Excersices in power system safety and protection

1. KKKZ 4014 POWER SYSTEM PROTECTION AND SAFETY
1. A 132 kV system is connected with 10 MVA, 132/11 kV transformer as shown in
Figure 1.
The transformer percentage reactance is 10%. The breaking capacity of the switchgear
to be installed.
Calculate
a) Short circuit MVA on the LV side of the transformer
b) Fault current
c) Source impedance
Solution:
We know, short circuit MVA =
%
100
X
P
,
=
10
10100
= 100 MVA
b) fault current at short circuit MVA on the LV side, IF =
kV3
Acircuit MVshort
IF = kA
kV
MVA
248.5
113
100


c) source impedance =
FI
kV
3
= 

21.1
248.53
11
kA
kV
Transformer rated power, P = 10 MVA
Transformer percentage reactance, X% = 10%
Short circuit MVA = 100 MVA
kV = 11 kV

2. 2. A 132 kV system is connected with 10 μVA, 132/11 kV transformer as shown in
Figure 2. The percentage reactance of the transformer is 10%.
Calculate
a) Short circuit MVA on the LV side
b) Fault current, IF
c) Source impedance
d) Fault current downstream at the particular distance far from the transformer with
the line impedance of 1Ω
Solution:
a) Short circuit MVA =
10
10100
%
100 VA
X
P 

= 100 μVA
b) Fault current, IF = kA
kV
248.5
113
100
3
Acircuit MVshort



c) Source impedance = 



21.1
248.53
11
3 kI
kV
F
d) Fault current, IF =
 
kA
kV
874.2
)21.2(3
11
121.13


New total source impedance => (1.21+1) = 2.21Ω
Source impedance = kA
I
kV
F
874.2
21.23
11
3




IF = kA
kV
874.2
21.23
11
)impedancesource(3



P = 10 μVA
X = 10%

3. 3. The impedance of a 1250 kVA transformer is 5.5% of 11 kV system. The high
voltage fault of the 11 kV system is 2.5 kA as shown in Figure 3.
Calculate
a) Total impedance in Ω at 1,000 V
b) Fault current at 1,000 V
Solution:
Based on ohmic method
Z Ω source = 



54.2
5.23
11
currentfaultHV3
kV
Again in impedance of the transformer,
Z Ω transformer = 





324.5
25.1100
115.5
100
% 22
MVA
kVZ
Z Ω total = (Z Ω source + Z Ω transformer) = 2.54 + 5.324 = 7.86 Ω
a) total impedance at LV = 2
2
HV
x LVHVinZtotal
= 

065.0
121
86.7
11
186.7
2
2
b) Fault current at 1000 V
IF 1000V = kA883.8
065.03
1
at LVtotalz3
LV




Data given
kV = 11 kV
HV fault current = 2.5 kA
kV = 11 kV
MVA = 1.250 MVA
Z = 5.5%

4. 4. The impedance of a 1250 kVA transformer is 5.5% of 11kV system. The high
voltage fault current of the 11kV system is 2.5 kA. Using percentage reactance
method
Determine
a) Total impedance in Ω at 1,000 V
b) Fault current at 1,000 V
Solution:
Based on percentage reactance method
Fault MVA at system = 3 x kV x IF
= 3 x kV x 2.5
= 47.63 MVA
Again source percentage impedance of the base MVA
= %
systematFault MVA
ratingx MVA1000
= %
63.47
250.1100
= 2.624%
a) total percentage impedance at 1000 V
= transformer percentage impedance + source percentage impedance kV
= 5.5% + 2.624% = 8.124%
b) fault MVA at 1000V = kA
kV
883.8
1013
10386.15
3
Fault MVA
3
6





Fault MVA at the transformer
= MVA386.15
124.8
250.1100
impedancepercentageTotal
Ratingx MVA100



