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CLIENT CONSULTANT CONTRACTOR PROJECT: PACKAGE-II : DOC. NAME : DESIGN CALCULATION SHEET 51NS CORE FOR NEUTRAL CT OF Converter duty TRANSFORMER OF SS2 DOCUMENT NO. 1.1
1.2
1.3
1.4
REV.-00
Dated : 03 /07/2014
System basic Input Data Transformer rating
=
630 KVA
C T ratios
=
600/1A
Design
=
3 PH 4 WIRE
Voltage
=
690
V AC
Frequency
=
50
Hz
Ratio
=
300
/
Accuracy class for Core‐II ( 51NS)
=
5P
15
Accuracy class for Core‐I
=
PS
Selected Burden ( VA ) for core‐I (PS)
=
10
VA
Primary Voltage (v)
=
11
kV
Secondary Voltage ( V )
=
690
V
Transformer rating
=
0.630
Impedence of transformer
=
4
% impedance variation on negative side
=
0
Basic Input Data of Current Transformer: 1
A
Basic Input Data of Distribution Transformer:
MVA % ( no negative tolerance‐ for calculation purpose)
Distribution System Fault Current: Full load current,IFL
=
(MVA X 106) (√ 3 X Sec. Voltage) 6
Min Transformer Impedance, ZM
=
(0.63x 10) (1.7322x690)
=
527.17
= = =
Maximum through fault current on LV side of transfromer = , MVA, IF = IF Max.through Fault current at CT Secondary Side Ifs
= = =
1 of 4
A
(% impedance x (100‐% tolerance))/100 4 x ( 100‐0))/100 4 % (IF x 100)/ ZM (527.17 x 100) /4) 13179.3 Amp IF / CTR 43.94 Amp
CT (CoreI) sizing calculation for 51NS Protection For 0.63 MVA Converter Trafo SS2 Core I: 2.0
CT sizing calculation for IDMT Earth Fault protection Relay(51NS) = Number of Identical C T S Location of C TS
2.1
2.2
2.3
=
1 NOS
Core _1 of NCT on Trafo or seperate CT
IDMT Earth Fault protection Relay(51NS) Input data Relay Module
=
E/F Setting
=
10
%
Current I/P burden, VA
=
0.1
VA
Relaty operating Current, IR
=
0.1
Amp
CoreI CT Input data Ratio
=
300
/ 1 A
Selected Burden ( VA )
=
10
VA
Accuracy Class
=
5P
15
Calculated of CT Burdens : =
50
m
Lead resistance at 70ºC for 2.5mm copper ( Ω/km )
=
8.87
Ω/km
Lead Resistance ( RL) (Ω)
=
0.89
(Ω)
Lead Burden ( VA )
=
0.89
VA
Total Meter Burden ( VA )
=
0.1
VA
=
0.99
VA
Cable Length l ( m ) 2
Total Burden on CT ( VA ) 0
CT Secondary resistance at 75 C (Rct) (Due to Design limitation of CT manufacturer,Rct) Selected burden is more than required. 2.4
= ≤
Ω
HENCE SAFE.
When CT,CoreI is used with ratio 600/1A,In case of phase to phase fault maximum primary current to be measured by the relay for operation 20%If = Fault current at Secondary
If
=
Voltage developed by CT Vct
CT Vct 2.5
4
13.2
kA
=
If X (Rct + Rl) Pri.CT Ratio
=
13.2 X 1000 X (4+0.887)/300)
=
215.03
V
Accordingly minimum VA burden of considered CT Rating with ALF 15 ALF
=
15
=
((Vct/ALF)xCT sec current) ‐ (Rct x CT sec Current)
=
((215.03/ 15 )X 1)‐(4 X1)
=
14.34
=
10.34
‐
4
Hence 10VA CT having 5P15 accuracy is adequate Hence NCT core 1 or seperate CT on 0.63 MVA Converter trafo CT 300/1A,5P15,10VA, RCT =< 4ohm
