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Session 16; Page 1/
ECE 525: Homework
Solution
138 kV10 0M VAZS
= j0.1 pu ZS
= Z
S
ZS
=Z
S Vs
= 1.0pu
Bus
Bus
ZL
= j0.1 pu
B
ZL
=Z
L
Bus
138:12 .47kV 50 MV AX = 10%R = 0D-Y
g
ZL
= j0.3 pu
B
on 100MVAand 1 38kV
B
Load3 MWunity
R1 pf
Load3 MWunity
R2 pf
Load3 MWunity
pf
Each
of the 3 feeder
sections has:
Zfeeder
=Z
feeder
= j0.1pu
Zfeeder
= j0.3pu
Bus
Bus
Bus
Assume the following:
Set the coordinating time interval at 6 cycles for each device (relay/recloser control)
The worst case zero sequence imbalance for each load current on the distribution feeder is 20%
The transformer follows the ANSI/IEEE standard phase shift
The fault currents by location and fault type are as follows (note that you will need refer currents
across the transformer for some coordination cases. See spreadsheet on web pageDefine constants:
MVA
1000kW
pu
MW
MVA
a^
j 120e ^
deg
Summary settings on page 25
A
012
(^12) a a
1 a^2 a
Session 16; Page 2/
Set base quantities:
S^ B
100MVA
V
B_HV
138kV
V
B_LV
V^
B_HV
12.47kV 138kV
V
B_LV
12.47 kV
IB_HV
S^ B
3 V
B_HV
IB_HV
418.37 A
^
IB_LV
S^ B
3 V
B_LV
IB_LV
4629.91 A
Load Currents:
- Each load on feeder:
IloadLV
3MW3 V
B_LV
IloadLV
138.9 A
^
pf
LV
- Load on transmission line (set based on transformer loaded at 40MVA):
Iload_HV
50MVA3 V
B_HV
Iload_HV
167.35 A
^
pf
HV
Fault Currents:
At Bus 1:
Three phase fault:
SLG fault:
LL fault:
IA3phB
2091.85A e
j (^) (^90)
deg
IASLGB
1568.9A e
j (^) (^90)
deg
IALLB
0A e
j 0^
deg
IB3phB
2091.85A e
j (^) (^210)
deg
IBSLGB
0A e
j (^) (^0) deg
IBLLB
1811.6A e
j (^) (^180)
deg
IC3phB
2091.85A e
j 30^
deg
ICSLGB
0A e
j 0^
deg
ICLLB
1811.6A e
j 0^
deg
Session 16; Page 4/
1. Determine relay settings for feeder relay (B3) and reclosers (R1 and R2). Coordinate settings between theprotective devices for phase faults (phase to phase and 3 phase) and with the transformer damage curve. Include aninstantaneous element for B3 to improve response time.
Determine CTR for B3, R1 and R2:
IloadR
IloadLV
IloadR
138.9 A
CTR
R
IloadR
2I
loadLV
IloadR
277.79 A
CTR
R
IloadB
3I
loadLV
IloadB
416.69 A
CTR
B
- Note, you probably will not be able to find anything between 400/5 and 450/
Phase element settings for recloser R2:
IloadR2_sec
IloadR2CTR
R
IloadR2_sec
4.63 A
Imin_f_R2_sec
IBLLB5CTR
R
Imin_f_R2_sec
190.93 A
^
Note that I'm using the phase to phase fault. The phase
element will be a backup to the ground element forSLG faults
0.5 I
^ min_f_R2_sec
95.47 A
2 I
^ loadR2_sec
9.26 A
Note that there is a big difference between these.
Set pick up for the phase element at 2X load current.
Session 16; Page 5/
IpuP_R
2 I
loadR2_sec
IpuP_R
9.26 A
Another option that is acceptable is to split the difference between the load current and the
minimum fault current. This does give better margin for inrush currents, but less resistive faultcoverage.
