Protection Class CT design

For high impedance relays (differential or restricted earth fault relays), 'Class X' current transformers are recommended to be used. As I understand from your query, you need to size the CT to be installed on the neutral point of the secondary side of 11/0.415 kV transformer, as well as for the 415 V phase CTs for restricted earth fault application. Please note that both CTs (neutral & phase) shall have the same characteristics. The following is an example to size the CT:

Input data:

- 11/0.415 kV Power transformer capacity: 2500 kVA (given);

- Transformer % impedance (Z): 6% (assumed);

- Length of cable from neutral CT to the relay located at LVSP: 200 m (given);

- Cross section of CT cable to be used: 6 mm² -copper (assumed);

- CT cable resistance: 0.0032 Ω/m (from manufacturer data).

Step # 1: Calculation of CT Rated Primary Current

I = kVA/ (0.415x1.732) = 2500/(0.415x1.732) = 3478.11 A, CT with primary current of 4000 A to be selected.

You may select the secondary current of the CT 1 or 5 A. I suggest selecting 1 A secondary current, as the cross section and length of pilot wires can have a significant effect on the required knee voltage of the CT and therefore the size and cost of the CT. When the relay is located some distance from the CT, the burden is increased by the resistance of the pilot wires.

Step # 2: Calculation of max. Fault Current

Ift = kVA/ (0.415x1.732x Z)

Ift = 2500/ (0.415x1.732x0.06) = 57968.59 A (rounded to 58000 A)

Step # 3: Calculation of the Knee Voltage of the CT (Vkp)

Vkp = (2x Iftx (Rct+Rw)/CT transformation ratio)

Where:

Rct: is the CT resistance (to be given by the manufacturer), for the purpose of illustration, we will use a typical Rct value in this example (1.02 Ω) given by one of the CT manufacturers.

Rw: total CT cable resistance= 2x cable length (200 m) x wire resistance= 2x200x0.0032= 1.28 Ω

CT transformation ratio = CT Primary Current/CT Secondary Current

CT transformation ratio = 4000/5= 800 A, for CT with 5 A secondary current; or,

CT transformation ratio = 4000/1= 4000 A, for CT with 1 A secondary current. We will use 1 A in this example.

Vkp = (2x58000x (1.02+1.28)/4000)= 66.7 V

The Vkp of the CT should be higher than the setting of relay stability voltage (Vs), to ensure stability of the protection during max. through fault current. To calculate the stability voltage, you should follow the related formula given by the relay manufacturer, as each relay manufacturer has its own formula.

You may calculate the Vkp as above using a CT with secondary current of 5 A, and you will notice the difference in the Vkp.

I hope the above could help you.

Regards,