Long Protection CT Leads

[EddyWirbelstrom]

It is intended to run protection CT secondary leads a distance of 1300 meters to a back-up earth fault relay.
The CT is installed on a transformer neutral with a 250A Neutral Earth Resistor.
CT ratio = 250 : 1
CT class = 5VA Class 5 P10 ( to IEC 185 )
CT class = 5P 50 F10 ( to AS 1675 )
Relay = Alstom type MCSU, In = 1A
Relay burden at max setting of 0.152A = 0.0007 VA
Therefore relay impedance at max setting = 0.0007 / 0.152^2 = 0.03 Ohm.
Impedance of CT secondary lead loop = 21 Ohm.
The 5VA CT could push the relay set current of 0.152A through an impedance of R = 5 / 0.152^2 = 216 Ohm.
Therefore the CT can deliver the set current of 0.152A through the burden impedance of 21.03 Ohm.
Question:
What other considerations are necessary for this installation ?

 

[jbartos]

Suggestion: The longer CT secondary run will exhibit: XL=2xpixfxL, inductive reactance and its effect,
XC=1/(2xpixfxC), capacitive reactance and its effect
beside the noticeably higher resistance, R.
Therefore, the CT secondary run conductors impedance
Zc=R+j(XL-XC), in Ohms will have to be added to the CT loop burden. The burden shall be formulated as Zb=Rb+jXb, in Ohms. The transient behavior of the longer CT secondary run will also surface.

 

[EddyWirbelstrom]

The CT lead impedance was based on the published R+jX values for 1300m of 2.5mm2, Cu, PVC/PVC, 0.6/1kV, multicore cable.
Only inductive reactance is included in the published cable data. Any suggestions for the uF/km between the two cores of such a cable ?

 

[stevenal]

Where's the 0.152A come from? Isn't your fault current 250A primary > 1A secondary?

 

[EddyWirbelstrom]

The 0.152A comes from the maximum available setting current of the MCSU Sensitive Earth Fault relay which has a setting range of 0.5% to 15.2% of rated current.
At the maximum earth fault current of 250A, the 250:1 ratio, 5VA Class 5P10 CT can only push 1A through a burden of 5 / 1^2 = 5 Ohm. The burden of CT leads + relay is 21.36 Ohm. The concern is that the CT will begin to saturate and may not deliver sufficient volts to drive the set current of 0.152A through the 21.36 Ohm burden.

 

[jbartos]

Suggestion:
5VA is CT rating
Ictsec=sqrt(5VA/21.36Ohm)=0.48A instead of 1A
The 21.36Ohm would limit CT to:
0.48 x 250/1=120A instead of 250A max earth fault

 

[EddyWirbelstrom]

Yes, the 250 : 1, 5VA Class 5 P10 CT can continuously deliver 0.48 A through a load burden of 21.36 Ohm.
However the CT "Accuracy Limit Factor" of 10 (P10)indicates that the CT can deliver (for a short time) 10 x its rated current while its ratio error remains below 5%.
For a CT with rated 5VA, 1A secondary, the rated load burden is R = VA / I^2 = 5 Ohm. Therefore the CT can (for a short time) deliver 10A x 5 Ohm = 50V which would drive 2.34A through a load burden of 21.36 Ohm.

 

[jbartos]

Suggestion to the previous posting: The short time rating of CT results in heavily distorted waves of current and voltage. This short time rating is just for the CT capability to withstand it without any physical damage such as burned winding insulation or physical damage due to electromagnetic forces. Therefore, the regular Ohms relationship should be compared to oscilloscope traces of waveforms.

 

[stevenal]

Your concern that saturation will prevent operation is valid. I would choose to not operate in this region. At least find a way to accurately digitally simulate the system, or to test it before relying on it.

 

[jghrist]

See The Impact of High Fault Current and CT Rating Limits on Overcurrent Protection, by Gabriel Benmouyal and Stanley E. Zocholl for an analysis of CT saturation with microprocessor o/c relays.

http://www.selinc.com/techpprs/6142.pdf

 

[stevenal]

Have you considered using a loading resistor to convert the current to voltage, and relaying off the voltage? Use a value that won't overburden your CT and keep the current loop short. Run the voltage signal the 1300 meters instead of current. Check voltage drop.

Jim,

If remember right, Mr. Zocholl simulated ANSI CTs. Accurate simulation will also require knowledge of the protection algorithm used.

 

[wilkin]

Your calculations are well informed and accurate i would add the following comments
The MCSU burden at 8% setting is 0.03 ohms the pilots are 21 ohms the voltage developed is .08 x 1.9 x 21.03 = 3.2 volts
5VA CT 5P10 will be acurate to 1% of 1amp at 5 volts ie 0.01 amps(the actual current could be obtaind from a mag curve)
The relay setting would be setting + ct mag current or 0.152+1 =0.252amps
POC would be 0.152 x 250 = 38 amps.
The ct will provide without saturation 50 volts
Therefore if 250 amps were to flow to earth the ct would produce 1 x 21.03 = 21.03 volts.
No saturation of the ct would occur and no additional harmonics generated to disturbe the relay.
The capacitance of the pilots will cause a time delay adding to the delay setting of the relay but this should be negligble

 

[wilkin]

Sorry for maths error corrections as follows
The MCSU burden at 8% setting is 0.03 ohms.
The pilots are 21 ohms.
The voltage developed is .08 x 1.9 x 21.03 = 3.2 volts
5VA CT 5P10 will be acurate to 1% of 1amp at 5 volts ie 0.01 amps(the actual current could be obtaind from a mag curve)
The relay setting would be setting + ct mag current or 0.152+0.01 =0.162amps
POC would be 0.162 x 250 = 40.5 amps.
The ct will provide without saturation 5 x 10 = 50 volts
Therefore if 250 amps were to flow to earth the ct would produce 1 x 21.03 = 21.03 volts.
No saturation of the ct would occur and no additional harmonics generated to disturbe the relay.
The capacitance of the pilots will cause a time delay adding to the delay setting of the relay but this should be negligble

 

[xabproject]

dear mnewman,
2.5sqmm copper for 1300m run!
I do not think anything less than 16sqmm is a choice.
Pl think twice before going ahead.
Best regards

 

[electricpete]

xab - shouldn't one consider the max current (250A primary) along with the resistance, as shown by wilkin?

 

[xabproject]

Electricpete,
I do not see the need. But then, is this all required?!