IpuP_R2_Alt
2 I
^ loadR2_sec
0.5 I
min_f_R2_sec
^
IpuP_R2_Alt
52.36 A
Since this recloser does not coordinate with anything else, set TD for the phase element at the minimum
TD
R2_P
Phase element settings for recloser R1:
IloadR1_sec
IloadR1CTR
R
IloadR1_sec
4.63 A
Imin_f_R1_sec
IBLLB5CTR
R
I'm using the Bus5 fault since I want to ensure detection
Also note that I'm using the phase to phase fault. The
phase element will be a backup to the ground elementfor SLG faults
Imin_f_R1_sec
95.47 A
0.5 I
^ min_f_R1_sec
47.73 A
2 I
^ loadR1_sec
9.26 A
Again, there is a big difference, just use 2X the load current for the pickup
IpuP_R
2 I
loadR1_sec
IpuP_R
9.26 A
Session 16; Page 7/
Now find M for recloser 1 for this fault:
M
R1P_If_max
IA3phB4CTR
R
IpuP_R
M
R1P_If_max
Now express TD as a function of time:
TD
VI
t req
M
^
t^ reqsec 3.88^2 M
Now for the case at hand:
TD
R1P_calc
TD
VI
t pu_R1P_desired
M
R1P_If_max
^
TD
R1P_calc
Set TD for the phase element at:
TD
R1_P
Session 16; Page 8/
Phase element settings for relay controlling breaker B3:
IloadB3_sec
IloadB3CTR
B
IloadB3_sec
4.63 A
Imin_f_B3_sec
IBLLB5CTR
B
I'm using the Bus5 fault since I want to ensure detection
Also note that I'm using the phase to phase fault. The
phase element will be a backup to the ground elementfor SLG faults
Imin_f_B3_sec
63.64 A
0.5 I
^ min_f_B3_sec
31.82 A
2 I
^ loadB3_sec
9.26 A
Again, there is a big difference, just use 2X the load current for the pickup
IpuP_B
2 I
loadB3_sec
IpuP_B
9.26 A
Another option that is acceptable is to split the difference between the load current and the
minimum fault current. This does give better margin for inrush currents, but less resistive faultcoverage.
IpuP_B3_Alt
2 I
^ loadB3_sec
0.5 I
min_f_B3_sec
^
IpuP_B3_Alt
20.54 A
Determine M for R1 and breaker B3 for the largest fault right at the downstream side of R1:
M
R1P_If_max
IA3phB3CTR
R
IpuP_R
M
R1P_If_max
Session 16; Page 10/
M
B3LL_B
IBLLB5CTR
B
IpuP_B
M
R1LL_B
IBLLB5CTR
R
IpuP_R
M
R2LL_B
IBLLB5CTR
R
IpuP_R
tpu_R2P_LLBus
tVI
TD
R2_P
M
R2LL_B
^
tpu_R2P_LLBus
51.508 ms
tpu_R1P_LLBus
tVI
TD
R1_P
M
R1LL_B
^
tpu_R1P_LLBus
160.297 ms
tpu_B3P_LLBus
tVI
TD
B3_P
M
B3LL_B
^
tpu_B3P_LLBus
315.19 ms
tpu_R1P_LLBus
tpu_R2P_LLBus ^
0.11 s ^
ok coordination
tpu_B3P_LLBus
tpu_R1P_LLBus ^
0.15 s ^
ok coordination
tpu_R2P_LLBus
tVI
TD
R2_P
M
R2LL_B
^
tpu_R2P_LLBus
52.724 ms
tpu_R1P_LLBus
tVI
TD
R1_P
M
R1LL_B
^
tpu_R1P_LLBus
173.1 ms
tpu_B3P_LLBus
tVI
TD
B3_P
M
B3LL_B
^
tpu_B3P_LLBus
360.42 ms
tpu_R1P_LLBus
tpu_R2P_LLBus ^
0.12 s ^
ok coordination
tpu_B3P_LLBus
tpu_R1P_LLBus ^
0.19 s ^
ok coordination
Session 16; Page 11/
Instantaneous Element for B3: ^
Current for a three phase at 60% from Bus 2 to Bus 3
If3phase_0.6FBus2_3pu
j 0.1^
j 0.1 ^
j 0.2 ^
0.6 j
^
If3phase_0.6FBus2_3pu
2.174i
pu
If3phase_0.6FBus2_
IB_LV
I f3phase_0.6FBus2_3pu
If3phase_0.6FBus2_
10.07i
kA
IB3_sec
If3phase_0.6FBus2_
CTR
B
IB3_sec
111.83 A
^
IB3_inst_set
112A
As a comparison, here is the time the 51P element at Bus 3 would take to respond to this fault
tVI
TD
B3_P
IB3_sec IpuP_B
^
0.25 s ^
So big difference in response time.
