Scroll Top
10 Old Grimsbury Rd, Banbury OX16 3HG, UK

MHH623556 Power Systems Analysis and Protection – Tri B – 2022 / 2023MHH623556

Coursework 2: Protection Study of a Radial Power System   

Tasks:

Using appropriate design rules (reference source(s)) set up the relays to achieve the desired discrimination and back-up. This can be achieved through adjusting every relay parameter as explained in the lecture and during the CW introduction.

Test your system by running appropriate protection studies (for three phase balanced faults only)

For the CW2 Submit a report consisting of solutions for questions in part 1 and 2:

Part 1: ERACS modelling

  1. The system parameters are as follows: Grid short circuit level is 950 MVA. Transformer 115kV/13.2 kV, impedance = 6% pu. Transmission line impedance is 70.5 ohm at 115 kV, and is 0.93 ohm at 13.2 kV. Calculate using the concepts you have learned, the nominal and short circuit current of every relay unit.             [30]
  2. A short summary of the adopted design rules and how they were applied including all essential calculations based on load flow simulations.                                              [20]
  3. A printout from ERACS of your relay setting.                                                      [20]
  4. A combined i-t graph for the 4 relays, also extracted from ERACS                   [10]
  5. A summary of your protection study, illustrating correct operation of the relay [20]

           

Part 2: Digitalization of substation

Substation digitalization involves introducing electronics and communication network into the operation of different apparatus in a substation.

  1. Explain which aspects of substation performance can benefit from this technology and how? (in terms of protection devices, data transfer, monitoring, etc.)                                                                                                             [10]
  2. Explain the differences between the conventional and digital protection system.

                                                                                                                  [10]

Note: please answer the part 2 less than 1 page for both parts together. Include at least two scientific references that you have reviewed for your research on this topic.

Solution

Report on Protection Study of a Radial Power System

  1. Introduction

The network comprises of a Grid infeed at 115kV connecting a radial distribution system via a 30 MVA, 115 kV/13.2 kV transformer, an OHL, and a total load of 23.6 MVA as shown in Figure-1.

The objective of this report is to study the protection of the radial power system mentioned above, using non-directional IDMT relays and to achieve desired discrimination and backup through current and time grading.

The network schematic and the input data are shown in the figure below. Relay-1 is connected to one of the outgoing load feeders at Bus C, Relay-2 is connected to the overhead line at Bus B, Relay-3 is connected to the secondary of the transformer at Bus B, and Relay-4 is connected to the primary of the transformer at Bus A.

Figure-1: Radial network schematic

  1. Part 1: ERACS Modelling
  2. Relay Nominal and Short circuit calculation
  3. Nominal Current

Starting with Relay-1, the nominal current through the relay is the load current as calculated by,

Inc1=Sload/(√3xVbusC) = 4.2×106/(√3×13.2×103) = 183.7A

For Relay-2 at Bus B, the nominal current is the total load current feeding all the loads connected at Bus C. From Figure-1, there are three equal loads rated at 4.2MVA each are connected at Bus-C. Hence, the nominal current through Relay-2 is given by

Inc2=3*Sload/(√3xVbusC) = 3×4.2×106/(√3×13.2×103) = 551.1A

For Relay-3, the nominal current is the transformer secondary current at 13.2kV side, calculated by

Inc3=Strafo/(√3xVsec) = 30×106/(√3×13.2×103) = 1312.16A

At Bus B, a load of 11MVA is connected and the load current is given by,

IloadB= SloadB/(√3xVbusB) = 11×106/(√3×13.2×103) = 481.13A

The total load current at Bus B = 551.1+481.13 = 1032.23A which is less than the rated secondary current of transformer of 1312.16A. Hence, the relay nominal current is calculated as transformer secondary current to protect the transformer.

For Relay-4, the nominal current is the transformer primary current at 115kV side, calculated by

Inc4=Strafo/(√3xVsec) = 30×106/(√3x115x103) = 150.61A

  • Short Circuit Current

To calculate the short circuit through the relays, it is required to derive the positive sequence  impedances (reactance) for the radial network in Figure-1. The calculation is shown below.

