Download - FAULT Analysis presentation Armstrong
Presented by Armstrong Okai Ababio
1Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Goal of the Presentation-1
To Appreciate the use of Ohmic and MVA Methods for Symmetrical Short Circuit Calculation.
A symmetrical or balanced fault affects each of the three-phases equally (Three phase fault)
2Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Goal of the Presentation-2To appreciate the use of Symmetrical Components
for Asymmetrical Short Circuit Calculation.An asymmetric or unbalanced fault does not
affect each of the three phases equally.Common types of asymmetric faults:
line-to-lineline-to-grounddouble line-to-ground
The analysis of this type of fault is often simplified by using methods such as symmetrical components.
3Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
IntroductionThe operation of a power system departs from normal after
the occurrence of a fault. The most common faults in any electrical installation are short circuits, i.e. a breakdown of insullation between conductive parts which normally are at different potential.
Fault give rise to abnormal operating conditions such as excessive current and voltages at certain points in the network.
The magnitude of fault current depends on the generated power in the system, the distance to these sources, fault resistance and for ground fault it depends also on the neutral point treatment of the system.
4Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Effect of Faults in a 3 phase Network A fault will cause voltage and current disturbances, including a
complete voltage collapse near the fault location.
The voltage gradient in the ground and across the earthing resistance create dangerous step and touch voltages near the fault.
The dynamic forces of the fault current, the electric arc at the fault location and the thermal effects of the current on the network elements in the current path all cause damage to the plant equipment and personnel.
The voltage and current disturbances interrupt the transmission of power and thus also affect customers.
5Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Effect of Faults-2 A large amount of power is dissipated at the fault point, which means lost
of revenue for the utility.
Failure to remove a fault will usually result in rapid expansion of damage
to the system.
Various protection equipment are used in guarding against such faults
condition. An idea of the magnitudes of such fault currents gives the
Engineer the current settings of the various protection devices and
ratings of circuit breakers in the network.
6Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Effect of Faults-3 In the light of the above discussed effects of faults, FAULT or SHORT
CIRCUIT ANALYSIS is highly needed.
Performing short-circuit calculations requires an understanding of various system components and their interaction.
Components of Power System:
Generators, Transformers, Transmission lines, Protection Systems(control and monitoring system), Switch gear, Load.
7Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Types of Fault/ Short Circuits in a 3 phase network
Three-phase fault with or without earth (5%)
Phase-to- phase clear of ground (10 – 15%)
Two-phase-to-earth fault (10 – 20%)
Phase-to-earth fault (65 – 70%) Fault Incident:
85% of faults are overhead line.
50% of these are due to lightning strikes.
L1L2L3
L1L2L3
L1L2L3
L1L2L3
8Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
3 PHASE FAULTS-1 A 3-ph fault affects the three-phase network equally (symmetrical fault).
They rarely occur. All three conductors are equally involved and carry the same rms short- circuit current. There is the need to use only one conductor for the calculation.
It is valid because system is maintained in a balanced state during fault. Voltages are equal and 120° apart. Currents are equal and 120° apart. Power system plant symmetrical:- Phase impedance equal Mutual impedance equal Shunt admittances equal Causes: System energization with maintenance earthing clamps still connected. 1ø faults developing into 3ø faults. Etc.
9Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
3 PHASE FAULTS-2 A 3-ph fault affects the three-phase network symmetrically. They
rarely occur. All three conductors are equally involved and carry the same rms short- circuit current. There is the need to use only one phase (conductor ) for the calculation.
The Per unit system or MVA method is used for the analysis of Three phase fault.
10Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
3 PHASE FAULTS
11Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
3 PHASE FAULTS
12Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Over View of Per Unit System Per- Unit Notations. It is used to simplify calculations on system with more
than two voltages. ( The √ are eliminated) It can be seen by inspection of any power system diagram that: a. several voltage levels exist in a system b. it is common practice to refer to plant MVA in terms of per unit or
percentage values c. transmission line and cable constants are given in ohms/km
Before any system calculations can take place, the system parameters must be referred to 'base quantities' and represented as a unified system of impedances in either, percentage, or per unit values.
