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Electrical Power Network of Funen Analyses, calculations, dimensioning, and protections. Blanca Naudín Aparicio Supervisors HeikoSteenbuchVester Arne FaaborgPoulsen Electrical Power Engineering Spring Semester 2011 SDU University of Southern Denmark 1 Electrical Power Network of Funen Sworn statement I hereby solemnly declare that I have personally and independently prepared this report. All quotations in the text have been marked as such, and the report or considerable parts of it have not previously been subject to any examination or assessment. Odense, on 29th of May of 2011. Blanca Naudín Aparicio 2 Electrical Power Network of Funen Table of Contents Introduction .................................................................................................................................. 5  Synopsis ............................................................................................................................. 5  Preface............................................................................................................................... 6 400-150 kV (High Voltage) Network of Funen .............................................................................. 8  Assumptions ...................................................................................................................... 9 I. Lines ............................................................................................................................... 9 II. Pylons ............................................................................................................................. 9 III. Feeders ........................................................................................................................ 11  Load Flow Calculations .................................................................................................... 11 I. Line FGD3-SVB3 ........................................................................................................... 11 a. Assumptions ............................................................................................................ 12 b. Mathematical calculations ...................................................................................... 13 c. NEPLAN .................................................................................................................... 17 II. Transformers................................................................................................................ 19 a. Considerations ......................................................................................................... 19 b. Calculations and Configurations .............................................................................. 19 III. Generators in Fynsværket............................................................................................ 22 a. Unit 3. FVO3 ............................................................................................................ 22 b. Unit 7. FVO5 ............................................................................................................ 24 c. Configurations and results ...................................................................................... 26 IV. Analyses. ...................................................................................................................... 27 V. Critical Operation Conditions ...................................................................................... 32 a. Network feeders ...................................................................................................... 32 b. Lines ......................................................................................................................... 33 c. Transformers ........................................................................................................... 37 d. Generators............................................................................................................... 40  Distance Protection ......................................................................................................... 43 I. Theoretical review ....................................................................................................... 43 a. Basis for dimensioning ............................................................................................ 43 b. Applied methods and practice ................................................................................ 47 II. Distance Protection of a string of the 150 kV network of Funen. ............................... 50 a. Maximum and minimum short circuit currents ...................................................... 50 b. Distance relays ........................................................................................................ 53 3 Electrical Power Network of Funen c. Conclusion ............................................................................................................... 57 60 kV (High Voltage) Network of Funen ...................................................................................... 58  Abildskov ......................................................................................................................... 58 I. Load data – 60 kV busbars ........................................................................................... 58 a. Power generation .................................................................................................... 58 b. Power losses ............................................................................................................ 59 c. Load profiles ............................................................................................................ 59 d. Load duration curve ................................................................................................ 60 e. Assumptions for 60 kV network calculations .......................................................... 60 I. Energy transferred in 150/60 kV transformers in Abildskov in 2010 .......................... 61 a. Load transformers ................................................................................................... 61 b. No-load transformers .............................................................................................. 66 II. Energy transferred in 150/60 kV two transformers in Abildskov - critical condition .. 66 a. Lossesduration curve ............................................................................................... 66 III. Economic aspects or conclusions ................................................................................ 67 a. Examples of duration curves analysis ..................................................................... 67 b. Power losses in transformers .................................................................................. 69 c. Power losses in network ......................................................................................... 70 IV. Conclusion .................................................................................................................... 70 10-0,4 kV (Low Voltage) Network of Funen ................................................................................ 71  Protection in Low Voltage Installations .......................................................................... 71 I. Introduction ................................................................................................................. 71 a. Protection of gear and cables against overload currents ....................................... 72 b. Protection of gear and cables against short circuit currents .................................. 73 c. Selectivity in electrical installations ........................................................................ 77 d. Protection from electric shock, directly and indirectly ........................................... 78 II. Fåborg .......................................................................................................................... 84 a. Cable dimension ...................................................................................................... 84 a. Swith gear for overload ........................................................................................... 93 b. Short circuit protection ........................................................................................... 99 c. Indirect contact protection ................................................................................... 105 d. FBY1-Q. Length of the 10 kV cable ........................................................................ 106 Conclusion ................................................................................................................................. 108 Literature ................................................................................................................................... 110 4 Electrical Power Network of Funen Appendix I. Abbreviations ........................................................................................................ 111 Appendix II. Values .................................................................................................................... 112 Appendix III. Low Voltage 1 ....................................................................................................... 114 Appendix IV. Low Voltage 2 ...................................................................................................... 117 Appendix V. LowVoltage 3 ........................................................................................................ 119 5 Electrical Power Network of Funen Introduction  Synopsis The overall purpose of this project is to achieve a high degree of understanding of electrical power systems, from low voltage to high voltage networks. This project is dealing with different topics about electrical power systems, and they could be load flow analyses, calculation of electrical line parameters, load profiles analyses, dimensioning and protection of low voltage installations, distance protection in high voltage networks, etc. There will be a demonstration both theoretical insight as well as the capabilities of using the theory in a practicable project. Among the subjects is the modeling of complex networks in NEPLAN. Finally, the purpose of the project is to communicate the results, background, applied theory, applied methods and the conclusions in a report. 6 Electrical Power Network of Funen  Preface1 Denmark’s transmission and distribution systems are built to connect not only its power systems, but also to supply or be supplied with other power plants abroad. There are denominated electricity motorways (transmission lines) which supplied electricity to consumer by generation nodes. These complex transmission and distribution systems consist on facilities with different rated voltages: 400, 150, and 132 kV. It is also divided into two different areas, which have obviously different sizes and which are, at the same time, non-synchronous. It means they have to be managed differently:  Jutland and Funen belong to the western Denmark, which forms part of the synchronous area of the European continent (UCTE). Here, the transmission grid is operated at 400 kV with a combination of ring connections and radial structure, and at 150 kV as a parallel grid. The ring connection is the most useful tool when finding failures on the network elements.. It is connected to the UCTE synchronous area at the German border via 400 kV, 220 kV and 150 kV AC lines (1200 import/800 export MW). The Western Danish system is also connected to the Nordel synchronous area, which includes Sweden, Norway and Finland, via HVDC links to Norway (1000 MW) and Sweden (720 MW), which makes exchange of energy of energy possible without it being synchronized.  The eastern Denmark (Zealand) represents a part of another synchronous area, the Nordel. The eastern transmission network is composed by a 400 kV radial grid and a 132 kV ring connected grid. It is connected to Sweden via AC lines (1700 import/1300 export MW) and through a HVDC connection to Germany (600 MW). Energinet.dk owns the 400 kV installations and the international connections, whereas the 150/132 kV installations are owned by the regional transmission companies, which make the 150/132 kV grids available to Energinet.dk. However, Energinet.dk owns the 132 kV grid in northern Zealand. There are several elements on the system, which are:  Electricity consumption - the use of electricity at home and at work.  