Transmission Lines and Power Flow Analysis

Transcription

1 (SOE) University of St Thomas (UST) Transmission Lines and Power Flow Analysis Dr. Greg Mowry Annie Sebastian Marian Mohamed 1

2 The UST Mission Inspired by Catholic intellectual tradition, the University of St. Thomas educates students to be morally responsible leaders who think critically, act wisely, and work skillfully to advance the common good. Since civilization depends on power, power is for the common good!! 2

3 Outline I. Background material II. Transmission Lines (TLs) III. Power Flow Analysis (PFA) 3

4 I. Background Material 4

5 Networks & Power Systems In a network (power system) there are 6 basic electrical quantities of interest: 1. Current i t = dq/dt 2. Voltage v t = dφ/dt (FL) 3. Power p t = v t i t = dw = de dt dt 5

6 Networks & Power Systems 6 basic electrical quantities continued: t 4. Energy (work) w t = t p τ dτ = v τ i τ dτ t 5. Charge (q or Q) q t = i τ dτ t 6. Flux φ t = v τ dτ 6

7 A few comments: Networks & Power Systems Voltage (potential) has potential-energy characteristics and therefore needs a reference to be meaningful; e.g. ground 0 volts I will assume that pretty much everything we discuss & analyze today is linear until we get to PFA 7

8 Networks & Power Systems Linear Time Invariant (LTI) systems: Linearity: If y 1 = f x 1 & y 2 = f(x 2 ) Then α y 1 + β y 2 = f(α x 1 + β x 2 ) 8

9 Networks & Power Systems Implications of LTI systems: Scalability: if y = f(x), then αy = f(αx) Superposition & scalability of inputs holds Frequency invariance: ω input = ω output 9

10 Networks & Power Systems ACSS and LTI systems: Suppose: v in t = V in cos ωt + θ Since input = output, the only thing a linear system can do to an input is: Change the input amplitude: V in V out Change the input phase: in out 10

11 Networks & Power Systems ACSS and LTI systems: These LTI characteristics along with Euler s theorem e jωt = cos ωt + j sin ωt, allow v in t = V i cos ωt + θ to be represented as v in t = V i cos ωt + θ V i e jθ = V i θ 11

12 Networks & Power Systems In honor of Star Trek, V i e jθ = V i θ is called a Phasor 12

13 V i e jθ = V i θ 13

14 Networks & Power Systems ACSS and LTI systems: Since time does not explicitly appear in a phasor, an LTI system in ACSS can be treated like a DC system on steroids with j ACSS also requires impedances; i.e. AC resistance 14

15 Comments: Networks & Power Systems Little, if anything, in ACSS analysis is gained by thinking of j = 1 (that is algebra talk) Rather, since Euler Theorem looks similar to a 2D vector e jωt = cos ωt + j sin ωt u = a x + b y 15

16 e j90 = cos 90 + j sin 90 = j Hence in ACSS analysis, think of j as a 90 rotation 16

17 Operators v(t) V V dv dt vdt j V V j 17

18 Ohm s Approximation (Law) Summary of voltage-current relationship Element Time domain Frequency domain R v Ri V RI L v L di dt V j LI C i C dv dt V I j C 18

19 Impedance & Admittance Impedances and admittances of passive elements Element Impedance Admittance R Z R Y 1 R L Z j L Y 1 j L C Z 1 j C Y j C 18

20 The Impedance Triangle Z = R + j X 18

21 The Power Triangle (ind. reactance) S = P + j Q 18

22 Triangles Comments: Power Triangle = Impedance Triangle The cos( ) defines the Power Factor (pf) 22

23 Reminder of Useful Network Theorems KVL Energy conservation v Loop = 0 KCL Charge conservation i Node or bus = 0 Ohm s Approximation (Law) Passive sign Convention (PSC) 23

24 Passive Sign Convection Power System 24

25 PSC Source Load or Circuit Element P Source = vi P Load = vi P Source = P Load 25

26 Reminder of Useful Network Theorems Pointing Theorem S = V I Thevenin Equivalent Any linear circuit may be represented by a voltage source and series Thevenin impedance Norton Equivalent Any linear circuit my be represented by a current source and a shunt Thevenin impedance 26 V TH = I N Z TH

27 (SOE) University of St Thomas (UST) Transmission Lines and Power Flow Analysis Dr. Greg Mowry Annie Sebastian Marian Mohamed 1

28 II. Transmission Lines 2

29 Outline General Aspects of TLs TL Parameters: R, L, C, G 2-Port Analysis (a brief interlude) Maxwell s Equations & the Telegraph equations Solutions to the TL wave equations 3

30 4

31 5

32 Transmission Lines (TLs) A TL is a major component of an electrical power system. The function of a TL is to transport power from sources to loads with minimal loss. 6

33 Transmission Lines In an AC power system, a TL will operate at some frequency f; e.g. 60 Hz. Consequently, λ = c/f The characteristics of a TL manifest themselves when λ ~ L of the system. 7

