Rodmcm
Electrical
- May 11, 2004
- 259
Hi
I am going around in circles with CT burden calculations… Can someone assist
Say we have the following
Ip= 100A, Is=5A, Type=5P, ALF=10, VA=15 and say Rct=0.2 ohm, Rl=0.45 Ohm (50 metres of 4mm2) and Relay Resistance Rr= 0.004 ohm and the fault level is 900 A and the maximum running current 80A.
We know that the impedance at rated burden and current is 15/(5*5) = 0.6 Ohm and hence we know that the CT must be capable of developing an internal voltage of (5 x 10 x 0.6) = 30V to deliver Is x ALF at the rated burden. This is as I understand it is the minimum or roughly appox knee voltage.
We know that at the fault level the voltage at the CT terminals to drive the secondary fault current through the cable and protection burden is 900*5*(0.45+0.004)/100 = 20.4V
I have read that the saturation voltage must be greater than ALF*Is*(Rct+Rl+Rr) = 10*5*(0.2+0.454) = 32 V
How is the actual saturation voltage calculated or assessed? How does the internal knee voltage of the CT calculated above relate to the required terminal voltage or the saturation voltage ? Is the saturation voltage calculation an approximation as it does not include reactive impedance of the CT or is it negligible enough to neglect? What is the pass/fail criteria for these calculations? Obviously AFL to fault level is one and Ip to load is another. Will this CT do the job?
Toddlers quide anyone!
I am going around in circles with CT burden calculations… Can someone assist
Say we have the following
Ip= 100A, Is=5A, Type=5P, ALF=10, VA=15 and say Rct=0.2 ohm, Rl=0.45 Ohm (50 metres of 4mm2) and Relay Resistance Rr= 0.004 ohm and the fault level is 900 A and the maximum running current 80A.
We know that the impedance at rated burden and current is 15/(5*5) = 0.6 Ohm and hence we know that the CT must be capable of developing an internal voltage of (5 x 10 x 0.6) = 30V to deliver Is x ALF at the rated burden. This is as I understand it is the minimum or roughly appox knee voltage.
We know that at the fault level the voltage at the CT terminals to drive the secondary fault current through the cable and protection burden is 900*5*(0.45+0.004)/100 = 20.4V
I have read that the saturation voltage must be greater than ALF*Is*(Rct+Rl+Rr) = 10*5*(0.2+0.454) = 32 V
How is the actual saturation voltage calculated or assessed? How does the internal knee voltage of the CT calculated above relate to the required terminal voltage or the saturation voltage ? Is the saturation voltage calculation an approximation as it does not include reactive impedance of the CT or is it negligible enough to neglect? What is the pass/fail criteria for these calculations? Obviously AFL to fault level is one and Ip to load is another. Will this CT do the job?
Toddlers quide anyone!