FAULT LEVEL CALCULATION, ENGINEERING
PROJECTS, ELECTRICAL EQUIPMENT RATING
DESIGN
DESIGN FACTORS CONSIDERED
A)
B)
The tolerance on 6.6kV/0.433kV Transformer impedance is taken as 5%
at principal tap. Actual minimum impedance at the lowest tap of other
transformers is considered.
Impedance of cables is neglected.
C)
First cycle asymmetrical peak short circuit (S/C) current (motor
contribution as well as grid incomer)= 2.55 symmetrical RMS first cycle
S/C current.
D)
A factor of 1.1 is taken into account for the increase in source fault
feeding voltage. A factor of 1.06 is also taken into account for persistent
system overvoltage during fault conditions for 6.6kV motor contribution.
E)
Subtransient Impedance of motors, Z d is calculated on the following
basis:
VOLTAGE LEVEL
MOTOR ID
6.6KV
ALL MOTORS
415V
Lumped motor Load (having
rating >= 37KW
SUBTRANSIENT IMPEDANCE (Zd)
F)
The tolerance on generator reactance (Xd) is taken as -15% as per IEC34.
G)
Power factor is taken as 0.85 for 6.6kV drives and 0.8 for 415V drives.
H)
Efficiency is taken as 0.9 for 6.6kV motors and 0.85 for 415V motors..
I)
The values of Z m o m (momentary) & Zintt(interrupting) for motors are
derived as under
Z m o m = 1 . 2 x Z d ( F o r m o t o r 7 5 0 k W a n d b e l o w )
Zintt = 3 . 0 x Z d ( F o r m o t o r 7 5 0 k W a n d b e l o w )
Z m o m = 1 x Z d ( F o r m o t o r a b o v e 7 5 0 k W )
Zintt = 1 x Z d ( F o r m o t o r a b o v e 7 5 0 k W )
415V FAULT LEVEL:
Basic Considerations:
1.
The fault level calculations for three-phase fault and single line to
ground faults are carried out for all the three types of 6.6KV/0.433KV LT
Transformers in service at XYZ PLANT
2MVA,
1.6MVA,
1MVA
2.
A negative tolerance of 5% is taken on transformer impedance at
nominal tap for lowest tap imp.value for fault current calculations.
3.
X/R values as per transformer test report.
4.
Fault is considered on the 415V Bus fed from HT board connected to
Level1 6.6KV Board HT212 as this is the strongest source to the 41V
faults.
5.
Motor contribution is calculated for largest lumped motor load for
each transformer rating for motor ratings equal to or greater than 37KW
fed from the switchboard and downstream MCCs connected to the
switchboard.
6.
Onl y case 1 is considered as in all the cases the fault current will
not change considerabl y.
7.
Impedance of cable and bus duct connections is neglected.
8.
Source over-voltage factor of 1.06 is considered.
Base MVA
=10
Base KV
=6.6KV/0.433KV
Base Current
=10/(3x0.433)
=13.33KA
INPUT D ATA
Transformer Data:
1. 2MVA, 6.33% imp,-5% for lowest tap ;X/R=7.52
2. 1.6MVA,6.19% imp,-5% for lowest tap;X/R=6.33
3. 1MVA, 4.95% imp ,-5% for lowest tap;X/R=6 (assumed)
Source:
1. S ymmetrical momentary fault current=35.86KA
Short circuit MVA
=3x6.6x35.86
2.
3.
S ymmetrical interrupting fault current
Short circuit MVA
=410MVA
=29.35KA
=3x6.6x29.35
=335MVA
=15 (assumed)
X/R Ratio
Lumped motor data:
1.
Largest Lumped motor Load fed from 2MVA transformer to
board PC-331-40B assuming that incomer to the adjacent bus PC-33141M is off and all motor load is fed from the same incomer through the
coupler.
BOARD NAME
SL#
FEEDER NAME
RATED KW
PC-331-40B(SS8)
9-PAM-004B
110
PC-331-40B(SS8)
9-PAM-015B
90
PC-331-40B(SS8)
COOLING TOWER FAN-A
90
PC-331-40B(SS8)
COOLING TOWER FAN-C
90
PC-331-41MB(SS8)
9-KAAM-003
90
PC-331-40B(SS8)
43-PAM-CF-006A
75
PC-331-40B(SS8)
43-PAM-CF-001A
75
PC-331-41MB(SS8)
43-PAM-CF-006B
75
PC-331-41MB(SS8)
43-PAM-CF-001B
75
10
PC-331-41MB(SS8)
9-PAM-005A
37
TOTAL(KW)
Rating
=807KW
Motor MVA =807/(0.8x0.85x1000)
Whereas, pf=0.8 and efficiency=0.85
=1.187
Zd
=10/(6x1.187)
=1.404pu
Zdmom
=1.2x1.404
=1.685pu
Zdintt
=3x1.404
=4.212pu
X/R
= 6(assumed)
Rmom
=1.685/[(6 2 +1 2 )]
=0.277pu
Xdmom
=0.277x6
=1.66pu
Rintt
=4.212/[(6 2 +1 2 )]
Xdintt
=0.693pu
=0.693x6
807
=4.156pu
Largest Lumped motor Load fed from 1.6MVA transformer to
board PC-306-11Bassuming that incomer to the adjacent bus PC-306-11B
is off and all motor load is fed from the same incomer through the coupler
1.
