CT Sizing Calculation for HT/LV Transformer Feeder...
Description
CLIENT CONSULTANT CONTRACTOR PROJECT: PACKAGE-II : DOWN HILL CONVEYOR SYSTEM DOC. NAME : DESIGN CALCULATION SHEET 51NS CORE FOR NEUTRAL CT OF DISTRIBUTION TRANSFORMER OF SS1 AND SS2 DOCUMENT NO. 1.1
1.2
1.3
1.4
REV.-01
Dated :03 /07/2014
System basic Input Data Transformer rating
=
1250 KVA
C T ratios
=
800/1A
Design
=
3 PH 4 WIRE
Voltage
=
415
V AC
Frequency
=
50
Hz
Ratio
=
800
/
Accuracy class for Core‐II ( 51NS)
=
5P
15
Accuracy class for Core‐I
=
PS
Selected Burden ( VA ) for core‐I (PS)
=
15
VA
Primary Voltage (v)
=
6.6
kV
Secondary Voltage ( V )
=
415
V
Transformer rating
=
1.250
Impedence of transformer
=
5
% impedance variation on negative side
=
0
Basic Input Data of Current Transformer: 1
A
Basic Input Data of Distribution Transformer: 11Kv
MVA % ( no negative tolerance‐ for calculation purpose)
Distribution System Fault Current: =
Full load current,IFL
For SS2
(MVA X 106) (√ 3 X Sec. Voltage) 6
= = Min Transformer Impedance, ZM
= = =
Maximum through fault current on LV side of transfromer , = MVA, IF = IF = Max.through Fault current at CT Secondary Side I fs
1 of 2
= =
(1.25x 10) (1.7322x415) 1739.06
A
(% impedance x (100‐% tolerance))/100 5 x ( 100‐0))/100 5 % (IF x 100)/ ZM (1739.06 x 100) /5) 34781.2 Amp IF / CTR 43.48 Amp
CT (CoreI) sizing calculation for 51NS Protection For 1.25 MVA Distribution Trafo SS1/ 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
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
=
800
/ 1 A
Selected Burden ( VA )
=
15
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
Total Burden on CT ( VA )
=
0.99
VA
CT Secondary resistance at 75 0C (Rct) (Due to Design limitation of CT manufacturer,R ct)
= ≤
Cable Length l ( m ) 2
Selected burden is more than required. 2.4
4
Ω
HENCE SAFE.
When CT,CoreI is used with ratio 2500/1A,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
1 NOS
34.8
kA
=
If X (Rct + Rl) Pri.CT Ratio
=
34.8 X 1000 X (4+0.887)/800)
=
212.59
V
Accordingly minimum VA burden of considered CT Rating with ALF 15 ALF VA
= =
15 ((Vct/ALF)xCT sec current) ‐ (Rct x CT sec Current)
=
((212.59/ 15 )X 1)‐(4 X1)
=
14.17
‐
4
VA = 10.17 Hence 10VA CT having 5P15 accuracy is adequate Hence NCT core ‐1 or seperate CT on 1.25 MVA Distribution trafo CT ‐800/1A,5P15,15VA, RCT =
Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.