Electric power production - is composed of many different types of production units, such as large central power plants, small CHP plants and renewable energy sources, mainly wind power.  National transmission system - including the largest high-voltage lines and cables in Denmark, which operated between 132,000 and 400,000 volts.  International connections - connecting Denmark with my neighbors.  Elm arched - electricity markets ensure that energy is sold at the right price.  Love, frameworks and rules of operation of the electrical system.  Tools - requires many computer systems for information system power and optimal utilization. 1On wind power integration into electrical power system: Spain vs. Denmark http://www.energinet.dk 7 Electrical Power Network of Funen The overall objective of system operation is to secure high and do this as efficiently as possible. Security of supply - there is always power in the plug - is an essential prerequisite for a functioning society. Power is perishable is the ultimate global product and should be used in the same second, it is produced. As the company responsible for system operators (TSOs) Energinet.dk is responsible for ensuring that there is always a balance between consumption and production. That task is growing steadily in scope, the development of wind energy. 8 Electrical Power Network of Funen 400-150 kV (High Voltage) Network of Funen The 400-150 kV (High Voltage) network of Funen and the south-eastern part of Jutland is shown in the Figure 1. As I can see, there are different lines supplied with different voltages:  Red line: 400 kV  Black line: 150 kV  Blue dotted line: HVDC connection to Zealand  Green Line: 400 kV Connection to Tyskland (Germany) Figure 1.400-150 kV network of Funen and the south-eastern part of Jutland. In this report, you will find some abbreviations for the name of the different cities, that is, for each place, I will have 3 capital letters, followed by a number (1, 2, 3, or 5), which depends on the rated voltage level. There are few examples bellow:  LAG5: Landerupgård, with 400 kV like nominal voltage (~410kV)  FGD3: Fraugde, with 150 kV like nominal voltage (~165kV)  SØN2: Sønderborg, with 60 kV like nominal voltage (~66kV)  FBY1: Fåborg, with 10 kV like nominal voltage (~12kV) You will find a description for all the abbreviations of the cities at the Appendix I. Abbreviations 15 Electrical Power Network of Funen frequency effect is that it reduces the effective cross-section area used by the current, and the effective resistance increases. 𝑅𝐴𝐶=𝑅𝐴𝐶·𝑘= 1′61 · 1′02 = 1′64Ω 𝑘=1’02 for 60 Hz frequency.  Inductance of transposed three-phase transmission lines With the actual transmission lines technology and because of the construction considerations, the phase conductors cannot handle symmetrical arrangement along the length. If the spacing of the length is different, then the inductance will be different for each phase, also the voltage is unbalanced in each conductor. In a transmission line I can assume a symmetrical arrangement in it by transposing the phase conductors. Each cable occupies the location of the other two phases for one third of the total line length. For my calculations I have to know the GMD, which means the average distance geometrical mean distance substitutes distance D. Figure 3. Distance geometrical mean distance description. In a transmission line the inductance per phase per unit length is: 𝐿𝑝𝑕𝑎𝑠𝑒=µ0 2𝜋𝑙𝑛 𝐺𝑀𝐷 𝐺𝑀𝑅𝑝𝑕𝑎𝑠𝑒 However, before I have to calculate the GMD, and the GMR. To calculate the GMD I need the drawing below. 16 Electrical Power Network of Funen Figure 4. 150 kV pylon description. We calculated the distances with the simple Pythagoras theorem: 𝐷𝐴−𝐵= 27502+62502=6828,25 𝑚𝑚 𝐷𝐵−𝐶= 27502+77502=8385,44 𝑚𝑚 𝐷𝐴−𝐶= 15002+55002=5700,88 𝑚𝑚 𝐺𝑀𝐷= 𝐷𝐴−𝐵.𝐷𝐵−𝐶.𝐷𝐴−𝐶 3= 6828 ·8385 ·5700 3=6885,34 And the GMR is easy to find in the table below: 17 Electrical Power Network of Funen Table 6.GMR for Martin type conductor for overhead lines. Finally I can calculate the line inductance per phase: 𝐿𝑝𝑕𝑎𝑠𝑒=µ0 2𝜋𝑙𝑛 𝐺𝑀𝐷 𝐺𝑀𝑅𝑝𝑕𝑎𝑠𝑒 =4𝜋10−7 2𝜋𝑙𝑛 6884,34 14′62 = 1′23 µ𝐻𝑚 Afterwards, I can calculate theinductive reactance per unit length the inductance per phase, which will be related to the line inductance per phase in this way: 𝑋𝐿𝑝𝑕𝑎𝑠𝑒 = 2𝜋𝑓𝐿𝑝𝑕𝑎𝑠𝑒= 2𝜋.50.1′23.10−6= 0′386.10−3Ω𝑚 c. NEPLAN For calculating the line FGD3-SVB3 by using NEPLAN, I have to know the data for the different kind of conductor, like its dimension, its material properties, or even the pylon characteristics. The phase conductors are steel-aluminum, STAL (ACSR), type Martin, simplex (one conductor per phase). The earth conductor is steel-aluminum, STAL (ACSR), type Partridge. And as I assumed before, a conductor temperature of 20:C is assumed. So the conductor data are: 18 Electrical Power Network of Funen Table 7. Conductor selected data for FGD3-SVB3 line. The parameters I will have to introduce for the phase conductor will be: Table 8. Phase conductor configuration for the Line FGD3-SVB3. Phase conductor Conductors per bundle Distance 2*GMR R20 Sag (cm) (cm) (Ω/Km) (m) Martin 1 0 2,924 0,0423 2,0125 The distance considered is equal to zero, due to I will have only one conductor per line. The sag I have considered for the line will be at least the 7%. So I have to calculate the 7% of the higher height of all the lines, what be the height of the L1 (28,75 m). 𝑆𝑎𝑔 𝑚 = 7% · 𝐿1𝑦−𝑎𝑥𝑖𝑠=7 100 ·28,75 = 2,0125 (𝑚) The data for the earth conductor I will have to introduce will be: Table 9.Earth conductor configuration for the Line FGD3-SVB3. Earth conductor x y R 2*GMR (m) (m) (Ω/Km) (cm) Partridge 3,125 35,5 0,214 1,3214 The calculated values I have obtained: Table 10.NEPLAN calculation for the Line FGD3-SVB3. Line Lenght 𝑹𝟏 𝑿𝟏 𝑪𝟏 𝑩𝟏 𝑹𝟎 𝑿𝟎 𝑪𝟎 𝑩𝟎 (Km) (Ω/Km) (Ω/Km) (uF/Km) (uS/Km) (Ω/Km) (Ω/Km) (uS/Km) (uS/Km) FGD3- SVB3 38 0,04235 0,38631 0,0091 2,853 0,19034 1,3125 0,00461 1,448 19 Electrical Power Network of Funen II. Transformers a. Considerations All 400/150 kV transformers are direct earthed YNyn0 autotransformers. All 150/60 kV transformers are YNd transformers. Earthing of the 150/60 kV transformers depend on the 𝑍0𝑍1 ratio from faults in the 150 kV network. Typically, at least one transformer in each 150/60 kV station is direct earthed. b. Calculations and Configurations  Voltage, and X/R ratios When talking about unit transformers, I will have to calculate the X/R ratio value. This value will be important for the short circuit calculations, due to determining the total impedance of the circuit will be the key element of it. This impedance is represented like complex impedance, with a module and an angle, where the module is just the hypotenuse of the right angle which you can see below. The angle of this complex impedance will be represented with the phi symbol(𝜑). 𝑡𝑎𝑛−1 𝑋𝑅 =𝜑 𝑠𝑖𝑛 𝜑 ·𝑍=𝑋 𝑐𝑜𝑠 𝜑 ·𝑍=𝑅 𝑍= 𝑅2+𝑍2 Figure 5. Impedance representation A very common assumption is to use an X/R ratio between 12 and 15. I will see how the values for all the transformers (400-150 kV, and 150-60 kV) will be around 30-50, what will be consider like low-medium value in a NEPLAN scale, where the ratio values are: Low value (~35), Medium value (~40), and High value (~60). For the calculation of the transformers, I have to follow the next steps with the example FGD5- FGD3: 𝑈𝑅 1 = 0.23% 𝑈𝑋 1 =12.5% 𝑈𝐾 1 = 𝑈𝑅 1 2+𝑈𝑋 1 2= 0.232+12.52→𝑈𝐾 1 =12.5% We could check the values for the positive sequence impedance: 𝑅 1 =𝑈𝑅 1 100 ·𝑈2 2 𝑆𝑟=0.23 100 ·1682 400 = 0.162Ω ; 𝑋 1 =𝑈𝑋(1) 100 ·𝑈2 2 𝑆𝑟=12.5 100 ·1682 400 = 8.82Ω We could check the values for the zero-sequence also 𝑅 0 =𝑈𝑅(0) 100 ·𝑈2 2 𝑆𝑟=0.2 100 ·1682 400 = 0.141Ω ; 𝑋 0 =𝑈𝑋(0) 100 ·𝑈2 2 𝑆𝑟=11.9 100 ·1682 400 = 8.39Ω 20 Electrical Power Network of Funen And the ratio between the inductance and the resistance, I will have: 𝑋(1) 𝑅(1) =8.82 0.162 =54.44 𝑋(0) 𝑅(0) =8.39 0.141 =59.50 The next table shows the results of the short-circuit voltage for the positive and the zero sequences [𝑈𝐾 1 , 𝑈𝐾 0 ], and the ratios between the inductance and the resistance for the different sequences for all the transformers: Table 11.Transformer data.Voltage, and inductance/resistance ratios. We will have to decide which kind of configuration I will want to have in my transformers. I already know that all the 150-60 kV transformers are YNd, but they could be YNd1, YNd5 or YNd11. The number at the end depends on that for different kind of connections, I will have different phase between the voltage in the primary side and the voltage on the secondary side. This value represents how the secondary side is retarded respect to the primary side, and it will be a multiple of 30°. The different configuration descriptions are:  YNd1. The secondary side is 30° backward, or 330° forward  YNd5. The secondary side is 150° backward, or 210° forward  YNd11. The secondary side is 30° forward, or -330° backward Trafo. Trafo nr. X(1)/R(1) X(0)/R(0) From-to MVA kV kV % % % % % % FGD5-FGD3 1 400 410 168 0,23 12,5 12,50 0,2 11,90 11,90 54,35 59,50 FGD5-FGD3 2 400 410 168 0,23 12,5 12,50 0,2 11,90 11,90 54,35 59,50 KIN5-KIN3 400 410 168 0,23 12,5 12,50 0,2 11,90 11,90 54,35 59,50 ABS3-ABS2 A 125 165 66 0,37 13,5 13,51 0 12,30 12,30 36,49 - ABS3-ABS2 B 125 165 66 0,32 13,5 13,50 0 12,30 12,30 42,19 - FGD3-FGD2 A 125 165 67 0,30 12,7 12,70 0 10,60 10,60 42,33 - FGD3-FGD2 B 125 165 66 0,37 13,5 13,51 0 12,30 12,30 36,49 - FVO3-FVB2 1 150 160 66 0,34 13,6 13,60 0 13,50 13,50 40,00 - FVO3-FVA2 2 150 160 66 0,34 13,6 13,60 0 13,50 13,50 40,00 - FVO3-FVA2 3 180 160 66 0,31 12,7 12,70 0 12,60 12,60 40,97 - GRP3-GRP2 A 125 165 66 0,37 13,5 13,51 0 12,30 12,30 36,49 - GRP3-GRP2 B 75 158 66 0,40 12,8 12,81 0 10,60 10,60 32,00 - OSØ3-OSØ2 125 165 66 0,37 13,5 13,51 0 12,30 12,30 36,49 - SVB3-SFV2 A 125 165 66 0,37 13,5 13,51 0 12,30 12,30 36,49 - SVB3-SFV2 B 125 165 66 0,37 13,5 13,51 0 12,30 12,30 36,49 - SØN3-SØN2 1 75 158 66 0,36 11,7 11,71 0 9,50 9,50 32,50 - SØN3-SØN2 2 125 165 67 0,28 11,8 11,80 0 11,60 11,60 42,14 - FBY2-FBY1 1 30 67 10 0,31 10,4 10,40 0,31 9,80 9,80 33,55 31,61 FBY2-FBY1 2 30 67 10 0,31 10,4 10,40 0,31 9,80 9,80 33,55 31,61 𝑆 𝑈 𝑈 𝑈 𝑈 𝑈 𝑈𝑅 𝑈 𝑈 21 Electrical Power Network of Funen Figure 6.YNd transformer representation. In this point, I can assume which kind of connection I are going to use by knowing that I will consider my transformer like ideal, with symmetrical load, and in a direct sequence. 𝑈 𝑅𝑆,1 =𝑈 𝑈,1 −𝑈 𝑉,1 𝑈 𝑅𝑆,2 =𝑈 𝑈,2 𝑈 𝑅𝑆,1 𝑈 𝑅𝑆,2 = 𝑈 1 1∠0−1∠−120 𝑈 2 ∠0= 3 𝑈 1 ∠30 𝑈 2 ∠0= 3𝑁1 𝑁2∠30 So now I can say that I will use the configuration YNd1, where the secondary side is 30° backward, or 330° forward respect to the primary side.  Regulation When working with NEPLAN, I have to introduce the values for the different parameters of each transformer. I will have to make also a regulation, and it will depend on the secondary side, which I consider the controlled node is located. This way, I will have to introduce the minimum (𝑈2,𝑚𝑖𝑛), and the maximum (𝑈2,𝑚𝑎𝑥) values which I could measure in the secondary side. We will have to configure the “tap position”, where I will choose the different voltages in a total of 21 different positions. The main positions are:  Position 1. Minimum voltage value 𝑈2,𝑚𝑖𝑛  Position 11. Rated voltage value 𝑈2  Position 21. Maximum voltage value 𝑈2,𝑚𝑎𝑥 The next table shows the values I have to introduce in NEPLAN for having well configured transformers. For a normal Load Flow Calculation, I used tap 11, what means nominal voltage values. 