04B-Tutorial CT Selection

Tutorial Current Transformer Requirements for Protection

1. A 400/5A bar primary CT with a Stalloy core supplies a 3VA CDG11 set at 10% of nominal current. It is required to cater for a current of 10 times relay setting, neglecting the CT resistance and any non linearity in the relay impedance. Consider a conductor diameter of 2 cm’s and allow 2 cm’s on each side of the core for the secondary winding and the insulation. System frequency is 50Hz. Find (i) The minimum knee point voltage. (ii) The cross sectional area of the core. (iii) A possible set of CT dimensions, ensuring that the CT will fit in the available housing space of 20 cm’s diameter and 15 cm’s deep. (iv) The Kv factor in volts/tesla (v) The Ki factor in mm/turn Note* Take Bm – Max flux density of stalloy = 1Wb/m

2

CDG11 Rct 5 Amp Relay Set at 10% Ip/N

Xct

CT Ratio = 400/5A

∴ N = 80 Turns RCT is ignored in this example.

3VA

2. Three CROSS (Cold Rolled Grain Orientated Silicon Steel) 400/5A Bar Primary current transformers have the following characteristics: Secondary resistance = 0.2 Ω Kv = 20V/tesla Ki = 4mm/turn The CT’s are connected to give phase and E/F protection using a CDG31 50-200% overcurrent relay set to 125%, and a CDG11 20-80% earth fault relay set at 40%. What is: The minimum primary current for operation of the phase fault relays for a 3 phase fault  b.) The minimum primary current for operation of the earth fault relay for a single phase to ground fault. a.)

 Assume relay burden at setting is 3VA for the phase fault relays and 2.4VA at setting for earth fault relays. Use the magnetising curves provided to estimate the magnetising currents of both the active and idle CT’s.

3. An earthed 132kV transformer is protected by Restricted Earth Fault protection using 4 CT’s and an MCAG14 relay plus stabilising resistor. Relay setting range 0.1-0.4A, set at 0.4A. The system fault level is 3500MVA, and the loop impedance from relay to the CT’s = 2.0 Ω The CT details are as follows: Ratio = 500/1

Material = CROSS

Kv = 272 V/tesla Ki = 2.08 mm/turn R = 0.7 Ω Find the primary fault setting and the minimum stabilising resistor value. Assume relay burden is 1VA. Use magnetising curve supplied, scale the Kv and Ki factors to calculate the magnetising current. 500/1

Rs 500/1 MCAG14

Solutions: 1.

I.

Setting of relay = 10% of 5A = 0.5 Amps CDG11 consumes constant VA at setting, therefore 3VA at 0.5  A must demand 6 V at the relay terminals (VA = V x I). If the CT is to be capable of driving 10 x setting current through the re;ay, then the CT must have a minimum knee point voltage of 10 x 6 V.

∴ Minimum Vk = 60 V. II. Calculate the cross sectional area of the core using the formula ES = 4.44 f B m A N (all metric units)  All factors but A are known so, 60 = 4.44 x 50 x 1 x A x 80

∴ A =

60 222 × 80

= 0.003378 m 2 2

= 33.78 cm  (area of Stalloy Core) III. Calculation of a possible set of CT dimensions With 2 cm’s diameter primary conductor and 4 cm’s for winding and insulation, minimum internal diameter = 6 cm’s.  Assume core to be approximately square e.g. 33.8 cm2 = 5 cm x 6.76 cm

Proposed size for core. 6.76 cm

16 cm

6 cm

Mean magnetic diameter =

6 + 16 2

= 11cm

Overall dimensions, including windings and insulation (2 cm)

10.76 cm

20 cm

2 cm

This CT will fit within the available space of 20cm x 15cm. IV. Calculation of Kv Factor Employing metric units with core area in cm K V  =

 AN 

1

× 10 −4 =

4.44 × 50

∴ Kv = 60 V/ tesla

AN 

4.5

=

33.78 × 80 45

= 60

2

V. Calculation of Ki factor Employing metric units with mean magneetic length in mm Mean magnetic length = πD = π x 11 x 10 = 346 mm Ki = Mean magnetic length/ N =

346 80

= 4.32mm / Turn

2. 400/5

O/C

Earth Fault

Relay Impedances: Phase Overcurrent = 3 VA at 125% of 5 A

∴ Z = 3/6.252 = 0.077 Ω E/F = 2.4 VA at 40% of 5 A

∴ Z = 2.4/22 = 0.6 Ω Full equivalent circuit, including magnetising impedances O/C 0.077 Ohms

Rct 0.2 Ohms

Z mag 0.077 Ohms

0.2 Ohms

0.077 Ohms

0.2 Ohms

E/F 0.6 Ohms

a.)