RATED KW
SL#
BOARD NAME
PC-306-11B(SS4)
23-PM-001A
110.00
PC-306-11B(SS4)
23-PM-001C
110.00
PC-306-12B(SS4)
23-PA-CF-001B
110.00
PC-306-12B(SS4)
23-PA-CF-001D
110.00
PC-306-11B(SS4)
47-PM-111A
45.00
PC-306-11B(SS4)
90-PM-101A
45.00
PC-306-11B(SS4)
90-PM-101B
45.00
PC-306-12B(SS4)
90-PM-101C
45.00
PC-306-12B(SS4)
47-PM-111B
45.00
TOTAL(KW)
FEEDER NAME
665
Zdmom
=2.045pu
Zdintt
=5.113pu
X/R
= 6(assumed)
Rmom
=0.336pu
Xdmom
=2.017pu
Rintt
=0.841pu
Xdintt
=5.043pu
1.
Largest Lumped motor Load fed from 1MVA transformer to board PMCC341-1B assuming that incomer to the adjacent bus PMCC-341is off and all
motor load is fed from the same incomer through the coupler
SL#
BOARD NAME
PMCC-341
47-PM-102A
125
PMCC-341-1B
47-PM-103A
125
TOTAL(KW)
Zdmom
=5.44pu
Zdintt
=13.6pu
X/R
= 6(assumed)
Rmom
=0.894pu
Xdmom
=5.366pu
Rintt
=2.24pu
Xdintt
=13.41pu
Source Imp:
Zs
Rs
=10/410
=0.0244pu
=0.0244/[(15 2 +1 2 )]
=0.0016pu
FEEDER NAME
RATED KVA/KW
250
Xs
=15x0.0016
=0.0243pu
Transformer imp.
1. 2MVA Transformer (-5% tolerance)
ZT1
=0.06331x0.95x10/2
=0.3007pu
R T 1 =0.3007/[(7.52 2 +1 2 )]
=0.0396pu
XT1
=7.52x0.0396
=0.2981pu
R T 0 =90% of R T 1
=0.0356pu
X T 0 =90% of X T 1
=0.268pu
2. 1.6MVA Transformer (-5% tolerance)
ZT1
=0.0619x0.95x10/1.6
=0.3868pu
R T 1 =0.3868/[(6.33 2 +1 2 )]
=0.06pu
XT1
=6.33x0.06
=0.3821pu
R T 0 =90% of R T 1
=0.054pu
X T 0 =90% of X T 1
=0.344pu
3. 1MVA Transformer (-5% tolerance)
ZT1
RT1
XT1
RT0
XT0
=0.0495x0.95x10/1
=0.47pu
=0.47/[(6 2 +1 2 )]
=0.077pu
=6x0.077
=0.464pu
=90% of R T 1
=0.069pu
=90% of X T 1
=0.418pu
Equivalent circuit calculations
3- PH ASE FAULT:
The equivalent reactance and resistance diagram is shown in sh. 26 of
diagram " Fault Level Calculations"
1.
2MVA Transformer:
Momentary:
Equivalent Resistance, Req
=0.0359pu
Equivalent Reactance, Xeq
=0.2501pu
Equivalent imedance for 3 phase fault:
Zeq
=(0.2501 2 +0.0359 2 )
=0.2724pu
Three phase short Circuit current in K Amps with over voltage factor of
1.06::
If
=1.06x(1/0.2605)x13.334
=51.9Karms
However, this value is not expected to go be yond desined value of
50Kamp(1 sec) due to following reasons:
as the calculated value will reduce after taking into account cable
impedance.
All level transformers are not expected to work at the least impedance
tap, resulting into higher impedance during fault.
Peak Value of short circuit current is as per following formula:
Ipeak= 2xImomx(1+e - t / )
= R/L;t=10ms
X/R
=0.2501/0.0359
=6.966
L/R
=6.966/(2f)
=0.0239
Ipeak
2.
=2x51.9x(1+e - 0 . 0 1 / 0 . 0 2 3 9 )
=121.7K Amp Peak
1.6MVA Transformer:
Momentary:
Equivalent Resistance, Req
Equivalent Reactance, Xeq
=0.0323pu
=0.3383pu
Equivalent imedance for 3 phase fault:
Zeq
=0.342pu
Three phase short Circuit current in K Amps with over voltage factor of
1.06:
If
=41.3K Arms
This is found OK as the design Value is 50K A.
Peak Value of short circuit current
L/R
=0.0206
Ipeak
1.
=2x41.3x(1+e - 0 . 0 1 / 0 . 0 2 0 6 )
=94K Amps Peak
1MVA Transformer:
Momentary:
Equivalent Resistance, Req
=0.0725pu
Equivalent Reactance, Xeq
=0.4475pu
Equivalent imedance for 3 phase fault:
Zeq
=0.4533pu
Three phase short Circuit current in K Amps with over voltage factor of
1.06:
If
=31.2K Arms
This is found OK, as the design Value is 50K A.
Peak Value of short circuit current
L/R
=0.0196
Ipeak
=2x31.2x(1+e - 0 . 0 1 / 0 . 0 1 9 6 )
=70K Amps Peak
SINGLE PH ASE TO E ARTH FAULT:
The impedance diagram for single phase to Earth fault is shown in sh. 27
of diagram " Fault Level Calculations". After network reduction, the current
calculations are as follows:
1.
2MVA transformer
Equivalent impedance to single line to Ground Fault:
Zeq
hence,FaultMVA
Fault Current
= 0.9208pu
=1.06x(3x(1/0.9208))x10
=34.5
=34.5/(3x0.415)
=46K A
1.
1.6MVA transformer
Equivalent impedance to single line to Ground Fault:
Zeq
= 1.171pu
hence,FaultMVA =1.06x(3x(1/1.171))x10
=27.2
Fault Current
=27.2/(3x0.415)
=36K A
2.
1MVA transformer
Equivalent impedance to single line to Ground Fault:
Zeq
= 1.484pu
hence,FaultMVA =1.06x(3x(1/1.484))x10
=21.4
Fault Current
=21.4/(3x0.415)
=28.6K A
