? so all current will comes to magnetizing part of CTs which there would be very few current flowing to 50/51 relay so relay will not operate for fault. Anyone could explain me ?

While selecting CTs with 30,000A/20 = 1500/5 A for this size of motor is very unusual, how to select suitable CT ?

Thank you

Schwatzernov

 

[davidbeach]

Some people don't do saturation calculations.

Some instantaneous relays can respond prior to CT saturation.

Some instances there is still enough current from the saturated CT to pick up the instantaneous element.

Sometimes the degree of saturation has to decline, which it will, before enough current is available for the instantaneous element.

Sometimes something burns up.

 

[Alex68]

Every time a small motor is fed by Medium Voltage, the difference between the rated current and the fault current is huge.
In case of fault, the saturation of the CT used for "normal" currents is required and desired to avoid the destruction of the protection by the extremely big secondary currents (3000 A in your case).
Thus you have to use two different protective systems: one for "normal" currents (rated current and start up current) and one for fault conditions.

Usually for small motors, two different soultions are possible:
1) CT and relay for normal conditions + fuse for fault conditions (old but cheaper solution)
2) CT for normal conditions + CT for fault conditions, both connected to a protective multifunction relay.

In both cases, the "CT for normal conditions" shall be able to measure up to start up currents but shall strongly saturate in case of fault.
In your example, the first CT could be 50/1A 5P5 5VA and the second one could be 5000/1 10P10 5VA.

Regards
Alex68

 

[slavag]

Hello.
David great explanation, shortly and fully.
Alex, 5000.1A is too big CT.
In lot of cases we dont pay attention on the short circuit of system. Usually , you calculate CT according to the nominal current and start-up current, and instan. setting you put according to start-up current and you need check that CT not saturated in time of start-up.
Your CT 5P20, 15VA, 50.5A seems as very good choice.
Regards.
Slava

 

[AMBMI]

Hi I agree with Slavag.
The check to perform should be that the CT is saturating for a current greater than twice the relay instantaneous threshold.
So this motor should have about 33 A of rated current (eff 0.95, cos phi =0.9). The starting current should be about 196 A (we assume 6 pu) so we put the inst threshold at about 390 A. I'm pretty sure that such CT isn't saturating at 780 A.
To perform the exact calculation we need the secondary resistance of the CT (1 ohm in this case?), length of the CT-relay cable and its diameter.

 

[AMBMI]

Uhmmmm I see only now that the CT is a 50/5 and not 50/1 so probably the Rct is about 0.2 and not about 1 ohm....

 

[oldfieldguy]

It motor controls, it is not unusual to "wrap" a larger-ratio CT to get a smaller ratio.

In other words, if you need a 50/5 ratio, and the saturation voltage on a 50:5 is too low for your application, you can use a 150:5 with the primary conductor wrapped through the window of the CT three times. In effect the ratio of the CT is divided by the number of primary turns.

This is NOT an uncommon practice.

old field guy

 

[slavag]

Old field guy.
You are right, its common practic too, in old application with EM relays. But always need check this according to protective relay.Not once, I see problems with thermal protection of motors.It operated according to FLC or Itheta or Itherm.
You must check setting of this parameter, for example
Itheta min setting eql 0.5Inom, in this case 150A div 2 eql 75A, is not correct setting for this motor where 35A is nominal current.
Regards.
Slava

 

[oldfieldguy]

slava

You're correct. the EE setting up the system needs to determine the proper ratio for system protection. He can then achieve this ratio by 'wrapping' a higher ratio CT. I have seen two, three or four turns through the window of a CT for this application. A 150:5 CT would need three turns on the primary for FLA=35.

old field guy