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L18update

Course: ECE 524, Spring 2012
School: Idaho
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524: Transients ECE in Power Systems Session 18; Page 1/19 Spring 2012 ECE 524: Lecture 18 Three phase TRV transients (see update on page 8 and PSCAD implementation at end) 6 MVAR 1000kW s 10 sec Consider the three phase system shown below: ATP Implementation UI LOAD U VS 12 MVAR VLL 34.5kV Lph 2 VLL 3 Vm QL 12MVAR XL XL VLL Vm 28.17 kV 2 QL XL 99.19 Lph 263.1 mH 2 60Hz Cph 2nF 1...

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524: Transients ECE in Power Systems Session 18; Page 1/19 Spring 2012 ECE 524: Lecture 18 Three phase TRV transients (see update on page 8 and PSCAD implementation at end) 6 MVAR 1000kW s 10 sec Consider the three phase system shown below: ATP Implementation UI LOAD U VS 12 MVAR VLL 34.5kV Lph 2 VLL 3 Vm QL 12MVAR XL XL VLL Vm 28.17 kV 2 QL XL 99.19 Lph 263.1 mH 2 60Hz Cph 2nF 1 1 2 Lph Cph fn Tn 1 fn 2nF UI 34.5kV NEU fn 6938.13 Tn 144.13 s Choose simulation time step: Tn 30 4.8 s 1 s 5 Zs j 10 ohm ECE 524: Transients in Power Systems Case 1: Session 18; Page 2/19 Spring 2012 Zn ohm Breaker currents 300 [A] 200 100 0 -100 -200 -300 0.00 0.02 0.04 (f ile Example.pl4; x-v ar t) c :VSWA -LOADA Neutral to ground voltage: 0.06 c :VSWB -LOADB 0.08 [s] 0.10 c :VSWC -LOADC 15.0 [kV] 7.5 0.0 -7.5 -15.0 -22.5 -30.0 49.5 50.5 51.5 (f ile Example.pl4; x-v ar t) v :NEU Phase voltages 52.5 53.5 [ms] - 50.0 [kV] 37.5 25.0 12.5 0.0 -12.5 -25.0 -37.5 -50.0 0.00 0.02 (f ile Example.pl4; x-v ar t) v :LOADA -NEU 0.04 v :LOADB -NEU 0.06 v :LOADC -NEU 0.08 [s] 0.10 54.5 ECE 524: Transients in Power Systems Session 18; Page 3/19 Spring 2012 Phase voltage after first switch clears 50.0 [kV] 37.5 25.0 12.5 0.0 -12.5 -25.0 -37.5 -50.0 49 50 51 (f ile Example.pl4; x-v ar t) v :LOADA -NEU 52 v :LOADB -NEU 53 54 [ms] v :LOADC -NEU Phase voltage after remaining switches clear: 50.0 [kV] 37.5 25.0 12.5 0.0 -12.5 -25.0 -37.5 -50.0 53.0 53.5 54.0 (f ile Example.pl4; x-v ar t) v :LOADA -NEU 54.5 v :LOADB -NEU 55.0 v :LOADC -NEU 55.5 [ms] 56.0 55 ECE 524: Transients in Power Systems Session 18; Page 4/19 Spring 2012 Breaker voltages: 90 [kV] 60 30 0 -30 -60 48 52 56 (f ile Example.pl4; x-v ar t) v :VSWA -LOADA VmaxA 85.586kV Case 2: Repeat with Breaker currents: 60 v :VSWB -LOADB v :VSWC -LOADC 64 [ms] 3Vm Zn 0ohm 300 [A] 200 100 0 -100 -200 -300 0.00 0.02 (f ile Example.pl4; x-v ar t) c :VSWA -LOADA 0.04 c :VSWB -LOADB 0.06 c :VSWC -LOADC 0.08 [s] 0.10 68 ECE 524: Transients in Power Systems Neutral current Session 18; Page 5/19 Spring 2012 50 [A] 0 -50 -100 -150 -200 -250 0.00 0.02 (f ile Example.pl4; x-v ar t) c :NEU Phase voltages: 0.04 0.06 0.08 [s] 0.10 - 40 [kV] 30 20 10 0 -10 -20 -30 -40 48 50 52 (f ile Example.pl4; x-v ar t) v :LOADA -NEU Breaker voltages: 54 v :LOADB -NEU 56 [ms] v :LOADC -NEU 60 [kV] Vmax 57.295kV 40 20 2 Vm 0 -20 -40 -60 49 50 51 (f ile Example.pl4; x-v ar t) v :VSWA -LOADA 52 v :VSWB -LOADB 53 54 v :VSWC -LOADC 55 [ms] 56 58 ECE 524: Transients in Power Systems Session 18; Page 6/19 Spring 2012 Case 3: Repeat with Cn 3 Cph Breaker currents: Cn 6 nF 300 [A] 200 100 0 -100 -200 -300 0.00 0.02 0.04 (f ile Example.pl4; x-v ar t) c :VSWA -LOADA 0.06 c :VSWB -LOADB 0.08 c :VSWC -LOADC Neutral current after first phase clears: 5.00 [A] 3.75 2.50 1.25 0.00 -1.25 -2.50 -3.75 -5.00 49.8 50.0 50.2 (f ile ExampleL17.pl4; x-v ar t) c :NEU 50.4 50.6 50.8 51.0 51.2 [ms] 51.4 - Neutral current when all phases clear: 5.00 [A ] 3.75 2.50 1.25 0.00 -1.25 -2.50 -3.75 -5.00 5 3.5 53.8 ( f ile E x a m p leL 1 7 . p l4 ; x -v a r t ) c : N E U 54.1 - 54.4 54.7 [ms ] 55.0 [s] 0.10 ECE 524: Transients in Power Systems Session 18; Page 7/19 Spring 2012 Neutral voltage: 20 [kV ] 10 0 -10 -20 -30 -40 44 47 ( f ile E x a m p le L 1 7 . p l4 ; x -v a r t ) v : N E U 50 53 56 59 [ms ] 62 - Phase voltages: 50.0 [kV] 37.5 25.0 12.5 0.0 -12.5 -25.0 -37.5 -50.0 48.0 49.5 51.0 (f ile ExampleL17.pl4; x-v ar t) v :LOADA -NEU 52.5 v :LOADB -NEU 54.0 v :LOADC -NEU 55.5 [ms] 57.0 ECE 524: Transients in Power Systems Session 18; Page 8/19 Spring 2012 Breaker voltages: 80 [kV] 50 20 -10 -40 -70 0.0330 0.0464 0.0598 (f ile ExampleL17.pl4; x-v ar t) v :VSA -LOADA 0.0732 v :VSB Vmax Vmax 78.846kV Vm -LOADB v :VSC 0.0866 [s] 0.1000 -LOADC 2.8 Case 4: high resistance ground: 3 Rgr = Xc0 Xcph Rgr 1 2 60Hz Cph Xcph 6 Xcph 1.33 10 Rgr 442.1 k 3 90 *10 3 Vmax 84.462kV 70 50 Vmax Vm 3 30 10 -10 -30 -50 -70 0.0330 0.0464 (file ExampleL18.pl4; x-var t) v:VSA -LOADA 0.0598 v:VSB -LOADB 0.0732 v:VSC -LOADC 0.0866 0.1000 ECE 524: Transients in Power Systems Session 18; Page 9/19 Spring 2012 Case 5: Add grounding resistance (high resistance ground--limit In to 20A): Inmax 20A Breaker voltages: VLL Rgr Rgr 995.93 3 Inmax use 1000 ohm 70 [kV] 44 Vmax 60.018kV 18 Vmax Vm 2.13 -8 -34 -60 0.0330 0.0464 (f ile ExampleL17.pl4; x-v ar t) v :VSA 0.0598 -LOADA v :VSB 0.0732 -LOADB v :VSC 0.0866 [s] 0.1000 -LOADC Case 6: Now try a low resistance ground (In = 500A) Inmax 500A Rgr VLL 3 Inmax Rgr 39.84 use 40ohm 60 Vmax 56.771kV Vmax Vm 2.02 [kV] 40 20 0 -20 -40 -60 0.0330 0.0464 (f ile ExampleL17.pl4; x-v ar t) v :VSA 0.0598 -LOADA v :VSB 0.0732 -LOADB v :VSC 0.0866 -LOADC [s] 0.1000 ECE 524: Transients in Power Systems Session 18; Page 10/19 Spring 2012 Case 7: Now try a low inductance ground (In = 500A) Inmax 500A Lgr magZgr magZgr VLL 3 Inmax magZgr 39.84 Lgr 105.67 mH 2 60Hz 80 [kV] Vmax 73.874kV 60 40 Vmax Vm 2.62 20 0 On first phase to clear A little higher after subsequent phases clear it damping neglected -20 -40 -60 -80 0.0330 0.0464 (f ExampleL17.pl4; ile x-v ar