22 Electrical Power Network of Funen Table 12.Description of transformer tap positions. As far as I know, the minimum value will be less than the nominal value, and it will be around the 80% of this rated value. On the other hand, the maximum value will be higher than the nominal value, and it will be up to 120% of this value. The tap position for the transformers is not a relevant value for my analyses. However, I decided to calculate and configure them on NEPLAN. III. Generators in Fynsværket ‘’Today more than 85,000 households and about 7,000 institutions (including large-scale greenhouses) and companies are supplied with district heating from Fyn Power Station. It supplies more than 98% of the total district heating available in the heating network – about 78% is generated at Units 3 and 7 and about 20% at Odense CHP Plant. Fyn Power Station consists of three plant units and the Odense CHP Plant. Unit 3 was built in 1974 and decommissioned on 1 April 2010. Unit 7 was built in 1991 and uses coal and oil as fuel. The two large-scale plant units at Fyn Power Station, Units 3 and 7, burn about 800,000 tons of coal annually, most of which is brought to the power station on coal barges. Oil is basically only used for starting up the plants, while consumption of natural gas depends on supply and price, which vary substantially’’2 a. Unit 3. FVO3 The generator on Fynsværket, Unit 3, supplies power to the 150 kV busbar FVO3 via a YNd11 coupled unit transformer. The configuration for the generator and the unit transformer are: Table 13.Generator and unit transformer data, Unit 3 on Fynsværket. Generator Unit transformer, YNd11 SG = 300 MVA xd” = 23 % ST = 335 MVA xk(1) = 11,5% UG = 18,0 kV xd´ = 34 % U1 = 18,0 kV xk(0) = 9,8% U2 = 170 kV xyn = 0,0% 2Fyn Power Station, Vattenfall A/S. Trafo. Trafo nr. From-to kV kV kV FGD5-FGD3 1 168 135 194 10,05 9,56 12,50 11,90 14,44 13,74 FGD5-FGD3 2 168 135 194 10,05 9,56 12,50 11,90 14,44 13,74 KIN5-KIN3 168 135 194 10,05 9,56 12,50 11,90 14,44 13,74 ABS3-ABS2 A 66 54 78 11,05 10,06 13,51 12,30 15,96 14,54 ABS3-ABS2 B 66 54 78 11,05 10,06 13,50 12,30 15,96 14,54 FGD3-FGD2 A 67 59 78 11,19 9,33 12,70 10,60 14,79 12,34 FGD3-FGD2 B 66 54 78 11,05 10,06 13,51 12,30 15,96 14,54 FVO3-FVB2 1 66 54 78 11,13 11,05 13,60 13,50 16,08 15,95 FVO3-FVA2 2 66 54 78 11,13 11,05 13,60 13,50 16,08 15,95 FVO3-FVA2 3 66 54 78 10,39 10,31 12,70 12,60 15,01 14,89 GRP3-GRP2 A 66 54 78 11,05 10,06 13,51 12,30 15,96 14,54 GRP3-GRP2 B 66 54 78 10,48 8,67 12,81 10,60 15,13 12,53 OSØ3-OSØ2 66 54 78 11,05 10,06 13,51 12,30 15,96 14,54 SVB3-SFV2 A 66 54 78 11,05 10,06 13,51 12,30 15,96 14,54 SVB3-SFV2 B 66 54 78 11,05 10,06 13,51 12,30 15,96 14,54 SØN3-SØN2 1 66 58 75 10,29 8,35 11,71 9,50 13,30 10,80 SØN3-SØN2 2 67 59 76 10,39 10,21 11,80 11,60 13,39 13,16 FBY2-FBY1 1 10 8,5 11,5 8,84 8,33 10,40 9,80 11,97 11,28 FBY2-FBY1 2 10 8,5 11,5 8,84 8,33 10,40 9,80 11,97 11,28 Position 11 (Tap r) Position 1 (Tap min) Position 21 (Tap max) 𝑈 𝑈 𝑈 𝑈 (%) 𝑈 (%) 𝑈 (%) 𝑈 (%) 𝑈 (%) 𝑈 (%) 23 Electrical Power Network of Funen We could calculate now the different parameters of the generator which I will need for calculating the short-circuit currents in the different lines. The parameters I have to calculate in this point are: the saturated synchronous reactance (xd sat), the negative sequence reactance (x(2)), and the zero sequence reactance (x(0)) of the synchronous machine. If I consider the saturated transient reactance, the recommended values for the different kind of asynchronous machine are: Turbo-SM: (1.4,1.7)*xd", Salient pole with amortisseur (damper) winding: 20-45%. The provided value is xd’=34%, which is between the 20-45%, and also it is 1,4783 times the saturated subtransient reactance, what represents the 26.1% in this interval. By using this percent, I will calculate the rest of the parameters.  Saturated synchronous reactance (xd sat). NEPLAN recommends values for the different kind of generator, and these values are: Turbo-SM: 120-270, Salient pole-SM: 70-130. I could consider a value between the two intervals, what could be xd sat=120%.  Negative sequence reactancex(2). The recommended value for NEPLAN is x(2) = xd", so I will consider x(2)=23%  Zero sequence reactancex(0). The recommended value for NEPLAN is x(0) = (0.4,0.8)*xd". If I use the 26.1%, the obtained value will be x(0)=10.44% As I can see, the value in the secondary side is a maximum value. I talked before about the tap position, and I will have to use it again in this point, so I will make the next assumptions:  Position 21. Maximum voltage value ~170 kV  Position 11. Rated voltage value ~150 kV  Position 1. Minimum value ~130 kV Now I can calculate of the impedance and the resistance, and as I saw in the transformer calculations, I have to calculate the values for this unit transformer the same way I did before: 𝑈𝑅 1 =? % 𝑈𝑋 1 =? % 𝑈𝑅 1 = 0.07𝑈𝑋 1 𝑈𝐾 1 =11.5% 𝑈𝐾 1 = 𝑈𝑅 1 2+𝑈𝑋 1 2= 0.07 + 1 𝑈𝑋 1 2 𝑈𝐾 1 2= (0.07 + 1)𝑈𝑋 1 2 𝑈𝑋 1 2=𝑈𝐾 1 2 0.07 + 1 →𝑈𝑋 1 = 𝑈𝐾 1 2 (0.07 + 1) →𝑈𝑋 1 = 11.5 100 2 (0.07 + 1) 𝑼𝑿 𝟏 =𝟏𝟏.𝟏𝟏%→𝑼𝑹 𝟏 =𝟎.𝟎𝟕𝑼𝑿 𝟏 →𝑼𝑹 𝟏 =𝟎.𝟕𝟖% Now I could calculate the values for the positive sequence impedance: 𝑅 1 =𝑈𝑅 1 100 ·𝑈2 2 𝑆𝑟=0.78 100 ·1502 335 = 0.52Ω ; 𝑋 1 =𝑈𝑋(1) 100 ·𝑈2 2 𝑆𝑟=11.11 100 ·1502 335 = 7.46Ω We could calculate the values for the zero-sequence also: 24 Electrical Power Network of Funen 𝑈𝑅 0 =? % 𝑈𝑋 0 =? % 𝑈𝑅 0 = 0.07𝑈𝑋 0 𝑈𝐾 0 = 9.8% 𝑈𝐾 0 = 𝑈𝑅 0 2+𝑈𝑋 0 2= 0.07 + 1 𝑈𝑋 0 2 𝑈𝑋 0 = 9.8 100 2 (0.07 + 1) 𝑼𝑿 𝟎 =𝟗.𝟒𝟕%→𝑼𝑹 𝟎 =𝟎.𝟎𝟕𝑼𝑿 𝟎 →𝑼𝑹 𝟎 =𝟎.𝟔𝟔% 𝑅 0 =𝑈𝑅(0) 100 ·𝑈2 2 𝑆𝑟=0.66 100 ·1502 335 = 0.44Ω ; 𝑋 0 =𝑈𝑋 0 100 ·𝑈2 2 𝑆𝑟=9.47 100 ·1502 335 = 6.36Ω And the ratio between the inductance and the resistance, I will have: 𝑋(1) 𝑅(1) =7.46 0.52 =14.29 𝑋(0) 𝑅(0) =6.36 0.44 =14.45 In order to follow my calculations, I have decided to apply the calculated values, although it is less than the half of the regular low value in NEPLAN (~35).  Power configuration When working with node type for Load Flow Calculation (LF), I have chosen the “PQ” value, due to I have the assumed power factor, and the apparent power (Sir), so I could calculate the real power (P), and the reactive power (Q) by using the power triangle. We will assume a pf = 0.8, so I can calculate the real power to introduce this value on NEPLAN: 𝑆=𝑃+𝑗𝑄 𝑃=𝑆·𝑐𝑜𝑠 𝜃𝑣−𝜃𝑣 =𝑆·𝑝𝑓 𝑆=300𝑀𝑉𝐴 𝑝𝑓= 0.8 𝑃=300 · 0.8 = 240 𝑀𝑊 𝑄=𝑆·𝑠𝑖𝑛 𝜃𝑣−𝜃𝑣 𝑄=300 · 0.6 = 180 𝑀𝑉𝐴𝑟 Figure 7. Powertriangledescription b. Unit 7. FVO5 The generator on Fynsværket, Unit 7, supplies power to the 400 kV busbar FVO5 via a YN/auto/d coupled unit transformer. The configuration for the generator and the unit transformer are: 31 Electrical Power Network of Funen Table25. Possiblecriticaltransformers.  FVO3-FVB2 (1), FVO3-FVA2 (2). These transformers have different values from the others: Sr=150 MVA, U1=160 kV. This is because the generator on Fynsværket, Unit 3, which supply power to the 150 kV busbar FVO3.  FVO3-FVA2 (3). As I can see, the power that this transformer can handle is up to 180 MVA. It means that this one is the main transformer between FVO3 and FVA2.  GRP3-GRP2 (B), SØN3-SØN2 (1). I are in the same conditions: Sr=75MVA, U1=158 kV. Both have a partner transformer with the next values: Sr=125 MVA, U1=165 kV. As I know, the transformers I are studying won’t be able to handle the entire load if the “Transformer A” between the nodes is not working. I will have to check about the voltage out, because is less than the nominal value Table 26.Network feeder data of 400-150 kV Network of Funen for normal conditions. These results have been obtained, like I have said before at the Feeder Assumptions, by using a 50% Slack in each feeder Table 27.Asynchronous machine data of 400-150 kV Network of Funen for normal conditions. The angle of the load data just shows me the YNd1 configuration of the transformers on these places. Table 28. Load data of 400-150 kV Network of Funen for normal conditions. Trafo. Trafo nr. From-to MVA kV kV FGD5-FGD3 1 400 410 168 FGD5-FGD3 2 400 410 168 KIN5-KIN3 400 410 168 ABS3-ABS2 A 125 165 66 ABS3-ABS2 B 125 165 66 FGD3-FGD2 A 125 165 67 FGD3-FGD2 B 125 165 66 FVO3-FVB2 1 150 160 66 FVO3-FVA2 2 150 160 66 FVO3-FVA2 3 180 160 66 GRP3-GRP2 A 125 165 66 GRP3-GRP2 B 75 158 66 OSØ3-OSØ2 125 165 66 SVB3-SFV2 A 125 165 66 SVB3-SFV2 B 125 165 66 SØN3-SØN2 1 75 158 66 SØN3-SØN2 2 125 165 67 FBY2-FBY1 1 30 67 10 FBY2-FBY1 2 30 67 10 𝑆 𝑈 𝑈 Node Element P Q I Angle I Name Name MW MVar kA ° LAG5 N1 235,517 202,617 0,437 -40,7 SHE3 N2 -51,176 -3,35 0,179 176,3 Node Element P Q I Angle I Name Name MW MVar kA ° Unit 3 Unit3 -240 -180 8,141 146,5 Unit 7 Unit7 -391,2 -293,4 12,207 150,8 Node Element P Q I Angle I Name Name MW MVar kA ° ABS2 ABS2 58 29 0,574 -29,8 FGD2 FGD2 58 29 0,557 -28,3 FVA2 FVA2 77 39 0,723 -28,5 FVB2 FVB2 53 27 0,503 -29,6 GRP2 GRP2 47 24 0,451 -29 OSØ2 OSØ2 45 22 0,438 -28,9 SFV2 SFV2 58 29 0,573 -29,9 SØN2 SØN2 47 24 0,459 -29,7 32 Electrical Power Network of Funen V. Critical Operation Conditions We will consider some situations, with different operation conditions of the network where some of the lines, transformers or generators are disconnected due to maintenance or faults. These situations are considered realistic, what means that all the 150kV busbars are still supplied with power. In this point I will have to work with the Normal Operation and (N-1)- Security Check. ‘’There is a difference between the optimization of the system in the normal operation case and in the case of line failures. Several load flow calculations are performed in order to estimate the ranges of values of the node voltages and the element loadings. The results are the ranges of values of the node voltages and the maximum loadings of the elements (lines, cables, transformers). The (N-1)-security check simulates a failure for each element of the considered voltage level. If some part of the system is out of service because of this failure, the procedure changes the network topology in order to resupply this part and to minimize losses (optimal separation point procedures). An evaluation of the new network state is done after each failure. The results are the ranges of values of the node voltages and the maximum loadings of the elements (lines, cables, transformers)’’.3 a. Network feeders We will study the two possible situations, which one of them is not completely probable. That is when the 410 kV network feeder (N1) will be disconnected. However, I will consider it. These will be the possible situations due to having both feeders switched off, it will be something that cannot happen. The network has to be supplied.  Network Feeder N1 is disconnected Like I mentioned before, the disconnection of the network feeder N1 will not happen. It will mean a 400 kV broken underground cable. I could check how the power losses increased for this situation, and it was around 3.5 times bigger than for normal conditions. Table 29.400/150 kV Network of Funen values for disconnected Network Feeder N1. Table 30. Power and Losses data of 400-150 kV Network of Funen for disconnected Network Feeder N1. 3 Normal Operation and (N-1)-Security Check, NEPLAN Tutorial From P Loss Q Loss P Imp Q Imp P Gen Q Gen P Load Q Load Area/Zone MW MVar MW MVar MW MVar MW MVar Network 26,679 72,634 -161,521 -177,766 631,2 473,4 604,521 400,766 Area 1 26,679 72,634 0 0 631,2 473,4 604,521 400,766 Zone 1 26,679 72,634 0 0 631,2 473,4 604,521 400,766 Un P Loss Line Q Loss Line kV MW MVar 165 24,894 47,88 410 0,288 -54,783 0,586 P Loss Transformer Q Loss Transformer MW MVar 26,807 52,73 0,911 33 Electrical Power Network of Funen Table 31. Node and line overload data of 400-150 kV Network of Funen for disconnected Network Feeder N1. As I can see, this situation will be a complete disaster. All the nodes are violating the upper voltage limits, except from 2, which have an overvoltage of 107,83%, and 109,26%. The rest of the nodes, are around 130% up, what means about 550kV at the 410kV nodes, practically impossible to manage.  Network Feeder N2 is disconnected This case could be more possible than the circumstance I studied before. In this situation, the network works perfectly, like for normal conditions delivering power to the grid, and even the network feeder for the 400-150 kV side (N1) will be delivering power to the grid. Table 32.400/150 kV Network of Funen values for disconnected Network Feeder N2. Table 33. Power and Losses data of 400-150 kV Network of Funen for disconnected Network Feeder N2. The voltage drops on the nodes are around 97,63% to 104,12%, what meets the ±10% rule I have considered. b. Lines As I know, for realistic considerations, all the 150kV busbars are still supplied with power, and this is the definition of n-1 elements. I cannot work when one node is not supplied. We will analyze only the really extreme situations, due to there is a really large number of plausible states. After trying a few numbers of possible problems, I will show only the problematic ones, where there will some overloaded elements or violated upper or lower voltage limits, what means around ±10% of the rated voltage in each element. The different values for the different nodes can be seen below. FVA2 136,59 LAG5 134,65 GRP2 136,5 KIN5 134,54 FVB2 135,86 FGD3 134,1 KIN3 135,65 FGD2 133,9 GRP3 135,43 OSØ3 133,69 FVO5 135,25 FVO3 133,62 FGD5 134,69 OSØ2 131,73 ABS3-SØN3 222,27 ABS3-FVO3 112,41 Overloads Nodes (upper) % Lines From P Loss Q Loss P Imp Q Imp P Gen Q Gen P Load Q Load Area/Zone MW MVar MW MVar MW MVar MW MVar Network 4,534 56,786 -183,666 -193,614 631,2 473,4 626,666 416,614 Area 1 4,534 56,786 0 0 631,2 473,4 626,666 416,614 Zone 1 4,534 56,786 0 0 631,2 473,4 626,666 416,614 Un P Loss Line Q Loss Line kV MW MVar 165 1,612 -30,454 410 0,88 -21,996 1,135 67,042 0,906 42,194 P Loss Transformer Q Loss Transformer MW MVar 34 Electrical Power Network of Funen Table 34. ±10% limits for the 410kV, 165kV, and 66kV nodes. Few possible situations are shown on the Appendix Excel 400-150 kV Network, but I will study here only some of them.  410 kV lines are disconnected In this case, I have checked the possible situations were one or two 410kV lines are disconnected, and I have seen that it works without any problem. These examples are shown in Appendix Excel 400-150 kV Network  ABS3-FVO3, and FG3-SVB3 are disconnected For this condition, I will work with two different networks due to one of them will be formed with the 150kV busbars in Enstedværket, Sønderborg, Abildskov, and Svendborg. It will mean that I won’t have one ring, I will have two radial networks instead. Figure 8. 