Per-unit analysis is based on "normalized" representations of the electrical quantities (i.e., voltage, current, impedance, etc.). The per-unit equivalent of any electrical quantity is dimensionless. It is defined as the ratio of the actual quantity in units (i.e., volts, amperes, ohms, etc.) to an appropriate base value of the electrical quantity. Per Unit (p.u.) value = Actual value / Base value
13Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Referring Impedances
14Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana
Can be derived from V1/V2 = I2/I1
Which translates into V1/V2=(V2/R2)/ (V1/R1)
9/9/2016
Formulae used in Per-Unit Calculations The base quantities are: three-phase power in MVA (MVA rating of largest item or 100MVA) It is
constant at all voltages . The line voltage in kV. Fixed at one part and it is transferred through
transformers to obtain base voltages of the other part of the system. The following Equations are used in Per- Unit Calculation: - Ip.u = Iactual / Ibase - Sp.u = Sactual / S base
- Vp.u = Vactual / Vbase - Base Current (kA), Ib = MVAb (3ø)/ {1.732 x kVb (L-L)} - Base Impedance (Ohm) Zb = (kV)2
(L-L) / MVAb 3ø
- Zp.u(new) = Z p.u
(old) x[MVAb (N) / MVAb
(old)] x[kVb(old) / kVb
(new)]2
15Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Per Unit Impedance
16Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Per Unit Impedance
17Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Percentage Impedance
18Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Percentage Impedance
19Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Percentage Impedance
20Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Percentage Impedance
21Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Transformer Base Voltage Selection
22Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Example
23Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Example
24Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
MVA METHOD FOR 3 PHASE SHORT CIRCUIT CALCULATION
The MVA Method is recognized and widely acceptable by industry in calculating power system short circuits where the reactance of all circuit components far exceeds resistance producing a consistently high X/R ratio throughout the system.
Combined MVA of components connected in series and parallel are calculated using the following formulas:
series: MVA1, 2 = MVA1 X MVA2 / (MVA1 + MVA2) parallel: MVA1, 2 = MVA1 + MVA2
As can be seen from the formulas above, series MVA’s are being calculated same as resistances in parallel. Parallel MVA combinations are done same as resistances in series.
MVA diagram undergoes same reduction process as impedance diagram, only that MVA values are used instead of per unit impedances or reactances.
25Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
MVA METHOD FOR 3 PHASE SHORT CIRCUIT CALCULATION
The MVA method is a modification of the Ohmic method where the impedance of a circuit equals the sum of the impedances of components constituting the circuit.
In practice, the MVA method is used by separating the circuit into components and calculating each component with its own infinite bus as shown in figures 1 and 2 below:
M15 MVAXd = 0.2
20MVA33 / 11kV X = 10%
T2161 / 33 kVX = 10%
161 kV T1 50MVA
Line 33kV
X = 4.0Ω
Sk’’ = 700MVA
Figure 1-One Line Diagram
700 500 272.25
200
75
3ø F.
700 / 1 50 / 0.1 (33)2 / 4
15 / 0.2
20 / 0.1
3ø F.
Figure 2- MVA Diagram
1 2 34
5
11kV
G
103.8 MVA
26Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
MVA METHOD FOR 3 PHASE SHORT CIRCUIT CALCULATION In our example: MVA’s 1& 2 are in series - MVA (1&2) = (700 x 500) / (700 + 500) = 291.67
MVA’s (1&2) & 3 are in series - MVA = (291.67 x 272.25) / (291.67 + 272.25) = 140.81
MVA’s (1,2&3) // MVA 4 - MVA at the point = 140.81 + 75 = 215.81
MVA’s (1,2,3&4) in series with MVA 5 MVA at the Fault point = (251.81 x 200) /(251.81 + 200) = 103.802 Considering Voltage Factor of 1.1, MVA = 103.802 x 1.1 = 114.18
- 3Ø ISC =MVA /(1.732 x Un) =114.18 /(1.732 x 11) = 5.99 kA
Per Unit System is more accurate than the MVA method in the analysis of three (3) phase faults.
27Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
D.C. Transient and Offsets
28Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
END OF PART 1
WAGE OPENER FOR PART 2
RIDDLE!!!
29Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Goal of the Presentation-2To appreciate the use of Symmetrical Components
for Asymmetrical Short Circuit Calculation.An asymmetric or unbalanced fault does not
affect each of the three phases equally.Common types of asymmetric faults:
line-to-lineline-to-grounddouble line-to-ground
The analysis of this type of fault is often simplified by using methods such as symmetrical components.
30Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
ASYMMETRICAL OR UNBALANCED FAULTS
CALCULATIONS
Developed by Charles LeGeyt Fortescue (1876–1936) in 1918. Fortescue was an Electrical Engineer and he worked for Westinghouse Corporation at East Pittsburgh, Pennsylvania. He developed the method of Symmetrical Components to resolve unbalance fault conditions.
31Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
UNBALANCED FAULTS-1Unbalanced Faults may be classified into SHUNT
FAULTS and SERIES FAULTSSHUNT FAULTSLine to GroundLine to LineLine to Line to GroundThe above faults are described as single shunt
faultsbecause they occur at one location and involve aconnection between one phase and another or to
earth.32
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
UNBALANCED FAULTS-2Unbalanced Faults may be classified into
SHUNT FAULTS and SERIES FAULTS
OPEN CIRCUIT/ SERIES FAULTSSingle phase open circuitDouble phase open circuit
33Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
CAUSES OF UNBALANCE FAULTSCAUSES OF SHUNT FAULTS Insulation BreakdownLightning Discharges Mechanical DamageCAUSES OF OPEN CIRCUIT FAULTSBroken ConductorOperation of FusesMal-operation of single phase CBDuring Unbalanced Faults, symmetry of system
is lost. Hence single phase representation is no longer valid for the fault analysis.
34Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Analysis of Unbalanced FaultsUnbalanced faults are analysed using :-Symmetrical ComponentsEquivalent Sequence Networks of the Power
SystemConnection of Sequence Networks
appropriate to the Type of Fault
35Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Sequence ComponentsThe Symmetrical Components consist of
three subsystems for an unbalanced three-phase system. They are;
Positive Sequence Subsystem-consisting of three phasers of equal magnitude and 120° phase displacement, and having the same phase sequence as the original balanced system of phasers.
36Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Sequence ComponentsThe second subsystem is termed the
'negative sequence' system, consisting of three phasers of equal magnitude and 120° phase displacement, and having a phase sequence which is the reverse of the original balanced system of phasers.
The final subsystem is termed the 'zero sequence' system, consisting of three phasers of equal magnitude and zero phase displacement.
37Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
ILLustration of Symmetrical Components
Orientation of the symmetrical components are as follows:V A1
V B1V C1V C2
V A2 VAo
V B2
VBo
120° 240°V Co
Zero Sequence PhasorsNegative Sequence PhasorsPositive Sequence Phasors
38Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Sequence Components-Three Phase Fault
39Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Sequence Components PHASE TO PHASE FAULT
ONLY +VE AND –VE SEQUENCE COMPNENTS EXISTS
E
Z1 Z2
I a11 I a22 ZoI a0 0 = 0= 0
Sequence Network Interconnection for phase to phase fault
40Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Sequence Components-Single Phase-Earth Fault. All sequence components are present
41Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
By definition, the magnitude and phase angle of any phasor in an unbalanced three-phase system is equal to the vector addition of the symmetrical components from the respective sequence subsystems.
Symmetrical Components
42Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Symmetrical Components
43Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Symmetrical ComponentsTo provide a mathematical perspective on the
preceding discussion of phasor magnitudes and phase displacements, it is necessary to define a phase displacement operator 'a'.
a = 1<120° = -0.5 + j0.866
Successive applications of operator 'a' to a given vector will result in rotation of that vector through 120°, 240° and 360°, degrees, as shown below.
44Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Symmetrical Componentsa = 1<120° = 1ej2π/3 = -0.5 +j0.866 a2 = 1<240° = 1ej4π/3 = -0.5 -j0.866 a3 = 1<360° = 1e]2π = 1Therefore we have;
45Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Converting from phase value to Converting from phase value to sequence componentssequence components
46Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Converting from phase value to sequence components
47Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Symmetrical Components-Similarly Equations for sequence Currents can be derived
48Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana
It can be deduced thatIn= Ia + Ib + Ic = 3I0
9/9/2016
Phase Sequence Equivalent Circuits
E
a2E
aE
I
a2 I
a I
P Q
P1 Q1
Z1 = E/I
Positive Sequence Impedance
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 49
Phase Sequence Equivalent Circuits
E
aE
a2E
I
a2 I
a I P Q
P2 Q2
Z2 = E/I
Negative Sequence Impedance For static non-rotating plant :- Z2= Z1
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 50
Phase Sequence Equivalent Circuits
E
I
I
I P Q
P0 Q0
Z0 = E/I
3I
Zero Sequence Impedance
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 51
Sequence Networks
+ve, -ve and zero sequence networks are drawn for a ‘reference’ phase. This is usually taken as the ‘A’ phase.
Faults are selected to be ‘balanced’ relative to the reference ‘A’ phase.
e.g. For Ø/E faults, we consider an A-E fault For Ø/Ø faults, we consider a B-C fault Sequence network interconnection is the simplest
for the reference phase.
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 52
1. Start with neutral point N1 -All generator and load neutrals are connected to
N1 2. Include all source voltages:- Phase-neutral
voltage 3. Impedance network:- Positive sequence
impedance per phase 4. Diagram finishes at fault point F1
Positive Sequence Diagram
N1E1
Z1F1
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 53
Positive Sequence Diagram G
T Line F
R
N
E
N1
E1 Z G1 Z T1 Z L1 I1 F1
N1
V1
V1=Positive sequence PH-N voltage at fault pointI1=Positive sequence phase current flowing into F1
V1=E1 –I1 (ZG1+ ZT1 + ZL1)
System Single Line Diagram
Positive Sequence Diagram
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 54
Negative Sequence Diagram
Start with neutral point N2
-All generator and load neutrals are connected to N2
No voltages included -No negative sequence voltage is generated! Impedance network -Negative sequence impedance per phase Diagram finishes at fault point F2
N2
Z2
F2
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 55
Negative Sequence Diagram G
T Line F
R
N
E
N2
Z G2 Z T2 Z L2 I2 F2
N2
V2
V2=Negative sequence PH-N voltage at fault pointI2=Negative sequence phase current flowing into F2
V2= –I2 (ZG2+ ZT2 + ZL2)
System Single Line Diagram
Negative Sequence Diagram
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 56
Zero Sequence Diagram
For “In Phase” (Zero Phase Sequence) currents to flow in each phase of the system, there must be a fourth connection (this is typically the neutral or earth connection).
IAo+ IBo+ICo=3IAo
IAo
IBo
ICo
N
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 57
Zero Sequence Diagram
N
E
R3IAo
Resistance Earthed System :-
Zero sequence voltage between N & E is given byVo= 3IAo * RZero sequence impedance of neutral to earth pathZo= Vo/ IAo = 3R
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 58
Zero Sequence Equivalent “D y” Transformer”
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 59
Zero Sequence Equivalent “D y” Transformer
Thus, Equivalent single phase zero sequence diagram is as shown:-
Side terminal Z ToI o
Y sideterminal
N o(Eo)
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 60
Zero Sequence Equivalent Circuits
9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 61
Equations Defining Shunt Fault Conditions
It should be noted that for any type of fault there are three equations that define the fault conditions. They are as follows:
