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# Electric Utility Solutions: Voltage Regulation

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This is a systems engineering and analysis presentation from Milsoft's 2009 User Conference. It was originally presented by Bill Kersting. The Milsoft Electric Utility Solutions Users Conference is the premier event for our users and the vendors who provide interoperable solutions or services that enhance Milsoft Smart Grid Solutions. If you’d like to be on our mailing list, just email: missy.brooks@milsoft.com.

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### Electric Utility Solutions: Voltage Regulation

1. 1. Voltage Regulation Bill Kersting
2. 2. What is to be presented • • • • ANSI Voltage Standards Methods of Voltage Regulation Example of Regulator Settings Example of Placement of Regulators
3. 3. The ANSI Voltage Standards • Range A – – – – Nominal Utilization Voltage = 115 volts Maximum Utilization Voltage = 126 volts Minimum Service Voltage = 114 volts Minimum Utilization Voltage = 110 volts • Range B – – – – Nominal Utilization Voltage = 115 volts Maximum Utilization Voltage = 127 volts Minimum Service Voltage = 110 volts Minimum Utilization Voltage = 107 volts
4. 4. Tools for Voltage Regulation • Shunt Capacitors • Step-Voltage Regulators • Substation Load Tap Changing Transformers
5. 5. Distribution Line Voltage Drop R jX + V S + VL Load - I - VS Im(ZI) ZI Real(ZI) VL 0 jXI RI I Vdrop = VS − Vr Vdrop ≈ Real ( Z ⋅ I L ) = Real ( R ⋅ I L + jX ⋅ I L )
6. 6. Vdrop = Real(ZIL) • Impedance (Z) and current (I) must be computed as accurately as possible. • Impedance best computed using Carson’s Equations • Current is a function of “load.” • If Z and I are not computed accurately, all bets are off on the calculated system voltages.
7. 7. Capacitor Voltage Rise R + X + I L + IC VS VL Load IC _ _ ' VS IC jXI C VS δ ZI L VL θ RI L RI C Im ( ZI L ) Real ( ZI L ) jXI L IL Vrise = ' VS − VS Vrise ≈ Real ( Z ⋅ I C ) = Real ( R ⋅ I C + jX ⋅ I C )
8. 8. ANSI Range A Critical Voltages Laterals Sub Reg 128 126 124 122 120 118 116 114 Reg Output First Customer Last Customer
9. 9. Voltage Drop Assumptions • 1 Volt drop on the service drop • 2 Volt drop on the secondary • 3 Volt drop through the transformer • Minimum Voltage at the Transformer Primary Terminals will be 120 volts.
10. 10. Voltage Profiles Laterals Sub Reg 128 Max Load 126 124 122 Min Load 120 118 116 114 Reg Output First Customer Reg. Point Last Xfm Last Customer
11. 11. Step Voltage Regulator
12. 12. Type B Step Voltage Regulator Preventive Autotransformer N1 IS R S + Reversing Switch L Series Winding IL VS N2 Control CT Shunt Winding Control PT - + VL SL L
13. 13. The Step Voltage Regulator Model = VL 1 ⋅ VS aR I= L aR ⋅ I S where: aR = 1  N2 N1 = 1  0.00625 ⋅ Tap aR One tap change = 0.75 V change on 120 V base
14. 14. Three Phase Voltage Regulator Model [ I S ]abc [VS ]abc = [VS ]abc = [ I S ]abc = [VL ]abc [ I L ]abc aR ⋅ [VL ]abc d R ⋅ [ I L ]abc AR ⋅ [VS ]abc [VL ]abc
15. 15. Voltage Regulator Model Matrix  aR _ a  [ aR ] =  0   0  0 aR _ b 0 dR ] AR ] [= [ = [ aR ]−1   0   aR _ c   0
16. 16. Compensator Circuit MVA rating Iline R line + jX line CTp − CTs Ic R c + jX c kVLL hi − kVLLlow Reg. Point 1:1 + Vdrop + N PT :1 Vreg - + VR - Voltage Relay