150-60 kV Network of Funen when ABS3-FVO3, and FG3-SVB3 are disconnected. Table 35. 150-60 kV Network of Funen when ABS3-FVO3, and FG3-SVB3 are disconnected. Table 36. Power and Losses data of 150-60 kV Network of Funen when ABS3-FVO3, and FG3-SVB3 are disconnected. node plus 10% minus 10% kV kV kV 410 451 369 165 181,5 148,5 66 72,6 59,4 From P Loss Q Loss P Imp Q Imp P Gen Q Gen P Load Q Load Area/Zone MW MVar MW MVar MW MVar MW MVar Network 11,65 85,433 -176,55 -164,967 802,281 588,462 790,631 503,029 Area 1 11,65 85,433 0 0 802,281 588,462 790,631 503,029 Zone 1 11,65 85,433 0 0 802,281 588,462 790,631 503,029 Un P Loss Line Q Loss Line kV MW MVar 165 8,159 -4,714 410 1,522 -14,39 0,931 42,801 1,038 61,736 P Loss Transformer Q Loss Transformer MW MVar 35 Electrical Power Network of Funen Table 37.Line overload data of 150-60 kV Network of Funen when ABS3-FVO3, and FG3-SVB3 are disconnected. Table 38. Line data of 150-60 kV Network of Funen when ABS3-FVO3, and FG3-SVB3 are disconnected. ABS3-SØN3 119,9 Overloads Line Node Element P Q I Angle I Loading P Fe P Comp Name Name MW MVar kA ° % MW MW ABS3 ABS3-FVO3 0 0 0 90 0 0,0001 -2,7414 ABS3 ABS3-SVB3 58,547 34,603 0,273 -38,3 27,58 0,4555 2,2622 ABS3 ABS3-SØN3 -116,627 -66,736 0,54 142,5 119,9 4,4778 7,8484 FGD3 FGD3-OSØ3 -8,816 -21,73 0,079 113,1 6,78 0,0064 -0,222 FGD3 FGD3-FVO3 -17,66 -31,393 0,121 120,4 15,9 0,0226 -6,5966 FGD3 FGD3-SVB3 0 -3,21 0,011 91 0,78 0,0001 -3,2095 FGD5 FGD5-LAG5 176,89 129,312 0,302 -34,9 18,9 0,5453 -8,265 FGD5 FGD5-FVO5 -389,736 -229,822 0,624 150,8 39,03 0,4309 2,1358 FGD5 FGD5-KIN5 181,263 125,392 0,304 -33,4 19,01 0,2976 -4,6786 FVO3 FGD3-FVO3 17,683 24,796 0,102 -53,4 13,4 0,0226 -6,5966 FVO3 FVO3-OSØ3 53,935 39,766 0,224 -35,3 31,13 0,0415 -6,3536 FVO3 FVO3-GRP3 37,883 25,27 0,152 -32,6 21,16 0,2364 -1,7003 FVO3 ABS3-FVO3 0 -2,741 0,009 91,1 1,21 0,0001 -2,7414 FVO5 FGD5-FVO5 390,167 231,958 0,622 -28,9 38,9 0,4309 2,1358 GRP3 FVO3-GRP3 -37,646 -26,97 0,157 144,8 21,84 0,2364 -1,7003 GRP3 GRP3-KIN3 -9,428 0,495 0,032 -176,6 4,22 0,0013 -8,3551 KIN3 GRP3-KIN3 9,429 -8,85 0,044 43,6 5,78 0,0013 -8,3551 KIN5 KIN5-LAG5 171,535 138,87 0,308 -38,4 19,24 0,2484 -3,5818 KIN5 FGD5-KIN5 -180,965 -130,071 0,311 144,9 19,43 0,2976 -4,6786 LAG5 FGD5-LAG5 -176,344 -137,577 0,315 142 19,68 0,5453 -8,265 LAG5 KIN5-LAG5 -171,287 -142,451 0,314 140,3 19,61 0,2484 -3,5818 OSØ3 FGD3-OSØ3 8,822 21,508 0,078 -66,7 6,71 0,0064 -0,222 OSØ3 FVO3-OSØ3 -53,894 -46,119 0,238 140,5 33 0,0415 -6,3536 SHE3 SHE3-SØN3 171,081 115,062 0,721 -33,9 84,87 2,917 14,3536 SØN3 ABS3-SØN3 121,105 74,584 0,526 -35 116,81 4,4778 7,8484 SØN3 SHE3-SØN3 -168,164 -100,708 0,724 145,7 85,22 2,917 14,3536 SVB3 ABS3-SVB3 -58,092 -32,34 0,275 140,9 27,79 0,4555 2,2622 SVB3 FGD3-SVB3 0 0 0 90 0 0,0001 -3,2095 36 Electrical Power Network of Funen Table 39.Node data of 150-60 kV Network of Funen when ABS3-FVO3, and FG3-SVB3 are disconnected.  GRP3-FVO3, FGD3-ØSØ3, and FGD3-FVO3 are disconnected Like I had before, in this case I will make the configuration of 2 separated networks. The part of the whole network which I will study will be the portion formed with the 150kV busbarson Enstedværket, Sønderborg, Abildskov, Svendborg, Odense, and both sides on Fynsværket (150kV, and 60 kV). We have checked that everything works in an almostperfect way, so I will not show the results here. They are shown on the Appendix Excel 400-150 kV Network.  ABS3-FVO3, FVO3-ØSØ3, and FGD3-FVO3 are disconnected In this case, I will found high voltage values at Fynsværket, and some busbarswill not meet the ±10% concerning the rated voltage. Node U u U ang P Load Q Load P Gen Q Gen kV % º MW MVAr MW MVAr ABS2 56,187 85,13 -10,1 58 29 0 0 ABS3 143,784 87,14 -7,7 0 0 0 0 FGD2 68,284 103,46 -0,6 58 29 0 0 FGD3 172,039 104,27 1 0 0 0 0 FGD5 418,326 102,03 1,3 0 0 0 0 FVA2 70,167 106,31 -0,4 77 39 0 0 FVB2 69,528 105,35 -1,3 53 27 0 0 FVO3 172,603 104,61 1,1 0 0 0 0 FVO5 421,09 102,7 1,8 0 0 0 0 GRP2 68,093 103,17 -1,2 47 24 0 0 GRP3 170,066 103,07 0,4 0 0 0 0 KIN3 170,062 103,07 0,4 0 0 0 0 KIN5 413,936 100,96 0,6 0 0 0 0 LAG5 410 100 0 347,631 280,029 0 0 OSØ2 67,249 101,89 -1,5 45 22 0 0 OSØ3 172,363 104,46 1 0 0 0 0 SFV2 54,426 82,46 -12,5 58 29 0 0 SHE3 165 100 0 0 0 171,081 115,062 SØN2 63,061 95,55 -5,1 47 24 0 0 SØN3 156,222 94,68 -3,4 0 0 0 0 SVB3 139,508 84,55 -10 0 0 0 0 Unit 3 21,623 120,13 4,5 0 0 240 180 Unit 7 23,235 110,64 8 0 0 391,2 293,4 37 Electrical Power Network of Funen Table 40.Node data of 150-60 kV Network of Funen when ABS3-FVO3, FVO3-ØSØ3, and FG3-FVO3 are disconnected. c. Transformers We want to know if the transformers will be able to work when, for example, one of them won’t be working (in the case of two transformers between two nodes), or even if one of them could handle the load through itself. And also I would like to know the behavior of the network when some of them are off at the same time. I will also make a simulation of “breaking” the ring, and splitting it into two different areas (400 kV, and 150-60 kV) to know what could happen in these conditions. The different operation conditions for transformers I have considered are shown on the Appendix Excel 400-150 kV Network. Like I will see, the network is so perfectly calculated that it will work for the different combinations: when I will have only one disconnected, when two are disconnected too, or even when I will have seven transformers off. We have found one possible problematic situation, and I are going to study it. I will see how this example shows how the network will behave when all the 400-150 kV transformers will be disconnected. Like I mentioned before, it will represent again the division of the whole network in two different networks, the 410kV and the 150-60kV sides. The schematic and the obtained results are: Node U u U ang P Load Q Load P Gen Q Gen kV % º MW MVAr MW MVAr ABS2 62,877 95,27 -5,2 58 29 0 0 ABS3 160,134 97,05 -3,3 0 0 0 0 FGD2 66,35 100,53 -2,4 58 29 0 0 FGD3 167,321 101,41 -0,7 0 0 0 0 FGD5 415,22 101,27 0,9 0 0 0 0 FVA2 74,101 112,27 2 77 39 0 0 FVB2 73,501 111,37 1,2 53 27 0 0 FVO3 182 110,3 3,3 0 0 0 0 FVO5 417,996 101,95 1,4 0 0 0 0 GRP2 69,591 105,44 0 47 24 0 0 GRP3 173,687 105,26 1,6 0 0 0 0 KIN3 173,224 104,98 1,5 0 0 0 0 KIN5 413,501 100,85 0,5 0 0 0 0 LAG5 410 100 0 270,435 215,835 0 0 OSØ2 64,986 98,46 -3,7 45 22 0 0 OSØ3 166,864 101,13 -0,9 0 0 0 0 SFV2 63,66 96,45 -4,7 58 29 0 0 SHE3 165 100 0 0 0 89,014 30,447 SØN2 65,54 99,3 -3,4 47 24 0 0 SØN3 162,156 98,28 -1,8 0 0 0 0 SVB3 162,063 98,22 -2,9 0 0 0 0 Unit 3 22,711 126,17 6,4 0 0 240 180 Unit 7 23,086 109,93 7,6 0 0 391,2 293,4 38 Electrical Power Network of Funen Figure 9.400-150 kV Network of Funen when KIN5-KIN3, FGD5/FGD3 1, and FGD5/FGD3 2 are disconnected. Table 41.Transformer data of 400-150 kV Network of Funen when KIN5-KIN3, FGD5-FGD3 1, and FGD5-FGD3 2 are disconnected. Concerning to problems at the nodes, I have noticed that because of having two different networks, the 150-60 kV part is not properly supplied. In fact, the majority of them are violating the lower voltage limits. This is because of even when working in normal conditions, the 150 kV part of the whole network is supplied by the grid, and like I will see later, the Power Node Element P Q I Angle I P Loss Q Loss Tap Name Name MW MVar kA °MW MVar ABS2 ABS3-ABS2 A -28,984 -14,559 0,341 140 0,0388 1,6374 11 ABS2 ABS3-ABS2 B -29,016 -14,441 0,34 140,2 0,0448 1,636 11 ABS3 ABS3-ABS2 A 29,023 16,196 0,136 -40 0,0388 1,6374 11 ABS3 ABS3-ABS2 B 29,061 16,077 0,136 -39,8 0,0448 1,636 11 FGD2 FGD3-FGD2 B -28,225 -9,261 0,314 146,5 0,0381 1,3899 11 FGD2 FGD3-FGD2 A -29,775 -19,739 0,377 131,2 0,046 1,9475 11 FGD3 FGD5-FGD3 2 0 0 0 90 0 0 11 FGD3 FGD5-FGD3 1 0 0 0 90 0 0 11 FGD3 FGD3-FGD2 B 28,263 10,651 0,126 -33,5 0,0381 1,3899 11 FGD3 FGD3-FGD2 A 29,821 21,687 0,153 -48,8 0,046 1,9475 11 FGD5 FGD5-FGD3 2 0 0 0 90 0 0 11 FGD5 FGD5-FGD3 1 0 0 0 90 0 0 11 FVA2 FVO3-FVA2 3 -33,703 -17,056 0,387 138,3 0,0445 1,7779 11 FVA2 FVO3-FVA2 2 -43,297 -21,944 0,498 138,3 0,0558 2,2847 11 FVB2 FVO3-FVB2 1 -53 -27 0,619 136,8 0,1136 4,5422 11 FVO3 FVO3-FVB2 1 53,114 31,542 0,255 -43,2 0,1136 4,5422 11 FVO3 Unit 3 -239,468 -148,541 1,165 135,7 0,5322 31,4595 11 FVO3 FVO3-FVA2 3 33,748 18,834 0,16 -41,7 0,0445 1,7779 11 FVO3 FVO3-FVA2 2 43,353 24,228 0,205 -41,7 0,0558 2,2847 11 FVO5 Unit 7 -390,164 -231,769 0,623 151,3 1,0361 61,6309 11 GRP2 GRP3-GRP2 A -19,035 -20,026 0,294 117 0,0604 1,9339 11 GRP2 GRP3-GRP B -27,965 -3,974 0,301 155,4 0,035 1,2791 11 GRP3 GRP3-GRP2 A 28 5,253 0,12 -24,6 0,035 1,2791 11 GRP3 GRP3-GRP B 19,096 21,96 0,123 -63 0,0604 1,9339 11 KIN3 KIN5-KIN3 0 0 0 90 0 0 11 KIN5 KIN5-KIN3 0 0 0 90 0 0 11 OSØ2 OSØ3-OSØ2 -45 -22 0,54 137,3 0,1129 4,1191 11 OSØ3 OSØ3-OSØ2 45,113 26,119 0,216 -42,7 0,1129 4,1191 11 SFV2 SVB3-SFV2 A -29 -14,5 0,349 137,8 0,047 1,7147 11 SFV2 SVB3-SFV2 B -29 -14,5 0,349 137,8 0,047 1,7147 11 SØN2 SØN3-SØN2 1 -28,32 -4,635 0,265 164,6 0,0212 0,8912 11 SØN2 SØN3-SØN2 2 -18,68 -19,365 0,248 127,8 0,0387 1,2572 11 SØN3 SØN3-SØN2 1 28,341 5,526 0,108 -15,4 0,0212 0,8912 11 SØN3 SØN3-SØN2 2 18,719 20,622 0,104 -52,2 0,0387 1,2572 11 SVB3 SVB3-SFV2 B 29,047 16,215 0,139 -42,2 0,047 1,7147 11 SVB3 SVB3-SFV2 A 29,047 16,215 0,139 -42,2 0,047 1,7147 11 Unit 3 Unit 3 240 180 9,711 -44,3 0,5322 31,4595 11 Unit 7 Unit 7 391,2 293,4 12,169 -28,7 1,0361 61,6309 11 39 Electrical Power Network of Funen Plant Unit 3 will manage the voltage control on the 150kV part of this network, it will get the approximately 100% of the rated voltage in these nodes. Table 42. Violation upper voltage limits of 400-150 kV Network of Funen when KIN5-KIN3, FGD5/FGD3 1, and FGD5/FGD3 2 are disconnected. The 400kV network will be delivering power to the grid, however, the 150-60kV network, because of being lower supplied, will need to be supplied by the grid. Table 43.400-150 kV Network of Funen when KIN5-KIN3, FGD5/FGD3 1, and FGD5/FGD3 2 are disconnected. Table 44.Power and Losses data of 400-150 kV Network of Funen when KIN5-KIN3, FGD5/FGD3 1, and FGD5/FGD3 2 are disconnected. Table 45. Node and line overload data of 400-150 kV Network of Funen when KIN5-KIN3, FGD5/FGD3 1, and FGD5/FGD3 2 are disconnected. FVA2 85,29 FGD2 82,82 FVO3 84,62 KIN3 82,81 OSØ3 84,42 GRP3 82,8 FGD3 84,21 GRP2 82,09 FVB2 84,02 OSØ2 81,12 ABS3-SØN3 154,99 SHE3-SØN3 104,34 Nodes (lower) % Overloads Lines % From P Loss Q Loss P Imp Q Imp P Gen Q Gen P Load Q Load Area/Zone MW MVar MW MVar MW MVar MW MVar Network 17,023 133,966 -171,177 -116,434 848,597 602,228 831,574 468,262 Area 1 17,023 133,966 0 0 848,597 602,228 831,574 468,262 Zone 1 17,023 133,966 0 0 848,597 602,228 831,574 468,262 Un P Loss Line Q Loss Line kV MW MVar 165 13,121 26,243 410 1,59 -13,493 P Loss Transformer Q Loss Transformer MW MVar 1,276 1,036 59,585 61,631 Node U u U ang P Load Q Load P Gen Q Gen kV % º MW MVAr MW MVAr ABS2 54,971 83,29 -13,4 58 29 0 0 ABS3 140,822 85,35 -10,9 0 0 0 0 FGD2 54,661 82,82 -15,3 58 29 0 0 FGD3 138,944 84,21 -12,8 0 0 0 0 FGD5 417,585 101,85 1,4 0 0 0 0 FVA2 56,29 85,29 -14,8 77 39 0 0 FVB2 55,456 84,02 -16,2 53 27 0 0 FVO3 139,62 84,62 -12,5 0 0 0 0 FVO5 420,352 102,52 2 0 0 0 0 GRP2 54,18 82,09 -16,5 47 24 0 0 GRP3 136,612 82,8 -14 0 0 0 0 KIN3 136,629 82,81 -14 0 0 0 0 KIN5 413,484 100,85 0,7 0 0 0 0 LAG5 410 100 0 388,574 245,262 0 0 OSØ2 53,54 81,12 -16,6 45 22 0 0 OSØ3 139,299 84,42 -12,6 0 0 0 0 SFV2 53,714 81,38 -15,6 58 29 0 0 SHE3 165 100 0 0 0 217,397 128,828 SØN2 62,567 94,8 -6,1 47 24 0 0 SØN3 155,039 93,96 -4,4 0 0 0 0 SVB3 137,778 83,5 -13 0 0 0 0 Unit 3 17,837 99,09 -7,5 0 0 240 180 Unit 7 23,2 110,47 8,1 0 0 391,2 293,4 40 Electrical Power Network of Funen This situation could remind me of the critical condition when the network feeder of the 400 kV side (N1) was disconnected. Actually, is the same state, due to the 150-60 kV side is not being supplied perfectly, only with its network feeder (N2) and the Unit 3, and Unit 7. d. Generators  Unit 3 disconnected This situation will be consider similar than the situation for normal conditions. The Unit 3 is anuseful tool to get the ~100% of the voltage on all the nodes. When disconnecting it, all the 150-60 kV nodes will not meet this condition,they will be around 98% of the rated voltage. The real power losses will increase a 12,68% the power losses for the steady state, although the