Impedance calculation referred to 115kV side:

Zsource = V2/SCMVAsource = (115×103/√3)2/(950×106/3) = 13.92 Ω

ZTF30 = ZtfpuxZTFMVA = 0.06x(115×103)2/(30×106) = 26.45 Ω

Zline =  70.5 Ω

The source is connected to Bus A with an impedance of 13.92 Ω and as the network is radial, the faults at Bus A, B and C are fed only from the source. Hence, the short circuit currents for the relays shown in Figure-1 are calculated as below.

Relay 4, ISC-A = Vprefault/Zeq = (115×103/√3)/ 13.92 = 4769.78 A

Relay 3 and 2, ISC-B = Vprefault/Zeq = (115×103/√3)/ (13.92+26.45) = 1644.67 A

Relay 1, ISC-C = Vprefault/Zeq = (115×103/√3)/ (13.92+26.45+70.5) = 598.86 A

The above short circuit currents are as referred to 115kV. The short circuit currents when referred to 13.2kV are given as,

Relay 3 and 2, ISC-B = 1644.67×115/13.2 = 14328.56 A (referred to 13.2kV)

Relay 1, ISC-C = 598.86×115/13.2 = 5217.34 A (referred to 13.2kV)

Relay/BusCalculated Nominal current (A)Calculated Short Circuit Current (A)
Relay-1/ Bus C183.7598.86 (115kV) / 5217.34 (13.2kV)
Relay-2/ Bus B551.71644.67 (115kV) / 14328.56 (13.2kV)
Relay-3/ Bus B1312.161644.67 (115kV) / 14328.56 (13.2kV)
Relay-4/ Bus A150.614769.78 (115kV)

The results from ERACS simulation are given in the table below and the ERACS results are included in the Appendices A and B.

Relay/BusERACS LF current (A)ERACS 3Ph Short Circuit Current (A)
Relay-1/ Bus C1975226 (13.2kV)
Relay-2/ Bus B59214476 (13.2kV)
Relay-3/ Bus B108814476 (13.2kV)
Relay-4/ Bus A1254769 (115kV)
  1. CT Ratio

The CT ratios for the Relays in the radial network is calculated in this section.

The CT ratio is n:1, a margin of 10% is considered on the highest current between Inom and ISC to decide the CT ratio. The CT saturation is considered as 20p.u.

 Relay/BusNominal current, Inom (A)Calculated Short Circuit Current, ISC(A)ISC/20 (A)With 10% margin on highest current (Inom, ISC), (A)CT Ratio chosen
Relay-1/ Bus C183.75217.34 (13.2kV)260.87286.96300/1
Relay-2/ Bus B551.714328.56 (13.2kV)716.43788.07800/1
Relay-3/ Bus B1312.1614328.56 (13.2kV)716.431443.371500/1
Relay-4/ Bus A150.614769.78238.49262.34300/1

ii. Design Rules and Relay Settings Calculation

The design rules adopted for the settings of the relays is as below:

  • The overload factor for transformer primary (Relay 4)  is considered as 125% of Inom, whereas the instantaneous pickup is considered as 130%  of the SC current on LV side referred to HV side. The overload factor for transformer secondary (Relay 3)  is considered as 125% and the instantaneous pickup is considered as 8 times the rated current.
  • The overload factor for 13.2kV line (Relay 2) is considered as 120% and the instantaneous pickup is 6 times the relay nominal current.
  • The overload factor for load feeder (Relay 1)  is considered as 120% and the instantaneous pickup is 6 times the relay nominal current.
  • The time discrimination margin between the relays is considered as 250ms.
  • The relay curve selected is Normal Inverse (NI)

The relay settings calculation is explained below:

Relay 1: 13.2kV Load feeder

Overload Pickup Setting: (I >):

  • PUR1 (I>) = OLFx Inc1/(CT Ratio) = 1.2×183.7/(300/1) = 0.735, set to 0.75 in ERACS for Relay-1.

