CT Calculator

Calculation of CT Parameter Description

Parameter

Remark

From generator Data sheet

From generator Data sheet

(i) Generator parameter Rated Voltage(kV) Rated Current(Amp) Sync. Reactance Xd is 1/SCR Subtransient Reactance Xd"(Saturated) Transient Reactance Xd' (Saturated) Rated Voltage at -5% FL current at -5%of rated Voltage=

19 9116.3 200.0 12.0% 20.0% 18.05 9596.1

Voltage at + 5% FL current at +5% of rated Voltage=

19.95 8682.2

Choose CT ratio = 11000/1 105% of rated current at (-)5% of rated voltage

10075.94

Choose CT ratio = 11000/1A

11000

This gives an over load capability

1.21

% O/L Capability of Full Load Current

120.66

1) Calculation of CT parameter for generator differential protection (87G)

Relay setting VR = If (RCT + 2RL + Rrelay) where , Where If = fault current on CT sec, RCT = CT secondary resistance, Rrelay = Relay resistance in ohms, RL = Resistance of cable in ohms, RL in  for 100m of 2.5mm2

0.887

(Considering 100 m of 2.5 sq.mm Cu cond. cable between busduct CT & GRP the resistance of 2.5 sq.mm Cu cond cable at 70 oC in  / km (from KEI catalogue)

8.87

Relay resitance/phase Assumed Rrelay from SEC

0.1

Fault contribution from Gen in MVA= Taking 30% - Ve tol on Xd” as per IEC-34 part –I

3571.43

Fault contribution from system in MVA considering 40KA system Impedence on 220KV Side. Source impedence in Percentage at 300 Base

2024.10

Taking higher of the above two values =

3571.43

1.97%

Fault Current in KA Taking higher value of the fault contribution from i.e. fault contribution from system / fault contribution from gen

108.52

Secondary Fault Current

9.87

VR = 9.87 (RCT +2x0.887+0.1)

E47*(RcT+2*E31+E35)

Vk = 2 VR

2*9.87* (RCT +2x0.887+0.1)

Expressing VK term of +40V

19.74Rct+36.99

Vk = 20RCT + 37 Im  15 mA at Vk/2

Page 1 of 20

CT with group-1/group-2 protection 11000/1A, CLPS, Vk  20RCT + 36V Im  15 mA at Vk/2 (ii) METERING CORE CT chosen 11000/1A, 30VA, CL 0.2 ISF  5 Typical Meter burden in ohm (from catalogue of Conzerv model) Typical Anolgue Meter burden in ohm (from catalogue of AE/Rishab) & Transducer

0.4 7 Total 10Nos

Analogue meter connected & max 4 nos of transducer per CT

2

Max. Burden margin Consider for future use Total Burden on CT = Lead Burden+ Meter Burden Lead Burden for lead length of 100M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 15 ohm for 11000/1Amp Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use 30 VA, CL-0.2

=Sqr(1)*(Rct+2Rl) =Rct+1.774 16.774 26.174 As meter used is Multifunction type, hence 30 VA considered is more than Sufficient also Lead burden is less than above assumed buden as Meter & CT are at the same Switchgear.

Calculation of CT parameter for (87- 0A) on 19 kV side Same as generator differential protection CT core 11000/1A, CL-PS, Vk ≥ 20Rct + 36V, Im ≤ 15mA at Vk/2 2 CT on 220 KV side of Generator Step up transformer Bay i) Core –1 (87- 0A) 512.40

GT HT side current (current on 400 kV side) = = 513 A RL = 8.87x0.27 (considering 270m)= 2.3949 Rrelay = 0.1 CT ratio chosen = 600/1A VK = 40xIx(RCT+2RL+Rrelay)

Fault current on 400KV side taken as 40KA

= 40x1 (RCT +2x2.3949+0.1) =40 RCT+195.592 = VK  40 RCT+200V CT ratio 600/1A CLPS , VK  40 RCT+200V Im  30 mA at VK/2 ii) Core –2 (50 LBB, 51GT) R relay= 0.1 (from relay catalogue of ABB) 600/1A Total Burden on CT = Lead Burden+ Relay Burden Lead Burden for lead length of 270M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 6 ohm Total Burden on CT = Lead Burden+ Relay Burden