2. Determine relay settings for feeder relay (B3) and reclosers (R1 and R2). Coordinatesettings between the protective devices for ground faults (SLG) and with the transformer.
Zero sequence load currents
IloadB3_
20% I
loadB
IloadB3_
83.34 A
IloadR1_
20% I
loadR
IloadR1_
55.56 A
IloadR2_
20% I
loadR
IloadR2_
27.78 A
Assume that the ground element currents in the relays are calcuated using measurements from the phase CTs rather than
having separate neutral CTs.Therefore the CTRs do not change from the previous part of the problem
Session 16; Page 13/
Ground element settings for recloser R1:
IloadR10_sec
IloadR1_0CTR
R
IloadR10_sec
0.93 A
Imin_f_R1_3I0_sec
IASLGB5CTR
R
Imin_f_R2_3I0_sec
185.2 A
^
I'm using the Bus5 fault since I want to
ensure detection
0.5 I
^ min_f_R2_3I0_sec
92.6 A
2 I
^ loadR10_sec
1.85 A
Again, there is a big difference, just use 2X the load current for the pickup
IpuG_R
2 I
loadR10_sec
IpuG_R
1.85 A
Another option that is acceptable is to split the difference between the load current and the
minimum fault current. This does give better margin for inrush currents, but less resistive faultcoverage.
IpuG_R1_Alt
2 I
^ loadR10_sec
0.5 I
min_f_R1_3I0_sec
^
IpuG_R1_Alt
24.08 A
Now the time dial setting will need to account for the coordinating time interval.
CTI
60Hz
CTI
0.1 s ^
Rather short, but was specified in the problem
Again choose the Very Inverse Curve:
Session 16; Page 14/
tVI
TD M
(^
)^
TD
3.88^2 M
^
sec
Determine M for R2 and R1 for the largest fault right at the downstream side of R2:
M
R2G_If_max
IASLGB4CTR
R
IpuG_R
M
R2G_If_max
tpu_R2G_If_max
tVI
TD
R2_G
M
R2G_If_max
^
tpu_R2G_If_max
0.05 s
Minimum pickup time for R1 for the same fault: ^ t
pu_R1G_desired
tpu_R2G_If_max
CTI
tpu_R1G_desired
0.15 s
Now find M for recloser 1 for this fault:
M
R1G_If_max
IASLGB4CTR
R
IpuG_R
M
R1G_If_max
Now express TD as a function of time (repeated from above):
TD
VI
t req
M
^
t^ reqsec 3.88^2 M
Session 16; Page 16/
Determine M for R1 and breaker B3 for the largest fault right at the downstream side of R1:
M
R1G_If_max
IASLGB3CTR
R
IpuG_R
M
R1G_If_max
tpu_R1G_If_max
tVI
TD
R1_G
M
R1G_If_max
^
tpu_R1G_If_max
0.15 s
Minimum pickup time for R1 for the same fault: ^ t
pu_B3G_desired
tpu_R1G_If_max
CTI
tpu_B3G_desired
0.25 s
Now find M for recloser 1 for this fault:
M
B3G_If_max
IASLGB3CTR
B
IpuG_B
M
B3G_If_max
Now determine time dial for B3 phase
TD
B3G_calc
TD
VI
t pu_B3G_desired
M
B3G_If_max
^
TD
