CT Calculator

(1)

Calculation of CT Parameter Parameter Remark

Description From generator Data sheet

(i) Generator parameter

Rated Voltage(kV) 19

Rated Current(Amp) 9116.3

Sync. Reactance Xd is 1/SCR 200.0

Subtransient Reactance Xd"(Saturated) 12.0%

Transient Reactance Xd' (Saturated) 20.0%

Rated Voltage at -5% 18.05

FL current at -5%of rated Voltage= 9596.1

Voltage at + 5% 19.95

FL current at +5% of rated Voltage= 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)

0.887 8.87

Relay resitance/phase Assumed Rrelay from SEC 0.1

Fault contribution from Gen in MVA= 3571.43

Taking 30% - Ve tol on Xd” as per IEC-34 part –I

2024.10

Source impedence in Percentage at 300 Base 1.97%

Taking higher of the above two values = 3571.43

108.52

Secondary Fault Current 9.87

E47*(RcT+2*E31+E35) 2*9.87* (RCT +2x0.887+0.1)

Expressing VK term of +40V 19.74Rct+36.99

From generator Data sheet

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

(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 70o_{C in}__{/ km (from} KEI catalogue)

Fault contribution from system in MVA considering 40KA system Impedence on 220KV Side.

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

V_R = 9.87 (R_CT +2x0.887+0.1) Vk = 2 VR

V_k = 20R_CT + 37 Im  15 mA at Vk/2

(2)

CT with group-1/group-2 protection

(ii) METERING CORE

Typical Meter burden in ohm (from catalogue of Conzerv model)

0.4

7

Max. Burden margin Consider for future use

2

Total Burden on CT = Lead Burden+ Meter Burden

Rct Assumed for CT Sec.1A is 15 ohm for 11000/1Amp 16.774

26.174 30 VA, CL-0.2

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)

GT HT side current (current on 400 kV side) = 512.40

= 513 A

Rrelay = 0.1 CT ratio chosen = 600/1A

CT ratio 600/1A

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

Rct Assumed for CT Sec.1A is 6 ohm 10.79

Total Burden on CT = Lead Burden+ Relay Burden 10.89

11000/1A, CLPS, Vk  20RCT + 36V Im  15 mA at Vk/2

CT chosen 11000/1A, 30VA, CL 0.2 ISF  5

Typical Anolgue Meter burden in ohm (from catalogue of AE/Rishab) &

Transducer Total 10Nos Analogue meter

connected & max 4 nos of transducer per CT

Lead Burden for lead length of 100M= I2

_x(R

ct

+2R

L

)

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

=Rct+1.774 Total Burden on CT = Lead Burden+ Meter Burden+burden margin

Consider for future use

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.

R_L = 8.87x0.27 (considering 270m)= 2.3949

V_K = 40xIx(R_CT+2R_L+Rrelay) Fault current on 400KV side taken

as 40KA = 40x1 (R_CT +2x2.3949+0.1) =40 R_CT+195.592

= V_K 40 R_CT+200V

CLPS , VK  40 R_CT+200V Im  30 mA at VK/2

R relay= 0.1 (from relay catalogue of ABB)

Lead Burden for lead length of 270M= I2_x(R

ct+2RL) =Sqr(1)*(Rct+2Rl)

(3)

15 VA, CL5P20

iii) Core –3 (metering) 600/1A

Typical Meter burden in ohm (from catalogue of Conzerv model) 0.4

4.5

Max. Burden margin Consider for future use 2

17.69

20 VA, CL-0.2

iv) Core –4 (87 B1/B2)

Fault level of switchyard = 40 kA for 1 sec. CT ratio chosen 600/1A

v) Core –5

Spare core . Parameters same as core-4,

A GT Bushing CTs

i) Core-1 Phase CT for REF protection (64 GT)

% Impedence of GT is 14.50%

GT full load current on HV side =

512.40 Setting relay burden =5.4 VA

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

Cater the Load burden of Numerical Relay

Typical Anolgue Meter burden in ohm (from catalogue of AE/Rishab) &

Transducer Total 5Nos Analogue meter

connected & max 4 nos of transducer per CT

_x(R

ct

+2R

L

)

