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Solved Problems on Load Faults Current - Homework 3 | ECE 525, Study notes of Electrical and Electronics Engineering

Homework 3 Material Type: Notes; Professor: Johnson; Class: Power System Protection and Relaying; Subject: Electrical & Computer Engr; University: University of Idaho; Term: Fall 2012;

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ECE 525:
Power Systems Protection and Relaying Session 16; Page 1/29
Fall 2012
ECE 525: Homework #3
Solution
138kV
100M VA
ZS1 = j0.1 pu
ZS0 = Z S1
ZS2=ZS1
Vs = 1.0pu
Bus1 Bus2
ZL1 = j0.1 pu
B1
ZL2=ZL1
Bus0
138:12 .47kV
50 MV A
X = 10%
R = 0
D-Yg
ZL0 = j0.3 pu
B2
on 100MVA
and 1 38kV
B3
Load
3 MW
unity pf
R1
Load
3 MW
unity pf
R2
Load
3 MW
unity pf
Each of the 3 feeder sections has:
Zfeeder1 =Zfeeder2 = j0.1pu
Zfeeder0 = j0.3pu
Bus3 Bus4 Bus5
Assume the following:
Set the coordinating time interval at 6 cycles for each device (relay/recloser control)1. The worst case zero sequence imbalance for each load current on the distribution feeder is 20% 2. The transformer follows the ANSI/IEEE standard phase shift3. The fault currents by location and fault type are as follows (note that you will need refer currents4. across the transformer for some coordination cases. See spreadsheet on web page
Define constants:
MVA 1000kW pu 1 MW MVA
ae
j 120deg
 Summary settings on page 25
A012
1
1
1
1
a2
a
1
a
a2

pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
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On special offer

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Download Solved Problems on Load Faults Current - Homework 3 | ECE 525 and more Study notes Electrical and Electronics Engineering in PDF only on Docsity!

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: 

  1. Each load on feeder:

IloadLV

3MW3 V

B_LV 



IloadLV

138.9 A

^

pf

LV

  1. 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