Time Dial Setting (TDS) (t >):

Plug setting multiplier for Relay-1, PSMR1 = Fault current/ (PUR1xCT Ratio) = 5226/(0.75×300)  = 23.23

(The PSM of 23.23 is greater than CT PU saturation of 20p.u., hence, the PSMR1 = 20)

Assuming TDS of Relay-1, TDSR1 (t>) = 0.05 (From ERACS Relay curve)

Operating time of relay for normal inverse curve, tR1 = 0.14/((PSM^0.02)-1) = 0.14/(( 20^0.02)-1) = 2.267s

Actual operating time of relay, topR1 = operating time of relay*TDS = 2.267×0.05 = 0.1133s (113.3ms)

Instantaneous Pickup Setting (I >>):

Instantaneous pickup, INSTR1(I>>) = 6xInc1/(CT Ratio) = 6×183.7/(300/1) = 3.67, set to 3.5 in ERACS

Instantaneous Time Dial Setting (TDS) (t >>):

Setting the TDS for instantaneous unit at 0.05s, i.e., TDS (t>>) = 0.05 (From ERACS Relay curve)

Relay 2: 13.2kV Distribution feeder (To coordinate with Relay-1 for 3Ph fault at Load, fault current of 5226A)

For fault at Load connected to Bus C, fault current of 5226A and with above settings calculated for Relay-1, the operating time is 0.113s.

Now, Relay-2 should act as backup to Relay-1 with discrimination margin of 0.25s for the same fault current of 5226A.

With a discrimination margin of 0.25s, the Relay-2 should operate at topR2 = 0.113+0.25 = 0.363s on NI.

Overload Pickup Setting: (I >):

PUR2 (I>) = OLFx Inc2/(CT Ratio) = 1.2×551.7/(800/1) = 0.827, set to 0.85 in ERACS for Relay 2.

Time Dial Setting (TDS) (t >):

Plug setting multiplier for Relay-2, PSMR2 = Fault current/ (PUR2xCT Ratio) = 5226/(0.85×800) = 7.685

Calculating the TDS (t>) for Relay-2,

TDSR2 = topR2x((PSM^0.02)-1)/0.14 = 0.363x( 7.68^0.02)-1)/0.14 = 0.108

Setting the TDSR2 to the next available setting in ERACS, i.e., TDSR2 (t>) = 0.125

Now with Pickup setting of 0.85 and TDS of 0.125, the time of operation for remote fault current of 5226A is topR2 = TDS x 0.14/((PSM^0.02)-1) = 0.125x 0.14/((7.685^0.02)-1) = 0.42s

Also, with Pickup setting of 0.85 and TDS of 0.125, the time of operation for close-in fault current of 14476A is topR2 = TDS x 0.14/((PSM^0.02)-1) = 0.125x 0.14/((20^0.02)-1) = 0.283s

(The PSM of 21.28 is greater than CT PU saturation of 20p.u. for fault current of 14476A, hence, the PSMR2 = 20)

Instantaneous Pickup Setting (I >>):

Instantaneous pickup, INSTR2(I>>) = 6xInc2/(CT Ratio) = 6×551.7/(800/1) = 4.13, set to 4.5 in ERACS

Instantaneous Time Dial Setting (TDS) (t >>):

Coordinating with Relay-1 TDS, the TDS for Relay-2 with 0.25s discrimination margin, TDSR2 (t>>)= 0.05+0.25 = 0.3s

Relay 3: Transformer Secondary 13.2kV (To coordinate with Relay-2 for 3Ph fault on 13.2kV line, fault current of 14476A)

For fault at line sending end, fault current of 14476A and with above settings calculated for Relay-2, the operating time, topR2 is 0.283s.

Now, Relay-3 should act as backup to Relay-2 with discrimination margin of 0.25s for the same fault current of 14476A.

With a discrimination margin of 0.25s, the Relay-3 should operate at topR3 = 0.283+0.25 = 0.533s on NI.

Overload Pickup Setting: (I >):

PUR3 (I>) = OLFx Inc3/(CT Ratio) = 1.25×1312.16/(1500/1) = 1.093, set to 1.1 in ERACS for Relay 3.