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

20 VA arrived at considering all the meter connected

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 R_CT + 199.6

Knee point voltage selected is 40 RCT + 200V CT ratio chosen 600/1A

CLPS , VK  40 R_CT + 200V, Im  15 mA at VK/2

VK  40 R_CT + 150V Im  15 mA at VK/2

Considering SPAE 10 relay with 200V 227 mA

(4)

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

CT parameters chosen CT ratio 600/1A

ii)

CT chosen 600/1A 20 VA, CL 5P20

Typical Relay burden in ohm (from ABB catlogue)

0.2

Rct Assumed for CT Sec.1A is 6 ohm for 600/1Amp

8.4836

8.6836

10 VA, CL-5P20

B 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 355

Generator Xd” sat From Generator data sheet 0.120

Zg with 30% neg tolerance

Zg (pu) = 0.155X0.7= 10.85% 0.084

Zg (pu) with 355MVA base= 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 = 1.15% 1.152

Zgt with 10% negative tolerance 13.05% 13.05

V_R = I_f (R_CT + 2R_L + R relay) = 6.01 ( R_CT + 2 x 1.242 + 5.4/1)

V_K ` = 6.01 R_CT + 47.38 2.5 mm2_{Cu cond, 140 m length}

V_Kread = atleast 2 VR V_K = 2x(6.01 R_CT + 47.38)V V_K = 12.5 R_CT + 100V

VK  13 RCT + 100 V Im < 30 mA at VK/2 Same CT parameter is applicable for neutral REF CT

Core - 2 Neutral CT-GT inverse time ground O/C relay (51NGT)

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

_x(R

ct

+2R

L

)

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

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

(5)

System contribution

= Base MVA / (Zspu+Zgt) 2499.49

= 2499.49 MVA

6701.17 = 5752.40 MVA

Fault current = 166.06 KA 203.63

CT Chosen = 8500/1A Cl 5P20, 10VA 10181.38

Ratio of fault current to CT ratio < 20 times 23.96

Vdesin=BurdenxALF/CT Secondary

(ii) Core – 2 (51 UAT – HV side) CT chosen – 800/1A 10VA, CL 5P20

Typical Relay burden in ohm (from ABB catlogue) 0.1

9.3175 10 VA, CL-5P20

(iii) Core – 3 (87 UAT-HT side) UAT rating – 20 MVA Current on 20 kV side = 577.35

607.74 At the lowest tap

800/1A i.e – 5%, CT ratio will be

CT ratio choose CT = 800/1A

2x8.87x100/1000

C UAT LV side CTs

i) Core –1 (87- 0A) (UAT-LV)

18451

1614.04

CT chosen – 2000/1A

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

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.

_x(R

ct

+2R

L

)

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

As relay used is Numerical type, hence 10 VA considered is more than Sufficient. 800/1A , CL PS , Im  15 mA at VK/2 V_K = 40xIx(R_CT + 2R_L + R relay) = 40x1x(RCT +2x8.87x0.125 +0.1) VK >40RCT+92.7 V_K 40R_CT+92.7V I_m 15mA AT VK/2

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

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

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

(6)

1.774

74.96

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)

iii) Core –3 (51 UAT - LV) 2000/1A, 10VA, CL 5P20

0.1 Total Burden on CT = Lead Burden+ Relay Burden

10 VA, CL5P20

iv) Core –4 (Metering)

Burden margin Consider for future use 2

15.087 20 VA, CL5P20

v) 51 N – UAT- LV.

V_K = 40 x I (R_CT + 2R_L + R relay)

2 RL = 2 x 8.87 x 0.1 (1 run 2.5 mm2 cable / phase, L = 100 meters) = 2.661 V_K = 40x1(R_CT + 1.774 + 0.1) = 40R_CT+74.96 V_K selected is  40R_CT+80V CT chosen 2000/1A, CLPS, Vk  40RCT+80V Im  15 mA at Vk/2 = 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.

R relay in ohm (from relay catalogue of ABB model SPAJ) Lead Burden for lead length of 50M= I2_x(R

=Rct+0.887

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.

