Introduction: - Institute of Electrical and Electronics ...



APPENDIX AFAULT CALCULATION METHOD USING SYMMETRICAL COMPONENT Introduction:This Appendix can be used to calculate the three phase and phase to ground faults at different points of the station service system as required. Using the calculated faults currents, the engineer can select the appropriate equipment ratings and the required protective device to insure a reliable and safe station service system. This Appendix consists of the following:Review of per unit quantities.Review of symmetrical components Fault calculation process Review of Per Unit Quantities:Per unit system is used to help simplify the process of calculating the currents and voltages at different points of the power system when system analysis such as load flow or fault calculation is performed. Per unit quantity is defined by the following equation:Per unit quantity (pu)=Actual quantityBase quantityBase values for the system voltage, current, Power and impedance must be defined. Normally once the system voltage and power are defined, the remaining base quantities can be defined. The base values for single phase system and three phase system are calculated as shown in the table below:Table 1 Base Quantities SelectionBase QuantitySingle Phase SystemThree Phase SystemMVA??33.33MVA 100 MVAVoltage?L-N System Nominal VoltageL-L System Nominal VoltageCurrentI=base, VA1φbase voltage VLNI=base, VA3φ3 base voltage VLlImpedanceZ=VBASE2VA1φBASEZ=VBASE2VA3φBASEImpedance New BaseZnew=ZoldVBASEold2VBASEnew2*VAnewVAoldZnew=ZoldVBASEold2VBASEnew2*VAnewVAold? for voltage base values use the system nominal voltage.?? for power base a 100 MVA has been used for three phase. For single phase 33.33MVA can be usedExample 1:Figure SEQ Figure \* ARABIC 1- Example 1 Station Service One Line Diagram The above figure represents an example of the substation AC auxiliary system. The source at the 12.47 kV bus is represented by equivalent impedance. A three phase transformer is used to step the voltage down to 480V. A 480V panel is connected to the transformer low side and feeds a second 480V and a single phase transformer which is connected to another panel. It is required to determine the per unit quantities for this figure.Solution:Step 1:As a first step we need to determine the base values for current, voltage, MVA and impedance for the above system. This can be accomplished by dividing the system into areas. Each area includes one voltage level as shown in the figure below:Figure SEQ Figure \* ARABIC 2 Example One Base Values DeterminationsStep2:Create a base table as shown below:Table 2 Base Quantities SelectionBase QuantitySection 1Section 2Section 3Section 4MVA??10010033.333.3Voltage?12.47.48277120Current “A”4630120281119133277750Impedance1.555.002304.0023040.000432Using the above table and the following equation, the PU values for the above figure can be calculated as shown on the figure below:Per unit quantity (pu)=Actual quantityBase quantityFigure SEQ Figure \* ARABIC 3 Example One - Per Unit CalculationsThe above diagram can be represented by a single phase diagram as shown below. This diagram can be used to calculate the faults at different busses.Figure- SEQ Figure \* ARABIC 4 Example 1 Impedance per Unit RepresentationsSymmetrical Component Review: In 1918, Dr. C.L Fortescue an American scientist showed that any unbalanced system of 3- phase currents (or Voltages) may be regarded as being composed of three sets of balanced vectors which has come to be known as symmetrical component and is defined below:Positive Phase Sequence Components: A balanced system of three phase currents having a positive or normal phase rotation.Negative Phase Sequence Components: A balanced system of three phase currents having a negative or opposite to the positive sequence currents phase rotation.Zero Phase Sequence Current: a system of three phase currents equal in magnitude and having zero phase displacements.For a three phase unbalanced system shown in the figure below:Figure SEQ Figure \* ARABIC 5- Three Phase Unbalance Phaser RepresentationsThe positive, negative and zero sequence components of the above unbalanced three phase system can be represented as shown in the figure below:Figure SEQ Figure \* ARABIC 6- Positive, Negative and Zero Sequence Representations of