A presentation explaining how to calculate fault currents for 3-phase or 1-phase faults in power grid. Particularly useful for engineers working in electrical power transmission company.
Since the loads having the trends towards growing density. This requires the better appearance, rugged construction, greater service reliability and increased safety. An underground cable essentially consists of one or more conductors covered with suitable insulation and surrounded by a protecting cover. The interference from external disturbances like storms, lightening, ice, trees etc. should be reduced to achieve trouble free service. The cables may be buried directly in the ground, or may be installed in ducts buried in the ground.
It covers protection of low voltage and medium voltage motors through the use of contactors, switches, fuses, MCC, reduced voltage motor starting techniques and motor protection relays.
Since the loads having the trends towards growing density. This requires the better appearance, rugged construction, greater service reliability and increased safety. An underground cable essentially consists of one or more conductors covered with suitable insulation and surrounded by a protecting cover. The interference from external disturbances like storms, lightening, ice, trees etc. should be reduced to achieve trouble free service. The cables may be buried directly in the ground, or may be installed in ducts buried in the ground.
It covers protection of low voltage and medium voltage motors through the use of contactors, switches, fuses, MCC, reduced voltage motor starting techniques and motor protection relays.
This directional over current relay employs the principle of actuation of the relay....It has a metallic disc free to rotate between the poles of two...
Why do Transformers Fail?
�The electrical windings and the magnetic core in a transformer are subject to a number of different forces during operation, for example:
�Expansion and contraction due to thermal cycling
�Vibration
�Local heating due to magnetic flux
�Impact forces due to through-fault current
�Excessive heating due to overloading or inadequate cooling
Installation, Testing and Troubleshooting of TransformersLiving Online
This will provide you with practical knowledge (including tips, tricks and tools) covering the fundamentals of power transformers and their testing. It will greatly assist you in communicating more effectively with your electrical engineering colleagues. At the end of this workshop, participants will be familiar with the importance of transformer testing and their purpose, the different kinds of transformer tests and their procedures and the practical applications of principals applied in transformer operation and maintenance.
WHO SHOULD ATTEND?
This workshop will be appropriate for the following professionals:
Electrical engineers
Maintenance engineers
Maintenance supervisors
Power electricians
Power engineers
MORE INFORMATION: http://www.idc-online.com/content/installation-testing-and-troubleshooting-transformers-24
How is power transformer protected??? This provides a basic understanding of power transformer. Furthermore, the protective relay application on power transformer is included.
• modelling of the electric railway system including locomotives;
• influence of the electric railway system on power quality in the transmission system simulations and power quality measurements);
• modelling of reactive power compensation for electric railway systems and analysis of switching transients;
• influence of the electric railway system on pipelines and telecommunication cables.
This directional over current relay employs the principle of actuation of the relay....It has a metallic disc free to rotate between the poles of two...
Why do Transformers Fail?
�The electrical windings and the magnetic core in a transformer are subject to a number of different forces during operation, for example:
�Expansion and contraction due to thermal cycling
�Vibration
�Local heating due to magnetic flux
�Impact forces due to through-fault current
�Excessive heating due to overloading or inadequate cooling
Installation, Testing and Troubleshooting of TransformersLiving Online
This will provide you with practical knowledge (including tips, tricks and tools) covering the fundamentals of power transformers and their testing. It will greatly assist you in communicating more effectively with your electrical engineering colleagues. At the end of this workshop, participants will be familiar with the importance of transformer testing and their purpose, the different kinds of transformer tests and their procedures and the practical applications of principals applied in transformer operation and maintenance.
WHO SHOULD ATTEND?
This workshop will be appropriate for the following professionals:
Electrical engineers
Maintenance engineers
Maintenance supervisors
Power electricians
Power engineers
MORE INFORMATION: http://www.idc-online.com/content/installation-testing-and-troubleshooting-transformers-24
How is power transformer protected??? This provides a basic understanding of power transformer. Furthermore, the protective relay application on power transformer is included.
• modelling of the electric railway system including locomotives;
• influence of the electric railway system on power quality in the transmission system simulations and power quality measurements);
• modelling of reactive power compensation for electric railway systems and analysis of switching transients;
• influence of the electric railway system on pipelines and telecommunication cables.
