EE 51 POWER SYSTEM 1 I. General Background of an Electric Power System (Students enrolling this course are presumed to have comprehensive knowledge in Mathematics, Circuit theories and theories and principles of operation of different Electrical Machineries) An electric power system is a network of electrical components deployed to supply, transfer, and use electric power. An example of an electric power system is the grid that provides power to an extended area. An electrical grid power system can be broadly divided into the generators that supply the power, the transmission system that carries the power from the generating station to the load, and the distribution system that feeds the power to nearby homes and industries. Learning Outcomes: At the end of this unit, the learner will be able to: ● Identify and determine parameters involve in power system ● Employ related equations in calculating power and relationship between Voltages and currents Pre-Test: Instructions: Answer the each question/problem quietly and encircle the letter of the correct answer. 1. At a lower temperature, transmission cables may be coated with ice, what will happen to its resistance? (a) increases (b) decreases (c) the same (d) increase slightly 2. For short transmission line where length is up to 50 miles, the parameters considered are usually _____________. (a) capacitance and inductance, the resistance is neglected. (b) resistance and capacitance, inductance is neglected. (c) resistance, inductance and capacitance (d) Resistance and inductance, capacitance is neglected. 3. In a balance 3-phase star-connected system line voltage is Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 (a) (b) (c) (d) the phasor difference of the two phase voltages the phasor sum of the two phase voltages 0.707 times the phase voltages 1.414 times the phase voltages 4. In transmission line conductors, maximum tension occurs in ______. (a) the lowest point (b) the supports (c) the horizontal (d) the vertical 5. Is the process of alternating the positions of each phase conductor of a line in equal intervals so that each phase will have the same inductance. (a) superposition (b) interpolation (c) transition (d) transposition 6. When a unit weight is uniformly distributed along the horizontal or along its span, a cable assumes the form of? (a) hyperbola (b) catenary (c) parabola (d) line 7. The following are reasons why grounded wires are installed on top of the towers or poles of a line, except one. (a) to shield the line from direct lightning strokes (b) to reduce the amount of voltage induced on the line due to electrostatic fields produced by charged clouds. (c) to give provisions for neutral (d) to give provisions for discharging surge current by means of lightning arrester. 8. A line has GMD of 10.33-ft, GMR of 0.3732-inch and length of 96 km. What is the admittance? (a) 3.59 x 10-4 mho (c) 2889.04 mhos (b) 9.34 x 10-7 mho (d) 42.26 mhos 9. In a delta connected system the line current is (a) (b) (c) (d) 1.414 times the phase current phasor sum of the two phase currents equal to the phase current 1.732 times the phase current 10. A (n) _______ stores and returns energy to a circuit while a(n) ________ dissipates energy. (a) resistance, impedance (c) inductor, resistor (b) resistor, inductor (d) inductor, reactance 11. For an RL circuit, the power factor cannot be less than ______ or greater than _______. (a) 1, 0 (b) 0, -1 (c) 0, 1 (d) -1, 0 Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 12. The advantage of using static capacitor to improve the power factor is because they (a) (b) (c) (d) are not variable are almost loss free provide continuous charge of power factor none of these 13. The advantage of the star-connections over the delta connections for the same phase voltage is that it gives (a) step-down current (b) extra step-up voltage (c) extra step-up current (d) extra step-up power 14. The inductance is responsible for the current to _____ the applied voltage. (a) lag (b) be in phase with (c) lead (d) be opposite 15. A line has GMD of 10.33-ft, GMR of 0.3732-inch and length of 96 km. What is the line reactance? (a) 96 Ohms (c) 42 Ohms (b) 26 Ohms (d) 69 Ohms Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Content: The power system engineer - the power system engineer should know the methods of load making studies, fault analysis and stability studies and principles of economic load dispatch because such studies affect the design and operation of the system and the selection of the apparatus for its control. The general principal divisions of an electric power system: 1. Generating Stations 2. Transmission Lines 3. Distribution Systems 4. Utilization Figure 1.1 Structure of a power system and its equivalent one-line-diagram. Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 System modeling SYSTEM MODEL 1. Circuit Model for synchronous machine d. Fuse Ra Xs e. Current Transformer E Vt f. Power Circuit Breaker 2. Equivalent Circuit for Transformer R1 g. Air Circuit Breaker X1 h. Three-Phase Three-Wire Delta-Connection V1 a V2 i. Three-Phase Three-Wire Y – neutral ungrounded i. Three-Phase Three-Wire Y – Neutral-Grounded 3. The One-Line Diagram a. Machine or Rotating Armature j. Ammeter and Voltmeter A b. Two-Winding Transformer k. Potential Transformer or V c. Three-Winding Transformer Transmission lines are the connecting links between the generating stations and the distribution systems and back to the other power systems over interconnections. A distribution system connects all the individual loads to the transmission lines at substations which perform voltage transformation and switching functions. Voltage of large modern generators usually is in the range of 18 to 24 kV. - Generator voltage is stepped-up to transmission levels in the range of 115 to 765 kV. Standard HV EHV UHV : 115, 138 and 230 kV : 345, 500 and 756 kV : levels of 1000 to 1500 kV Load study - is the determination of the voltage, current, power, and power factor or reactive power at various points in an electric network under existing or contemplated conditions of normal operation. It is essential in planning the development of the system. Economic dispatch – is the name given to the process of apportioning the total load on a system between the various generating plants to achieve the greatest economy of operation. Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Transmission and Distribution system Line Parameters: A. Resistance (Series) - The ac resistance is equal to Ridc resistance of the conductor only if the distribution of current throughout the conductor is uniform. At frequencies of 60 Hz and below, the difference between ac resistance and the dc resistance is less than 1% for conductor less than 350 MCM in cross section. An increase in frequency causes non-uniform current density. This phenomenon is called skin effect. Skin Effect – the tendency of current to move outward the surface of conductor. Rac = K Rdc K Eq. 1.1 1 1 F2 2 For:: Copper Aluminum F = 0.0106 d²f F = 0.0063 d²f where; Rac = ac resistance; Rdc = dc resistance of each of the line f = frequency in Hz ; d = diameter of conductor in inch B. Inductance (L) GMD H/mi L 0.7411x 10 3 log GMR phase (For English Units) [Henry per mile per conductor (per phase)] Eq. 1.2a GMD H/m L 2 x 10 7 ln GMR phase (For S.I. Units) [Henry per meter per conductor (per phase) Eq. 1.2b Where: GMD = Geometric mean distance; GMR = Geometer mean radius; L = Series inductance Transposition – is the alternating of position of each phase of the line in equal intervals so that each phase has the same inductance. C. Capacitance (Shunt) Cn 0.0388 x 10 6 GMD log GMR F/mi phase (For English Units) [Farad per mile per phase (neutral)] Eq. 1.3a Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 (For S.I. Units) 2 π x 10 12 F/m [Farad per meter per phase (neutral)] phase GMD ln Eq. 1.3b GMR Where: Cn – shunt capacitance of the line Cn Note: GMD and GMR and r must have the same units. 2 d12 d23 1 GMD Figure 1.2 D. d31 3 3 d12 d23 d31 Eq. 1.4 Conductors of the three phases spaced d1, d2 and d3 between centers respectively of a power system. For solid conductors, GMR = r (radius) Conductance – (shunt) negligible - It is mainly due to corona and leakage to insulators. Corona is not likely to occur when the diameter of the conductor is small compared with the distance between conductors. High voltage small wires and close spacing contribute to a high voltage gradient which may induce corona. A rough and dirty surface on a conductor increases the probability of the occurrence of the corona, and damp weather increases the loss. - In general it is due to the leakage current at the insulators of overhead lines and through the insulations of cables and is taken to be negligible, thereby conductance between conductors of an overhead line is assumed to be zero. Corona –A luminous discharge caused by ionization of the gasses around the high voltage conductor usually accompanied by a hissing sound and characterized by the presence of ozone. Reasons why grounded wires are installed on top of the transmission line towers or poles 1. To shield the transmission line from direct lightning strokes. 