17. 17. Control Panel
18. 18. Control Circuit Line Current Control Current Transformer Line D rop Compensator Control Potential Transformer V oltage Relay Time Delay Motor Operating Circuit
19. 19. Regulator Control Settings • Voltage Level – voltage to hold at the regulation point • R and X setting (volts) – Equivalent impedance from the regulator to the regulation point • Time Delay – time after a tap change required before the tap is changed • Bandwidth – allowed deviation from the set voltage level
20. 20. Equivalent Line Impedance For i = a, b, c Vregi − Vreg _ pti Zlinei Iregi Ω where: Vreg = actual line-to-neutral voltage output of regulator Vreg _ pt = actual line-to-neutral voltage at the regulation point Ireg = actual line current leaving the regulator
21. 21. Compensator Impedance Zcomp = Zline ⋅ CT N pt Volts where: Zline = equivalent line impedance in Ohms CT = current transformer primary rating VLN rated N pt = potential transformer ratio = 120
22. 22. Voltage Level Bandwidth 123 122 121 Bandwidth = 2 V ∆T
23. 23. Modified IEEE 13 Node Test Feeder 1 2 9 8 3 13 12 4 14 5 6 10 7 11
24. 24. Modifications • • • • • Line 4-12 changed to phases B-C Transformer 6-7 changed to Ungrounded Wye – Delta Load at Node 7 converted to Delta-PQ Load at Node 8 converted to Delta-PQ Load at Node 14 changed to phase B with constant Z load • Load added at Node 5 phase c: 300 + j145.3 kVA • Interchange phase a and c distributed loads on line 3-4
25. 25. Step 1 • Select regulation point to be Node 4. • Turn off regulator in Analysis Manager. • Run power-flow with source set to 126 volts (IEEE 13 Node Test Feeder Start.wm). • Display Voltage Profile. • Compute compensator impedance.
26. 26. Step 1 Voltage Profile 135 135 132.5 Node Voltage 130 V.a 127.5 125 V.b 122.5 V.c 120 117.5 115 110 112.5 110 1 1 2 3 Node 4 5 5
27. 27. Voltages and Currents from Power-Flow Run eg. V2 := j ⋅0   2521.87⋅ e    − j ⋅120 ⋅deg  2521.87⋅ e    2521.87⋅ ej ⋅120 ⋅deg    Ireg := V4 :=  2310⋅ e− j ⋅3.5 ⋅deg     − j ⋅124.6 ⋅deg  2377.5⋅ e   j ⋅116.1 ⋅deg    2284.2⋅ e   590.8⋅ e− j ⋅34.4 ⋅deg     − j ⋅150.5 ⋅deg  632.5⋅ e    651.9⋅ ej ⋅81.4 ⋅deg   
28. 28. Compensator R and X Setting CT p = 700 Zlinei := N p t = 20 V2 − V4 i i Ireg i Z avg := mean( Zline) Z set := Z avg⋅ CT p Np t  0.1671 + 0.4037j    Zline =  0.0541 + 0.3817j   0.1426 + 0.4188j    Z avg = 0.1212 + 0.4014j Z set = 4.2 + 14j volts
29. 29. IEEE 13 with Regulator Set • Set source voltage to 120 V. • Set regulator control. – R and X = 4.2 + j14 – Set voltage output (level) to 121. • Analysis Manager – Set regulator to step. • Run voltage drop. – Show results – Show profile
30. 30. IEEE 13 with Regulator Set
31. 31. Full Load with Regs, no Caps 135 135 133 Node Voltage 131 V.a 129 127 V.b 125 V.c 123 121 119 115 117 115 1 1 2 3 Node 4 5 5
32. 32. Use WindMil “Set Regulation” • Select Voltage Drop. – Analysis Manager • Set regulators to infinite. • Set source to 126 volts. • Select Set Regulation. – Analysis Manager • • • • • Select substation regulator. Select Node 4 as load center. Most desirable voltage = 121 Tolerance 2% Unbalanced study
33. 33. Set Voltage Regulation
34. 34. WindMil R and X settings
35. 35. WindMil with R 4.8, X = 14.4 and no feeder caps • Set source voltages to 120. • Set voltage level (output voltage) to 121 V. • Run Voltage Drop.