t) v :VSA 0.0598 -LOADA v :VSB 0.0732 -LOADB v :VSC 0.0866 -LOADC [s] 0.1000 ECE 524: Transients in Power Systems Session 18; Page 11/19 Spring 2012 EMTDC Implementation Timed Breaker Logic BRKA Closed@t0 Switches set to have Rclosed=0 Timed Breaker Logic BRKB Closed@t0 Timed Breaker Logic BRKC Closed@t0 BRKA Vswita A A BRKB VS B Ib Vswitb R=0 B Ia C C Vswitc Vswitb Ic Vswitc A Ib In 1.E-8 [ohm] Ia Vswita Timed Breaker Logic Open@t0 BRKN In VLDC 0.002 [uF] Zground 0.002 [uF] Vn B C VLDA 0.002 [uF] VLDB Vn VS BRKN VLDA VLDB 0.2631 [H] 0.2631 [H] VLDC Ic A B C 0.2631 [H] BRKC Main : Line Currents Case 1: 0.30 Zn ohm Ia Ib Ic 0.20 Breaker currents y (kA) 0.10 0.00 -0.10 -0.20 -0.30 0.0600 0.0650 0.0700 0.0750 0.0800 0.0850 0.0900 0.0950 0.1000 0.1050 ECE 524: Transients in Power Systems Session 18; Page 12/19 Spring 2012 Main : V_n Neutral to ground voltage: Vn 15.0 10.0 5.0 y (kV) 0.0 -5.0 -10.0 -15.0 -20.0 -25.0 -30.0 0.0860 VLD A 40 Phase voltage after first switch clears 0.0870 0.0880 0.0890 0.0900 VLD B 0.0910 0.0920 0.0930 VLD C 30 20 y (kV) 10 0 -10 -20 -30 -40 Main : Graphs 40 Phase voltage after remaining switches clear: VLDA VLDB VLDC 30 20 y (kV) 10 0 -10 -20 -30 -40 0.0910 0.0915 0.0920 0.0925 0.0930 0.0935 ECE 524: Transients in Power Systems Session 18; Page 13/19 Spring 2012 Breaker voltages: Main : Switch Voltages 100 Vswita Vswitb Vswitc 80 60 y (kV) 40 20 0 -20 -40 -60 0.0860 0.0880 0.0900 VmaxA 84.472kV 0.0920 0.0940 0.0960 0.0980 0.1000 0.1020 0.1040 3Vm 84.51 kV Zn 0ohm Case 2: Repeat with Breaker currents: Main : Line Currents 0.30 Ia Ib Ic 0.20 y (kA) 0.10 0.00 -0.10 -0.20 -0.30 0.0775 0.0800 0.0825 0.0850 0.0875 0.0900 0.0925 0.0950 0.0975 0.1000 ECE 524: Transients in Power Systems Session 18; Page 14/19 Spring 2012 Neutral current Main : Graphs 0.050 In 0.000 y (kA) -0.050 -0.100 -0.150 -0.200 -0.250 0.0750 0.0800 0.0850 0.0900 0.0950 0.1000 0.1050 0.1100 Phase voltages: Main : Graphs 30 VLDA VLDB VLDC 20 y (kV) 10 0 - 10 - 20 - 30 0.0870 0.0880 0.0890 0.0900 0.0910 0.0920 0.0930 0.0940 M : Switch Voltages ain Breaker voltages: 60 Vswita Vswitb Vswitc 40 2 Vm 56.34 kV y (kV) 20 0 -20 Vmax 56.314kV -40 -60 0.0860 0.0880 0.0900 0.0920 0.0940 0.0960 0.0980 0.1000 0.1020 ECE 524: Transients in Power Systems Session 18; Page 15/19 Spring 2012 Case 3: Repeat with Cn 3 Cph Cn 6 nF Main : Line Currents Breaker currents: Ia 0.30 Ib Ic 0.20 y (kA) 0.10 0.00 -0.10 -0.20 -0.30 0.0775 0.0800 0.0825 0.0850 0.0875 0.0900 0.0925 0.0950 0.0975 0.0900 0.0905 0.0910 0.0915 0.1000 0.0920 Neutral current after first phase clears: Main : Graphs 0.0050 In 0.0040 0.0030 0.0020 y (kA) 0.0010 0.0000 -0.0010 -0.0020 -0.0030 -0.0040 -0.0050 0.0870 0.0875 0.0880 0.0885 0.0890 0.0895 Main : Gra p hs Neutral current when all phases clear: 0 .0050 In 0 .0040 0 .0030 0 .0020 y (kA) 0 .0010 0 .0000 - 0.0010 - 0.0020 - 0.0030 - 0.0040 - 0.0050 0 .0910 0 .0912 0 .0914 0 .0916 0 .0918 0.0920 0.0922 0.0924 0.0926 0.0928 0.0930 ECE 524: Transients in Power Systems Session 18; Page 16/19 Spring 2012 Neutral voltage: Main : V _n 20 Vn 10 y (kV) 0 -10 -20 -30 -40 0.0870 0.0880 0.0890 0.0900 0.0910 0.0920 0.0930 Phase voltages: Main : Graphs 50 VLDA VLDB VLDC 40 30 20 y (kV) 10 0 -10 -20 -30 -40 -50 0.0870 0.0880 0.0890 0.0900 0.0910 0.0920 0.0930 0.0940 ECE 524: Transients in Power Systems Session 18; Page 17/19 Spring 2012 Breaker voltages: Main : Switch Voltages 80 Vswita Vswitb Vswitc 60 40 y (kV) 20 0 -20 -40 -60 -80 0.0840 0.0860 0.0880 0.0900 0.0920 Vmax Vmax 78.833kV Vm Xcph 0.0960 0.0980 0.1000 0.1020 0.1040 2.8 3 Rgr = Xc0 Case 4: high resistance ground: 1 Xcph 2 60Hz Cph Rgr 0.0940 6 Xcph 1.33 10 Rgr 442.1 k 3 M : Switch Voltages ain 100 Vmax 84.449kV Vswita Vswitb Vswitc 80 Vm 60 3 40 y (kV) Vmax 20 0 -20 -40 -60 0.0850 0.0875 0.0900 0.0925 0.0950 0.0975 0.1000 0.1025 0.1050 0.1075 ECE 524: Transients in Power Systems Session 18; Page 18/19 Spring 2012 Case 5: Add grounding resistance (high resistance ground--limit In to 20A): Inmax 20A Rgr VLL Rgr 995.93 3 Inmax Breaker voltages: use 1000 ohm M : Switch Voltages ain 80 Vswita Vswitb Vswitc 60 Vmax 60.007kV 40 Vm 20 2.13 y (kV) Vmax 0 -20 -40 -60 0.0825 0.0850 0.0875 0.0900 0.0925 0.0950 0.0975 0.1000 0.1025 0.1050 0.1075 Case 6: Now try a low resistance ground (In = 500A) Inmax 500A Rgr VLL Rgr 39.84 3 Inmax use 40ohm M : Switch Voltages ain 60 Vmax 56.758kV Vm Vswitb Vswitc 40 2.01 20 y (kV) Vmax Vswita 0 -20 -40 -60 0.0850 0.0900 0.0950 0.1000 0.1050 0.1100 0.1150 0.1200 ECE 524: Transients in Power Systems Session 18; Page 19/19 Spring 2012 Case 7: Now try a low inductance ground (In = 500A) Inmax 500A Lgr magZgr magZgr 2 60Hz Vm 3 Inmax magZgr 39.84 Lgr 105.67 mH Vmax Vmax 73.795kV VLL 2.62 On first phase to clear A little higher after subsequent phases clear it damping neglected M :S ain witch Voltages 80 Vswita Vswitb Vswitc 60 40 y (kV) 20 0 -20 -40 -60 -80 0.0850 0.0900 0.0950 0.1000 0.1050 0.1100 0.1150 0.1200
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Chyna GoldenSeptember 29, 2011African-American HistoryPaper #1In todays world, we take for granted the many freedoms that we have been given. Wehave the ability to go and get a job and for the most part not be discriminated upon. We donthave to worr
Pittsburgh - COMMRC - 0500
Chyna GoldenJanuary 11, 2012ArgumentVideo Games ArgumentDoes NotConsider how the violent games are being considered, being should specific games andusers be considered or just in generalPaucity-scarcityResearch has tended to vary with social dista
Boise State - GEOS - 100
2/23/2012GEOS 100 Fundamentals of GeologyThursday, 23 February 2012Announcements:Reading Quiz 5 due Tuesday 28 Feb, 9:15 amLab this week: Presentation work timeLab Exam 1 coming up 5 9 March (week after next)!If you need a grade check, please give