B3G_calc
Set TD for the phase element at:
TD
B3_G
Session 16; Page 17/
Now verify coordination for all phase elements at B3, R1 and R2 (look at faults at Bus 4 and Bus 5):
M
B3SLG_B
IASLGB4CTR
B
IpuG_B
M
R1SLG_B
IASLGB4CTR
R
IpuG_R
M
R2SLG_B
IASLGB4CTR
R
IpuG_R
M
B3SLG_B
IASLGB5CTR
B
IpuG_B
M
R1SLG_B
IASLGB5CTR
R
IpuG_R
M
R2SLG_B
IASLGB5CTR
R
IpuG_R
tpu_R2G_SLGBus
tVI
TD
R2_G
M
R2SLG_B
^
tpu_R2G_SLGBus
48.274 ms
tpu_R1G_SLGBus
tVI
TD
R1_G
M
R1SLG_B
^
tpu_R1G_SLGBus
155.67 ms
tpu_B3G_SLGBus
tVI
TD
B3_G
M
B3SLG_B
^
tpu_B3G_SLGBus
266.048 ms
tpu_R1G_SLGBus
tpu_R2G_SLGBus ^
CTI
^
0.01 s ^
OK coordination
tpu_B3G_SLGBus
tpu_R1G_SLGBus ^
CTI
^
0.01 s ^
OK coordination
tpu_R2G_SLGBus
tVI
TD
R2_G
M
R2SLG_B
^
tpu_R2G_SLGBus
48.344 ms
tpu_R1G_SLGBus
tVI
TD
R1_G
M
R1SLG_B
^
tpu_R1G_SLGBus
156.56 ms
tpu_B3G_SLGBus
tVI
TD
B3_G
M
B3SLG_B
^
tpu_B3G_SLGBus
269.45 ms
tpu_R1G_SLGBus
tpu_R2G_SLGBus ^
CTI
^
0.01 s ^
OK coordination
tpu_B3G_SLGBus
tpu_R1G_SLGBus ^
CTI
^
0.01 s ^
OK coordination
Session 16; Page 19/
Fall 2012
I012LL_bus
A
012
(^1)
IALLB2 IBLLB2 ICLLB2
I012LL_bus
^ A
^
arg I
012LL_bus
^
deg
I0LL_bus2_HV
I1LL_bus2_HV
12.47 ^138
^
I012LL_bus
1
^
j 30e ^
deg
I1LL_bus2_HV
522.96 A
^
arg I
1LL_bus2_HV
^
^
deg
I2LL_bus2_HV
12.47 ^138
^
I012LL_bus
2
^
e^
j 30 ^
deg
I2LL_bus2_HV
522.96 A
^
arg I
2LL_bus2_HV
^
^
60 deg
IABC_LL_Bus2_HV
A^
012
I0LL_bus2_HV I1LL_bus2_HV I2LL_bus2_HV
IABC_LL_Bus2_HV
^ A
^
arg I
ABC_LL_Bus2_HV
^
deg
In this case the largest phase has about the same current as the 3 phase fault at Bus 2 when referred to the primary.
Determine CTR for B1 and B2, assuming 40MVA loading on transformer is worst case.:
IloadB
Iload_HV
IloadB
167.35 A
CTR
B
Session 16; Page 20/
IloadB
IloadB
IloadB
167.35 A
CTR
B
IloadB2_sec
IloadB2CTR
B
IloadB2_sec
4.65 A
^
I'm using the Bus3 fault since I want to ensure detection
Also note that I'm using the SLG fault since that has the
lowest phase currents on the HV side.
Use fault current referred to HV side
Imin_f_B2_sec
IABC_SLG_Bus2_HV
0
CTR
B
Imin_f_B2_sec
20.13 A
0.5 I
^ min_f_B2_sec
10.06 A
2 I
^ loadB2_sec
9.3 A
Sufficient difference, use 2X the load current for the pickup
IpuP_B
2 I
loadB2_sec
IpuP_B
9.3 A
Another option that is acceptable is to split the difference between the load current and the
minimum fault current. This does give better margin for inrush currents, but less resistive faultcoverage.
IpuP_B2_Alt
2 I
^ loadB2_sec
0.5 I
min_f_B2_sec
^
IpuP_B2_Alt
9.68 A