R relay= 0.1 (from relay catalogue of ABB)

=Sqr(1)*(Rct+2Rl) =Rct+4.79 10.79 10.89

Page 2 of 20

15 VA, CL5P20

Cater the Load burden of Numerical Relay

iii) Core –3 (metering) 600/1A Typical Meter burden in ohm (from catalogue of Conzerv model) Typical Anolgue Meter burden in ohm (from catalogue of AE/Rishab) & Transducer

0.4 4.5 Total 5Nos Analogue meter connected & max 4 nos of transducer per CT

Max. Burden margin Consider for future use Total Burden on CT = Lead Burden+ Meter Burden Lead Burden for lead length of 270M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 6 ohm for 600/1Amp Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use 20 VA, CL-0.2

iv)

2 =Sqr(1)*(Rct+2Rl) =Rct+4.79 10.79

17.69 20 VA arrived at considering all the meter connected

Core –4 (87 B1/B2) Fault level of switchyard = 40 kA for 1 sec. CT ratio chosen 600/1A CT sec lead resistance with 2.5 mm2 Cu cond cable 270 m length (max) RL = 8.87x0.27 = 2.3949 VK = If (RCT + 2 Rlead + Rrelay)

RCT value from CT manufacturer Bus bar protection is a numerical relay

VK = 40x1 (RCT +4.79+0.2) = 40 RCT + 199.6 Knee point voltage selected is 40 RCT + 200V CT ratio chosen 600/1A CLPS , VK  40 RCT + 200V, Im  15 mA at VK/2 v)

Core –5 Spare core . Parameters same as core-4, VK  40 RCT + 150V Im  15 mA at VK/2

A

GT Bushing CTs

i)

Core-1 Phase CT for REF protection (64 GT) % Impedence of GT is GT full load current on HV side = 512.40

14.50% Considering SPAE 10 relay with 200V mAburden =5.4 VA Setting227 relay

Considering the Source impedance based on 40kA, max. earth fault current I = F.L current / % Z x 0.9 ( -ve tol of 10%) 3457.15 = 3607.70 as GT impedance Zt = 14.5% Choosing bushing CT of 600/1 on GT HV side & neutral side

Page 3 of 20

CT sec. equiv. thro'fault current = 3607.70 x1/600 5.76

∴

VR = If (RCT + 2RL + R relay) = 6.01 ( RCT + 2 x 1.242 + 5.4/1)

VK ` = 6.01 RCT + 47.38

2.5 mm2 Cu cond, 140 m length

VK read = atleast 2 VR VK = 2x(6.01 RCT + 47.38)V VK = 12.5 RCT + 100V CT parameters chosen CT ratio 600/1A VK  13 RCT + 100 V Im < 30 mA at VK/2

ii)

Same CT parameter is applicable for neutral REF CT

Core - 2 Neutral CT-GT inverse time ground O/C relay (51NGT) CT chosen 600/1A 20 VA, CL 5P20

P20 because Chracteristic will get smoothen after 20 times of rated current

0.2

Typical Relay burden in ohm (from ABB catlogue) Total Burden on CT = Lead Burden+ Meter Burden Lead Burden for lead length of 140M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 6 ohm for 600/1Amp Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use 10 VA, CL-5P20

B

=Sqr(1)*(Rct+2Rl) =Rct+2.4836

8.4836 8.6836 As relay used is Numerical type, hence 10 VA considered is more than Sufficient.

UAT BUS DUCT MOUNTED CTS Formula used / reference / design assumption (87UAT) (HV side)

i)

Core –1 (50 UAT – HV side) For a fault at UAT HV terminals Fault contribution from generator Base-MVA Generator Xd” sat From Generator data sheet

355 0.120

Zg with 30% neg tolerance Zg (pu) = 0.155X0.7= 10.85% Zg (pu) with 355MVA base=

0.084 0.0845

Fault contribution from generator is Base MVA/Fault MVA

4201.68

Fault contribution from 400KV system in MVA

27712.8

Zspu with 10% negative tolerance = Zgt with 10% negative tolerance

1.15%

1.152

13.05%

13.05

Page 4 of 20

System contribution = Base MVA / (Zspu+Zgt)

2499.49

= 2499.49 MVA Total fault MVA(Contribution from Generator & GT system) = 3252.91+2499.49 = 5752.40 MVA Fault current