Boise State - GEOS - 100
2/28/2012GEOS 100 Fundamentals of GeologyTuesday, 28 February 2012Announcements:Reading Quiz 5 due today; RQ 6 due 9:15 am Tu 6 MarchLab Exam 1 next week!How do earth materials change from one type to anotherthrough the Rock Cycle?How is sediment
Boise State - COMM - 101
Name_Activity Report #9Day 1: InformativeStudent&TopicWhat was the Link-the statement that made the topicrelevant to you?Student&TopicWhat was the AttentionGetter?What was the Thesisstatement?What Visual Aiddid you find mosteffective?W
Boise State - COMM - 101
Group One: ListeningFirst, discuss and review the concepts in chapter 4. What is the processof listening, what are the obstacles, and how can we listen for specific goals?Second, find 4-5 interviews to watch and analyze. (television, Youtube)These can
Boise State - COMM - 101
Group Presentation Grading RubricDid the group show comprehension of text concepts assigned in project?12345678910Did the group define 1.the purpose of the project, 2. how the projected was completed, and 3.its results?12345678910Di
Boise State - COMM - 101
Boise State - COMM - 101
Informative Grading Form(I will fill this out during your speech)Name:_Introduction: 1pt each-Arouses attention-Adapts to this audience-Establishes credibility-Provides a short, precise thesis statement-Provides a preview of main points.12345
Boise State - COMM - 101
Keyword Outline 25 pts-Follows standard outline format. Turning in an essay or bulleted items will result in a zero onthe assignment. 3pts_-Uses proper outline symbols to show levels of subordination. 3pts_-Utilizes rules of coordination and subordina
Boise State - COMM - 101
Group ProjectTeam MemberDirections:Fill out one of these forms for each member of your team. Make sure that you assignpoints for each category of performance. Make comments about the team membersperformance in each category. Be specific, describe beh
Boise State - COMM - 101
Speech Assessments-Criteria is the same for both the peer and self assessments-Due on the class day after speeches are completed (check schedule for date).-About 2 pages in length, essentially 3 paragraphs following criteria below-Standard formatting
Boise State - COMM - 101
Visual Aid ChecklistChecklist for preparation of visual aidso Easilyvisible for entire audience: large and dark?o Emphasizeso Balancedo Accurateo Limitedo Meetscentral or key ideas?and pleasing to the eye?and sharp?in scope (only necessary inf
Boise State - COMM - 101
Ricardo Ruiz-Gonzalez Jr.Comm 101-008 (11:40am)1/23/2012Communication Activity #1When I was done reading chapter two I came out with a different way of perceivingpeople. I applied the Sharpen Your Skill activity of Perceiving Others to a recent meeti