Time Dial Setting (TDS) (t >):

Plug setting multiplier for Relay-3, PSMR3 = Fault current/ (PUR3xCT Ratio) = 14476/(1.1×1500) = 8.773

Calculating the TDS (t>) for Relay-3,

TDSR3 = topR3x((PSM^0.02)-1)/0.14 = 0.533x( 8.773^0.02)-1)/0.14 = 0.169

Setting the TDSR3 to the next available setting in ERACS, i.e., TDSR3 (t>) = 0.175

Now with Pickup setting of 1.1 and TDS of 0.175, the time of operation for remote and close in fault current of 14476A is topR3 = TDS x 0.14/((PSM^0.02)-1) = 0.175x 0.14/((8.773^0.02)-1) = 0.54s

Instantaneous Pickup Setting (I >>):

Instantaneous pickup, INSTR3(I>>) = 8xInc3/(CT Ratio) = 6×1312.16/(1500/1) = 5.24, set to 5.5 in ERACS

Instantaneous Time Dial Setting (TDS) (t >>):

Coordinating with Relay-2 TDS, the TDS for Relay-3 with 0.25s discrimination margin, TDSR3 (t>>)= 0.3+0.25 = 0.55s

Relay 4: Transformer Primary 115kV (To coordinate with Relay-3 for fault on transformer secondary, fault current of 14476A at 13.2kV)

For fault at transformer secondary, fault current of 14476A and with above settings calculated for Relay-3, the operating time topR3 is 0.54s.

Now, Relay-4 should act as backup to Relay-3 with discrimination margin of 0.25s for the same fault current of 14476A at 13.2kV which corresponds to 1661.6A at 115kV.

With a discrimination margin of 0.25s, the Relay-4 should operate at topR4 = 0.54+0.25 = 0.79s on NI.

Overload Pickup Setting: (I >):

PUR4 (I>) = OLFx Inc4/(CT Ratio) = 1.25×150.61/(300/1) = 0.62, set to 0.65 in ERACS for Relay 4.

Time Dial Setting (TDS) (t >):

Plug setting multiplier for Relay-4, PSMR4 = Fault current/ (PUR4xCT Ratio) = 1661.6/(0.65×300) = 8.52

Calculating the TDS (t>) for Relay-4,

TDSR4 = topR4x((PSM^0.02)-1)/0.14 = 0.79x( 8.52^0.02)-1)/0.14 = 0.247

Setting the TDSR4 to the next available setting in ERACS, i.e., TDSR4 (t>) = 0.275

Now with Pickup setting of 0.65 and TDS of 0.275, the time of operation for remote fault current of 1661.6A is topR4 = TDS x 0.14/((PSM^0.02)-1) = 0.275x 0.14/((8.52^0.02)-1) = 0.87s

Now with Pickup setting of 0.65 and TDS of 0.275, the time of operation for close in fault current of 4769A is topR4 = TDS x 0.14/((PSM^0.02)-1) = 0.275x 0.14/((20^0.02)-1) = 0.623s

(The PSM of 24.45 is greater than CT PU saturation of 20p.u. for fault current of 4769A, hence, the PSMR4 = 20)

Instantaneous Pickup Setting (I >>):

Instantaneous pickup, INSTR4(I>>) =1.3xFault current at LV referred to HV/(CT Ratio) = 1.3×1661.6/(300/1) = 7.2, set to 7.5 in ERACS.

Instantaneous Time Dial Setting (TDS) (t >>):

This setting for Relay-4 is set at TDS (t>>) is 0.1 as this function will not operate for fault on transformer secondary on DT.

iii. Relay Setting Print from ERACS

Relay (Curve)OCDT
Relay-1 (NI)I>0.75I>>3.5
t>0.05t>>0.1
Relay-2 (NI)I>0.85I>>4.5
t>0.125t>>0.3
Relay-3 (NI)I>1.1I>>5.5
 t>0.175t>>0.55
Relay-4 (NI)I>0.65I>>7.5
 t>0.275t>>0.1

               The relay settings print out from ERACS software is included in Appendix-C.

iv. Combined i-t Graph from ERACS

The combined i-t graph from the ERACS software for all the 4 Relays in the radial network is included in Appendix D.

v. Protection Study Summary

The operation of relays for faults at various location in the radial network is explained below. The coordination graphs are included in Appendix-E.