Unbalanced SystemThe current in any phase is equal to sum of the positive, negative and zero sequence current. As shown in the figure 7 below:Figure SEQ Figure \* ARABIC 7- Symmetrical Components Current RepresentationsNotes Related to symmetrical components:The positive sequence currents (Ia1,Ib1,Ic1), negative sequence currents (Ia2,Ib2,Ic2) and zero sequence currents (Ia0,Ib0,Ic0) separately form a balanced system of currents. They are defined as the symmetrical components.The symmetrical components theory applies equally to three phase voltages and currents.The symmetrical components have no physical existence they are mathematical components of unbalanced currents and (or voltages) which actually flow in the system.In a balanced three phase system the negative and zero sequence currents are zero.Operator a:When an operator “a” is multiplied by a vector it rotates the vector through 1200 in the counter clockwise direction. The properties of the vector “a” are shown below:a = 1∠120 oa2= 1∠240 oa3= 1∠360 o =11+a+ a 2 =0a+ a 2 =-11+a=1∠60o1+ a 2 =1∠-60oa-a2=j3a2-a-=-j31-a-=3∠-30 o1-a2=3∠30 oUsing the operator “a” above, the relationship between the different phase components are shown in the table below:Positive sequenceIa1=Ia11∠0o= Ia1Ib1=Ia11∠240o= a2Ia1Ic1=Ia11∠120o= aIa1Negative sequenceIa2=Ia21∠0o= Ia2Ib2=Ia21∠120o= a2Ia2Ic2=Ia21∠240o= a2Ia2Zero sequenceIao=Ibo=IcoThe unbalanced currents of, Ia, Ib, Ic, can be expressed in terms of phase “a” component and by using operator “a” as follows:Ia=Ia0+Ia1+Ia2Ib=Ia0+a2 Ia1+aIa2Ic=Ia0+a Ia1+a2 Ia2In matrix form, the phase Ia, Ib, Ic can be expressed as follows:IaIbIc=IaoIa1Ia21111a2a1aa2The above equation can be expressed in a single matrix equation as follows:Iabc =A I012 Where A is known as the symmetrical component transformation matrix which transforms the currents Iabc to Ia012 and is given by the following matrix:A=1111a2a1aa2Since a2=a* then A-1=13A*The sequence quantities can be expressed in terms of the phase quantities by finding the inverse of the matrix as follows:I0aI1aI2a=131111aa21a2aIaIbIcI0a=13Ia+Ib+IcI1a=13Ia+aIb+a2IcI2a=13Ia+a2Ib+aIcFor a balanced three phase system, I0a=0 since Ia+Ib+IC=0The above equations apply to voltages also.Sequence representations of power system elements:ELEMENTSYMBOLPOSITIVE SEQNEGATIVE SEQGENERATOR/SOURCELINETRANSFORMERFault Calculations:Using the station power one line diagram presented earlier as shown below, and the associated per unit diagram, calculate the three phase fault and the phase to ground fault at points A,B,C,D,E,F. Figure SEQ Figure \* ARABIC 8 Station Power Example One Line DiagramFigure SEQ Figure \* ARABIC 9 Station Power Example per unit Diagram3.1Three Phase Fault Calculations:Since for three phase faults all currents are equal in magnitude and 1200 apart, the system is balanced and can be represented by the positive sequence impedance only.Three Phase Fault at Point A:If= 1@00j.3871=-j2.583PU Three Phase Fault at Point B:0-89600If= 1@00j2.9594=-j.3379PUThree Phase Fault at Point C:If= 1@00j12.9594=-j.0771PUThree Phase Fault at Point D:-2208689161700If= 1@00j881=-j.0011PUThree Phase Fault at Point E:If= 1@00j3051=-j.000328PU3.2Phase to Ground Fault Calculations:At Point A:I0= 1@00j1.9318=-j.51765@PUIf= 3I0=3*-j.51765PU=-j1.553PUAt Point B:I0= 1@00j10.9344=-j.09145PUIf= 3I0=3*-j.09145PU=-j.2744PUAt Point C:I0= 1@0035.94@89.80=-j.0278 PUIf= 3I0=3*-j.0278PU=-j.08347PU027495500At Point D:I0= 1@00j3508=-j.000285PUIf= 3I0=3*-j.000285PU=-j.000855 PU At Point E:0-63500I0= 1@00j8716=-j.0001147PUIf= 3I0=3*j.0001147PU=-j.000341PUFault At Point F:A single phase fault at point F is treated as a single phase fault at point D with the leakage impedance of the single phase transformer between F and D appearing as the fault impedance (therefore the fault impedance will be Zt x 3. See below.0-127000Fault Summary:I0= 1@00j3517=-j.000284333PUIf= 3I0=3*j.000284333PU=-j.000853PUCalculation Summary:Fault Location3-Phase Fault “PU”Phase-Ground Fault “PU”Base Current Amp3-Phase Fault “AMPSPhase to Ground Fault “Amps”A-j2.583-j1.5534630119597190.4B-j.3379-j.274446301564.51270C-j.0771-j.08347120281927310039D-j.0011-j.000855120281132.3102.84E-j.000328-j.00034112028139.4541F.000853277750237?? Line to Neutral Fault ................
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