This paper proposes fault location model for underground power cable using microcontroller. The aim of this project is to determine the di stance of underground cable fault from base station in kilometers. This project uses the simple c oncept of ohm�s law.When any fault like short circuit occurs,voltage drop will vary depending on the length of fault in cable,since the current varies. A set of resistors are ther efore used to represen t the cable and a dc vol tage is fed at one end and the fault is detected by detecting the change in voltage using a analog to voltage converter and a microcontroller is used to make the necessary calculations so that the fault distance is displayed on the LCD display.
Voltage Profile Improvement using Switched Capacitors: Case of Single Wire Ea...IJMERJOURNAL
ABSTRACT: Most rural areas in Africa are characterized by scattered villages with a very low demand in electricity. Due to improper planning and lack of knowledge on low cost technologies, the cost of extending the grid to supply these area is very high relative to the returns. Rural electrification by means of extending the main grid and distributing power using a single wire with earth return (SWER) has shown to be the least expensive rural electrification method in remote area where loads are light and scattered.This paper presents a developed model of Single wire earth return distribution network and a voltage profile of the network using backward and forward sweep method load flow algorithm. And finally presents the analysis of the effect of shunt capacitors on the voltage profile of the network using Maximum power saving method for the sizing and placement of the capacitor.
Partial discharge monitoring of High Voltage Assets - TATA Steel Scunthorpe Mikolaj Kukawski
Extensive Presentation with 14 Case Studies from High Voltage Assets Preventative Maintenance by Non-intrusive testing using Partial Discharge detection and monitoring equipment. This Presentation contains: Author's Key Facts, Integrated Steelworks Plants brief overview of processes and Biggest Private HV Network in the UK Single Line Diagram - to highlight importance of HV Assets importance to the whole site uninterrupted operations, 14 case studies showing various High Voltage Switchgear and Transformers Partial Discharge issues detected, located and eliminated by prompt action.
EXPERIMENTAL DETERMINATION AND ANALYSIS OF TRANSMISSION LINE PERFORMANCEijiert bestjournal
It is necessary to calculate the voltage,current a nd power at any point on a transmission line provided the values at one point are known. We are aware that in three phase circuit problems it is sufficient to compute results in one phase and subsequently predict results in the other two phases by exploiting the three phase symm etry. Although the lines are not spaced equilaterally and not transposed,the resulting asy mmetry is slight and the phases are considered to be balanced. As such the transmission line calcu lations are also carried out on per phase basis. For that purpose in the transmission line demo pane l we will be designed to �To study the performance of the line.1)e.g. Relation between sen ding end quantity and receiving end quantity,Ferranti effect,efficiency of power line etc.2) To demonstrate fault clearing process using distance relay(Future Scope)
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
1. Today’s Session
• Fault Level Calculations
– Power-system Overview
– PU Concept
– 3 Ph. Fault Level Calculations
– 1 Ph Fault Level Calculations
• O/C E/F Relay Time Coordination
– General information about relay characteristic
– Worked out example
• Directional Relay
– What is MTA
– How to confirm direction of relay reach while in
service
What is ahead?
3. Fault Level Calculations 3
What is ahead?
• Overview of transmission network.
• What is mean by fault level
• PU Concept
• Network Modeling
• Fault Level Calculation & Worked out
example
– Three Phase Fault
– SLG Fault
7. Fault Level Calculations 7
400kV
220kV
220kV
132kV
33kV
33kV220kV
132kV
11kV33kV
S/S - A
S/S - F S/S - E
S/S - B
S/S - C
S/S - D
Local Network – or – Sub-Transmission System
12. Fault Level Calculations 12
PU Concept
• PU impedance values are calculated by
– For Transformers
– For Lines
%ImpedanceNew Base MVA = % ImpedanceName Plate
Base MVA
Capacity in MVA
%Impedance = Ohmic Value
Base MVA
kV 2
13. Fault Level Calculations 13
PU Concept
• PU impedance values are calculated by
– For Transformers
• %Impedance = Ohmic Value * ( Base MVA / Name Plate MVA )
– For Lines
• %Impedance = Ohmic Value * ( Base MVA / BasekV2
)
14. Fault Level Calculations 14
PU Concept
400/220 kV
15% 315 MVA
12 Ohm
Sec DSec CSec B
0.4
ohm/Km
200 Km
0.4
ohm/Km
50 Km220/132 kV
14% 200 MVA
132/33 kV
9.33% 25 MVA
0.4
ohm/Km
30 Km
Sec A
Base MVA = 200 MVA
400kV 33kV132kV220kV
T1 L1 T2 L2 T3 L3
17. Fault Level Calculations 17
Short Circuit Current
• 3 Phase Short Circuit MVA at fault point
Base MVA x 100
= ----------------------------------------------------
Percentage Impedance (from source to fault point)
21. Fault Level Calculations 21
Typical State Level Power System & our area of interest
4630
Base MVA=
Xs(pu)
MVA
100 Xs(pu)
4630
Xs (pu) = 0.0216
Or
Xs% = 2.16%
24. Fault Level Calculations 24
Unsymmetrical Faults-Sequence Network
• Up till now we have discussed 3 Phase
faults which are rear. Most of the fault in
power system are single line to ground
faults.