2. To reduce the amount of voltage induced upon the transmission line conductor due to the electrostatic field produced by a charged cloud. 3. High voltage will occur between wires during lightning, so it must be discharged to the ground by means of lightning arrester. Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Effect of temperature (surrounding temperature) on the resistance of conductor Recall: R2 R 1 [ 1 α1 (t 2 t1 )] Where: - R1 and R2 are resistances respectively at temperatures t1 and t2. (R1 = resistance of the conductors at ambient temperature) α1 1 /Co T t1 = Ohms temperature coefficient of resistance at temperature t1 T = inferred absolute zero temperature of the conductor (for copper T = - 234.5 ºC) Voltage and Current Relations - (For 3-phase and 1-phase AC transmission and distribution systems) 1. All loads connected at the receiving end (load side) are considered to be balanced. Likewise, it follows that the voltage and currents are also balanced. 2. All voltage equations must be done in per phase basis. 3. All vector / phasor diagrams must be done in a per phase basis. 4. Circuit representation (representation of a transmission lines) must be done in per neutral basis, or per phase basis, or per wire basis, or per conductor basis. 5. Line parameters (resistance, inductance, and capacitance – impedance) must be treated per wire, or per phase, or per conductor, per line basis (even if it is not specified to be). 6. For transmission (short) lines: up to 50 miles in length (use only resistance and inductance, neglect capacitance effect). 7. For medium length transmission lines: from 51 to 150 miles in length (use all the three parameters). 8. For long transmission lines– 151 miles up (use all the three parameters). Conventional symbols: Receiving end data: Load Side ERL ER = = Receiving end line to line voltage Receiving end line to neutral (per phase) voltage E - - - for 3-phase T.L. ER RL 3 Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 P.F.R = ERL - - - for 1-phase T.L. 2 Receiving end power factor ER Eq. 1.5a P.F.R Cos θR Eq. 1.5b Where: θR R.F.R IR PR QR SR = = = = = = Receiving end p.f. angle Receiving end reactive factor P.F.R Sin θR Eq. 1.5c Receiving end line current Receiving end real (true) power in kW, MW,GW Receiving end reactive power in kVARs, MVARs Receiving end apparent power in kVA, MVA Sending end data: Source ESL ES = = P.F.S = Sending end line to line voltage Sending end line to neutral (per phase) voltage E - - - for 3-phase T.L. ES SL 3 E - - - for 1-phase T.L. . ES SL 2 Sending end power factor P.F.S Cos θS Where: θ S R.F.S IS PS QS SS = = = = = Eq. 1.6a Eq. 1.6b = Sending end p.f. angle Sending end reactive factor P.F.S Sin θS Eq. 1.6c Sending end line current Sending end real (true) power in kW, MW,GW Sending end reactive power in kVARs, MVARs Sending end apparent power in kVA, MVA Power: Power triangle P S θ S -jQ jQ θ P Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Apparent Power (S): S P jQ in VA, kVA, or MVA - Where: (+) j Q - - - Capacitive VARs ( -) j Q - - - Inductive VARs Three-Phase: /S / 3 EL IL 1000 /S/ 3E I 1000 Eq. 1.7a Single-Phase: EL IL /S / 1000 - - kVA /S/ - - kVA - - kVA 2E I - - kVA 1000 Eq. 1.7b True (Real) Power (P): P S Cos θ - Three-Phase: P in W, kW, or MW 3 EL IL Cos θ 3 E I Cos θ 1000 1000 EL IL Cosθ 2 EI Cosθ P 1000 1000 Single-Phase: kW kW Eq. 1.7c Reactive (Imaginary) Power (Q): Q S Sin - Three-Phase: Q 3 EL IL Sin θ 1000 Single-Phase: Q EL IL Sin θ 1000 in VAR, kVAR, or MVAR 3 E I Sin θ 1000 2 EI Sin θ 1000 kVARs kVARs Eq. 1.7d Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Learning Activities Activity 1 Understanding Basic Concepts and Nomenclature of Power System Direction. Do research, study and understand the basic concept, operation, characteristics and technical features of the following. 1. 2. 3. 4. 