reactive power losses will decrease around 7%. Table 46.400-150 kV Network of Funen when Unit 3 in Fynsværket is disconnected. Table 47.Power and Losses data of 400-150 kV Network of Funen when Unit 3 in Fynsværket is disconnected. Table 48.Node data of 400-150 kV Network of Funen when Unit 3 in Fynsværket is disconnected. From P Loss Q Loss P Imp Q Imp P Gen Q Gen P Load Q Load Area/Zone MW MVar MW MVar MW MVar MW MVar Network 4,348 47,609 56,148 -22,791 478,16 318,256 473,812 270,646 Zone 1 4,348 47,609 0 0 478,16 318,256 473,812 270,646 Un P Loss Line Q Loss Line kV MW MVar 165 1,866 -29,153 410 0,52 -26,038 MVar 0,557 20,99 81,81 1,405 P Loss Transformer Q Loss Transformer MW Name U u U ang P Load Q Load P Gen Q Gen kV % º MW MVAr MW MVAr ABS2 63,287 95,89 -5,2 58 29 0 0 ABS3 161,138 97,66 -3,3 0 0 0 0 FGD2 65,136 98,69 -3,9 58 29 0 0 FGD3 164,363 99,61 -2,1 0 0 0 0 FGD5 411,768 100,43 0,3 0 0 0 0 FVA2 66,253 100,38 -4,3 77 39 0 0 FVB2 65,569 99,35 -5,3 53 27 0 0 FVO3 163,27 98,95 -2,6 0 0 0 0 FVO5 414,557 101,11 0,8 0 0 0 0 GRP2 66,209 100,32 -3,4 47 24 0 0 GRP3 165,516 100,31 -1,7 0 0 0 0 KIN3 165,916 100,56 -1,6 0 0 0 0 KIN5 410,218 100,05 0 0 0 0 0 LAG5 410 100 0 30,812 47,646 0 0 OSØ2 63,574 96,32 -5,4 45 22 0 0 OSØ3 163,442 99,06 -2,6 0 0 0 0 SFV2 63,26 95,85 -5,5 58 29 0 0 SHE3 165 100 0 0 0 86,96 24,856 SØN2 65,692 99,53 -3,4 47 24 0 0 SØN3 162,521 98,5 -1,8 0 0 0 0 SVB3 161,081 97,62 -3,6 0 0 0 0 Unit 3 0 0 0 0 0 0 0 Unit 7 22,92 109,14 7,2 0 0 391,2 293,4 47 Electrical Power Network of Funen For distance protection measurement the reactance (𝑋𝐹) of fault impedance is the only component that you can use to effectively determine the distance to fault, but the resistivity component can vary due to the indeterminate arc resistance at the fault location. The ideal reactance should be as flat as possible running parallel at the R-axis. Figure 15.Combined circle- and straight line characteristic.  Starting The principal mission of starting function is first to detect and classify a fault (short-circuit) in the power system line without any mistakes, especially in single phase faults to ensure selective phase tripping. The function of the drop-off and the pick-up of starting are to determine the beginning and the end of the fault. The starting function might for case trigger the zone timers and the fault recorder. b. Applied methods and practice There are several methods for calculating and dimensioning the distance protection.  Over-current starting: This method is the simplest and the fastest. It can be used in network with small line impedances and where enough large short-circuit current flows. In that way the smallest short-circuit current cannot be smaller than twice the maximum load current. Also the applied setting has to be around 1.3 times the maximum current in the phases and half time 𝐼𝑁 for the earth-current. In the case of parallel lines and when one line is off, the rest may carry twice the current, but just for a short time. So the setting of the phases must be double. If I want to check out the dependability of the fault detection I have to use a fault in two phases because the fault current is smaller than a current fault in three phases by a factor 3. 48 Electrical Power Network of Funen Figure 16. Reach of the overcurrent starter (for phase to phase faults). Figure 17.Voltage at the relay location during short-circuit.  Under-impedance starting (U< and I>) There are few reasons for why the current when short-circuit appears in the feeder may be too small for overcurrent starting. These reasons are: weak source or high source impedance, current splitting in parallel paths of a meshed system and earth-current limited by reactance or resistance in the transformer star-point. The source impedance and the fault impedance are the causes of the voltage at the relay location. In this starting configuration the current controls the voltage thresh-old, this is why the pick-up sensitivity of the voltage is increased as the current increases. When I>> corresponds to an overcurrent starter stage. Typical settings are: 49 Electrical Power Network of Funen 𝐼>=0.25𝐼𝑁 𝐼≫= 2.5. 𝐼𝑁 𝑈 𝐼> =70%. 𝑈𝑁 𝑈 𝐼≫ =90%. 𝑈𝑁  Effectively earthed system: The method below is required in this method to achieve phase-selective fault detection because normally simple overcurrent starting is not enough. In my case the short-circuit current has the possibility to carry on the healthy phases during earth-faults. The overcurrent starting has to be setting above these healthy phase-currents. Because of the difference between the positive and the zero sequences systems at the two lines and theses currents have to arise. In the picture above I can see what happens when there is an earthed transformer with no infeed at the end of one line. I can see, while the fault is just in a single-phase to ground, that the current at the three phases are the same. Figure 18.Short circuit in an effectively earthed system with unequal source and earthing conditions. 50 Electrical Power Network of Funen  Impedancestarting: An impedance characteristic is well suited to discriminate between fault and lad conditions as well. All the possibilities of faults loops that I can get in this case are uninterruptedly measured and controlled by several technologies. The optimization of the starting characteristics, when I are talking about conventional relays, was made with the implementation of the circle and straight line elements. The objectives of the optimization can be seen below: - Large reach in X-direction for the detection of remote faults. - Sufficient arc-compensation in the other direction (R-Axis) while maintaining secure margin against load encroachment. II. Distance Protection of a string of the 150 kV network of Funen. For the calculations of Distance Protection, the first step is selecting a string of the 150kV network on Funen. The one that I have chosen is the string between the nodes FGD3, and SHE3. The most important reasons for choosing this line is because it is large enough to make selectivity protection that I have to do at the end of this part of the project. To set the distance relay I have to calculate first the maximum and the minimum current of the string I chose. I have to make the calculations by hand first, and by NEPLAN afterwards, to check out if both values are equals or close enough. a. Maximum and minimum short circuit currents  By hand For the hand calculations I have to select first where I are going to place the short circuit fault, in my case it is going to be in the ABS3 node. It is actually a random selection but I chose it because it is just in the middle of the string. The values necessary to make the hand calculations are the impedances of each length of the line, the impedances of the feeders, and the line voltage. The c factor is shown in the table below, but for the NEPLAN calculations and for the hand calculations I chose 1,1 for maximum current and 1’00 for minimum current. Due of the fault location, in the middle of the string, I will obtain two different values of maximum current, one from each end of the line but I are only interested in the current that flows from SHE3, as I are only configuring distance protection from this side. Either I are going to have just one value of the minimum current because I have to calculate it at the end of the string. 51 Electrical Power Network of Funen Table 55. “c” factor for short circuit calculations.  Maximum Short-Circuit Current: For the maximum current I need a 3-phase fault in the line, and the definitive value is given by the formula: 𝐼𝐾′′=𝑐∙𝑈∆ 3∙𝑍𝑆𝐶=1´1.165000 3∙36′05 = 2′906𝑘𝐴 But first I need the value of the 𝑍𝑆𝐶.This value is the sum of the impedances in the line before the fault location. Figure 19.Equivalent circuit of the line from SHE3 to ABS3. - Impedance of thefeeder: 𝑍𝑛𝑄=𝑐𝑈𝑛𝑄2 𝑆𝑛𝑄 𝑛2= 1′11650002 4000 . 1 = 7′48Ω - Short circuitimpedance: 𝑍𝑆𝐶=10′3 + 18′27 + 7′48 =36′05Ω 52 Electrical Power Network of Funen  Minimum short-circuit current: For the minimum short-circuit current, the kind of fault that I need is a fault between 2 lines or a single line fault. So for this reason, I need to calculate first in both ways and then I will choose the lower value. - Single-phase-to-ground fault. 𝐼′′𝑘(1) =𝑐∙ 3∙𝑈∆ 2∙𝑍1+𝑍0 =1´0. 3. 165000 2∙10´3 + 28´03 = 6′46𝑘𝐴 - Line to line fault. 𝐼′′𝑘(2) = 3 2𝐼′′𝑘(3) = 3 2. 5´894 = 5´103𝑘𝐴 We have to calculate the 3-Phase current 𝐼′′𝑘(3)in SHE3-SØN3 line: 𝐼𝐾′′=𝑐∙𝑈∆ 3∙𝑍𝑆𝐶=1´1.165000 3∙ 10´3 + 7´48 = 5′894𝑘𝐴 As I can see in the results the minimum short circuit current for me is Line to Line fault because is the smaller of them.  NEPLAN As I did for the Load flow calculations, in NEPLAN I have to build the schema of my line, as well I have to determine the different values of cables, nodes, transformers and feeders.  Maximum Short-Circuit Current: For the maximum short-circuit current I are going to simulate a 3-Phase short circuit current placed at ABS3 node, but at the same time I have to disconnect the ABS3-SVB3 line as well. Table 56.3-Phase short circuit current values in ABS3. As I can see the values calculated by hand and by NEPLAN are close enough to say that they are pretty correct.  Minimum current: For the minimum short circuit current I have to choose one of the ends of the string and I chose the one at SØN3 node. To deciding witch of the both currents that I are going to calculate the criteria is just chose the smaller one of them. 3-phase Fault Un UL-E (RST) ∆U-LE (RST) Ik'' (RST) ∆ik'' (RST) Location kV kV kV kA kA 1 ABS3 165 104,789 180 2,915 -77,38 53 Electrical Power Network of Funen - Single-phase-to-ground fault Table 57. Single-phase-to-ground fault in SØN3 - Line to line fault Table 58.Line to line fault in SØN3. As I can see again the values are close enough tosay that they are equals. Finally I chose is Line to Line value because is the smaller of them. b. Distance relays  R/X characteristics The next step is set the distance relays by all over the line. The best option is to put two relays for each section of the string. This decision was made because I want to control the current in both ways because it is possible for this kind of networks. The distance relay that I chose for my project is the one called ABB REL 316. The principal characteristic of this relay is that it can work in 3 zones and I assume that these 3 zones are enough for a high protection level. Each relay has its own current transformer and its own voltage transformer. In the picture below you can see how the schema in NEPLAN looks with all the protection elements. Single- Fault To Node UL-E (RST) ∆U-LE (RST) Ik'' (RST) ∆ik'' (RST) phase Location kV kV kA kA 1 SØN3 Faulted 95,263 180 6,41 -79,46 2 15,721 -9,93 0 -90 3 15,721 -9,93 0 -90 4 0,888 94,74 5 0,888 94,74 6 0,68 94,74 7 0,68 94,74 8 1,568 -85,18 9 1,568 -85,18 Element Fault UL-E (RST) ∆U-LE (RST) Ik'' (RST) ∆ik'' (RST) Location kV kV kA kA 1379 SØN3 0 90 0 -90 2 82,5 90 4,773 191,95 3 82,5 270 4,773 11,95 4 0,037 -88,39 5 0,037 91,61 6 0,036 91,61 7 0,036 -88,39 8 4,772 -78,05 9 4,772 101,95 54 Electrical Power Network of Funen Table 59.400-150 kV distance relay settings.  Tripping schedules The tripping schedule is a graphic given by NEPLAN where I can study how the relays will behave when a fault appears. NEPLAN will give me two graphics, one for each direction of the current. The tripping schedule graphics is showed below 55 Electrical Power Network of Funen Table 60.SHE3 to FGD3 line tripping schedule. Table 61.FGD3 to SHE3 line tripping schedule. As I can see both graphic are not exactly equals like they should be, that is because I had some problems with the software. The graphic shows the relation between the impedance of the line, so I can detect a fault in the line, and the time that the relay will need to disconnect the line. 56 Electrical Power Network of Funen For example I are going to take the tripping schedule of the line from FGD to SHE, and make a short-circuit in the line from ABS to SØN. The system will detects that Z=31 (Ohm) for example, so as I can see in the graphic the relay called ABS3 DISREL-4903759 will disconnect the line 0,1 seconds after the fault appears. If this relay has problems and cannot disconnect the line, then the relay called SVB3 DISREL-4903750 will tries to disconnect the line 0,4 seconds after the fault appears, if after exactly 1 second the fault still there then the relay called FGD3 DISREL- 4903777 will be the last opportunity to disconnect the line. If after one second the line is not disconnect, it will mean that I will be situated in the critical zone, and provably the components of the line will be destroyed. That is one example when I have three opportunities to disconnect the line, but for example in others I only have one or two opportunities. Table 62.SHE3 to FGD3 line tripping schedule. 