IF = 2.5 kA
kV = 11 kV

5. 5. The impedance of a 1250 kVA transformer is 5.5% of 11kV system. The high
voltage fault current of the 11kV system is 2.5 kA. Using per unit method
Determine
a) Total impedance in Ω at 1,000 V
b) Fault current at 1,000 V
Solution:
Based on per unit method
Short circuit MVA = 3 x kV x IF
= 3 x kV x 2.5
= 47.63 MVA
Per unit source impedance = ..026.0%
63.47
25.1
Acircuit MVShort
MVABase
up
a) Total impedance at 1000 V = source impedance + transformer impedance
= 0.026 + 0.055
= 0.08124 p.u.
b) Fault current at 1000V = kA
upkV
883.8
1008124.013
1025.1
..3
MVABase
3
6





kV = 11 kV
IF = 2.5 kA
Base MVA = 1.25 MVA

6. 6. A 20 MVA generator is connected to a 20 MVA, 13.8/115 kV system as shown in
Figure 4. The 87 differential relays are connected to the generator neutral and the
circuit breaker. The CT ratio is 1000:5
Determine
a) Maxm
load current at 13.8 kV
b) Internal 3 phase fault current at F
c) 3 phase fault current in the 87 relay operating coil
d) 3 phase fault current before the generator synchronization to the power system
Solution:
a) Maxm
load current at 13.8 kV
IMAX LOAD 2ND = A74.836
108.133
1020
3
6



b) On a 20 MVA base, the equivalent reactance,
Xeq = ..04.0
100
2.020
MVAsystem
reactancesystemxMVA
up


Transformer reactance, XT = 10% = 0.1
Generator sub-transient reactance, X’’
d = 0.32
Total reactance to the fault, Xtotal = ..097.0
46.0
14.032.0
XXX
)X(XX
eqTd
''
eqTd
''
up




Per unit 3 phase fault at 20 MVA, I3PHASE = ..27.10
097.0
1
up
Internal 3 phase fault current, I3PHASE = A85935
108.133
102027.10
3
6



c) Maxm
load current at secondary of the CT is
IMAX LOAD 2ND =
1000
574.836 
= 4.18 A
Fault current following in the (87 operating coil)
I3PHASE 87 = 10.27 x 4.18 = 42.96
d) If 3 phase fault current occured before the generator synchronisation
I3PHASE = ..13.3
X
1
d
"
up I3PHASE before synchronisation = 99.2618
108.133
102013.3
3
6



at 13.8 kV
Current following in 87 operating coil
I3PHASE 87 = A1.13
1000
599.2618



7. 7. 60 MW distributed generator connected to 20 miles radial 69 kV lines as shown in
Figure 5. The generation output at transformer higher side 60 MW and 19.9 MVar, the
operating at 95% power factor. The system impedance is (3.8 + j14.4) ohms
Calculate
a) The voltage that exist on the utility line at generator location
b) Percentage of voltage rise across the line impedance
c) Give your reason/decision whether the power quality protection operate or not
Solution:
cos = 0.95
 = cos-1
0.95 = 18.2o
MVar = 60 MW x tan = 60 x tan 18.2o
= 19.7 MVar
Approximate current at transformer higher side
I = Ampj
j
HV
kVarjkW
)165502(
693
1970060000
3
107.191060 33





a) Voltage that exist in generator location
VG = 69000 + (3.8 + j14.4) x (502 - j165)
VG = 73 580 V
b) Voltage raise across the linear impedance
= %6.6%100
73580
6900073580
%100 




 





 
G
LG
V
VV
c) As we know, voltage regulation is considered about 5% according to the IEEE
standard. The 6.6% of voltage raise to is a little bit problem. However, if the load is
sensitive then raise voltage level could more problematic. So protection must need to
operate.