Single Phase-to-earth (A-E)
Ib= 0 (Ib= 0, Ic = 0, Because Phases B and C do not contribute to fault current)Ic = 0
Va = 0 (Va = 0, because voltage at the faulted phase decreases to Zero)
62Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Equations Defining Shunt Fault Conditions
Phase-phase (B-C)
Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)
Ib =-Ic
Vb = Vc
63Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Equations Defining Shunt Fault Conditions
Phase-phase-to-earth (B-C-E)
Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)
Ib + Ic = In
Vb = Vc=0 (Because voltages at the faulted phases decrease to Zero)
64Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Equations Defining Shunt Fault Conditions
Three Phase fault (A-B-C or A-B-C-E)
Ia + Ib + Ic = 0 (Because the system is balanced and hence In= Ia + Ib + Ic =0) Va = Vc
Vb = Vc
65Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Single Phase to Earth Fault-1We recall that for Single Phase to ground fault:
Ib=Ic = 0
Va = 0 and therefore we can write;
I1=1/3(Ia + aIb + a2Ic)=1/3Ia
I2=1/3(Ia + a2Ib + aIc)=1/3Ia Therefore I0=I1=I2= 1/3Ia
I0=1/3(Ia + Ib + Ic)=1/3Ia
66Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Single Phase to Earth Fault-2Also
Va = 0 and therefore we can write;
V1 +V2 +V0 =0 (Since Va =V1 +V2 +V0 =0)But
V1 =V-I1 Z1 , V2 =-I2 Z2 and V0=-I0Z0
Substituting we have;
V-I1 Z1 -I2 Z2-I0Z0=0 since I1=I2=I0, we haveV=I1 (Z1 + Z2 + Z0)
67Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Single Phase to Earth Fault-3The above analysis indicate that the
equivalent circuit for the fault is obtained by connecting the sequence networks in series as shown below;
68Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Single Phase to Earth Fault-4It also follows that;
I1 = E/(Z1 + Z2 + Z0)
From the equation described earlier, I0=I1=I2= 1/3Ia and there fore
Ia = 3I1 = 3E/(Z1 + Z2 + Z0)
69Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Single Phase to Earth Fault-5In the more general case, with nonzero fault
resistance, the equality of I1; I2 and I0 is maintained, and 3I1 flows through the fault resistance. Therefore, it is necessary that 3Zf exist in series with the zero sequence subsystem to achieve the required effect. The generalized equation is as follows:
Ia = 3I1 = 3E/(Z1 + Z2 + Z0 + 3Zf)
70Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016
Phase to Phase Fault (B-C)-1We recall that for a phase-phase fault,
Ia = 0, implying
I1=1/3(Ia + aIb + a2Ic)=1/3(aIb + a2Ic)
I2=1/3(Ia + a2Ib + aIc)=1/3(a2Ib + aIc)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 719/9/2016
Phase to Phase Fault (B-C)-2Also,
Ib= -Ic I1=1/3(aIb + a2Ic) = 1/3(alb-a2lb) = l/3Ib (a-a2)
I2=1/3(a2Ib + aIc) = 1/3(a2lb-alb) = l/3Ib (a2-a) The above equations illustrate that;I 1 = - I 2
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 729/9/2016
Phase to Phase Fault (B-C)-3By inspection, the equations for this fault
condition are as follows:
I1=E/(Z1 + Z2)
I2=-E/(Z1 + Z2)
I0 = 0It must be noted that no zero sequence current
exist since there is no connection to ground.Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 739/9/2016
Phase to Phase Fault (B-C)-4From the equation below;
I1=1/3(aIb + a2Ic) = 1/3(alb-a2lb) = l/3Ib (a-a2), Since Ib=-Ic
But
a = 1<120° = 1ej2π/3 = -0.5 +j0.866 a2 = 1<240° = 1ej4π/3 = -0.5 -j0.866
Therefore it follows that, a-a2 = j1.732= j√3 and I1= j√3/3Ib
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 749/9/2016
Phase to Phase Fault (B-C)-5Finally it implies that;
I1=1/3(aIb + a2Ic) = 1/3(alb-a2lb) = l/3Ib (a-a2) Since,
a-a2 = j1.732= j√3I1 = j √3 /3Ib = jIb/ √3 by rationalization.
hence Ib =-j √3 I1
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 759/9/2016
Phase to Phase Fault (B-C)-6Similarly;
I2=1/3(a2Ib + aIc) = 1/3(a2lb-alb) = l/3Ib (a2-a) And since
a2 -a = -j1.732= -j√3, we haveI2 = -j √3 /3Ib = -jIb/ √3 by rationalization.
