A Three-Phase Fault Currents Calculation Method Used · PDF fileCalculation Method Used For...

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A Three-Phase Fault Currents Calculation Method Used For Protection Coordination Analysis Khaled Saleh [email protected] Dr. Hatem Zeineldin [email protected] Dr. Amer Al-Hinai [email protected] 1 Paper ID: 14TD0269

Transcript of A Three-Phase Fault Currents Calculation Method Used · PDF fileCalculation Method Used For...

Page 1: A Three-Phase Fault Currents Calculation Method Used · PDF fileCalculation Method Used For Protection Coordination Analysis ... • The conventional short circuit calculation ...

A Three-Phase Fault Currents Calculation Method Used For

Protection Coordination Analysis Khaled Saleh – [email protected]

Dr. Hatem Zeineldin – [email protected]

Dr. Amer Al-Hinai – [email protected]

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Paper ID: 14TD0269

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Introduction

• A reliable fault current calculation method is required to perform precise protection coordination analysis.

• The conventional short circuit calculation method is not valid for distribution power systems that are equipped with Inverter-Based Distributed Generation (IBDG) units.

• Usually, a transient model of the power system equipped with IBDG units are developed using either PSCAD or MATLAB to compute the short circuit currents flowing through each protective device.

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Introduction • A simple method based on the superposition theorem is

proposed that calculates the three-phase fault currents for meshed distribution system in the presence of IBDG units.

• Most of the IBDGs are controlled using the current control scheme.

• During fault conditions, the fault contribution of IBDGs is bounded by their controller limiters.

• In this study IBDG during fault conditions is modeled as a constant current source injecting fault current equivalent to the upper boundary of its controller limiter.

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Fault Current Calculation Method

Bus 1

Bus 2 Bus 3SBDG IBDG

F4

R2 R5

R1

R6

R3

R4

F5

F6

Substation

4

Three-bus system example.

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Fault Current Calculation Method 5

Three-bus system superposition part 1.

R1

Bus 1

Bus 2

R3

R2 R5

Bus 3

R6 R4

F4

SBD

G

IBD

G

F6

F5Su

bstatio

n

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Fault Current Calculation Method 6

Three-bus system superposition part 2.

R1

Bus 1

Bus 2

R3

R2 R5

Bus 3

R6 R4

F4

SBD

G

IBD

G

F6

F5

Sub

station

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Fault Current Calculation Method 7

• In the first part of superposition, fault current passing through each DOCR for all fault locations on the system are computed.

• For the second part of superposition, fault current passing through each DOCR for all fault locations on the system are computed considering constant fault current magnitude with zero phase angle injected by the IBDGs.

• The final step towards implementing the proposed method is to sum the fault current contribution of the first and second part of superposition.

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Proposed Method Verification 8

R3

Bus 1

Bus 5

R1

R5

R4 R13

R7

Bus 2

R9

Bus6

R14 R15

R6

F10

F14

Bus 3

R11 R8Bus 4

R2 R12

Bus 7

R16 R10

60 MVA60 MVA

F11F9

F8 F13

F12

F15

2 MVA

2 MVA

2 MVA

Power Distribution System of the IEEE 14-bus system under study

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Proposed Method Verification 9

In order to be able to validate the proposed method, two simulation setups were developed:

1. Short circuit currents are computed based on the proposed method where each IBDG is set to inject fault current of 1.5 p.u of the DG capacity.

2. Short circuit currents are computed based on a transient model of the same test system developed on SIMULINK simpowersystems toolbox. – IBDG is a 2MVA photovoltaic plant with current controlled inverter

with controller limiters set to 1.5 p.u of the DG capacity.

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Proposed Method Verification 10

Fault location

DOCR Three-Phase Fault currents (p.u.) Percentage

error (%) Transient Model Proposed Method

F8

R1 2.345 2.335 0.426

R2 1.191 1.181 0.832

R4 0.073 0.070 3.409

R11 1.177 1.174 0.219

R6 0.233 0.229 1.659

F9

R5 2.669 2.655 0.518

R6 0.917 0.907 1.105

R2 0.662 0.657 0.819

R13 0.233 0.227 2.676

R4 0.214 0.218 2.154

R16 0.684 0.680 0.560

F10

R3 2.289 2.279 0.435

R4 0.699 0.696 0.444

R2 0.600 0.594 0.959

R14 0.687 0.683 0.553

R6 0.1931 0.1929 0.084

F11

R13 1.006 0.989 1.752

R14 1.747 1.735 0.701

R3 0.983 0.981 0.221

R5 1.150 1.143 0.614

R16 0.599 0.593 0.981

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Proposed Method Verification 11