• To find out the short circuit MVA due to
single phase to ground fault it is necessary
to use Sequence Network
• Let us have a closer look of sequence
network
26. Fault Level Calculations 26
Single Line to Ground Fault-Sequence Network Connections
Xg1XT11XL1XT21
XM21XM11
F
Xg2XT12XL2
XT22
XM22XM12
F
Xg0XT10XL0XT20
XM20
F
X1
X2
X0
Ia = 3 / (X1 + X2 + X0)
Typical state level power system is shown here.
This power system is consists of Generator, EHV lines, ICTs, Transformer, Distribution Feeders etc,
Power generated get injected into the system either at 400kV or at 220kV.
We can say that major portion of the power system is of 400kV Grid. (Shown violate)
Some portion is of 220kV Grid (Shown brown)
Generally most of engineers deals with small portion of the network shown inside. Which we may call as Sub-Transmission network.
Majority portion of sub-transmission network is of radial nature.
While dealing with such limited portion of the system from fault level calculation point of view we can replace the entire external grid system with a source and a thevinians equivalent impedance.
Here grid is replaced by thevinans equivalent impedance.
Or to be more specific, while calculating fault levels we can neglect resistance of the elements
Thus this impedance is considered as reactance
Then next obvious question is how to determine this equivalent reactance. It is explained afterword.
But before to proceed further let us have a closer look at Sub-Transmission network
And it is necessary to recall our concepts about pu calculations
Here sub-transmission network enlarged ( and rotated clockwise 900)
Though following points are un-important or known it is necessary to have common language while discussing some issues/problems regarding power system. Hence such terminology is reviewed here.
S/S voltage levels means highest voltage level at that S/S Hence
400kV S/S – A
200kVS/S – B
S/S – C
132kVS/S – D
S/S – E
S/S – F
Sub-Stations are also recognized as 400/220kV or 220/33 kV or 220/132kV or 132/33kV or 132/33-11kV (S/S-F) etc.
400kV and one 220kV line of S/S-A is in grid.
S/S-B and S/S-C get supplied through parallel lines.
S/S-D and S/S-E are connected in local loop network (S/S-D is having LILO arrangement).
S/S-F is of radial nature
There may have standby lines too.
We may define fault level as -
Fault level at any location is that value of current which is expected to flow upon occurrence of fault considering all other elements in network are in service and all generations are in service and generating their maximum output.
FOR FAULT LEVEL STUDY SELECTED LOCATION IS BUS IN THE SYSTEM
Every year such calculations are done for the entire network of the company at corporate level using computer programs and results circulated to field offices. Generally these fault levels are expressed in MVA or KA. When expressed in MVA it is also called as Short Circuit MVA.
Such a sample result is shown here. Generally while calculating fault levels, two type of faults are considered at every bus in the network (Grid).
3 Phase Fault
Single Line to Ground Fault (SLG)
Here 3 Ph fault is considered. Fault levels for S/S connected to generation are more (S/S-A). Also fault levels having more lines are more (S/S-B). Fault levels of S/S having higher rating transformers are also more
Let us have a closer look of S/S-B
During fault level study not only bus fault levels are given but also contribution of this fault by various elements connected to that bus are also given.