2000 / 1A , 10VA , CL 0.5 , ISF  5

Earth Fault current to be restricted to 300Amp. For neutral CT to be located in NGR.

(7)

300/1A, 15VA, CL 5P20 D ST LV side CTs i) Core –1 ( metering) 22292 1950.04 CT chosen – 2500/1A

18.087

ii) 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

4.435

Vk = 10.29x(Rct+4.435+0.1) 46.67

= 11RCT+46.67 V

iii) Core –3 (51 ST - LV)

2500/1A, 20VA, CL 5P20 From Above Full Load Current

R relay in ohm (from relay catalogue of ABB model SPAJ) 0.1

2500/1A

20 VA, CL5P20

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

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

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

2500 / 1A , 20VA , CL 0.5 , ISF  5 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.

2 R_L = 2 x 8.87 x 0.20 (1 run 2.5 mm2_{cable / phase, L = 250 meters) =} 4.435

= 10.29 R_CT + 46.67

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

=Rct+0.887

(8)

iv) Core –4 (87- ST- LV)

CT chosen – 2500/1A From Above Full Load Current

4.435 V_K = 40 x I (R_CT + 2R_L + R relay)

2 RL = 2 x 8.87 x 0.15 (1 run 2.5 mm2 cable / phase, L = 250 meters) = 2.661 V_K = 40x1(R_CT + 4.435 + 0.1) = 40R_CT+181.4 V_K selected is  40R_CT+200V CT chosen 2500/1A, CLPS, Vk  40RCT+115V Im  15 mA at Vk/2 ∴

(9)

CT sizing for 6.6 kV System

SNo. Type of switchboard CT ratio Remark

A

Motor Feeder

1 Boiler Feed Pump-1A, 1B 5400 600/1

i. For Differential Protection

Motor rated Current = 5400/Sqrt(3)/6.6/0.9 524.86

Vk>= 40Rct+70V, Im<=30mA at Vk/2

ii. For O/C & E/F Protection

10 VA, CL5P20

iii. For Metering Core

0.2 1

10.9096 15 VA, CL-0.5

2 Condensate Extraction Pump-1A, 1B 1100 150/1

Rct Assumed for CT Sec.1A is <=6 ohm

6.7096

Feeder Load rating (kW)

Rated current

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.

Differential CT for Motor Protection Vk = 40xIx(Rct+2RL+ Rrelay)

= 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

R relay in ohm (from relay catalogue of ABB model SPAJ)

_x(R

ct

+2R

L

)

=Sqr(1)*(Rct+2Rl)_=Rct+0.7096

Typical Anolgue Meter burden in ohm (from catalogue of AE/Rishab) & Transducer

_x(R

ct

+2R

L

)

=Sqr(1)*(Rct+2Rl)_=Rct+0.7096

Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use

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

_x(R

ct

+2R

L

)

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

(10)

10 VA, CL5P20

0.2

1

Burden margin Consider for future use

2 6.7096

9.9096

10 VA, CL-0.5

3 ID-1A, 1B 2240 300/1

6.7096

10 VA, CL5P20

0.2

1

2 6.7096

9.9096

10 VA, CL-0.5

4 FD-1A, 1B 800 87.48 100/1

i. For O/C & E/F Protection

_x(R

ct

+2R

L

)

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

=Rct+0.7096

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. Also Rct is max typical value for 750Amp Primary

= 40*1*(Rct+2x1.996+0.1)

= 40Rct+163.66 V RL = 8.87X0.225 (Considering 225m Length from Motor terminal to Relay)=1.996, R relay = 0.1

_x(R

ct

+2R

L

)

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

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

(11)

Rct Assumed for CT Sec.1A is <=4 ohm 6.7096

10 VA, CL5P20

ii. For Metering Core

0.2

1

2 4.7096

7.9096

10 VA, CL-0.5

5 PA-1A, 1B 1600 200/1

5.7096

5.8096

10 VA, CL5P20

0.2

1

2 5.7096

8.9096

10 VA, CL-0.5

6 Coal Mill-1A, 1B, 1C, 1D, 1E, 1F 400 43.74 50/1

0.1 Lead Burden for lead length of 40M= I2_x(R

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

= 40*1*(Rct+2x0.887+0.1) = 40Rct+74.96V

RL = 8.87X0.1 (Considering 100m Length from Motor terminal to Relay)=0.887 R relay = 0.1