  • 3Ph Fault at Load connected to Bus C – Coordination between Relay-1 and Relay-2

For a 3Ph fault at load connected to Bus-C, the fault current seen by the Relay-1 connected to the load feeder is 5226A which trips instantaneously at 0.05s. Also, if the Relay-1 fails to operate, then the Relay-2 which is a back-up to Relay-1 will operate at 0.3s to isolate the fault.

Further coordinating to upstream Relays for the same fault current, Relay-3 is operated at 1.021s and Relay-4 operates at 1.645s. This can be seen in the figure below.

From the figure, it is evident that all Relays operate with adequate margin for a fault in the far end of the radial network.

  • 3Ph Fault at 13.2kV line sending end – Coordination between Relay-2 and Relay-3

For a 3Ph fault at sending end of the 13.2kV line, close to Bus B, the Relay-2 operates at 0.3s for a fault current of 14350A and Relay-3 acts as a back up which operates in 0.55s as shown in the figure below.

  • 3Ph Fault at Transformer Secondary/Bus-B (13.2kV) – Coordination between Relay-3 and Relay-4

For a fault on transformer secondary side or Bus B, the Relay-3 operates at 0.55s clearing a fault current of 14476A and the Relay-4 acts as a back-up to Relay-3 which operates at 0.879s as shown in figure below.

III. Part 2: Digitalization of Substation

i. Substation Performance Benefits:

A digital substation benefits in being smaller in size, increased reliability, easy maintenance, less operational cost and easy to widen than a conventional one. It offers increased safety and is more efficient than its traditional equivalent.

  • By replacing the existing electromechanical/static protection with the numerical relays, the system availability is improved as these relays provide the advantage of connecting to the communication network in real time and further can be connected to the higher-level system such as substation automation, SCADA which allows monitoring of all secondary equipment connected to the system. 
  • The safety of the system and personnel is increased by the utilization of remote operation facility which is combined with rule-based access and authority and remote testing. 
  • The functionality of the system is improved with a fully distributed architecture. It is coupled with communication which is unrestricted, and it makes it possible to include advanced functions easily with minimal outage requirements because of the process capability. This benefits the user by providing additional safety and security to the  power system.
  • Using numerical/digital relays along with deploying IEC 61850 complaint solutions, interoperability with other manufacturer’s equipment can be achieved.

ii. Comparison between Conventional and Digital Protection:

Conventional ProtectionDigital Protection
Utilizes electromechanical and static relaysUtilizes digital and numerical relays
Relays consist of analogue electronic devices to create the relay characteristic or work based on actuating force created by electromagnetic interactionThese relays use analogue to digital conversion of all measured quantities and use microprocessor to implement protection algorithm
Limited functional settings, accuracy compared to digitalRelays have wide range of functions and settings and greater accuracy
Power system measurements not availableAble to monitor and measure power system parameters
No communication between the relaysCommunication between the relays to improve reliability and speed-up operation

Other advantages of digital relays over conventional relays are listed below.

  • Post-fault analysis of observed transient phenomena
  • Settings controllable by hierarchical systems
  • Post-fault calculation of a distance to a fault or fault locator
  • Inbuilt backup protections
  • Consistency of operation times – reduced grading margin
  • Circuit Breaker state and condition monitoring
  • User definable logic

References:

  • T. Werner, A. Meier, “Performance considerations in digital substation applications”, PAC World conference Glasgow, 2015
  • Bogdan Kasztenny, Eugeniusz Rosolowski, “A Digital Protective Relay as a Real-Time Microprocessor System”, Proceedings of the 1997 Workshop on Engineering of Computer-Based Systems (ECBS ’97) 0-8186-7889-5/97
  • Network Protection and Automation Guide – Protective Relays, Measurement and Control, Edition May 2011, Alstom Grid

Appendix-A: ERACS Load Flow Result

Appendix B: ERACS 3Ph SC Results

Appendix C: ERACS Relay Settings  Report

Appendix D: 4 Relays Combined i-t Graph

Appendix E: ERACS OC Relay Coordination Graphs

Related Posts

Leave a comment

× WhatsApp Us