Not that arrow marking is not necessary. It is only shown here to get the contribution more clear.
Such type of data can be given in SLD form as well as Table form too.
However if given in SLD form it is more understandable
================================================================================
Note :- we are using pu impedance and percentage impedance interchangeably throughout the discussion where
Percentage Impedance = Per-Unit Impedance * 100
================================================================================
While modeling a transmission network for fault level study following points are important.
Transmission system (or network) is made up of lines and transformers.
Line is characterized by inductance, resistance and line charging capacitance.
For small length lines, charging capacitance can be neglected while modeling the transmission system. Small length lines are generally of lower voltage. From transmission point of view that is 132kV and 220kV.
While modeling the 400kV line we have to consider line charging capacitance too.
For fault level calculations only it is necessary to consider reactive impedance of the line. We can exclude line charging and resistance.
Transformer impedances are always expressed in percentage (or PU ).
================================================================================
The per-unit impedance of the transformer is the same regardless of whether it is determined from ohmic values referred to high tension or low tension sides of the transformers.
For example consider a 132/33 kV 25 MVA transformer under short circuit test. That means about 3 phase 440 V (i.e. 440/1.732 = 254 V Ph-N) supply is given from 132kV side and 33kV side kept short circuited. This transformer short circuit current at 440 V will be about 3.9 amps. That is its leakage impedance is 254/3.9 = 65.12 Ohms
Its PU value (referred to 132kV side) is
65*25/(132*132) = 0.0933 ( or 9.33%)
If the leakage reactance had been measured on the low tension side, the value would be
65/16 = 4.0625 ( Where 16 is square of transformation ratio (132/33)2 )
Its PU value (referred to 33kV side) is
4.0625*25/(33*33) = 0.0933 (or 9.33%) same as before
Same case is about line modeling where base voltages for the lines connected on HV and LV sides of the transformers shall have same ratio as that of transformation ratio
In short while calculating per-unit impedance of the line it’s rated voltage is selected as base voltage and base MVA will be common for entire system.
The per-unit impedance of the transformer is the same regardless of whether it is determined from ohmic values referred to high tension or low tension sides of the transformers.
For example consider a 132/33 kV 25 MVA transformer under short circuit test. That means about 3 phase 440 V (i.e. 440/1.732 = 254 V Ph-N) supply is given from 132kV side and 33kV side kept short circuited. This transformer short circuit current at 440 V will be about 3.9 amps. That is its leakage impedance is 254/3.9 = 65.12 Ohms
Its PU value (referred to 132kV side) is
65*25/(132*132) = 0.0933 ( or 9.33%)
If the leakage reactance had been measured on the low tension side, the value would be
65/16 = 4.0625 ( Where 16 is square of transformation ratio (132/33)2 )
Its PU value (referred to 33kV side) is
4.0625*25/(33*33) = 0.0933 (or 9.33%) same as before
Same case is about line modeling where base voltages for the lines connected on HV and LV sides of the transformers shall have same ratio as that of transformation ratio
In short while calculating per-unit impedance of the line it’s rated voltage is selected as base voltage and base MVA will be common for entire system.
This will be clear by an example consider a simple system as shown above of 400/220kV T/F 220/132kV Transformer and 132/33kV Transformer.
Select base MVA of 200 MVA for entire system.
Sectionalize the system as per voltage levels
Section A is for 400kV
Transformer T1 will get included in this section as pu values of transformer whether referred to primary or secondary are same
Section B is for 220kV
Transformer T1 and also transformer T2 will get included in this section due to same reason as said above. That means all transformers get overlapped by two sections
Similarly section C is for 132kV and section D is for 33kV
Select 400kV as base voltage in section A
It will be converted into 220kV in section B, 132kV in section C and 33kV in section D
===========================================================================
By this way line rated voltage will be base voltage for that line
===========================================================================
Now it is possible to calculate per-unit values of transformer and lines with above said base mva and base kv
T1 Percentage Impedance = XT1% = 15*200/315 = 9.52%
L1 Percentage Impedance = XL1% = (X/Km)*L*( Base MVA / ( kV2 ) )*100
=0.4*200*(200/(220*220))*100 = 33.05%
T2 Percentage Impedance = XT2% = 14%
L2 Percentage Impedance = XL2% = 0.4*50*(200/(132*132))*100 = 22.96%
T3 Percentage Impedance = XT3% = 9.33 * 200/25 = 74.64%
L3 Percentage Impedance = XL3% = 0.4*30*(200/(33*33))*100 = 220.38%
Total percentage impedance up to source
220 + 74.64 + 22.96 + 14 + 33.05 + 9.52 = 374.17%
Short circuit MVA = 200*100/374.17 = 53.54 MVA
Short Circuit Current = 17.5*53.54 = 936 Amp
Same results we get by considering ohmic values. How ever every time it is necessary to take care of transformation ratio hence pu method is always preferable. More ever when we works with large transmission grid extensive use of matrix is required and there is no alternative to pu values. Still we will calculate same results by ohmic method just to realize simplicity of pu method.