63 Electrical Power Network of Funen Figure 25. Load profile of power losses over a day with low load.  Losses duration curve over a day Load profiles I present as load duration curves, summary of each type of load are present on Figure 26. Summary of losses duration curve over a day.We can compare how large are active power losses over a day for two transformers. Figure 26.Summary of losses duration curve over a day. We calculated value of power losses for two transformers, for high load, medium and low load day: 64 Electrical Power Network of Funen Table 67.Power losses data for Transformers in Abilsdkov. Real Power (P) [MWh] Reactive Power (Q) [MVAr] High load 3,41669 35,86433 Medium load 3,156226 26,55077 Low load 2,816142 14,43678 We noted that the total losses are the biggest, as I expected, over high load day. I can see on Figure 27. Annual losses duration,that almost 16 hours power losses are above value P(t)loss =0,140 [MWh]. After some calculations I made annual losses duration curve.  Annual losses duration curve How I get my annual losses duration curve. I assumed 91 days for high load, 91 days - low load and 183 for medium load. I used data from last subsection. I created season consist of 4 days – one high load and one low load, and two days with medium load. I get 96 values (hours). I sorted them and for approximation 8760 hours, I used scaling factor 1/96 on time axis. I got duration Power losses to percents. Relation which I got corresponds to the real annual values. Figure 27.Annual power losses duration curve. To calculate total power losses I multiplied by 91 each value of high load and low load day and multiplied by 183 values of medium load day. After summed them I got P(t)loss =1144,777[MWh] in all year. This is average P(t)loss =3,136 [MWh] each day. mediumlosslowlosshighlosslosstot tPtPtPtP ____ )(183)(91)(91)(  P(t)loss =1144,777 [MW] Q(t)loss =9436,189 [MW] 65 Electrical Power Network of Funen  Network losses In all network I have bigger power losses. NEPLAN generated the table with that data. Table68. Powerlossessummed Relations with values of power losses are similar to reported previously load profile analysis for single days. I get the biggest power losses in network during high load day. Power (active) losses in the network are bigger around 4,6 times, compare with power losses in two transformers, on low load day, around 7 times on medium load day and bigger 8,5 times on high load day. Reactive power losses were two times bigger in network for high and medium load days and almost the same for low load. Total power losses in network in 2010 year were: P(t)loss =7958,826 [MW] Q(t)loss =15616,731 [MW]  Energy transferred - load duration curve Figure 28.Annual load duration curve of energy transfer in Abildskow network in 2010. Load flow calculations: 24 24 24 Not converged load flow calculations: %1 000 Years from / to 2010 2010 2010 2010 2010 2010 Months from / to May May April April February February Days from / to 30 30 26 26 11 11 Time from / to / increment 0 0,989583 60 0 0,989583 60 0 0,989583 60 Network energy losses [MWh] 12,985 22,534 29,159 Network energy losses [Mvarh] 14,997 44,993 66,138 Areas Area 1 12,985 MWh 22,534 MWh 29,159 MWh Zones Zone 1 12,985 MWh 22,534 MWh 29,159 MWh 66 Electrical Power Network of Funen For making annual load duration curve, I used data created by NEPLAN. Procedure and steps for making graph were the same as described annual duration power losses. Total power demand was P(t)=34011 [MW] Total transferred power is a sum of power demand and power network losses: P(t) tot=P(t)+P(t)loss P(t) tot=34011 [MW]+1144,777 [MWh] P(t) tot=35155,777 [MWh] P(t)loss =1144,777 [MW] that is a big value of power losses, but lossesare only 2% ofthe transmittedenergy. b. No-load transformers Transformers are no-load, when I have open circuit on secondary side of transformers. I have very low no-load current on primary sides which causes current losses in cores.  No-load losses To calculate no load losses, I changed in NEPLAN, parameters of transformers (deleted 50 noload losses of each). After analysis NEPLAN generated some data for simulations. For two transformers in 2010 year, no –load losses were equal: P(t)loss =246,0446 [MW] Q(t)loss =8973,894 [MW] To remind, for full load transformers I had: P(t)loss =1144,777 [MWh] and Q(t)loss =9436,189 [Mvarh], and it was 4,65 times more active losses and about 462 Mvarh more reactive power losses. II. Energy transferred in 150/60 kV two transformers in Abildskov - critical condition We have to make the simulation assuming that transformer A has been disconnected for some reason in all the year 2010. a. Lossesduration curve For preparing annual losses duration curve for B transformer we: - Made simulation with load profile for high, medium and low load - Sorted data and calculated some ratios - Calculated total power losses in one year with only one working transformer We got: P(t)loss =940,5899 [MW] Q(t)loss =17965,61 [MW] 67 Electrical Power Network of Funen Figure 29.Annual losses duration curve of B transformer. III. Economic aspects or conclusions a. Examples of duration curves analysis Load profiles and duration curves I make for forecasting power demand or power losses on given period. If I want to know more details of power consumption or others, I can present my results in specific charts. For example I can compare number of hours in three types of day with different load profile. I made four areas of power consumption during single day: 1) P >50 [MWh] 2) P =(50-40) [MWh] 3) P =(40-30) [MWh] 4) P<30 [MWh] Figure30shows number of hours of energy consumption over three different load profiles days. I can see that the widest range of power consumption value has winter day with high load. Summer day with low load profile has consumption at a value P<40 [MWh]. 68 Electrical Power Network of Funen Figure 30.Number of hours of energy consumption over three load profiles days. We presented also a summ of that hours of enrgy consumption in one year on Figure 31 Figure 31.Number of hours of energy consumption in one year. 91 days with consumption above P=50 [MWh] represents a 10% of all annual energy consumption. 69,8% this is consumption in range P=(50-30)[MWh]. Average values of power consumed are present in Table 69. Average values of power consumption Calculations, for present power consumption in [%], I can make using duration curves or graphs as above. Energy companies plan generation and distribution of electrocity, after analysis of data like that. Information about power consumption are very important, when I have more than one kind of energy generators. I decide when I will be using which one to get maximumefficiencyat minimumcost. Making analysis I should also remember about losses in my generators and network. They describe in next subsections. 69 Electrical Power Network of Funen Table 69. Average values of power consumption. High load Medium load Low load Average power consumption per hour [MW] 46,088 39,794 29,6145 Power consumption per day [MW] 1106,125 955,057 710,746 Average power consumption per one day in year [MW] 38,82 Average power consumption per year (365 days) [MW] 931,81 b. Power losses in transformers We made power losses calculations with two working transformers (full-load and no-load) and in bad condition when only one transformer were working all year. Summaryof measurement data are present in table below: Table 70.Annual power losses on transformers. A&B transformers load A&B transformers noload B transformer load P(t)loss [MWh] 1144,777 246,0446 940,5899 Q(t)loss [Mvarh] 9436,189 8973,896 17965,61 As I expected, I get bigger power losses on full-load transformers as on no-load. It was bigger 4,6 times for active power losses. P=1144,777 [MWh] for full-load and P=246,0446[MWh] for no-load. Most interesting is compare annual power losses in two transformers and with one disconnected. I assumed that, the B transformer was working without break all year. Finally I got higher value of power losses for only one transformer. Lossdifferenceis P=694,5[MWh]. One working transformer is enough and it is cheaper solution for build network but it gives big value of power losses. I need second transformer also to ensure supply protection. If one of two transformers will breakdown or just will have any damage then second one can keep value of transferred power. So I got improved reliability of supply. Other advantage is a longer life of each transformer, when they are working together. Transformers do not have to work on maximum performance all time so they can longer operated. It Is also worth noting,that power losses could be smaller for two transformers work, than for one only. In that case could be only one disadvantage. On the beginning, there are higher costs of purchase andinstallation two transformers, than only one. However, I can conclude that definitely it is better to have two transformers. 70 Electrical Power Network of Funen c. Power losses in network On the end I present the sum of network losses. That data generated NEPLAN. As I can see on Table 71, power losses on the network are quite high. Mostly it is because of losses on lines (we have 10 lines) and connection losses. Table 71.Network energy losses. IV. Conclusion If electricity cannot be stored, this is very important to know loading curve. It helps to keep the correct balance between supply and power demand. “Energy supplied must always be equal to consumed.” So it is also necessary to know how much power should be generate in real time, under all system operating conditions. When I have large power consume at given time I can determine is it because of damage line or it is just high load. Using duration curve I can plan power demand in future and plan power supply. I calculate total power transferred and total power losses also I can try to minimize power losses. I can use it for make some statistics or to find out the solutions for economical questions. high load med. Load low load high load med. Load low load med. Load low load [MWh] 29,15922 22,53363 12,9852 26,82537 18,03196 10,6647 19,85174 12,21519 Total /year [Mvarh] 66,38756 44,99312 14,99724 66,33004 38,39739 14,96054 101,222 28,84281 Total /year Q(t) losses 61,86177 15639,7577 14424,16515 23156,60162 Network energy losses Load 2 transformers No-load 2 transformers Load 1 transformer high load P(t) losses 28,96524 7958,7965 6711,44505 7380,28755 71 Electrical Power Network of Funen 10-0,4 kV (Low Voltage) Network of Funen  Protection in Low Voltage Installations I. Introduction In Fåborg town, there is a company that has its own 10/0.4 kV transformer outside the building. The transformer is fed by a 10 kV line, which belongs to the Medium Voltage Installation that has been described above. Although more cables are shown in the schema of the low voltage installation, only L1, L2 and L3 are considered during calculating. Figure 32.Entry part of a low voltage installation at a company in Fåborg town. Cables connect the transformer switchboard (T.Sw) via cables to the main switchboard (M.Sw) inside the building. L1, L2 and L3 will be analyzed in order to obtain the cable dimension, the switch gear of overload, the short circuit protection and the protection from indirect contact. Figure 33.Diagram of a low voltage installation at a company in Fåborg town. 72 Electrical Power Network of Funen A brief description to different protection criteria for dimensioning low voltage installations is done below. The descriptions are obtained from the ReglamentoElectrotécnico de Baja Tensión (REBT). The REBT is the legislation for low voltage installations that is used in Spain. Due to the facility to find this legislation in Internet, in this project REBT is the chosen option to analyze the low voltage installation. “The objective of system protection is to detect faults and to selectively isolate faulted parts of the system. It must also permit short clearance time to limit the fault power and the effect of arcing faults.”4 a. Protection of gear and cables against overload currents In the REBT, GUÍA-BT-22, overload current protection explains that the circuits have to be protected from the overload currents, so the interruption of these circuits has to be done in a convenient time or the circuits have to be dimensioned considering the predictable overload currents. The overload current may be produced by: - The used devices or isolated defects due to big impedance. - Short circuits. - Atmospheric electrical discharges. The limit of the current carrying capacity in the cables has to be guaranteed by the used protection device. The protection device may be a circuit breaker with omnipolar cut or a calibrated fuse. The operating characteristics of the circuit breaker that protects a cable from overload currents have to satisfy the two following conditions: 1. 