NOMElec_FaultlevelCalc_20041005

8/3/2019 NOMElec_FaultlevelCalc_20041005

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Project: K. D. Malaviya National Oil Museum 1

K. D. MALAVIYA NATIONAL OIL MUSEUM

FAULT LEVEL CALCULATION

1 SCOPE :

Calculations for Source Impedance (Zs) :

The equivalent impedence for 500 MVA a fault level is

2.18 Ohms

500 MVA/ (3 x 33kV )

3 Calculations for cable A:

System : 33kV

At Supply point (I.e. outoging of State Electricity Board Circuit Breaker)8.75KA

sqrt 3 x 33 kV

Assuming conductor area to be A mm² and time duration of 1 sec

I sc = 0 .094 A/ sqrt t

8.75 = 0.094 A/ sqrt t

A = 8.75 x 1 /0.094 = 93.08mm²

The said cable should cater to normal full load current corresponding to 0.8MVA at 33kV

I.e 14.00Amp

From the cable manufacturer's data table 3C x 120 mm² cable is necessary to satisfy conditions

(a) and (b) for XLPE (E) cable :

4 Impedence of cable A ( Za) :

Length of Cable 'A' is about 20 Mtrs.

Impedence of cable A = (0.253 ) x 20 / 1000

Za = 0.0051 Ohms

Therefore total impedence at the owner's 33kV V.C.B panel is

= Za + Zs =

= 2.1780 + 0.0051 = 2.1805 Ohms

Selected size for Cable B is 3C X 120mm²

The building has one vertical electric shafts. Through there rising mains will be run for normal andemergency loads. The A.H.U.s and distribution boards on any floor will be connected to rising main withsuitable plug-in-units/breakers.

The sizing of L.T. cables has been done using the fact that circuit breakers will interrupt fault current in 8c.p.s ( I.e. 0.16 secs) and compensation required for temperature and bunching of cables in cable trays. Thsizing of H.T cables is done on the basis of available data of fult level at site (500 MVA) The block diagram

Zs = 33 kV/ 3

I = 500 MVA =

SupplyCo’sC.B

Owner’sVCB

MainLTPanel

A

B

BINCOMER 33kV

Cable

3 Panel 33kV

VCB Panel

33kV

Cable400kVA

X’mer











5Impedence of cable B ( Zb) :

Length of Cable 'B' is about 20 Mtrs.

Impedence of cable B = (0.253) x 20 / 1000Zb = 0.0051 Ohms

Therefore total impedence at the Transformer

Z t = (Za + Zs + Zb )

Z t = 2.1780 + 0.0025 +.0051 = 2.1856 Ohms

6 Impedence of Transformer ( Z tr) :

Assuming transformer impedence to be 5% is V² x X% , Where V is the secondary voltage

MVA

0.0215 Ohms

400 X1000 X100

(HV)²

(33000)²

Z eq = 0.00034 + 0.0215 = 0.0218 Ohms

Zeq = 0.0218 Ohms

7Fault level calculation :

Z eq

10.99KA

0.02Isc = 10.99 kA

So, we are considering that the maximum fault level as 11KA.

The impedence at the secondary side of the transformer is the sum of equivqlent impedence of StateElectricity Boards network, the cable 'A', the cable 'B' (all reflected on secondry side) plus impedence of thtransformer

Z tr = (V)² x Z% =

Z tr = (415)² x 5 =

Total impedence on secondarysideZ eq = (LV)² x.Z t + Z tr

Z eq = (415)² x2.1856+0.0215

Zeq = 0.00016 x 2.1856 +

0.0215

The fault current at the secondarycurrent of the transformer is

Isc = 415/sqrt 3

Isc = 415/sqrt 3 =



19

AHU RISER PIU 1 AHU-GRA-DB Gr-AHU Rm-A YFY 1 4C X 10

AHU RISER PIU 2 AHU-GRB-DB Gr-AHU Rm-B YFY 1 4C X 16

AHU RISER PIU 3 AHU-BANK-DB Gr-AHU Rm-Bank YFY 1 4C X 16

AHU RISER PIU 4 AHU-01A 01-AHU Rm-A YFY 1 4C X 10AHU RISER PIU 5 AHU-01B 01-AHU Rm-B YFY 1 4C X 10

AHU RISER PIU 6 AHU-02A 02-AHU Rm-A YFY 1 4C X 10

AHU RISER PIU 7 AHU-02B 02-AHU Rm-B YFY 1 4C X 10





AHU RISER PIU 12 AHU-04-DB 04-Elect. Rm-01 YFY 1 4C X 16


AHU RISER PIU 14 AHU-05B 05-AHU Rm-B YFY 1 4C X 10AHU RISER PIU 15 AHU-06A 06-AHU Rm-A YFY 1 4C X 10




AHU RISINGMAIN 400 A



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Sq.mm 2 8 SWG

Sq.mm 2 8 SWG

Sq.mm 2 8 SWG

Sq.mm 2 8 SWGSq.mm 2 8 SWG

Sq.mm 2 8 SWG

Sq.mm 2 8 SWG

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NOMElec_FaultlevelCalc_20041005