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CHAPTER 3 - Three Phase Network.ppt
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- 2. Single-Line Diagram B Y R 3-phase system single-phase system • One-linediagram is a simplified single-phase circuit diagram of a balanced three-phase electric power system. • It is indicated by a single line and standard apparatus symbols. 2
- 3. Single-Line Diagram Apparatus Symbolsof One-line Diagram Machine or rotating armature Two-winding power transformer Three-winding power transformer Power circuit breaker, oil/ liquid Air circuit breaker Load Or Or Or Or 3
- 4. Apparatus Symbols ofOne-line Diagram A V Or Current transformer Busbar Transmission line Fuse Potential transform er Three-phase, three-wire delta connection Three-phase wye, neutral ungrounded Three-phase wye, neutral grounded Ammeter Voltmeter Single-Line Diagram 4
- 5. Single-Line Diagram One-line Diagram Theinformation on a one-line diagram is vary according to the problem at hand and the practice of the particular company preparing the diagram. Example : 5
- 6. Single-Line Diagram Advantages ofOne-line Diagram Simplicity. One phase represents all three phases of the balanced system. The equivalent circuits of the components are replaced by their standard symbols. The completion of the circuit through the neutral is omitted. 6
- 7. Representation of ElectricPower System Impedance and Reactance Diagrams Impedance (Z = R + jX) diagram is converted from one-line diagram showing the equivalent circuit of each component of the system. It is needed in order to calculate the performance of a system under load conditions (Load flow studies) or upon the occurrence of a short circuit (fault analysis studies). Reactance (jX) diagram is further simplified from impedance diagram by omitting all static loads, all resistances, the magnetizing current of each transformer, and the capacitance of the transmission line. It is apply to fault calculations only, and not to load flow studies. Impedance and reactance diagrams sometimes called the Positive-sequence diagram. 7
- 8. Representation of ElectricPower System 1 2 3 Load B T2 T1 Load A Impedance and Reactance Diagrams Example : One-line diagram of an electric power system 8
- 9. Representation of ElectricPower System E1 E2 E3 Gen. 3 Load B Transformer T2 Transmission Line Transformer T1 Load A Generators 1 and 2 Impedance diagram corresponding to the previous one-line diagram Impedance and Reactance Diagrams 9
- 10. Representation of ElectricPower System Reactance diagram corresponding to the one-line diagram E1 E2 E1 Generators 1 and 2 Transmission Line Transformer T2 Gen. 3 Transformer T1 Impedance and Reactance Diagrams 10
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- 59. 59 Per-Unit Representation The primaryadvantages of per-unit system in power system analysis are: 1. The per-unit values for the transformer impedance, voltage and current are identical when referred to the primary and secondary (no need to reflect impedances from one side of the transformer to the other, the transformer is a single impedance) 2. The per-unit values for various components lie within a narrow range regardless of the equipment rating. 3. The use of √3 in three-phase calculation is reduced 4. Ideal for computer simulations
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- 70. 70 Per-Unit Representation Summary Commonquantities used in power system analysis are voltage (kV), current (kA), volt-amperes (kVA or MVA), and impedance (Ω). It is very cumbersome to convert currents to a different voltage level in a power system having two or more voltage levels. Per-unit representation is introduced such that the various physical quantities are expressed as a decimal fraction or multiples of base quantities and is defined as: quantity of value base quantity actual unit - per in Quantity
- 71. 71 Per-Unit Representation Summary Forsimplicity, per-unit is always written as pu. For single-phase systems: 1 1 1 1 1 2 LN LN LN 1 MVA base MW power, Base kVA base kW power, Base MVA base ) kV voltage, (base impedance Base A current, base V voltage, base impedance Base kV voltage, base kVA base A current, Base
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- 73. 73 Per-Unit Representation Summary Changing theBase of Per-unit Quantities The impedance of individual generators and transformers are generally in terms of % or pu quantities based on their own ratings (By manufacturer). For power system analysis, all impedances must be expressed in pu on a common system base. Thus, it is necessary to convert the pu impedances from one base to another (common base, for example: 100 MVA). Per-unit impedance of a circuit element is 2 kV) voltage, (base MVA) (base X ) impedance, (actual
- 74. 74 Per-Unit Representation Summary old new 2 new old old new MVA base MVA base kV base kV base unit Z - per unitZ - Per Changing the Base of Per-unit Quantities The equation shows that pu impedance is directly proportional to base MVA and inversely proportional to the square of the base voltage. Therefore, to change from old base pu impedance to new base pu impedance, the following equation applies:
- 75. 75 Per-Unit Representation Example : A30 MVA 13.8 kV three-phase generator has a subtransient reactance of 15%. The generator supplies two motors over a transmission line having transformers at both ends, as shown in the one-line diagram below. The motors have rated inputs of 20 MVA and 10 MVA, both 12.5 kV with x” = 20%. The three- phase transformer T1 is rated 35 MVA, 13.2Δ – 115Y kV with leakage reactance of 10%. Transformer T2 is composed of three single-phase transformers each rated at 10 MVA, 12.5Δ – 67Y kV with leakage reactance of 10%. Series reactance of the transmission line is 80 Ω. Draw the reactance diagram with all reactance marked in per unit. Select the generator rating as base in the generator circuit. Cont… 1 2 (13.8 kV) k n (120 kV) l m T1 T2 p r (12.9 kV)
- 76. 76 Per-Unit Representation Solution: The three-phaserating of transformer T2 is 3 X 10 MVA = 30 MVA. and its line-to-line voltage ratio is 12.5 – √3 X 67 = 12.5 – 116 kV. A base of 30 MVA, 13.8 kV in the generator circuit requires a 30 MVA base in all parts of the system and the following voltage bases: In transmission line: 13.8(115/13.2) = 120 kV. In motor circuit: 120(12.5/116) = 12.9 kV. The reactance of the transformers converted to the proper base are: Transformer T1: X = 0.1 (13.2/13.8)2(30/35) = 0.0784 pu. Transformer T2: X = 0.1(12.5/12.9)2(30/10) = 0.28 pu. The base impedance of the transmission line is (120 kV)2/30 MVA = 480 Ω and the reactance of the line is (80/480) = 0.167 pu. Reactance of motor 1 = 0.2 (12.5/12.9)2(30/20) = 0.282 pu. Reactance of motor 2 = 0.2 (12.5/12.9)2(30/10) = 0.563 pu. Cont…
- 77. 77 Per-Unit Representation Solution: Eg Em2 Em1 jo.15 j0.0784j0.167 j0.0940 j0.282 j0.563 p r n m l k Reactance diagram :