Chapter 10 Symmetrical Components and Unbalanced Faults Part II 10 4 Sequence Networks of a Loaded Generator In the figure to the right is a generator supplying a three phase load with neutral connected through impedance Z n to ground The generator generates three phase balanced voltages defined as 1 abc E a 2 Ea a The terminal voltage of the generator is found using Kirchoff s voltage law Va Ea Z s I a Z n I n Vb Eb Z s Ib Z n I n Zs Ea Zn Ec In Zs Ia Va Eb Zs Ib Ic V Vc b Vc Ec Z s I c Z n I n And since I n I a I b Ic we have Zn Zn Ia Va Ea Z s Z n Z s Zn Zn Ib Vb Eb Z n V E Z Zn Z s Z n I c c c n Or in compact form V abc E abc Z abc I abc This equation can be transformed to the 012 or symmetrical component form thus AVa012 AEa012 Z abc AI a012 Multiplying by A 1 we have Va012 E a012 A 1Z abc AI a012 Ea012 Z 012 I a012 Where Z 012 A 1Z abc A Zn Zn 1 1 1 1 1 Z s Z n 1 2 1 a a Zn Z s Zn Zn 1 a2 3 2 a Z n Zn Z s Z n 1 a 1 a Performing the above matrix multiplication we have 0 0 0 Z s 3Z n 0 0 Z Z 012 0 Z s 0 0 Z 1 0 0 0 Z s 0 0 Z 2 1 a a 2 It is important to note that the matrices above are diagonal and the equations are completely decoupled Expanding into three voltage equations we have Va0 0 Z 0 I a0 Va1 E a Z 1I a1 V 0 Z I 2 a These equations represent three separate circuits as shown to the right one for the positive sequence one for the negativesequence and one for the zero sequence networks 2 2 a Z1 I1 a Z2 I 2a Z0 I0 a Ea V 1a V2 a V0 a Positive sequence Negative sequence Zero sequence The following important observations are made The three sequences are independent The positive sequence network is the same one used in the one line diagram for studying balanced three phase currents and voltages Only the positive sequence network has a voltage source Thus the positivesequence voltage will generate only positive sequence currents There is no voltage source in the negative and zero sequence networks Negative and zero sequence currents cause only negative and zero sequence voltages The neutral of the system is the reference point for the positive and negativesequence networks The ground is the reference point for the zero sequence hence the zero sequence current cannot flow unless there is a connection to ground The grounding impedance is reflected in the zero sequence network as 3Z n Each of the three sequence networks can be solved separately on a per phase basis The phase currents and voltages can then be found by superposition by adding their symmetrical components of current and voltage respectively 10 5 Single Line To Ground Fault This is the most common fault on a three phase system It is illustrated using the following simple network 2 Here the fault occurs between line a and ground through a fault impedance Z f Assuming the generator was initially at no load the boundary conditions at the fault are Va Z f I a Zs Ia Ea Zn Ec Va Zf Eb I b 0 Zs In Zs Ib Ic 0 Ic 0 Using t he second equation in the symmetrical V Vc b components of currents we have I a0 1 1 1 Ia 1 1 2 Ia 3 1 a a 0 1 a2 a 0 I a2 1 Therefore I a0 I a1 I a2 I a Thus the three symmetrical components are equal and 3 each is equal to one third phase a current Phase a voltage in terms of symmetrical components is Va Va0 Va1 Va2 Using equation last page and the fact the three symmetrical components of current 1 are I a0 I a1 I a2 I a we have 3 Va Ea Z 0 Z 1 Z 2 I a0 Where Z 0 Z s 3Z n and Z 1 Z 2 Zs Recall that Va Z f I a and that I a 3I a0 the above equation becomes 3Z f I a0 Ea Z 1 Z 2 Z 0 I a0 Or Ea I a0 1 2 Z Z Z 0 3Z f And the fault current is 3E a I a 3I a0 1 2 Z Z Z 0 3Z f Using the symmetrical components of current in equation the symmetrical components of voltage are found hence by transformation the voltages during the fault Observing the equations and we notice that the sequence networks derived earlier may be connected as shown here which satisfies these two equations Z1 I1a Z2 I 2a Z0 I 0a Ea V1 a V2 a V0 a 3Z f 3 If the generator neutral is grounded then Z n 0 and Z 0 Z s and for a bolted fault Z f 0 10 6 Line To Line Fault This fault is shown in the figure to the right Note the neutral is not grounded The fault is shown between phases b and c It is assumed the generator is initially on no load The boundary conditions at the fault point are Vb Vc Z f I b Ib Ic 0 Zs I a 0 Ea Ec Eb Zs Zs Ia 0 Using I a 0 and I c I b the symmetrical components of currents become I a0 1 1 1 0 1 1 2 Ia 3 1 a a Ib 1 a2 a I I a2 b Expanding the matrix equation we have I a0 0 1 I 1a a a 2 I b x 3 1 I a2 a 2 a Ib 3 Ic Va Ib Zf V b Vc Note that I 1a I a2 from x above From the equations for phase voltages reproduced below we have Va Va0 Va1 Va2 Vb Va0 a 2Va1 aVa2 Vc Va0 aVa1 a 2Va2 Hence Vb Vc a 2 a Va1 Va2 Z f Ib Using the values of Va1 and Va2 from and note that I a2 I a1 we have a 2 a Ea Z 1 Z 2 I a1 Z f I b And substituting for I b from x above we have 4 Ea Z 1 Z 2 I a1 Z f 3I 1a a 2 a a a 2 Since a 2 a a a2 3 solving for I a1 we have I 1a Ea Z Z2 Z f 1 The phase currents are I a 1 1 2 I b 1 a I 1 a c The fault current is 1 0 a Ia1 a 2 I a1 I b Ic a 2 …