_per unit additional examples.pdf

Additional 42 CONTENTS examples The transformer equivalent impedance referred to the high-voltage side is µ ¶2 400 Ze2 =...

Additional 42 CONTENTS examples The transformer equivalent impedance referred to the high-voltage side is µ ¶2 400 Ze2 = (0.9 + j1.8) + (128 + j288) = 488 + j1008 Ω 20 The high-voltage base impedance is (400)2 ZB 2 = = 4000 Ω 40 488 + j1008 Zpu 2 = = 0.122 + j0.252 pu 4000 We note that the transformer per unit impedance has the same value regardless of whether it is referred to the primary or the secondary side. 3.13. Draw an impedance diagram for the electric power system shown in Figure 26 showing all impedances in per unit on a 100-MVA base. Choose 20 kV as the voltage base for generator. The three-phase power and line-line ratings are given below. G1 : 90 MVA 20 kV X = 9% T1 : 80 MVA 20/200 kV X = 16% T2 : 80 MVA 200/20 kV X = 20% G2 : 90 MVA 18 kV X = 9% Line: 200 kV X = 120 Ω Load: 200 kV S = 48 MW +j64 Mvar ¾»........T 1.......... 1 2 T2................... ¾»................ Line.............. G1........................................................... G2..... ½¼.................................. ½¼ Load ? FIGURE 26 One-line diagram for Problem 3.13 The base voltage VB G1 on the LV side of T1 is 20 kV. Hence the base on its HV side is µ ¶ 200 VB1 = 20 = 200 kV 20 This fixes the base on the HV side of T2 at VB2 = 200 kV, and on its LV side at µ ¶ 20 VB G2 = 200 = 20 kV 200  CONTENTS 43 The generator and transformer reactances in per unit on a 100 MVA base, from (3.69)and (3.70) are µ ¶ 100 G: X = 0.09 = 0.10 pu 90 µ ¶ 100 T1 : X = 0.16 = 0.20 pu 80 µ ¶ 100 T2 : X = 0.20 = 0.25 pu 80 µ ¶ µ ¶2 100 18 G2 : X = 0.09 = 0.081 pu 90 20 The base impedance for the transmission line is (200)2 ZBL = = 400 Ω 100 The per unit line reactance is µ ¶ 120 Line: X= = 0.30 pu 400 The load impedance in ohms is (VL−L )2 (200)2 ZL = ∗ = = 300 + j400 Ω SL(3φ) 48 − j64 The load impedance in per unit is 300 + j400 ZL(pu) = = 0.75 + j1.0 pu 400 The per unit equivalent circuit is shown in Figure 27. j0.2............... j0.3............... j0.25.............................................................................................. j0.1......... j0.081........ ¶³ 0.75 + j1.0 ¶³ G1 G2 µ´ µ´ FIGURE 27 Per unit impedance diagram for Problem 3.11. 44 CONTENTS........ T1............ T2.... 1................ 2 3................ 4........................... Line 1............................................................................ 220 kV..................... ¾» ¾» G........ T4 M.... ½¼....... ½¼..... T3....................................................... 5 6......................................... Line 2.............................................................................. 110 kV................................................ Load............ ? FIGURE 28 One-line diagram for Problem 3.14 3.14. The one-line diagram of a power system is shown in Figure 28. The three-phase power and line-line ratings are given below. G: 80 MVA 22 kV X = 24% T1 : 50 MVA 22/220 kV X = 10% T2 : 40 MVA 220/22 kV X = 6.0% T3 : 40 MVA 22/110 kV X = 6.4% Line 1: 220 kV X = 121 Ω Line 2: 110 kV X = 42.35 Ω M: 68.85 MVA 20 kV X = 22.5% Load: 10 Mvar 4 kV ∆-connected capacitors The three-phase ratings of the three-phase transformer are Primary: Y-connected 40MVA, 110 kV Secondary: Y-connected 40 MVA, 22 kV Tertiary: ∆-connected 15 MVA, 4 kV The per phase measured reactances at