6701.17

= 166.06 KA

203.63

CT Chosen = 8500/1A Cl 5P20, 10VA

10181.38

Ratio of fault current to CT ratio < 20 times However the ratio of rated burden of CT & the connected burden is about 10 times. Therefore the voltage, developed by the CT (200V) will be increased by that ratio, therefore the CT does not saturate and hence CT chosen is ok. (ii)

23.96 Vdesin=BurdenxALF/CT Secondary

Core – 2 (51 UAT – HV side) CT chosen – 800/1A 10VA, CL 5P20 Typical Relay burden in ohm (from ABB catlogue) Total Burden on CT = Lead Burden+ Relay Burden Lead Burden for lead length of 125M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 7 ohm for 600/1Amp Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use 10 VA, CL-5P20

(iii)

0.1 =Sqr(1)*(Rct+2Rl) =Rct+2.2175 9.2175 9.3175 As relay used is Numerical type, hence 10 VA considered is more than Sufficient.

Core – 3 (87 UAT-HT side) 800/1A , CL PS , Im  15 mA at VK/2 UAT rating – 20 MVA Current on 20 kV side = 577.35 607.74 800/1A

At the lowest tap i.e – 5%, CT ratio will be

VK = 40xIx(RCT + 2RL + R relay)

CT ratio choose CT = 800/1A

= 40x1x(RCT +2x8.87x0.125 +0.1) VK >40RCT+92.7 VK  40RCT+92.7V

2x8.87x100/1000

Im  15mA AT VK/2 C

UAT LV side CTs

i)

Core –1 (87- 0A) (UAT-LV) Total running load in KVA on the 6.6 kV bus ( higher of the two buses U1A and U1B taken from UAT sizing calculation)

18451

Total running load in Amp on the 6.6 kV bus ( higher of the two buses U1A and U1B taken from UAT sizing calculation)

1614.04

∴

CT chosen – 2000/1A

This also cater the installed Capacity of UAT i.e. 20MVA

Page 5 of 20

VK = 40 x I (RCT + 2RL + R relay) 2 RL = 2 x 8.87 x 0.1 (1 run 2.5 mm2 cable / phase, L = 100 meters) 2.661

=

1.774

VK = 40x1(RCT + 1.774 + 0.1) = 40RCT+74.96

74.96

VK selected is  40RCT+80V CT chosen 2000/1A, CLPS, Vk  40RCT+80V Im  15 mA at Vk/2 ii)

Core - 2 64-UAT - LV side & 87-UAT 20/16 MVA , 20/6.9 KV Rated Current of Unit Transformer at nominal Tap

1749.55

Fault Current limited by transformer impedance of 10% at -5%

18416.28 = 18.416 KA

Fault Current reflected on CT Secondary

9.21

VK = 9.21x(Rct+2x0.887+0.1) = 9.21 RCT + 17.26 =10Rct+20V Im  15mA at Vk/2 Since the phase side CTs for REF protection will be same as that used for 87 UAT, same parameters as that of core-2 will be provided for neutral CT to be located in NGR or Neutral Bushings of transformer. iii)

Core –3 (51 UAT - LV) 2000/1A, 10VA, CL 5P20 R relay in ohm (from relay catalogue of ABB model SPAJ) Total Burden on CT = Lead Burden+ Relay Burden Lead Burden for lead length of 50M= I2x(Rct+2RL)

0.1 =Sqr(1)*(Rct+2Rl) =Rct+0.887

Rct Assumed for CT Sec.1A is 6 ohm Total Burden on CT = Lead Burden+ Relay Burden 10 VA, CL5P20 As relay used is numerical(SPAJ-140C) type, hence 10 VA considered is more than Sufficient also Lead burden is less than above assumed buden as relay & Ct are at the same Switchgear. iv)

Core –4 (Metering) 2000 / 1A , 10VA , CL 0.5 , ISF  5 Typical Meter burden in ohm (from catalogue of Conzerv model) Burden margin Consider for future use Total Burden on CT = Lead Burden+ Meter Burden Lead Burden for lead length of 50M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 12 ohm Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use 20 VA, CL5P20

v)

6.887 6.987

0.2 2 =Sqr(1)*(Rct+2Rl) =Rct+0.887 12.887 15.087 As meter used is Multifunction type, hence 10 VA considered is more than Sufficient also Lead burden is less than above assumed buden as Meter & CT are at the same Switchgear.