𝐼𝐵≤𝐼𝑁≤𝐼𝑍 2. 𝐼2≤1.45 ∙𝐼𝑍 Where: IB is the current, with which the circuit has been designed according to the predictable load. IZ is the current carrying capacity. In is the rated current in the protection device. In the case of regulated protective devices. In is the regulated current selected. I2 is the current that ensures the protection of the device for a long time (tc). 4Electrical Installations Handbook. Part 1.3: System Protection. John Wiley &Sons.Third edition, 2000. 79 Electrical Power Network of Funen Table73. IP digits. 3. Protectionthroughobstacles. This action doesn't guarantee a complete protection and its application is limited, in a practical way, to the locals where electric services are done. In these locals, only electric workers have access to it. The obstacles have to prevent from: o a not deliberate physical approach to the active parts. o not deliberate contacts with active parts, in the case of the intervention in low voltage equipment while this is in service. 4. Protection by distance to the active parts. This action doesn't guarantee a complete protection and its application is limited, in a practical way, to the locals where electric services are done. In these locals, only electric workers have access to it. This protection only prevents from accidental contacts with the active parts. 80 Electrical Power Network of Funen Figure 38.Accessibility volume to the area S. 5. Additional protection by residual current devices. This action is intended for complementing other protection actions against direct contacts. Using current differential-residual devices, which value is equal or under 30 mA, is a complementary protection action in the case of failure of the other protection against direct contact or reckless of the users.  Protection against indirect electric shock. “Protection against indirect contact hazards can be achieved by automatic disconnection of the supply if the exposed-conductive-parts of equipment are properly earthed. Two levels of protective measures exist:  1st level: The earthing of all exposed-conductive-parts of electrical equipment in the installation and the constitution of an equipotential bonding network. 81 Electrical Power Network of Funen Figure 39.Illustration of the dangerous touch voltage Uc.  2nd level: Automatic disconnection of the supply of the section of the installation concerned, in such a way that the touch-voltage/time safety requirements are respected for any level of touch voltage Uc(1).”8 There are three methods of automatic disconnection depending on how the electric system is. - Automatic disconnection for TT system. “Automatic disconnection for TT system is achieved by RCD having a sensitivity of where RA is the resistance of the installation earth electrode.”9 Figure 40.Automatic disconnection of supply for TT system. - Automatic disconnection for TN systems. 8 http://www.electrical-installation.org/wiki/Measures_of_protection:_two_levels. 9 http://www.electrical-installation.org/wiki/Automatic_disconnection_for_TT_system 82 Electrical Power Network of Funen “The automatic disconnection for TN system is achieved by overcurrent protective devices or RCD’s.”10 Figure 41.Automatic disconnection in TN system. - Automatic disconnection on a second fault in an IT system. “In this type of system: o The installation is isolated from earth, or the neutral point of its powersupply source is connected to earth through a high impedance. o All exposed and extraneous-conductive-parts are earthed via an installation earth electrode. First fault situation: In IT system the first fault to earth should not cause any disconnection. 10 http://www.electrical-installation.org/wiki/Automatic_disconnection_for_TN_systems. 83 Electrical Power Network of Funen Figure 42. Fault current path for a first fault in IT system. Second fault situation: The simultaneous existence of two earth faults (if not both on the same phase) is dangerous, and rapid clearance by fuses or automatic circuit-breaker tripping depends on the type of earth-bonding scheme, and whether separate earthing electrodes are used or not, in the installation concerned.”11 Figure 43. Circuit-breaker tripping on double fault situation when exposed-conductive-parts are connected to a common protective conductor. 11 http://www.electricalinstallation.org/wiki/Automatic_disconnection_on_a_second_fault_in_an_IT_system. 84 Electrical Power Network of Funen II. Fåborg a. Cable dimension The first step when dimensioning the cables is to calculate the currents that flow through these cables. The description of the cables is the following one: “L3 supplies a 10 A load and is most of the way bundled along with five other cables on a wall, each loaded with 10 A. The load current of L2 is determined from the maximum load current of the six cables that are supplied from S.Sw multiplied with a simultaneity factor of 0.6. The load current of L1 is determined from the maximum load currents of L2 plus the two other cables that are supplied from M.Sw. A simultaneity factor of 0.6 is used to calculate the load current.”12 To start the calculation, it is assumed a cosφ = 1. A resistive load supplied through a – more or less – resistive cable gives the worst case for voltage drop. When calculating the currents, a simultaneity factor is considered according to the project description. The calculations have been made in Excel, and are shown in Appendix CD - 10 kV Network - Low Voltage Installation. 12Electrical Power Engineering Project. 85 Electrical Power Network of Funen Table 74.Calculation of L1, L2 and L3 currents. Load toS.Sw POLAR I phase R I phase S I phase T I phase N U phase (V) Factor RMS ANGLE RMS ANGLE RMS ANGLE RMS ANGLE Load 1 230,00 1,00 10,00 0,00 10,00 -120,00 10,00 120,00 0,00 0,00 Load 2 230,00 1,00 10,00 0,00 10,00 -120,00 10,00 120,00 0,00 0,00 L3 230,00 1,00 10,00 0,00 10,00 -120,00 10,00 120,00 0,00 0,00 Load 4 230,00 1,00 10,00 0,00 10,00 -120,00 10,00 120,00 0,00 0,00 Load 5 230,00 1,00 10,00 0,00 10,00 -120,00 10,00 120,00 0,00 0,00 Load 6 230,00 1,00 10,00 0,00 10,00 -120,00 10,00 120,00 0,00 0,00 From S.Sw to M.Sw POLAR I phase R I phase S I phase T I phase N U phase (V) Factor RMS ANGLE RMS ANGLE RMS ANGLE RMS ANGLE Line 1 230,00 1,00 100,00 0,00 100,00 -120,00 100,00 120,00 0,00 0,00 L2 230,00 0,60 36,00 0,00 36,00 -120,00 36,00 120,00 0,00 0,00 Line 3 230,00 1,00 50,00 0,00 50,00 -120,00 50,00 120,00 0,00 0,00 From M.Sw to T.Sw POLAR I phase R I phase S I phase T I phase N U phase (V) Factor RMS ANGLE RMS ANGLE RMS ANGLE RMS ANGLE L1 230,00 0,60 111,60 0,00 111,60 -120,00 111,60 120,00 0,00 0,00 86 Electrical Power Network of Funen After calculating the currents, the dimension of the cables is the next step. L1, L2 and L3 are studied using different laws from the REBT.  Assumptions In the Danish legislation, loaded cables loaded which exceed more than 75% are considered when applying the reduction factor depending on grouped cables. In the English legislation, loaded cables loaded which exceed more than 30% are considered when applying the reduction factor depending on grouped cables. In the Spanish legislation, this reduction factor is considered in all the cases. It doesn’t matter that the cables are not exceeding a specific per cent. 1. L1 “The main cable (L1) between T.Sw and M.Sw is placed in ground outside the building. It enters the building through a pipe in the concrete wall, and is fixed on a cable rack inside the building. L1 is grouped with two other cables – 95 mm2 Cu cables – both loaded more than 30 %. The distance between the cables in the ground is 15cm.”13 1.1 Thermal criteria. L1 has different sectors, so it has to be dimensioned according to three different laws: 1.1.1. Cables underground. In this case, tables and reduction factors are found in ITC-BT-07, from the REBT. The ITC-BT-07 is shown in theAppendix III. Low Voltage 1, at the end of this Project. The current which has been calculated above in L1 is 𝐼𝐿1=111.60 𝐴. The Table 3 (AIII) - Appendix III. Low Voltage 1shows the maximum permissible current. The maximum current has to be higher than 𝐼𝐿1. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=16 𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =115 𝐴. Reductionfactors in Appendix III. Low Voltage 1:  Table 4. Reduction factor depending on temperature. 𝑇𝐺𝑅𝑂𝑈𝑁𝐷=15º𝐶,𝑋𝐿𝑃𝐸 → 𝑓𝑇= 1.07  Table 5. Reduction factor depending on thermal resistivity to ground. 𝜌= 1.5 º𝐾∙𝑚𝑊 ,𝑇𝑕𝑟𝑒𝑒 𝑝𝑕𝑎𝑠𝑒𝑠 → 𝑓𝜌= 0.87 To get this value, an interpolation is needed. 13Electrical Power Engineering Project. 87 Electrical Power Network of Funen 𝑓𝜌= 0.89 +0.84 −0.89 1.65 −1.40 1.5 −1.4 = 0.87  Table 6. Reduction factor depending on grouped cables. 𝑇𝑕𝑟𝑒𝑒 𝑐𝑎𝑏𝑙𝑒𝑠,𝑑= 0.25𝑚 → 𝑓𝑔= 0.8 The formula to apply now is: 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥𝐼𝑡𝑎𝑏𝑙𝑒∙𝑓𝑇∙ 𝑓𝜌∙ 𝑓𝑔 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥115 ∙1.07 ∙0.87 ∙0.8 = 88.6428 𝐴 The obtained value is lower than 𝐼𝐿1=111.60 𝐴. So a bigger cross section is required from table 3. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=25𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =150 𝐴. 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡≥150 ∙1.07 ∙0.87 ∙0.8 = 111.7 𝐴 The difference between𝐼𝐿1 and 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒𝑐𝑢𝑟𝑟𝑒𝑛𝑡 is so low, that a bigger cross section is needed. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=35𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =180 𝐴. 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥180 ∙1.07 ∙0.87 ∙0.8 = 134.0496 𝐴 1.1.2. Cables underground in contact with concrete. In this case, tables and reduction factors are found in ITC-BT-07, in Appendix III. Low Voltage 1.Although tables and reduction factors are the same as the previous case, now the cables are in contact with a different material, concrete. The current L1 is 𝐼𝐿1=111.60 𝐴. The Table 3 (AIII) Appendix III. Low Voltage 1shows the maximum permissible current. The maximum current has to be higher than𝐼𝐿1. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=16 𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =115 𝐴. Reductionfactors in Appendix III. Low Voltage 1:  Table 4. Reduction factor depending on temperature. 𝑇𝑊𝐴𝐿𝐿=25º𝐶,𝑋𝐿𝑃𝐸 → 𝑓𝑇= 1  Table 5. Reduction factor depending on thermal resistivity to ground. 𝜌= 2.5 º𝐾∙𝑚𝑊 ,𝑇𝑕𝑟𝑒𝑒 𝑝𝑕𝑎𝑠𝑒𝑠 → 𝑓𝜌= 0.71 88 Electrical Power Network of Funen  Table 6. Reduction factor depending on grouped cables. 𝑇𝑕𝑟𝑒𝑒 𝑐𝑎𝑏𝑙𝑒𝑠,𝑑= 0.25𝑚 → 𝑓𝑔= 0.8 The formula to apply now is: 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥𝐼𝑡𝑎𝑏𝑙𝑒∙𝑓𝑇∙ 𝑓𝜌∙ 𝑓𝑔 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥115 ∙1∙0.71 ∙0.8 = 65.32 𝐴 The obtained value is lower than 𝐼𝐿1=111.60 𝐴. So a bigger cross section is required from table 3. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=25𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =150 𝐴. 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥150 ∙1∙0.71 ∙0.8 = 85.2 𝐴 So a higher cross section is needed. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=50𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =215𝐴. 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥215 ∙1∙0.71 ∙0.8 = 122.12𝐴 1.1.3. Cables inside a gallery. In this case, tables and reduction factors are found in ITC-BT-07, in Appendix IV. Low Voltage 2.Outdoorinstallation conditions. The current in L1is 𝐼𝐿1=111.60 𝐴. The Table 1- Appendix IV. Low Voltage 2 shows the maximum permissible current. The maximum current has to be higher than𝐼𝐿1. 3 × 𝑋𝐿𝑃𝐸 → 𝑠=35 𝑚𝑚2,𝐼𝑚𝑎𝑥,𝑡𝑎𝑏𝑙𝑒 =135 𝐴. Reduction factors:  Table 2. Reduction factor depending on the temperature of the air. 𝑇𝐴𝐼𝑅=35º𝐶,𝑋𝐿𝑃𝐸 → 𝑓𝑇= 1.05  Table 3. Reduction factor depending on how the cables are grouped. Considering a tray with holes (ventilation), with a distance D between cables. 𝐴 𝑡𝑟𝑎𝑦 , 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑎𝑏𝑙𝑒𝑠 → 𝑓𝑔= 1 The formula to apply now is: 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥𝐼𝑡𝑎𝑏𝑙𝑒∙𝑓𝑇∙ 𝑓𝑔 𝐼max ,𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ≥135 ∙1.05 ∙1 = 141.75 𝐴 95 Electrical Power Network of Funen Figure 45.K factor by UNE 20460-4-43. Although XLPE doesn’t appear in the table, the material PR/EPR has the same characteristic temperatures as the XLPE. The factor also depends on the material of the cable, in this case, copper. 4.5 ∙105 ≤1432∙502=51122500 So the chosen protection is OK.  