5. Single Line Diagram of Power Supply System Classification of Substations Difference Between Transmission and Distribution Line Difference Between Power Transformer and Distribution Transformer Different Layouts for Substation Activity 2 Application of Related Equations Employed in Power System Direction. Study and understand the following illustrative problems below. Illustrative Problem 1.1 A three-phase 60-Hz line has flat horizontal spacing. The conductors have a GMR of 0.0133m with 10m between adjacent conductors. Determine the inductive reactance per phase in ohms per kilometer. Solution: Given Data: 3-phase transmission line @ f=60 Hz Spacing: Flat horizontal (10 meters between adjacent conductors) d12 = 10m; d23 = 10m; d31 = 20m Conductor: GMR = 0.0133 inch = 3.38 x10-4 m Required: Inductive reactance (XL)per phase in ohms per kilometer. Recall: XL 2π f L Ohms Inductance (L) per conductor using Eq. 1.2b (For S.I. Units) GMD H/m 1000m L 2 x 10 7 ln x km GMR phase Where: GMD 3 101020 12.60m L 2.11 mH per km hence XL 2π 60 2.11x103 0.794 Ohms per km Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Illustrative Problem 1.2 A three-phase 60-Hz line spaced triangularly with distances 12ft, 12ft and 16ft respectively between the three phases. The conductors have an outside diameter of 3.28 cm. Determine the capacitive reactance to neutral in ohmmeters and the capacitive reactance of the line in ohms if its length is 125 mi. Solution: Given Data: 3-phase transmission line @ f=60 Hz Spacing: Triangular d12 = 12 ft ; d23 = 12 ft; d31 = 16 ft Conductor: D = 3.28 cm GMR (Assume solid conductor) = r = D/2 = 0.0164 m Length = 125 mi = 201,137.5m Required: Capacitive reactance (Cn) to neutral in ohm-meters and the capacitive reactance (XCn) of the line in ohms Capacitive reactance to neutral (XCn): 1 Recall: XCn Ohms 2ππXCn Capacitance per conductor to neutral using Eq. 1.3b for S.I. Units 2 π x 10 12 F/m x length GMD phase ln GMR 2 π x 10 12 F/m Cn phase 3 1 121216 3.28 ln 0.0164 F/m 12 Cn 1.126 x 10 phase Cn Hence, XCn 1 2,354,782,955 Ω meter 2π 60 1.126 x 1012 XCn 2,354,782,955 Ω meter 11,707.330 Ohms 201,137.50m Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Illustrative Problem 1.3 Given the following data for 3-phase transmission line; distance, 25 mile; power, 6,000 kW; power factor, 0.80 ,lagging current; voltage at load, 33,000 volts, 3-phase, 60-cycles; size wire, No. 00 AWG solid copper (diameter, 365 mils, ρ = 11 ohms per cir. mil-ft. @ 600C); spacing of wires, 48 in., triangular. Determine (a) AC resistance per conductor at 600C; (b) inductive reactance per conductor and capacitive reactance to neutral; (c) receiving end current. Solution: Given Data: 3-phase transmission line Load: PR =6,000kW P.F.R = 0.80 lagging ERL = 33,000 Volts at 60Hz Conductor: No. 00 AWG solid copper d = 365 mils ρ = 11 ohms per cir. mil-ft Length = 25 miles Spacing: Triangular d1 = d2 = d3 = 48 inches Required: (a) AC resistance per conductor at 600C Using Eq. 1.1 Rac = K Rdc Where: K FCopper 1 1 F2 2 0.0106d2 f Rdc ρ Rac Rac (b) L A 2 2 1 1 0.0106 365 60 1000 2 10.92 Ohms 25 x 5280 11 3652 inductive reactance per conductor and capacitive reactance to neutral Inductive reactance per conductor (XL) Recall: XL 2π f L Ohms Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Inductance per conductor (L) using Eq. 1.2a for English Units GMD H/mi L 0.7411x 10 3 log x length GMR phase Where: From Eq. 1.4 GMD 3 48 48 48 48 inches GMR r 365 0.1825 inch 1000 2 48 H/mi L 0.7411x 10 3 log x 25 mi 0.1825 phase L 44.84 mH Hence, 44.84 XL 2π 60 16.90 Ohms 1000 Capacitive reactance to neutral (XCn): 1 Recall: XCn Ohms 2ππXCn Capacitance per conductor to neutral using Eq. 1.3a for English Units 0.0388 x 10 6 48 log 0.1825 40.10x108 F F/mi x 25 mi phase Cn Cn Hence, XCn (c) 1 6.62 kΩ 2π 60 40.10 x 108 Receiving end current (IR) For three-phase system from Eq. 1.7c 3 ER IR CosθR PR 1000 Hence, PR x 1000 6,000x1000 IR 131.216 Amp 3 ER CosθR 33,000 3 0.80 3 Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 Activity 3 Self-Assessment: Practice Problems Direction. Analyze and identify data given on the following problems and employ related equations to determine the corresponding required data for each problem. (See Answer-Key to verify your answers) Practice Problem 1.1 A three-phase 60-Hz line spaced triangularly with distances 5.2m, 5.2m and 8.6m respectively between the three phases. The conductors have an outside diameter of 1.824-inch. Determine the capacitive reactance to neutral in gigaOhm-meters and the capacitive reactance of the line in Mega-Ohm s if its length is 160 km. Practice Problem 1.2 A three-phase 60-Hz line has flat horizontal spacing. The conductors have a GMR of 1.58cm