_x(R

ct

+2R

L

)

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

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

3C x 185 mm2

(12)

10 VA, CL5P20

0.2

1

2 1.7096

4.9096

10 VA, CL-0.5

7 CWP-1A, 1B 1400 200/1

5.7096

5.8096

10 VA, CL5P20

0.2

1

2 5.7096

8.9096

10 VA, CL-0.5

8 Main Incomer from Unit Transformer 2000/1

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

3C x 185 mm2

= 40*1*(Rct+2x3.548+0.1)

_x(R

ct

+2R

L

)

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

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

(13)

Lead Burden for lead length of 40M= I2x(Rct+2RL)

15 VA, CL5P20

0.2 1

12.7096 15.9096 20 VA, CL-0.5

9 Unit Service Transformer-1A & 1B 2500 219.79 300/1 i. For O/C & E/F Protection

10 VA, CL5P20

0.2

1

2 6.7096

9.9096

10 VA, CL-0.5

10 Spare motor feeder on all the Unit-1 buses 5400 590.47 600/1 CT Details same as Feedpump Motor feeder

11 Spare transformer feeder on all Unit-1 buses 2500 219.79 300/1

12 ESP Service Transformer Feeder-1A & 1B feeder 2500 219.79 300/1

13 Tie to C1A & C1B 2000/1

B

Motor Feeder

=Sqr(1)*(Rct+2Rl) =Rct+0.7096

=Rct+0.7096 Rct Assumed for CT Sec.1A is 12 ohm for

2000/1Amp

3C x 240 mm2

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

CT Details same as Unit service Trans. feeder

CT Details same as I/C feeder from Unit Transformer

6.6 kV Unit switchgear (U2A & U2B) comprising of the following panels with ratings and accessories as indicated in single line diagram, specification & data sheet.

(14)

1 Boiler Feed Pump-2A, 2B 5400 600/1

Rct Assumed for CT Sec.1A is 7 ohm

7.7096

7.8096

10 VA, CL5P20

0.2

1

2 7.7096

10.9096

15 VA, CL-0.5

2 Condensate Extraction Pump-2A, 2B 1100 150/1 i. For Differential Protection

10 VA, CL5P20

0.2

1

2

CT Details same as above Feedpump Motor feeder

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

_x(R

ct

+2R

L

)

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

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

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

=Rct+0.7096

_x(R

ct

+2R

L

)

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

(15)

6.7096

9.9096

10 VA, CL-0.5

3 ID-2A, 2B 2240 300/1

10 VA, CL5P20

0.2 1

Total Burden on CT = Lead Burden+ Meter Burden Lead Burden for lead length of 40M= I2x(Rct+2RL)

9.9096 10 VA, CL-0.5

4 FD-2A, 2B 800 87.48 100/1

6.7096

10 VA, CL5P20

0.2

1

2

Total Burden on CT = Lead Burden+ Meter Burden Total Burden on CT = Lead Burden+ Meter Burden+burden margin Consider for future use

= 40*1*(Rct+2x1.995+0.1)

=Rct+0.7096

=Sqr(1)*(Rct+2Rl) =Rct+0.7096

3C x 185 mm2

_x(R

ct

+2R

L

)

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

=Rct+0.7096

(16)

4.7096

7.9096

10 VA, CL-0.5

5 PA-2A, 2B 1600 200/1

10 VA, CL5P20

0.2

1

2 5.7096

8.9096

10 VA, CL-0.5

6 Coal Mill-2A, 2B, 2C, 2D, 2E, 2F 400 43.74 50/1

1.7096

1.8096

10 VA, CL5P20

0.2 1

_x(R

ct

+2R

L

)

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

=Rct+0.7096

= 40*1*(Rct+2x0.887+0.1)

=Sqr(1)*(Rct+2Rl) =Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

3C x 185 mm2

_x(R

ct

+2R

L

)

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

=Rct+0.7096

(17)