Calculate 3 phase fault primary setting The phase currents are balanced for this condition and no current will flow in the neutral. Setting current of overcurrent relays = 6.25 A

∴ Voltage which needs to be developed within the CT = 6.25 x total impedance.  Assume all impedances have the same angle for simplicity

∴ V at CT = 6.25(0.077 + 0.2) = 1.73 V 0.077

1.73 V

0.2 6.25 A

In = 0

To calculate expected magnetising current in CT: I.

Scale magnetising curve with the Kv and Ki factors given in the question ( as traced) 8V

4V

1.73 V

0.08 A

II.

Scale of 1.73 V – Result = 0.02 A Neglecting any phase angle differences Isec’y = 6.25 A + Imag = 6.27 Amps

∴ Primary sett ing for O/C relays = 6.27 x 80 = 501.6 A for 3 phase faults  b.)

Calculate earth fault primary setting Step A

Calculate voltage across E/F relay = Setting Current x E/F relay impedance = 2.0 x 0.6 Ω = 1.2 V

Step B

Calculate the magnetising current in the 2 idle sound phase CT’s, noting they have 1.2 V impressed across them Use magnetising curve – scaled by Kv and Ki –  as before 8V

4V

1.2 V

0.16 A

0.08 A

Step C

Draw the known parameters so far 0.077 Ohm

0.2 Ohm Faulted Phase

2.032 A

Sound Phase Imag = 0.016 A

Sound Phase Imag = 0.016 A

2A

Step D

1.2 V

0.6 Ohm

Calculate the voltage at the driving CT V at CT = 1.2 V + Total current x 0.277 Ohm = 1.2 V + 2.032 x 0.277 = 1.763 V

Step E

Calculate the magnetising current taken by the CT Use mag curve – Scaled as before

8V

4V

1.76 V

0.02 A

0.08 A

Step F

Construct the final diagram, assuming all currents are in phase. 2.052 A 0.077 Ohm

0.2 Ohm Faulted Phase

2.032 A

0.02 A Sound Phase Imag = 0.016 A

Sound Phase Imag = 0.016 A

2A

1.2 V

0.6 Ohm

Current in driving CT = 2.052 A sec’y = 164.15 A primary

∴ Primary E/F sensitivity = 164.2A

3. Step A

Calculate E/F rating MVA rating/phase = 3500/3 Phase/neutral voltage = 132/ √3

∴ Fault level =

Step B

3500

3 = 1166.67 = 15.3kA 132 76.2 3

Calculate CT secondary current (prospective)  I S  =

 I P n

=

15300 A 500

30.6 A

Step C

Draw detailed diagram, as seen from the CT on the neutral Rct

Rloop

0.7

2.0 Rstab Fault Is = 30.6 A If other CT's Saturate

Step D

Calculate recommended voltage setting of high impedance relay, ie, MCAG14 + R STAB V setting = If (sec’y) x resistance in one leg of supply. = 30.6 A x (0.7 + 2.0) = 82.6 V

Step E

Calculate the mag current taken by the CT – using curve supplied and scaled by the Kv and Ki factors given

108.8 V

82.6 V

54.4 V

0.025 A

Step F

0.0416 A

Calculate total current in driving CT (assuming currents are in phase) at setting.

Note:- Worst case scenario is one CT driving its own magnetising current plus 3 idle CT’s in parallel.

∴ Total magnetising = 4 x 0.025 A = 0.1 A

0.025

0.025

0.025

0.025

Setting Current = 0.4 A

CAG14 set 0.4 A

Total Is for setting = 4 x 0.025 A + 0.4 A = 0.5 A So Primary setting = 500 x 0.5 A = 250 A Step G

Calculate total Z of relay leg =

Re laySettingVoltage Re laySettingCurrent 

=

82.6 0.4

= 206.5Ω 2

Subtract CAG14 Ohms = 1 VA / 0.4 =6.25 Ω

∴ RSTAB = 206.5 – 6.25 = 200.25 Ω