36. 36. WindMil R and X setting with no feeder capacitors
37. 37. WindMil with regs, no caps 135 133 Node Voltage 131 Va 129 127 Vb 125 Vc 123 121 119 117 115 1 2 3 Node 4 5
38. 38. WindMil Voltage Profile
39. 39. Observations • Regulator taps – Phase a: 12 – Phase b: 13 – Phase c: 15 • Concern that Phase c is near maximum tap • Concern about high voltage at Node 2 • Need to add shunt capacitors
40. 40. Shunt Capacitors • Source reactive power – Phase A: 834 kVAr – Phase B: 805 kVAr – Phase C: 1040 kVAr • Install shunt capacitors – Node 3: 100 kVAr per phases a,b,c – Node 4: 300 kVAr per phases a,b,c – Node 4: Switched 300 kVAr per phases a,b,c
41. 41. WindMil R and X settings with capacitors
42. 42. Full Load with Regs and Caps
43. 43. Observations • Regulator taps – Phase a: 6 – Phase b: 6 – Phase c: 8 • Concern for voltage unbalance at Node 4
44. 44. Node 4 Voltage Unbalance  119.8  V4 :=  124     121.4  Vavg := mean V4 Dev :=  1.9333  Dev =  2.2667    0.3333   i ( ) V4 − Vavg i Vunbalance := max( Dev ) Vavg ⋅ 100 Vunbalance = 1.862 Vavg = 121.7333 %
45. 45. Minimum load of 50% • Analysis manager – Set load growth to -50% • Run voltage drop – Observe power factor at source – Switch 900 kVAr at Node 4
46. 46. 50% load reduction with all capacitors
47. 47. 50% load with all Capacitors
48. 48. 50% load with 900 kVAr at Node 4 switched off 135 133 Node Voltage 131 Va 129 127 Vb 125 Vc 123 121 119 117 115 1 2 3 Node 4 5
49. 49. 10% Growth with Original Capacitors
50. 50. 10% Load Growth • Analysis Manager – Set load growth to 10%. • Run voltage drop – Voltage profile – Check kVAr supplied by sub. – Install new shunt capacitors if necessary.
51. 51. 10% load growth with original capacitors
52. 52. 50% load reduction, switch off 900 kVAr at Node 4
53. 53. 10% Load Growth with original caps
54. 54. 10% Growth with 300 kVAr added at Node 10
55. 55. 10% Load Growth 100 kVAr per phase added at Node 10
56. 56. IEEE 34 Node Test Feeder • Will be used to: – Determine location of downstream step voltage regulators – Voltage level – R and X settings • My method • WindMil method
57. 57. Modified IEEE 34 Node Test Feeder http://ewh.ieee.org/soc/pes/dsacom/testfeeders.html 29 23 28 22 27 33 Sub 2 3 4 5 6 26 15 21 7 9 24 14 16 17 18 8 13 1 19 20 30 31 32 10 11 12 25
58. 58. To Start • • • • • System is very unbalanced. System is very long (35 miles). Voltage level is 24.9 kV. Set substation output voltage to 126 volts. Run power flow for the IEEE 34 node system with no regulators or shunt capacitors (IEEE 34 Node Bare Bones).
59. 59. IEEE 34 with no regulators and no capacitors
60. 60. IEEE 34 with no regulators or capacitors
61. 61. Install substation regulators • Install 3 Step Voltage Regulators connected in grounded Y in the substation to start the regulation process. • Potential transformer ratio = 14,400/120 • Current transformer ratio = 100/0.1 • Voltage level = 126 volts • Bandwidth = 2 volts • R and X = 0 • Run power flow.