51 N – UAT- LV. Earth Fault current to be restricted to 300Amp. For neutral CT to be located in NGR.

Page 6 of 20

300/1A, 15VA, CL 5P20 D

ST LV side CTs

i)

Core –1 ( metering) Total running load in KVA on the 6.6 kV bus ( higher of the two buses C1A and C1B taken from ST sizing calculation)

22292

Total running load in Amp on the 6.6 kV bus ( higher of the two buses C1A and C1B taken from ST sizing calculation) ∴

CT chosen – 2500/1A

1950.04

Taking into consideration 25MVA(Rated Installed Capacity) Current is 2187Amp

Typical Meter burden in ohm (from catalogue of Conzerv model) Burden margin Consider for future use Total Burden on CT = Lead Burden+ Meter Burden Lead Burden for lead length of 50M= I2x(Rct+2RL)

0.2 2 =Sqr(1)*(Rct+2Rl) =Rct+0.887

Rct Assumed for CT Sec.1A is 15 ohm Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use 2500 / 1A , 20VA , CL 0.5 , ISF  5

ii)

15.887 18.087 As meter used is Multifunction type, hence 20 VA considered is more than Sufficient also Lead burden is less than above assumed buden as Meter & CT are at the same Switchgear.

Core - 2 (ST - 64 RST - LV side) 25 MVA , 220/6.9 KV Rated Current of Station Transformer at nominal Tap

2186.93

Fault Current limited by transformer impedance of 14.5% at -15%

25728.62 = 25.73 KA

Fault Current reflected on CT Secondary

10.29

2 RL = 2 x 8.87 x 0.20 (1 run 2.5 mm2 cable / phase, L = 250 meters) = 4.435

4.435

Vk = 10.29x(Rct+4.435+0.1)

46.67

= 10.29 RCT + 46.67 = 11RCT+46.67 V i.e. CTR chosen - 2500/1A, 11RCT+50V, Im  15mA at Vk/2 iii)

Core –3 (51 ST - LV) 2500/1A, 20VA, CL 5P20 R relay in ohm (from relay catalogue of ABB model SPAJ) 2500/1A Total Burden on CT = Lead Burden+ Relay Burden Lead Burden for lead length of 50M= I2x(Rct+2RL) Rct Assumed for CT Sec.1A is 15 ohm Total Burden on CT = Lead Burden+ Relay Burden 20 VA, CL5P20 As relay used is numerical(SPAJ-140C) type, hence 20 VA considered is more than Sufficient also Lead burden is less than above assumed buden as relay & Ct are at the same Switchgear.

From Above Full Load Current 0.1

=Sqr(1)*(Rct+2Rl) =Rct+0.887 15.887 15.987

Page 7 of 20

iv)

Core –4 (87- ST- LV) ∴

CT chosen – 2500/1A

From Above Full Load Current

VK = 40 x I (RCT + 2RL + R relay) 2 RL = 2 x 8.87 x 0.15 (1 run 2.5 mm2 cable / phase, L = 250 meters) 2.661

=

4.435

VK = 40x1(RCT + 4.435 + 0.1) = 40RCT+181.4 VK selected is  40RCT+200V CT chosen 2500/1A, CLPS, Vk  40RCT+115V Im  15 mA at Vk/2

Page 8 of 20

CT sizing for 6.6 kV System SNo.

A

Type of switchboard

Feeder Rated Load current rating (kW)

CT ratio

Remark

6.6 kV Unit switchgear (U1A & U1B) comprising of the following panels with ratings and accessories as indicated in single line diagram, specification & data sheet. Motor Feeder

1 i.

Boiler Feed Pump-1A, 1B For Differential Protection Motor rated Current = 5400/Sqrt(3)/6.6/0.9 Differential CT for Motor Protection Vk = 40xIx(Rct+2RL+ Rrelay)

5400

600/1

524.86

= 40*1*(Rct+2x0.7983+0.1) = 40Rct+67.864V

RL = 8.87X0.09 (Considering 90m Length from Motor terminal to Relay)=0.7983 R relay = 0.1

Vk>= 40Rct+70V, Im= 40Rct+70V, Im