Fault location

DOCR Three-Phase Fault currents (p.u.) Percentage

error (%) Transient Model Proposed Method

F12

R15 1.583 1.574 0.576

R16 1.140 1.132 0.697

R5 1.261 1.255 0.440

R9 1.128 1.124 0.347

R13 0.329 0.325 1.269

F13

R11 1.881 1.874 0.358

R12 1.423 1.414 0.670

R7 1.881 1.874 0.358

R1 1.411 1.406 0.339

F14

R9 2.265 2.257 0.338

R10 0.863 0.855 0.871

R8 0.429 0.425 1.045

R15 0.850 0.847 0.390

F15

R7 2.836 2.829 0.233

R8 1.026 1.021 0.498

R10 0.448 0.453 0.960

R12 1.026 1.021 0.498

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Protection Coordination Problem 12

• One of the most popular solutions proposed in the literature to solve the protection coordination problem of interconnected systems is the use of optimization methods.

• The objective is to minimize the sum of primary and backup DOCRs operating time (T) while maintaining the conditions of protection coordination. Hence, the objective function is given as:

N

i

bx

ij

p

ij

M

j

ijttMinimizeT11

)(

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Protection Coordination Problem 13

• Conventional Relay Tripping Characteristic (IEC 255-3 standard):

• TDS: Time Dial Setting

• M: Multiple of pickup

currents= (Isc/Ip)

• Isc=Short circuit current

• Ip=Pickup current setting

• A=0.14, B=0.02

2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

2

2.5

Multiples of pickup currents (M)

trip

pin

g tim

e (

seconds)

TDS = 0.05

TDS = 0.1

TDS = 0.2

1

BM

ATDSt

Time-Current relay characteristic for different values of TDS

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Protection Coordination Problem 14

• The protection coordination optimization (PCO) model considers three sets of constraints that must be satisfied so that optimal feasible solution is achieved.

• Relay settings:

• Relay operating time:

• Relay Coordination:

iTDSTDSTDS iii maxmin

iIII pipipi maxmin

},{,, min xjittt ij

bx

ij

p

ij

CTItt p

j

bx

j },{ xj

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Simulation Results 15

Relay TDS (S) Ip (pu) Relay TDS (s) Ip (pu) 1 0.050 0.826 9 0.050 0.617

2 0.050 0.352 10 0.050 0.264

3 0.050 0.555 11 0.050 0.648

4 0.150 0.034 12 0.050 0.531

5 0.050 0.694 13 0.110 0.089

6 0.050 0.132 14 0.050 0.396

7 0.050 1.075 15 0.050 0.471

8 0.050 0.230 16 0.050 0.329

T(s)= 19.1675

Optimal TDS and Ip DOCR Settings

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Simulation Results 16

Optimal primary and backup relay operating times

Fault location

Operating times of relays (s) (p= primary, b= backup) P b1 b2

F8 R1: 0.3332 R2: 0.2856

R4: 1.4532 R11: 0.5856

R6: 0.6332 -

F9 R5: 0.2573 R6: 0.1783

R2: 0.5573 R13: 0.8145

R4: 0.5573 R16: 0.4783

F10 R3: 0.2444 R4: 0.3388

R2: 0.6652 R14:0.6388

R6: 0.9203 -

F11 R13: 0.3120 R14: 0.2335

R3: 0.6120 R5: 0.6979

- R16: 0.5908

F12 R15: 0.2868 R16: 0.2797

R5: 0.5868 R9: 0.5797

R13: 0.5868 -

F13 R11: 0.3262 R12: 0.3537

R7: 0.6262 R1: 0.6537

- -

F14 R9: 0.2663

R10: 0.2944

R8: 0.5663 R15: 0.5944

- -

F15 R7: 0.3582 R8:0.2313

R10: 0.6582 R12: 0.5313

- -

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Conclusion 17

• This paper proposes a method based on superposition theory that can be used to calculate the three-phase fault currents in a meshed distribution system equipped with IBDG units.

• The methodology is based on the fact that the current controlled IBDG units inject a specific amount of fault current that does not exceed 2p.u depending on the controller’s limiters (modelled as constant current source).

• The proposed method was tested and achieved reasonable accuracy when verified against a transient model.

• Finally, the fault currents computed using the proposed method were utilized by the CPC model developed to solve the protection coordination problem.

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Thank you for your Attention

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