Z n. 100 kv. 15 kv. pu := 1. MVA := 1000.kW. Transformer nameplate data: X T_pu := 0.1pu S T := 10MVA. V L := 15kV. V H := 100kV

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1 /9 j := pu := MVA :=.kw 7.. Three MVA, -5 kv transformers have nameplate impedances of % and are connected Δ-Y with the high voltage side Δ. Find the zero sequence equivalent circuit. kv Z n 5 kv Transformer nameplate data: T_pu :=.pu S T := MVA V L := 5kV V := kv Per-unit Base quantities: S B := 3S T S B = 3 MVA Total system complex power V BL := V L 3 V BL = kv Line-Line low side base voltage V B := V V B = kv Line-Line high side base voltage 2 V BL Z BL := S B 2 V B Z B := S B Z BL = 22.5 Ω Z B = Ω Low side impedance base igh side impedance base

2 2/9 f e := 6z The assumed electrical frequency j T 3*Z n Generalized Zero Sequence Equivalent Circuit (a) If the neutral is ungrounded: Z n = infinity j. Zero Sequence Equivalent Circuit for Ungrounded Neutral (b) If the neutral is grounded solidly: Z n_pu := pu Z _pu := j T_pu + 3Z n_pu Z _pu =.j pu Z _pu =. pu arg( Z _pu ) = 9 deg j. Zero Sequence Equivalent Circuit for Solidly Grounded Neutral

3 3/9 (c) If the neutral is grounded through a 5 Ω resistance: Z n_res := 5Ω Z n_res_pu := Z n_res Z BL Z n_res_pu =.222 pu 3Z n_res_pu =.667 pu Z _res_pu := j T_pu + 3Z n_res_pu Z _res_pu = j pu Z _res_pu =.674 pu arg( Z _res_pu ) = 8.53 deg j..667 Zero Sequence Equivalent Circuit for Neutral Grounded through 5 Ω Resistance (d) If the neutral is grounded through a 5μF capacitance: C n := 5 μf Z n_cap := j2 π f e C n C n = 5 3 F Z n_cap =.53j Ω Z n_cap_pu := Z n_cap Z BL Z n_cap_pu =.24j pu 3Z n_cap_pu =.7j pu Z _cap_pu := j T_pu + 3Z n_cap_pu Z _cap_pu =.29j pu Z _cap_pu =.29 pu arg( Z _cap_pu ) = 9 deg

4 4/9 j. -j.7 j.29 Zero Sequence Equivalent Circuit for Neutral Grounded through 5 μf Capacitance

5 5/9 j := pu := MVA :=.kw j 2 π 3 a:= e In the transformer connection shown below each single-phase transformer has a turns ratio of n = n /n 2 =. The low-voltage side carries an unbalanced load with sequence currents given as I a = -I a2 -3 A. Find the sequence currents and 2 as phasors. A B C 2 N n () n n n n 2 () I a I a2 a b c From the connection shown above, we have: I a = ni A ni C where: n = I a = n n := ( ) n n 2 Writing the above equation for positive sequence quantities (loosely following the development in Blackburn): I a = n ( ) and noting that

6 6/9 = ai A Therefore: I a = n ai A. ( ) I a = n ( a) We note: a =.5.866i a =.732 arg( a) = 3deg Thus: I a = n 3 e j 3deg or: = I a e j3 deg n 3 For the given positive sequence unbalanced load current: I a := e j 3deg := I a n ( a) = = arg( ) = deg

7 7/9 Writing similar equation for the negative sequence quantities: I a2 = n 2 2 ( ) and noting that 2 = a 2 2 Therefore: I a2 = n 2 a 2 2. ( ) I a2 = n a 2 2 We note: a 2 = i a 2 =.732 ( ) = 3 deg arg a 2 Thus: I a2 = n 3 2 e j3 deg or: I a2 e j 3deg 2 = n 3 For the given positive sequence unbalanced load current: I a2 := e j 3deg 2 := I a2 ( ) n a 2 2 = i 2 = arg( 2 ) = 2 deg

8 8/9 As an aid in visualizing the sequence currents on each side of the transformer bank and the impact of the Δ-Y connection (the / for I a and I a2 is to scale the quantities for plotting): I a I a arg,, arg I a, arg I a2 ( ) arg ( 2) ( ) ( ) Noodling (as in thinking about with one's noodle (head), not cat-fishing by hand) a bit more with this problem, I thought it would be mildly interesting to look at the phase currents resulting from these sequence currents. Using the synthesis equation here is what I came up with: A 2 := a 2 a a a 2 I a I b I c := A I 2 a I a = I b = i I c = i I a2

9 9/9 I b = arg I b ( ) = 2 I c = arg( I c ) = 6 deg deg It looks like the load is between only phases "b" and "c," and not involving "a" phase. The phase currents on the high-voltage side of the transformer: := A =.547 arg = i i i = The result on the high-voltage side of the transformer is a bit more challenging to classify. We have the phase "A" and "C" currents in-phase and equaling the phase "B" current; but at 8 degrees opposite. It's tempting to call it a very "ugly" and "unbalanced" 3-phase fault, definitely not bolted. This problem once again demonstrating the impact of Δ-Y transformation on sequence currents and the resulting phase currents. deg I b I c arg( I b ), arg ( I c), arg( ), arg( ), arg ( )

Chapter 8: Unsymmetrical Faults

Chapter 8: Unsymmetrical Faults Introduction The sequence circuits and the sequence networks developed in the previous chapter will now be used for finding out fault current during unsymmetrical faults.