L2 𝐿2→𝐹𝑢𝑠𝑒: 40 𝐴 In the catalogue NH- Fuse system (Low voltage) from the company Ferraz Shawmut, GROUPE CARBONE LORRAINE, the required information is found. The type of fuse is gG (general purpose protection of wires and cables). The model is NH-fuses, ~400 V gG. Table 78. L2 fuse model data. As in the previous case, the energy has to be evaluated following the literature from blackboard: The first step is to calculate the time of the short circuit. 𝑡= 𝑘∙𝑆 𝐼𝑘 2 96 Electrical Power Network of Funen The material of the cable is PVC and the conductor is made of copper. The factor “K” used is 115. Table 79.K factor by UNE 20460-4-43. In the case of the fuses, the lowest short circuit current is associated with the highest energy. Therefore, the minimum short circuit current is required to make this calculation. Further down, the minimum short circuit calculations are shown. In NEPLAN, the minimum values are obtained with ine ohase to ground fault. The minimum short circuit current at the node S.Sw is 0.967 kA. 𝑡= 115 ∙10 0.967 ∙103 2 = 1.4143 𝑠 The time is bigger than 0.1 seconds. The curve where the energy has to be found is below. From the figure 14, the short circuit current related to 1.4 s is around 350A. the energy will be checked using: 𝐼2𝑡 𝐶𝐵≤ 𝐼2𝑡 𝐶𝑎𝑏𝑙𝑒 =𝑘2∙𝑆2 35021.4143 𝐶𝐵=173251 ≤1152∙102=1322500 The chosen fuse is OK. 97 Electrical Power Network of Funen Table 80.The curve of the fuse.  L3. 𝐿3→ 𝑀𝐶𝐵: 10 𝐴 The maximum short circuit current that device has to protect is 2.105 kA. This current is calculated by NEPLAN, at the node S.Sw. The used catalogue is from Merlin Gerin, Multi 9 System, ProtectionMiniature Circuit Breakers, SCHNEIDER ELECTRIC. The model selected is circuit-breakers up to 63 A, C60N, 6kA, C curve, AS/NZS 4898. The MCB has three poles. The device protects the line, because the ultimate breaking capacity is 6 kA and the maximum short circuit current that the line withstands is 2.105 kA. 98 Electrical Power Network of Funen Figure 37. Data sheet for the MCB. Schneider Electric. The following curve shows the maximum short circuit current related to the energy. Table 81.Thermal stress limitation curve of the MCB. Curve C, model 60 N. 99 Electrical Power Network of Funen From the graphic, knowing that the maximum short circuit current is 2.105 kA at that point, and energy of 6x103 A2s is set. As the previous case, the factor “K” will depend on the type of conductor and cable (copper and PVC), so the value is 115. This data can be checked in table 7. Applying the formula that has being explained in the theory: 𝐼2𝑡 𝐶𝐵≤ 𝐼2𝑡 𝐶𝑎𝑏𝑙𝑒 =𝑘2∙𝑆2 6∙103≤1152∙2.52=82656 So the chosen device is OK. b. Short circuit protection The short circuit currents are calculated by hand and using NEPLAN. The maximum and minimum short circuit current helps me in order to choose the protection. To calculate the short circuit currents in NEPLAN, first the low voltage installation is configured in this software. Some of the parameters used when setting the elements, come from the calculation by hand. This is the case of the impedance in the feeder and the impedance in the transformer. However, the characteristics of the cables are gotten from the library of NEPLAN, with the cross sections which have been previously set. The impedances calculated by hand are explained below: - Impedance of thefeeder. 𝑍𝑛𝑄=𝑐𝑈𝑛𝑄2 𝑆𝑛𝑄 𝑛2 Data from the transformer: 𝑛=𝑈𝑟𝑇𝐻𝑉 𝑈𝑟𝑇𝐿𝑉 So the impedance is related to the low voltage side of the transformer. An approximation can be done to get the values of the resistance and the impedance. 𝑋𝑄= 0.995 ∙𝑍𝑛𝑄𝑅𝑄= 0.1 ∙𝑋𝑛𝑄 - Impedance of thetransformer. In the library of the NEPLAN there are transformers. One of those is chosen, with the rated voltage and the apparent power S already set in the project. 100 Electrical Power Network of Funen From NEPLAN, UK and UR are given in the data of the transformer. 𝑍𝑇=𝑈𝐾∙𝑈2𝐿2 100 ∙𝑆𝑛 𝑅𝑇=𝑈𝑅∙𝑈2𝐿2 100 ∙𝑆𝑛𝑋𝑇= 𝑍𝑇2− 𝑅2𝑇 The resistance and the impedance of L1, L2 and L3 cables are gotten from the data in the library of NEPLAN. Table 82.Values of the resistance and the impedance of all the elements. Scc (MVA) Un1 (kV) NETWORK MT 30,000 10 Sn transfo (kVA) Ucc (%) URc (%) Pcun (kW) Un1 (kV) Un2 (V) In1 (A) In2 (A) TRANSFO 1(Line) 400,000 4 1,150 6 10 400 23,094 577,350 TRANSFO 1(Phase) 400,000 4 1,150 10 231 13,333 577,350 R phase (mΩ) X phase (mΩ) R neutro (mΩ) X neutro (mΩ) R0(mΩ) X0 (mΩ) R0 neutro (mΩ) X0 neutro (mΩ) NETWORK MT 0,531 5,307 0,000 0,000 2,123 21,227 0,000 0,000 TRAFO 1 4,60000 15,324 0,000 0,000 4,600 15,324 0,000 0,000 L1 3P+N 15,560 3,320 28,960 3,440 45,920 3,320 68,040 44,600 L2 3P+N 90,500 4,700 90,500 4,700 133,050 4,700 133,050 4,700 L3 3P+N 218,400 3,300 218,400 3,300 229,320 3,300 229,320 3,300 101 Electrical Power Network of Funen In the tables above are shown the values for the positive and the negative impedance. The zero sequence in the feeder is calculated supposing that: 𝑍0𝑄𝑍1𝑄 ~ 4 The zero sequence in the transformer can be supposed as: 𝑍0𝑇𝑍1𝑇 ~ 1  Maximum short circuit current. The maximum short circuit current appears when there is a fault in the three phases. To calculate this current by hand, the next formula has to be used: 𝐼𝐾′′=𝑐∙𝑈∆ 3∙𝑍𝑆𝐶 The factor “c” is shown in the table below. Table 83. “c” factor for short circuit calculations. The factor “c” is set as 1.05 in NEPLAN and in the calculations by hand. The faults are done in four different points, in the nodes T.Sw, M.Sw, S.Sw and L3. It is only needed the accumulative impedance of one phase in the four different points. Below, the results that have been obtained in NEPLAN and the calculations by hand are shown. The values are very similar in both cases. These values are very reasonable. 102 Electrical Power Network of Funen Figure 46.Maximum short circuit currents in NEPLAN. Table 84. Maximum short circuit currents obtained with calculations by hand. Rsc (mΩ) Xsc (mΩ) Zsc (mΩ) I''sc (kA) Short circuit in T.Sw 5,131 20,631 21,260 11,406 Short circuit in M.Sw 20,691 23,951 31,651 7,661 Short circuit in S.Sw 111,191 28,651 114,823 2,112 Short circuit in the node L3 329,591 31,951 331,136 0,732 103 Electrical Power Network of Funen  Minimum short circuit current. The minimum short circuit currents have to be analyzed by two different faults. The calculations by hand are made using the following formulas: - Single-phase-to-ground fault. In this case, the impedance zero is used. 𝐼′′𝑘(1) =𝑐∙ 3∙𝑈∆ 2∙𝑍1+𝑍0 - Line to line fault. 𝐼′′𝑘(2) = 3 2𝐼′′𝑘(3) These calculations were made by EXCEL, but the obtained values were lower than the results made by NEPLAN. Therefore, these calculations are not shown because of a mistake in these. When using NEPLAN, A factor “c” of 0.95 is set. This value is obtained from the table 9. Below, the results from NEPLAN are shown. The lowest values are obtained in the case of one phase to ground. 104 Electrical Power Network of Funen - Single-phase-to-ground fault. Figure 47. NEPLAN - minimum short circuit current – one phase to ground fault. - Line to line fault. Figure 48.NEPLAN - minimum short circuit current – line to line fault. 111 Electrical Power Network of Funen Appendix I. Abbreviations LAG - Landerupgård FGD - Fraugde KIN - Kingstrup FVO - Fynsværket GRP - Graderup FVB - Fynsværket FVA - Fynsværket SHE - Endstedværket SØN - Sønderborg ABS - Abildskov SVB - Svendborg SFV - Svendborg FÆG - Fællinggård RAD - Radby SHP - SønderHøjrup KRI - Korinth ESP - Espe RIE - Ringe HNE - Horne FBV - Fåborg Vest FBY - Fåborg By OSØ - Odense SØ 112 Electrical Power Network of Funen Appendix II. Values Table 1 (AII).Line data from the Electrical Power Project specifications. Table 2 (AII).Transformer data from the Electrical Power Project specifications. 113 Electrical Power Network of Funen Table 3 (AII).Conductor data for overhead lines. 114 Electrical Power Network of Funen Appendix III. Low Voltage 1 ITC-BT-07 Redessubterráneasparadistribución en bajatensión, “Underground networks for low voltage distributions”. These conductors, which are used in underground networks, they will be made by copper or aluminium. These conductors must have corrosion protection and enough strength. The cables can have one or more conductors and its rated voltage won’t be less than 0.6/1 KV. The cross section of copper conductor won’t be less than 6 mm2 and the cross section of aluminium conductor won’t be less than 16 mm2. Depending on the number of conductors that are used in the distribution, the minimum cross section of the neutro conductor will be: a)Two or three conductors: The cross section is the same as the phase conductors. b) With 4 conductors, the minimum cross section of the neutro conductor depends on table1. Table 1 (AIII).Minimum neutro cross section depending on phase conductor cross section. 1. Maximum current carrying capacity. 1.1 Maximum permissible temperature. The maximum admissible currents depend on the permanent performance. In each case, the maximum temperature that the insulation can withstand without alteration of their electrical, mechanical or chemical properties. 115 Electrical Power Network of Funen Table 2 (AIII). Cables with dry isolation, ºC assigned to the conductor. 2. Conditions in underground system. A tripolar or tetrapolar cable or a set of three single-core cables in mutual contact, or a wire cable or two single-core cables in mutual contact, directly over its entire length buried in a trench 0.70 m deep in ground, resistivity thermal average of 1 Km / W and ground temperature at this depth, 25 ° C. 3. Tables that are used to dimension the cable. L1 conductors are made from copper. This decision is made due to the two other cables that are buried with L1. L1 is grouped with two other cables -95 mm2 Cu. Table 3 (AIII).Maximum permissible current,in amps, forcableswithcopper conductors in underground systems (permanent service). 4. Reduction factors. - Buried cables in ground, with a different temperature to 25ºC in the ground. 116 Electrical Power Network of Funen Table 4 (AIII).Reduction factor depending on temperature. - Buried cables in a ground with a thermal resistivity different to 1 ºK m /W. Table 5 (AIII).Reduction factor depending on thermal resistivity to ground. - Number of tripolar or tetrapolar cables which are grouped under ground. Table 6 (AIII).Reduction factor depending on grouped cables. - Cables buried in ground at different depths. Table 7 (AIII).Reduction factor depending on the depth of the cables. 117 Electrical Power Network of Funen Appendix IV. Low Voltage 2 ITC-BT-07 Redessubterráneasparadistribución en bajatensión, “Underground networks for low voltage distributions” 1. Outdoorinstallation conditions. A tripolar or tetrapolar single cable or a set of three single-core cables in contact with each other, with a placement that allows effective air renewal, with the environmental temperature of 40 ° C. For example, in cable trays placed on or attached to a wall, etc... Table 1 (AIV). Maximum permissible current,in amps, forcables with copper conductors in airy installations in ventilated gallery (permanent service). 2. Reduction factors. - Cablesinstalledinairy installations with a different temperaturethan40°C. Table 2 (AIV).Reduction factor depending on the temperature of the air. - Reduction factor of three-phase cables which are grouped in airy installations. 118 Electrical Power Network of Funen Table 3. Reduction factor depending on how the cables are grouped. 119 Electrical Power Network of Funen Appendix V. LowVoltage 3 ITC-BT-19 Instalaciones interiores o receptoras. Prescripcionesgenerales, “Indoors installations or receptors.General requirements”. 1. Cross section of the cables. Drop voltages. For industrial installations directly feed at high voltage through its own distribution transformer, it is considered that the interior installation of low voltage comes from the transformer output. In this case, the maximum permissible drop voltage will be 4.5% for lighting and 6.5% for other uses. In interior installations, to take account of harmonic currents due nonlinear loads and possible imbalances, unless justified by calculation, the section of the neutral conductor is at least equal to the section of the phases. 2. Maximum permissible currents. This law is based on Norma UNE 20.460. The followingtable showsthe rated current forambient air temperatureof40 °Cand fordifferentinstallation methods, groups and typesofcables. For temperatures, installation methods, groups and typesofcable, as wellasburied conductors, it is needed to seek reduction factors in the UNE20 460-5 to 523. 120 Electrical Power Network of Funen Table 1 (AV).Maximum admissible currents depending on assembly, number of conductors and type of isolation. Table 52- C20. 3. Reduction factors. - Reduction factor when the air temperature is different than 40ºC. Table 2 (AV). Reduction factors depending on temperature. UNE 20 460-5-523. - Reduction factor depending on the number of cables that are grouped together. L3