the terminal of a winding with the second one short-circuited and the third open-circuited are ZP S = 9.6% 40 MVA, 110 kV / 22 kV ZP T = 7.2% 40 MVA, 110 kV / 4 kV ZST = 12% 40 MVA, 22 kV / 4 kV Obtain the T-circuit equivalent impedances of the three-winding transformer to the common MVA base. Draw an impedance diagram showing all impedances in per unit on a 100-MVA base. Choose 22 kV as the voltage base for generator. The base voltage VB1 on the LV side of T1 is 22 kV. Hence the base on its HV side  CONTENTS 45 is µ ¶ 220 VB2 = 22 = 220 kV 22 This fixes the base on the HV side of T2 at VB3 = 220 kV, and on its LV side at µ ¶ 22 VB4 = 220 = 22 kV 220 Similarly, the voltage base at buses 5 and 6 are µ ¶ 110 VB5 = VB6 = 22 = 110 kV 22 Voltage base for the tertiary side of T4 is µ ¶ 4 VBT = 110 = 4 kV 110 The per unit impedances on a 100 MVA base are: µ ¶ 100 G: X = 0.24 = 0.30 pu 80 µ ¶ 100 T1 : X = 0.10 = 0.20 pu 50 µ ¶ 100 T2 : X = 0.06 = 0.15 pu 40 µ ¶ 100 T3 : X = 0.064 = 0.16 pu 40 The motor reactance is expressed on its nameplate rating of 68.85 MVA, and 20 kV. However, the base voltage at bus 4 for the motor is 22 kV, therefore µ ¶µ ¶2 100 20 M: X = 0.225 = 0.27 pu 68.85 22 Impedance bases for lines 1 and 2 are (220)2 ZB2 = = 484 Ω 100 (110)2 ZB5 = = 121 Ω 100 46 CONTENTS 1 j0.20 j0.25 j0.15 4............................................................................................................... -I...... j0.16............... j0.35............... j0.06............... j0.18............................................................................................................................. j0.3................................... j0.27................................ 0.12 Im ? ¶³..... ¶³ Eg G M E m µ´ −j10 µ´ FIGURE 29 Per unit impedance diagram for Problem 3.14. Line 1 and 2 per unit reactances are µ ¶ 121 Line1 : X= = 0.25 pu 484 µ ¶ 42.35 Line2 : X= = 0.35 pu 121 The load impedance in ohms is (VL−L )2 (4)2 ZL = ∗ = = −j1.6 Ω SL(3φ) j10 The base impedance for the load is (4)2 = 0.16 Ω ZBT = 100 Therefore, the load impedance in per unit is −j1.6 = −j10 pu ZL(pu) = 0.16 The three-winding impedances on a 100 MVA base are µ ¶ 100 ZP S = 0.096 = 0.24 pu 40 µ ¶ 100 ZP T = 0.072 = 0.18 pu 40 µ ¶ 100 ZST = 0.120 = 0.30 pu 40  CONTENTS 47 The equivalent T circuit impedances are 1 ZP = (j0.24 + j0.18 − j0.30) = j0.06 pu 2 1 ZS = (j0.24 + j0.30 − j0.18) = j0.18 pu 2 1 ZT = (j0.18 + j0.30 − j0.24) = j0.12 pu 2 The per unit equivalent circuit is shown in Figure 29. 3.15. The three-phase power and line-line ratings of the electric power system shown in Figure 30 are given below. T T2..... Vg........... 1........... ¾»...... 1 Line 2.................... Vm ¾».............................................. G.......................................... M ½¼.......................................... ½¼ FIGURE 30 One-line diagram for Problem 3.15 G1 : 60 MVA 20 kV X = 9% T1 : 50 MVA 20/200 kV X = 10% T2 : 50 MVA 200/20 kV X = 10% M: 43.2 MVA 18 kV X = 8% Line: 200 kV Z = 120 + j200 Ω (a) Draw an impedance diagram showing all impedances in per unit on a 100-MVA base. Choose 20 kV as the voltage base for generator. (b) The motor is drawing 45 MVA, 0.80 power factor lagging at a line-to-line ter- minal voltage of 18 kV. Determine the terminal voltage and the internal emf of the generator in per unit and in kV. The base voltage VB G1 on the LV side of T1 is 20 kV. Hence the base on its HV side is µ ¶ 200 VB1 = 20 = 