hence Ib =j √3 I2
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 769/9/2016
Phase to Phase Fault (B-C)-7By inspection, the equations for this fault
condition are as follows:
I1=E/(Z1 + Z2) but Ib =-j √3 I1 Therefore
Ib =-j √3 E/(Z1 + Z2)
Also, Ic =-Ib
Therefore Ic =j √3 E/(Z1 + Z2)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 779/9/2016
Phase to Phase Fault (B-C)-8In the more general case, with nonzero fault
resistance, the generalized equation, based on the fault impedance of Zf is expressed below:
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 78
fcb ZZZ
EjII
21
3
9/9/2016
Double Phase to Ground Fault (B-C-E)We recall that for double Phase to Ground Fault
(B-C-E), the following equations can be written:
Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)Ib + Ic = In
Vb = Vc=0 (Because voltages at the faulted phases decrease to Zero)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 799/9/2016
Double Phase to Ground Fault (B-C-E)From the general equation below, we have:
V-I1 Z1 = -I2 Z2 from(V1=V2=V0=Va/3) and Vb = Vc=0
Also, Since Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)
we have, I1 + I2 + I0 = 0
And it follows that;
I1 = -( I2 + I0)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 809/9/2016
Double Phase to Ground Fault (B-C-E)Vb = Vc=0 (Since voltages at the faulted phases
decrease to Zero)Also implies, V1 = V2 = V0 = 1/3Va
Given that all three sequence voltages are equal in phase and magnitude, and given that the sequence currents sum vectorially to zero, it is obvious that the three sequence networks are in parallel
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 819/9/2016
Double Phase to Ground Fault (B-C-E)From the fact that V1 = V2 = V0 = 1/3Va, we can write
V2 = V0, and hence I2 Z2 = I0Z0
The negative and zero sequence currents can be derived on the basis of the current divider principle as shown below:
I2=-I1 Z0/(Z2+ Z0), and I0=-I1 Z2/(Z2+ Z0)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 829/9/2016
Double Phase to Ground Fault (B-C-E)Equating V1 and V2, we have
V-I1 Z1 = -I2 Z2 or V = I1 Z1 - I2 Z2 , where V=E= phase voltage
Substituting I2, I2=-I1 Z0/(Z2+ Z0), into the above equation,
It gives, V = I1 [Z1 + Z0Z2 /(Z0 + Z2)] Therefore;
I1 = E(Z0 + Z2)/(Z1Z0 + Z1Z2 +Z0Z2)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 839/9/2016
Double Phase to Ground Fault (B-C-E)Finally,
Substituting I2 and I0 into the above equation,
I2=-I1 Z0/(Z2+ Z0) , I0=-I1 Z2/(Z2+ Z0)andI1 = E(Z0 + Z2)/(Z1Z0 + Z1Z2 +Z0Z2)
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 84
0212 IaIIaIb
9/9/2016
Double Phase to Ground Fault (B-C-E)We have;
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 85
020121
203ZZZZZZ
aZZEIb
9/9/2016
Three Phase Fault (A-B-C or A-B-C-E)We recall that for a three phase fault;
Ia + Ib + Ic = 0 (Because the system is balanced and hence In= Ia + Ib + Ic
=0) Va = Vc
Vb = Vc
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 869/9/2016
Three Phase Fault (A-B-C or A-B-C-E)It should be noted that, because this fault
type is completely balanced, there are no zero- or negative-sequence currents;
By inspection we can write;
I1= E/(Z1+ Zf) also I2 = I0 =0
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 879/9/2016
Three Phase Fault (A-B-C or A-B-C-E)
It finally follows that for a three phase fault;
Ia= Ib = Ic = E/(Z1+ Zf) also I2 = I0 =0
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 889/9/2016
CONCLUSION-1
It must be emphasized that in very large and complex networks, system computer programs are used for short circuit analysis.
In the Electricity Company of Ghana (ECG), ASPEN software is used for short circuit analysis.
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016 89
CONCLUSION-2 I would like to first of all thank God for His guidance and for
the wisdom and understanding He gave me to put this presentation together.
Secondly, I appreciate the efforts of Ing. Godfred Mensah/ SM/System Planning and Mr. Frank Osei Owusu of Protection Applications whom I understudied when preparing for this presentation.
Finally, I very much appreciate your presence for this presentation
THANK YOU ALL FOR COMING.!!!
Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016 90