12 ohm referred to 132kV = 12*(4*4) = 192 Ohms ( Where 4 is transformation ratio 132/33kV)
L3 + T3 + L2
65 + 192 + 20 = 277 ohms
This referred to 220kV = 277*(1.6666*1.6666) = 769 Ohms ( Where 1.6666 is transformation ratio 220/132kV)
(L3 + T3 + L2 + T2 + L1) Referred to 220kV =
769 + 34 + 80 = 883 Ohms
This referred to 400kV = 883*(1.8181*1.8181) = 2918 Ohms ( Where 1.8181 is transformation ratio 400/220kV)
(L3 + T3 + L2 + T2 + L1 + T1) = 2918 + 75 = 2993 Ohms
Current at 400kV level = (400000/1.732)/2993 = 77.16 Amp
Current at 33kV level = 77.16*400/33 = 935 Amp.
Let us again consider radial part of the part of the power system (Sub-Transmission Network) in which we are interested, Let we know the 3 phase short circuit MVA of the Root Bus. From this we can find out source impedance and from source MVA we can proceed further for calculating short circuit MVA of other busses in the radial Sub-Stations branched from Root Bus.
As we have just calculated
3 Phase Fault Level at a bus in the grid = Base MVA / Equivalent Per-Unit Impedance up to source
Hence
Equivalent Per-Unit Impedance up to source from a bus in the system = Base MVA / 3 Phase Fault Level at that bus
For the system under consideration its value is 2.16% which is considerable low as compared to other components pu impedances. Hence in such cases we may neglect it too without hampering further results much more
3 Ph Fault level calculations
Calculate source impedance for 132kV Bus Padegaon from given fault level
Consider Base MVA as 100
X1s = 100*100/2850 = 3.51%
b) Calculate % impedance of Padegaon-Harsool single circuit
XL1 = 0.4*8*(100*100)/(132*132) = 1.84% = XL2
c) Total impedance up to Harsool 132kV Bus
= 3.51 + ( 1.84 || 1.84 )
= 3.51 + 1.84/2
= 4.43%
d) 3 Phase Fault level at 132kV Harsool
= 100*100/4.43
= 2257 MVA
e) 132/33 kV 50 MVA Transformer percentage impedances at new base MVA
XT1 = 10.21*100/50 = 20.42%
XT2 = 10.88*100/50 = 21.76%
f) Total impedance from 132kV Harsool S/S up to source
X1s= 100*100/2257 = 4.43%
g) Total impedance up to Harsool 33kV Bus
= X1s + (XT1 || XT2)
= 4.43 + ( 20.42 || 21.76)
= 14.96%
h) Fault level at 33kV Harsool Bus
= 100*100/14.96 = 668 MVA
Fault level at 132kV Chikalthana Bus
XL1 = XL2 = 0.4*5*(100*100/(132*132)) = 1.15%
Xeq = 4.43 + 1.15/2 = 5.005
Fault level at 132kV Chikalthana = 100*100/5.005 = 1998 MVA
j) Fault level at 33 kV Chikalthana
XT1new = 10.03 * 100/50 = 20.06%
XT2new = 10.52*100/25 = 42.08
Xeq = 5.005 + ( 20.06 || 42.08 ) = 18.59%
Fault Level = 100*100/18.59 = 538 MVA
Like wise fault levels of other busses also can be calculated
Without going into much details of the sequence network concept studied in academic years now we will recall some of them.