8. 8. A three phase, 11 kV, 15 kW generator of power factor 0.8 is operating at full load
prior to fault. The reactance, transient reactance, sub-transient reactance and time
constant at nominal voltage 1 p.u. and constant 1 p.u. are as follows.
Xd = 2.0 p.u. T’
d = 5.0 sec
X’
d = 0.25 p.u. T’’
d = 0.15 sec
X’’
d = 0.12 p.u.
Determine
a) Fault current
b) EMF of the generator at reactance, transient reactance and sub-transient reactance
c) Current, transient current and sub-transient current (I, I’
, I”
)
Solution:
MVA of the power factor = MVA75.18
8.0
15
factorpower
MV

a) We know, fault current, IFC = A
kV
MVA
984
10113
1075.18
113 3
6





b) We know, EMF, E =    22
cossin  IXIXV dd  =    22
8.0126.0121 
= 2.72 p.u.
E’
=    2'2'
cossin  IXIXV dd  =    22
8.0125.06.0125.01 
= 1.67 p.u.
E’’
=    2''2''
cossin  IXIXV dd  =    22
8.0112.06.0112.01 
= 1.076 p.u.
c) Current I =
Reactance
EMF
x fault current
Reactance current, I =
2
72.2
 F
d
I
X
E
x 984 = 8823.23A
Transient current, I’
=
25.0
67.1
'
'
 F
d
I
X
E
x 984 = 4593.43A
Sub-transient current, I’’
=
12.0
076.1
''
''
 F
d
I
X
E
x 984 = 1338.29A
Data given;
V = 1 p.u. at nominal voltage
I = 1 p.u. at nominal voltage
Xd = 2.0 p.u.
cos = 0.8
sin = 0.6

9. 9. A 11 kV, 15 kW generator is running with 0.8 power factor. The generator has an
impedance of 1.61 Ω per phase
Determine
a) Fault MVA of the generator
b) Percentage impedance of the generator
Solution:
Phase voltage =
3
1011
3
voltagetime 3

 = 6350 V
Fault current =
61.1
6350
impedancephase
voltagephase
 = 3940 A
a) Fault MVA = 3 x kV x IF = 3 x 11 kV x 3940 = 75 MVA
b) We know, the percentage impedance, Z %
Z % =
 2
voltageline
ratingMVA
x phase impedance x 100%
MVA =
8.0
15
factorpower
MW
 = 18.75 MVA
Z % = 2
11
75.18
x 1.61 x 100% = 25%

10. 10. Primary fault current is 9 kA with 300:1 CT connected transformer restricted earth
fault protection system as shown in Figure 6. The secondary winding resistance, RCT
of CT is 3 Ω and the lead resistance, R1 is 1 Ω. The relay type is CAG 14, 1 A,
10-40%, burden 1.0 VA
Calculate
a) The stabilizing voltage
b) Minimum knee-point voltage
c) The stabilizing resistance of the restricted earth fault protection
Solution:
Secondary fault current =
300
1x9000
= 30 A
Relay operating current,
Choose 10% tap on CAG 14
I = 10% x 1 = 0.1 A
Relay operating voltage =
1.0
1

I
VA
= 10 V
a) Stabilizing voltage, V = I (RCT + R1) = 30 (3 + 1) = 120 V
b) We know current transformer minimum knee-point voltage is twice of the
stabilizing voltage
Knee-point voltage = 2 x stabilizing voltage
= 2 x 120
= 240 V
c) Stabilizing reactance =
resistortheroughcurrent th
resistoracrossvoltage
Voltage across the resistor = stabilizing voltage - relay voltage
= 120 - 10
= 110 V
Stabilizing reactance =
1.0
110
= 1100 Ω
Given
Relay voltage = 10 V
RCT = 3 Ω
R1 = 1 Ω
I = 30 A

11. 11. A distribution or similar facility served by a 7500 kVA, 115:12 kV transformer
with 7.8% of impedance as shown in Figure 7.
Determine
a) Secondary fault current
b) Fuse settling current
c) Justify your fuse design
Solution:
We know, short circuit MVA =
8.7
7500100
X%
100P 
 = 96.153 kVA
a) Fault current at secondary, IF = short circuit MVA = 4.626 A
b) Primary current, I primary =
1153
7500
1153 


kVA
= 37.65 A at 15k
3 phase fault current = 3 x I primary fault current
= 3 x 37.65
= 65 Ampere
Fuse current setting = 65 Ampere
c) Yes, the calculated fuse setting current is same with the fuse used. So, the design
parameter is satisfied.