with 32ft between adjacent conductors. Determine the inductive reactance per phase in ohms per mile. Practice Problem 1.3 Given the following data for 3-phase transmission line; distance, 120km; power, 16,000 kW; power factor, 0.84 ,lagging current; voltage at load, 69kV, 3-phase, 60-cycles; size wire, No. 000 AWG solid copper (diameter, 410 mils, ρ = 11 ohms per cir. mil-ft. @ 600C); spacing of wires, 4.8m., triangular. Determine (a) AC resistance per conductor at 600C; (b) inductive reactance per conductor and capacitive reactance to neutral; (c) receiving end current. Assessment Instructions: Answer the each question/problem quietly and encircle the letter of the correct answer. 1. In a balance 3-phase star-connected system line voltage is (a) the phasor sum of the two phase voltages (b) 1.414 times the phase voltages (c) 0.707 times the phase voltages (d) the phasor difference of the two phase voltages 2. In transmission line conductors, maximum tension occurs in ______. (a) the lowest point (b) the vertical (c) the horizontal (d) the supports 3. Is the process of alternating the positions of each phase conductor of a line in equal intervals so that each phase will have the same inductance. (a) superposition (b) transposition (c) transition (d) interpolation Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 4. When a unit weight is uniformly distributed along the horizontal or along its span, a cable assumes the form of? (a) parabola (b) catenary (c) hyperbola (d) line 5. The following are reasons why grounded wires are installed on top of the towers or poles of a line, except one. (a) to shield the line from direct lightning strokes (b) to give provisions for neutral (c) to reduce the amount of voltage induced on the line due to electrostatic fields produced by charged clouds. (d) to give provisions for discharging surge current by means of lightning arrester. 6. A (n) _______ stores and returns energy to a circuit while a(n) ________ dissipates energy. (a) resistance, impedance (c) resistor, inductor (b) inductor, resistor (d) inductor, reactance 7. For an RL circuit, the power factor cannot be less than ______ or greater than _______. (a) 0, 1 (b) 0, -1 (c) 1, 0 (d) -1, 0 8. The advantage of the star-connections over the delta connections for the same phase voltage is that it gives (a) step-down current (b) extra step-up voltage (c) extra step-up current (d) extra step-up power 9. The inductance is responsible for the current to _____ the applied voltage. (a) lag (b) be in phase with (c) lead (d) be opposite 10. A three-phase 60-Hz line spaced triangularly with distances 18-ft, 18-ft and 24ft respectively between the three phases. The conductors have an outside diameter of 765mils. Determine the capacitive reactance to neutral in gigaOhm-meters and the capacitive reactance of the line in Mega-Ohm s if its length is 100 mi. (a) 23.56 GΩ-meter; 14.727 MΩ (c) 2.716 GΩ-meter; 18.876 kΩ (b) (d) 2.716 GΩ-meter; 16.876 kΩ 23.56 GΩ-meter; 147.27 MΩ 11. A three-phase 60-Hz line has flat horizontal spacing. The conductors have a GMR of 0.840-inch with 9m between adjacent conductors. Determine the inductive reactance per phase in ohms per mile. (a) 0.808 Ohms per mile (b) 8.08 Ohms per mile (c) 0.761 Ohms per mile (d) 7.61 Ohms per mile 12. Given the following data for 3-phase transmission line; distance, 120mi; power, 20,000 kVA; power factor, 0.80 ,lagging current; voltage at load, 66kV, 3- Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT EE 51 POWER SYSTEM 1 phase, 60-cycles; size wire, No. 0000 AWG solid copper (diameter, 460 mils, ρ = 11 ohms per cir. mil-ft. @ 600C); spacing of wires, 20-ft., triangular. Determine (a) AC resistance per conductor at 600C; (b) inductive reactance per conductor and capacitive reactance to neutral; (c) receiving end current. (a) RAC = 34.011 Ω; XL = 101.23 Ω; XCn = 19.150 kΩ; IR = 139.968 A (b) RAC = 62.427 Ω; XL = 61.764 Ω; XCn = 24.015 kΩ; IR = 133.882 A (c) RAC = 34.011 Ω; XL = 101.23 Ω; XCn = 15.190 kΩ; IR = 139.968 A (d) RAC = 26.427 Ω; XL = 61.764 Ω; XCn = 24.015 kΩ; IR = 133.882 A Answer Key Pre-Test 1. 2. 3. 4. 5. b d a b d 6. 7. 8. 9. 10. c c c d c 11. 12. 13. 14. 15. c b b a c Activity 3 1. 2.356 GΩ-meter; 14.727 kΩ 2. 0.808 Ohms per mile 3. (a) (b) (c) RAC = 26.427 Ohms; XL = 61.764 Ohms ; XCn = 24.015 kΩ IR = 133.882 Amp. Compiled and Adapted by: JOY C. IMPERIAL, M.Eng’g, REE, LPT
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