Consider a SLD of stand alone system where a generator feeding 2 synchronous motor loads through a step up transformer, transmission line and a step down transformer.
Let there is a SLG fault at F
To find out the value of fault current first we will draw the equivalent sequence networks
+ Ve sequence network is with a source
If all values are expressed in pu this source voltage shall be obviously 1.0 pu
Subscript 1 in pu impedances of components stands for +Ve sequence impedances
-Ve sequence network is exactly like +Ve sequence network
Except there is no voltage source. As generated voltage (behind generator reactance Xg ) is always balance hence –Ve and zero sequence of supply voltage component is zero.
Subscript 2 in pu impedances of components stands for -Ve sequence impedances
NOTE THE REFERANCE BUS IN THE SEQUENCE NETWORK. It is referred to only that sequence network only.
While drawing Zero sequence network special care for earthing provided and transformer winding connections has to be taken.
Delta connected side get connected to Zero Bus directly. Where as star side get connected to component against it only if star point is earthed
If earthing is done through reactance X , then in zero sequence network it is replaced by 3X
For calculating single phase fault current we have to connect these three networks in series as shown in next slide.
After connecting sequence network in series ( with respect to fault point ). it is possible to reduce each sequence network into corresponding equivalent sequence impedance as shown above
Here +Ve sequence network caries +Ve sequence current –Ve sequence network caries –Ve sequence current and Zero sequence network caries zero sequence current.
Due to series configuration I1 = I2 = I0
And R Ph current involved in fault = I1 + I2 + I0 = 3I1
IR = 3/(X1+X2+X0)
Or
1 Phase Short Circuit MVA = 3 * ( Base MVA) / (X1 + X2 + X0) if X is expressed in pu
= 3*(BaseMVA)*100/(X1% + X2% + X0%)
Note : Positive and negative sequence reactance of network components are equal.
Zero sequence reactance for the transformer is same as that of +Ve sequence reactance
Zero sequence reactance for line are given in data sheet and it is about 2.5 to 3 times its +Ve sequence reactance
As said previously 3 Phase and 1 Phase fault levels of busses in grid are given by studying the system using computer.
With availability of this fault levels and formulae reviewed we can proceed further for calculation of single line to ground fault as below
1 Ph Fault level calculations
Calculate positive sequence source impedance for 132kV Bus Padegaon from given fault level as previous
Consider Base MVA as 100
X1s = 100*100/2850 = 3.51%
b) Now 1 Ph fault level = 3*BaseMVA*100/(X1 + X2 + X0)
But X1 = X2 = 3.51%
Hence 2230 = 3*100*100/(3.51+3.51+X0)
7.02 + X0 = 30000/2230
X0 = 13.45 – 7.02
X0 = 6.43%
As said previously 3 Phase and 1 Phase fault levels of busses in grid are given by studying the system using computer.
With availability of this fault levels and formulae reviewed we can proceed further for calculation of single line to ground fault as below
b) Calculate % impedance of Padegaon-Harsool single circuit
XL1 = 0.4*8*(100*100)/(132*132) = 1.84% = XL2
Let XL0 = 3*XL1 = 5.52%
d) 1 Phase Fault level at 132kV Harsool
= 3*100*100/(2*(XS1+(XL1||XL2)) + (XS0+XL0))
= 30000 / ( 2*(3.51 + ( 1.84 || 1.84 )) + ( 6.43 + ( 5.52 || 5.52 )) )
= 30000 / ( 8.86 + 9.19 )
= 30000 / 18.05
= 1662 MVA
e) 132/33 kV 50 MVA Transformer percentage impedances at new base MVA
XT1 = 10.21*100/50 = 20.42%
XT2 = 10.88*100/50 = 21.76%
h) 1 Phase Fault level at 33kV Harsool Bus
= 30000 / ( 2*( Xs1 + (XL11 || XL21) + ( XT11 || XT21) ) + (Xs0 + (XL10 || XL20) + ( XT10 || XT20) ) )
= 30000 / ( 2*(3.51 + 0.92 + 10.53) + (6.43 + 2.76 + 10.53) )
= 30000 / ( 2*14.96 + 19.72 )
= 30000 / 49.64
= 604MVA
Like wise you can proceed
Like wise fault levels of other busses also can be calculated