4.9096 10 VA, CL-0.5

7 CWP-2A, 2B 1400 200/1

10 VA, CL5P20

0.2

1

2 5.7096

8.9096

10 VA, CL-0.5

8 Main Incomer from Unit Aux transformer 2000/1

9 Unit Service Transformer-2A & 2B 2500 219.79 300/1

10 Spare motor feeder on Unit-2 buses 5400 590.47 600/1 CT Details same as Feedpump Motor feeder

11 Spare transformer feeder on Unit-2 buses 2500 219.79 300/1

12 ESP Service Transformer Feeder-2A & 2B feeder 2500 219.79 300/1

C

1 5400 600/1

=Sqr(1)*(Rct+2Rl) =Rct+0.7096

3C x 185 mm2

= 40*1*(Rct+2x2.4836+0.1)

Or CT details will be same as earlier Supplied CT.

=Sqr(1)*(Rct+2Rl) =Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

CT Details same as I/C feeder from Unit Transformer

CT Details same as Station Service transformer feeder to SST Transformer

CT Details same as Unit service Trans. feeder

6.6 kV Station switchgear (C1A & C1B) comprising of the following panels with ratings and accessories as indicated in single line diagram, specification & data sheet.

BFP-1C & 2C Motor Feeder From C1A & C2A Board Respct.

(18)

Rct Assumed for CT Sec.1A is 7 ohm

7.7096

7.8096

10 VA, CL5P20

0.2

1

2 7.7096

10.9096

15 VA, CL-0.5

2 Compressor Feeders From C1A & C2A Board 500 100/1

Motor rated Current = 5400/Sqrt(3)/6.6/0.9 54.67 i. For O/C & E/F Protection

10 VA, CL5P20

0.2

1

2 4.7096

7.9096

10 VA, CL-0.5

3 Tie to U1A 2000/1

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

_x(R

ct

+2R

L

)

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

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

=Sqr(1)*(Rct+2Rl) =Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

CT Details same as Unit Switchgear I/C feeder from Unit Aux. Transformer

(19)

4 SST-1 feeder From C1A Board 1600 140.67 300/1

5 CWP-1C From C1B Board 1400 153.09 200/1

10 VA, CL5P20

0.2

1

2 5.7096

8.9096

10 VA, CL-0.5

6 Tie to U1B 2000/1

7 Spare feeder 300/1

8 CWPT-1 feeder From C1B Board 2500 219.79 300/1

9 Incomer 2500/1

D

1 Compressor Feeder From C2A Board 500 54.67 100/1

2 CWPT-2 feeder From C2A Board 2500 219.79 300/1

3 Tie to U2A 2000/1

4 Incomer 2500/1

5 Spare feeder 300/1

6 Incomer 2500/1

7 Tie to U2B 2000/1

8 SST-2 feeder From C2B Board 1600 140.67 300/1

9 Compressor Feeder From C2B Board 500 54.67 100/1

10 BFP-2C Motor Feeder From C2B Board 5400 600/1

CT ratio is kept same as to maintain the uniformity with Station Service Transformer CT Details same as Other Station service Switchgear

= 40*1*(Rct+2x3.548+0.1)

=Rct+0.7096

_x(R

ct

+2R

L

)

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

=Rct+0.7096

CT Details same as Other Station service Switchgear

7Rx1Cx630 mm2

6.6 kV Station switchgear (C2A & C2B) comprising of the following panels with ratings and accessories as indicated in single line diagram, specification & data sheet.

CT Details same as above air compressor Feeder.

CT Details same as Other Station service Switchgear 5Rx1Cx630 mm2 7Rx1Cx630 mm2 3C x 240 mm2 7Rx1Cx630 mm2 5Rx1Cx630 mm2 3C x 240 mm2

CT Details same as above air compressor Feeder.

3Rx1Cx630 mm2

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

(20)

10 VA, CL5P20

0.2 1

10.9096 15 VA, CL-0.5

11 Spare feeder 600/1 CT Details same as earlier main Feeder R relay in ohm (from relay catalogue of ABB model

SPAJ)

=Rct+0.7096

=Sqr(1)*(Rct+2Rl) =Rct+0.7096