62. 62. Modified IEEE 34 Node Test Feeder http://ewh.ieee.org/soc/pes/dsacom/testfeeders.html 29 23 28 22 27 33 Sub 2 3 4 5 6 21 7 9 24 8 26 15 14 13 1 19 16 20 17 18 30 31 32 10 11 12 25
63. 63. IEEE 34 with Y connected sub regulators, Voltage Output (level) = 126, R and X = 0
64. 64. Voltage Profiles with Substation Regulators with Voltage Level = 126
65. 65. Observations and next step • Node 5 is the first node downstream where the voltage drops below 120. • Select Node 5 as the regulation point for the substation regulator. • Set regulators to infinite. • Run Set Regulator to compute R and X settings. • Set R and X on the sub regulator control. • Set voltage level on regulator to 120 volts. • Run power flow with regulators set as step.
66. 66. Sub Regulator set with R = 14.4 and X = 9.6 Voltage Output (level) = 120
67. 67. Install Regulators at Node 5 • Set voltage level = 126 • Regulator set to infinite • R and X = 0
68. 68. IEEE 34 Node Test Feeder http://ewh.ieee.org/soc/pes/dsacom/testfeeders.html 29 23 28 22 27 33 Sub 2 3 4 5 6 21 7 9 24 8 26 15 14 13 1 19 16 20 17 18 30 31 32 10 11 12 25
69. 69. Substation Regulators Set Regulators Installed at Node 5 Voltage Level = 126
70. 70. Observations • All voltages at Node 5 are between 119 and 121 volts. • The first downstream node where all of the voltages drop below 120 V is Node 11. • Set regulators to Infinite. • Run Regulation Set to compute R and X from Node 5 to Node 11. • Set regulators to step. • Run power flow.
71. 71. Sub and Node 5 (R=16.8, X = 7.2) Regulators Set
72. 72. Observations • • • • Install a regulator at Node 11. Set voltage level to 126. Set regulators to step. Run Voltage Drop.
73. 73. Regulator Installed at Node 5 29 23 28 27 22 33 Sub 2 3 4 5 6 21 7 9 24 8 26 15 14 13 1 19 16 20 17 18 30 31 32 10 11 12 25
74. 74. Reg at Node 11 Set to 126 Volts, R and X = 0
75. 75. Regulator at Node 5 set with V = 126 No R and X • With the regulator set at 126 volts, all of the downstream voltages in the main feeder are greater/equal to 120 volts. • No need to set R and X for this regulator • The only problems occur on the 4.16 kV line from 19 to 20.
76. 76. Profile including the 4.16 kV line
77. 77. Install regulator at secondary terminals of the transformer • • • • • • Potential Transformer Ratio = 2400/120 Primary CT Rating = 100 amps Calculate R and X. Set regulators to Infinite. Load center is Node 20. Run Set Regulation.
78. 78. System does not converge 4.16 Reg set with V = 122, R = 12, X = 7.2 • Use Set Regulation to compute R and X for Reg 11 and Reg 20 with voltage output = 122.
79. 79. Node 5: R = 9.6, X = 4.8; Voltage Output = 122 Node 20: R = 12, X = 9.6; Voltage Output = 122
80. 80. Set source voltage to 120 • Run with no capacitors. • Add capacitors.
81. 81. Source set to 120, no capacitors
82. 82. Correct feeder power factor to near 1 • Display P and Q on 4.16 kV line. – Install a three phase capacitor bank to supply most of the 4.16 kV kVAr load. – 75 kVAr/phase at Node 20 • Need to add 200 kVAr/phase – Node 16, 100 kVAr/phase – Node 822: 100 kVAr/phase A – Node 848: 100 kVAr/phase
83. 83. Final with capacitors
84. 84. Final Regulator Tap Positions • • • • Sub Regulator: 9, 8, 7 Node 812: 10, 6, 7 Node 830: 5, 6, 7 Node 888: 12, 12, 12
85. 85. Final kVAr supplied by source • Source power factor: – Phase a: 43 (PF = 99.8 %) – Phase b: 71 (PF = 99.5 %) – Phase c: 16 (PF = 99.9 %)
86. 86. To be continued by you • Minimum load – Which capacitors to switch • Load growth – Where and how big new capacitor banks