sizing CT protection by burden
sizing CT protection by burden
(OP)
1- Reference: Protection Techniques In Electrical Energy Systems – helmut ungrad & others
Chap 4- page 42
Where the actual secondary burden of a CT differs from the rated burden, the effective overcurrent factor Nnt varies according to equation:
nt = nn(( Sn + Se)/(St + Se)) (4.2)
nt – effective overcurrent factor
nn – rated overcurrent factor
Sn – rated CT power in VA
St - actual secondary burden (including lead burden) in VA
Se – burden of the CT itself in VA ( approx. 0.1 Sn)
It follows from equation 4.2 that the effective overcurrent factor nt can be increased by reducing the burden St. This fact is frequently used in practice to improve CT performance.
2-Application
A customer needs to install a system in which a CT rated current is 100A. Short circuit reaches a level of 4000A.
Equivalent load over the relay to the wiring transformer is 100 VA.(St)
For the TC protect the electrical system of this level of short circuit, there is a need to design a CT that does not saturate up to 20 times the rated current.
Assuming that the customer wants response has an error of less than 10%, specification will say:
• 200 / 5 A - 10P20 - 100 VA (10B400) solution 1
For better sensibility of protection we choose CT ratio = 100/5 A and Sn = 200 VA
Check:
nt = nn(( Sn + Se)/(St + Se))
nt – effective overcurrent factor
nn = 20
Sn = 200
St = 100
Se = 0
So nt = nn * Sn/St
nt = 20* 200/100 = 40;
• 100 / 5 A - 10P20 - 200 VA (10B800) solution 2
What do you think about solution 2?
Chap 4- page 42
Where the actual secondary burden of a CT differs from the rated burden, the effective overcurrent factor Nnt varies according to equation:
nt = nn(( Sn + Se)/(St + Se)) (4.2)
nt – effective overcurrent factor
nn – rated overcurrent factor
Sn – rated CT power in VA
St - actual secondary burden (including lead burden) in VA
Se – burden of the CT itself in VA ( approx. 0.1 Sn)
It follows from equation 4.2 that the effective overcurrent factor nt can be increased by reducing the burden St. This fact is frequently used in practice to improve CT performance.
2-Application
A customer needs to install a system in which a CT rated current is 100A. Short circuit reaches a level of 4000A.
Equivalent load over the relay to the wiring transformer is 100 VA.(St)
For the TC protect the electrical system of this level of short circuit, there is a need to design a CT that does not saturate up to 20 times the rated current.
Assuming that the customer wants response has an error of less than 10%, specification will say:
• 200 / 5 A - 10P20 - 100 VA (10B400) solution 1
For better sensibility of protection we choose CT ratio = 100/5 A and Sn = 200 VA
Check:
nt = nn(( Sn + Se)/(St + Se))
nt – effective overcurrent factor
nn = 20
Sn = 200
St = 100
Se = 0
So nt = nn * Sn/St
nt = 20* 200/100 = 40;
• 100 / 5 A - 10P20 - 200 VA (10B800) solution 2
What do you think about solution 2?






RE: sizing CT protection by burden
for newer relays 100VA, unpossible
10P20, take 5P30, don't increase burden, its not good solution,
change secondary to 1A.
from my point of view, 200VA is oversizing.
in additional, for protection you can use CT with 500/5 or 1000/5A, not 100/5A.
you must calculated your CT according to protective terminal manual, not according to good book, but its only book.
RE: sizing CT protection by burden
- high CT ratio for avoiding saturation
- low CT ratio for CT better sensibility of protection
RE: sizing CT protection by burden
Of course oversizing CT is possible too.
with newer relay n'>15-20 approx.
RE: sizing CT protection by burden
when we specify CT equipments we do not know yet the manufacturer of relays. You can review later, although not a good idea, there may be implications.
RE: sizing CT protection by burden
Your method seems quite reasonable. But it takes Se to be equal to Zero. It can be true only when the connected burden is quite high. Otherwise Se becomes quite comparable with your actual connected burden and significantly affects the effective overcurrent factor. Considering that the modern relays have small a burden, I feel it to be a safe bet to go by your 1st method.
In the 2nd method you are choosing a higher rating only so that you can put on an additional burden.
RE: sizing CT protection by burden
This is an example, not a real case.
RE: sizing CT protection by burden
for every protection function calculation is different.
for distance protection as example, you have about 3 factors and after calculation, additional factor : with or without autorecloser.
what ia transint condition, what is X/R of grid.
for HI its something other, for busbar other.
sensitivety of protection is not depend on the CT ratio.
saturation is not function of primary current.
Please read IEC60044-1, IEC60044-6(transient condition).
5P20 200VA, its something crazy today, put 5P30 or 5P40.
But, up to you.
RE: sizing CT protection by burden
There are about 5 years ago we started the project of a large hydroelectric plant. The 500kV substation was 4000MVA-SF6.
TCs were 6000 / 5 A, 10P20, multiple ratio,TPY type.
A manufacturer of CT had suggested adopting CT of 1A. We agree with 4000 / 1 A, 10P20. At that time we did not have clear information of remote interconnection substation.
3 years after it was specified remote substation, 500kV, outdoor installation. Another company did the project and adopted the same CT 1 A. Now we have problem because the measurement CT standard only admits CT 5 A!
Studies, projects and teams of substation equipment are fully decoupled from protection. The activities usually occur before 6 months of protection.
RE: sizing CT protection by burden
Now its clear.
Its GIS or AIS, try divide CT to two pcs, one for revenue meters and mesuarments with 5A, cl0.5FS5 and 0.2S FS5 and "right" ratio
but for protection use 1A secondary, and 5P30 as minimum.
btw, Im use now PX-TPS ,but TPY for distance protection with AR is best solution.
500kV, near to big generation, for calculation, as minimum: 63kA and X/R=40, that means 127ms.
RE: sizing CT protection by burden
I'm not sure I really see the advantage of using 1A secondary for protection. Most users scale the leads from CT to relay down based on 1A or 5A secondary and the result is not much difference.
The actual CT itself will have roughly the same core and copper cost for either 1A or 5A secondary rating.
RE: sizing CT protection by burden
maximum actual shortcircuit at 525 kV is 17 kA , X/R=133.5 , tau = 353 ms.
scottf
the big problem is that we are already familiar with 5A CT, 100VA, 6mm2. When you change for 1A, you do not have the feeling of burden , cable, etc. It is very dangerous this change because it involves people from different areas of the project
RE: sizing CT protection by burden
Now its more clear!!! 353ms tau, its really problem, big problem.
scottf, you are right, for protection is not importnat 1A or 5A, in newewr relay its parameter of setting only or jumper.
from CT point of view its no problem too.
only cable size, if its long distance, for 5A needed, a 10 or 16 mm^2, for 1A less. its all.
RE: sizing CT protection by burden
You can find sizing CT- 1 A (500kV) and 5A(13,8kV) in a same substation.
RE: sizing CT protection by burden
In fact manufacture circuitbreaker of 525 kV was submitted to the "SPECIAL TEST" with 120kA tau= 120 ms, a value that was used as a comparison to the real short circuit(= 17 kA and 353 ms).
The difference between the current peaks of asymmetric test values and peak asymmetric real current during minimum time of opening of the breaker + minimum of protection operation interrupton time (52ms)is:
Iassim peak special test = 46.47 kA
Iassim real Short = 44.73 kA
Difference = 1.74 kA
RE: sizing CT protection by burden
"SPECIAL TEST" with 20kA tau= 120 ms