200 kV 20 This fixes the base on the HV side of T2 at VB2 = 200 kV, and on its LV side at µ ¶ 20 VB m = 200 = 20 kV 200 48 CONTENTS The generator and transformer reactances in per unit on a 100 MVA base, from (3.69)and (3.70) are µ ¶ 100 G: X = 0.09 = 0.15 pu 60 µ ¶ 100 T1 : X = 0.10 = 0.20 pu 50 µ ¶ 100 T2 : X = 0.10 = 0.20 pu 50 µ ¶ µ ¶2 100 18 M: X = 0.08 = 0.15 pu 43.2 20 The base impedance for the transmission line is (200)2 ZBL = = 400 Ω 100 The per unit line impedance is 120 + j200 Line: Zline = ( ) = 0.30 + j0.5 pu 400 The per unit equivalent circuit is shown in Figure 31. j0.2...................... 0.3 + j0.5............................. j0.20.............................................................................................................. + +........................... j0.15............ Vg Vm............ j0.15............ ¶³ ¶³ Eg Em µ´ µ´ − − FIGURE 31 Per unit impedance diagram for Problem 3.15. (b) The motor complex power in per unit is 456 36.87◦ Sm = = 0.456 36.87◦ pu 100 and the motor terminal voltage is 186 0◦ Vm = = 0.906 0◦ pu 20  CONTENTS 49 0.456 −36.87◦ I= = 0.56 −36.87◦ pu 0.906 0◦ Vg = 0.906 0◦ + (0.3 + j0.9)(0.56 −36.87◦ = 1.317956 11.82◦ pu Thus, the generator line-to-line terminal voltage is Vg = (1.31795)(20) = 26.359 kV Eg = 0.906 0◦ + (0.3 + j1.05)(0.56 −36.87◦ = 1.3756 13.88◦ pu Thus, the generator line-to-line internal emf is Eg = (1.375)(20) = 27.5 kV 3.16. The one-line diagram of a three-phase power system is as shown in Figure 32. Impedances are marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW −j33.4 Mvar, and at bus 3 is S3 = 77 MW +j14 Mvar. It is required to hold the voltage at bus 3 at 4006 0◦ kV. Working in per unit, determine the voltage at buses 2 and 1. V1 V2 V3 j0.5 pu j0.4 pu ? ? S2 S3 FIGURE 32 One-line diagram for Problem 3.16 S2 = 15.93 MW − j33.4 Mvar = 0.1593 − j0.334 pu S3 = 77.00 MW + j14.0 Mvar = 0.7700 + j0.140 pu 4006 0◦ V3 = = 1.06 0◦ pu 400 S∗ 0.77 − j0.14 I3 = 3∗ = = 0.77 − j0.14 pu V3 1.06 0◦ V2 = 1.06 0◦ + (j0.4)(0.77 − j0.14) = 1.16 16.26◦ pu 50 CONTENTS Therefore, the line-to-line voltage at bus 2 is V2 = (400)(1.1) = 440 kV S∗ 0.1593 + j0.334 I2 = 2∗ = = 0.054 + j0.332 pu V2 1.16 −16.26◦ I12 = (0.77 − j0.14) + (0.054 + j0.332) = 0.824 + j0.192 pu V1 = 1.16 16.26◦ + (j0.5)(0.824 + j0.192) = 1.26 36.87◦ pu Therefore, the line-to-line voltage at bus 1 is V1 = (400)(1.2) = 480 kV 3.17. The one-line diagram of a three-phase power system is as shown in Figure 33. The transformer reactance is 20 percent on a base of 100-MVA, 23/115-kV and the line impedance is Z = j66.125Ω. The load at bus 2 is S2 = 184.8 MW +j6.6 Mar, and at bus 3 is S3 = 0 MW +j20 Mar. It is required to hold the voltage at bus 3 at 1156 0◦ kV. Working in per unit, determine the voltage at buses 2 and 1. V1................................................................................... V3 V2 j66.125 Ω ? S2 ? S3 FIGURE 33 One-line diagram for Problem 3.17 S2 = 184.8 MW + j6.6 Mvar = 1.848 + j0.066 pu S3 = 0 MW + j20.0 Mvar = 0 + j0.20 pu 1156 0◦ V3 = = 1.06 0◦ pu 115 S∗ −j0.2 I3 = 3∗ = = −j0.2 pu V3 1.06 0◦ V2 = 1.06 0◦ + (j0.5)(−j0.2) = 1.16 0◦ pu  CONTENTS 51 Therefore, the line-to-line voltage at bus 2 is V2 = (115)(1.1) = 126.5 kV S2∗ 1.848 − j0.066 I2 = ∗ = = 1.68 − j0.06 pu V2 1.16 0◦ I12 = (1.68 − j0.06) + (−j0.2) = 1.68 − j0.26 pu V1 = 1.16 0◦ + (j0.2)(1.68 − j0.26) = 1.26 16.26◦ pu Therefore, the line-to-line voltage at bus 1 is V1 = (23)(1.2) = 27.6 kV