34 USA example The USA is ~ 2700 mi. wide & 1600 mi. from north-to-south (4.35e6 x 2.57e6 m). For 60 Hz, λ = 3e8 60 = 5e6 m Thus λ 60 Hz ~ L USA and consequently the continental USA transmission system will exhibit TL characteristics 8

35 Overhead TL Components 9

36 Overhead ACSR TL Cables 10

37 Some Types of Overhead ACSR TL Cables 11

38 TL Parameters 12

39 Model of an Infinitesimal TL Section 13

40 General TL Parameters a. Series Resistance accounts for Ohmic (I 2 R losses) b. Series Impedance accounts for series voltage drops Resistive Inductive reactance c. Shunt Capacitance accounts for Line-Charging Currents d. Shunt Conductance accounts for V 2 G losses due to leakage currents between conductors or between conductors and ground. 14

41 Power TL Parameters Series Resistance: related to the physical structure of the TL conductor over some temperature range. Series Inductance & Shunt Capacitance: produced by magnetic and electric fields around the conductor and affected by their geometrical arrangement. Shunt Conductance: typically very small so neglected. 15

42 TL Design Considerations Design Considerations Responsibilities Electrical Factors 1. Dictates the size, type and number of bundle conductors per phase. 2. Responsible for number of insulator discs, vertical or v-shaped arrangement, phase to phase clearance and phase to tower clearance to be used 3. Number, Type and location of shield wires to intercept lightning strokes. 4. Conductor Spacing's, Types and sizes Mechanical Factors Focuses on Strength of the conductors, insulator strings and support structures Environmental Factors Include Land usage, and visual impact Economic Factors Technical Design criteria at lowest overall cost 16

43 Transmission Line Components Components Made of Types Conductors Insulators Shield wires Support Structures Aluminum replaced copper Porcelain, Toughened glass and polymer Steel, Alumoweld and ACSR Lattice steel Tower, Wood Frame ACSR - Aluminum Conductor Steel Reinforced AAC - All Aluminum Conductor AAAC - All Aluminum Alloy Conductor ACAR - Aluminum Conductor Alloy Reinforced Alumoweld - Aluminum clad Steel Conductor ACSS - Aluminum Conductor Steel Supported GTZTACSR - Gap-Type ZT Aluminum Conductor ACFR - Aluminum Conductor Carbon Reinforced ACCR - Aluminum Conductor Composite Reinforced Pin Type Insulator Suspension Type Insulator Strain Type Insulator Smaller Cross section compared to 3 phase conductors Wooden Poles Reinforced Concrete Poles Steel Poles Lattice Steel towers. 17

44 ACSR Conductor Characteristic Name Overall Dia DC Resistance Current Capacity (mm) (ohms/km) (Amp) 75 C Mole Squirrel Weasel Rabbit Racoon Dog Wolf Lynx Panther Zebra Dear Moose Bersimis

45 TL Parameters - Resistance 19

46 TL conductor resistance depends on factors such as: Conductor geometry The frequency of the AC current Conductor proximity to other current-carrying conductors Temperature 20

47 Resistance The DC resistance of a conductor at a temperature T is given by: where T = conductor resistivity at temperature T l = the length of the conductor A = the current-carry cross-sectional area of the conductor R(T) = ρ T l A The AC resistance of a conductor is given by: R ac = P loss I 2 where P loss real power dissipated in the conductor in watts I rms conductor current 21

48 TL Conductor Resistance depends on: Spiraling: The purpose of introducing a steel core inside the stranded aluminum conductors is to obtain a high strength-to-weight ratio. A stranded conductor offers more flexibility and easier to manufacture than a solid large conductor. However, the total resistance is increased because the outside strands are larger than the inside strands on account of the spiraling. 22

49 The layer resistance-per-length of each spirally wound conductor depends on its total length as follows: where R cond = ρ A (π ) 2 p cond Ω/m R cond - resistance of the wound conductor (Ω) 1 + (π 1 P cond ) 2 - Length of the wound conductor (m) p cond = l turn 2rlaye r - relative pitch of the wound conductor (m) l turn - length of one turn of the spiral (m) 2rlaye r - Diameter of the layer (m) 23

50 Frequency: When voltages and currents change in time, current flow (i.e. current density) is not uniform across the diameter of a conductor. Thus the effective current-carrying cross-section of a conductor with AC is less than that for DC. This phenomenon is known as skin effect. As frequency increases, the current density decreases from that at the surface of the conductor to that at the center of the conductor. 24

51 Frequency cont. The spatial distribution of the current density in a particular conductor is additionally altered (similar to the skin effect) due to currents in adjacent current-carrying conductors. This is called the Proximity Effect and is smaller than the skin-effect. Using the DC resistance as a starting point, the effect of AC on the overall cable resistance may be accounted for by a correction factor k. k is determined by E&M analysis. For 60 Hz, k is estimated around 1.02 R AC = k R DC 25

52 Skin Effect 26

53 Temperature: The resistivity of conductors is a function of temperature. For common conductors (Al & Cu), the conductor resistance increases ~ linearly with temperature ρ T = ρ α T T o the temperature coefficient of resistivity 27

54 TL Parameters - Conductance 28

55 Conductance Conductance is associated with power losses between the conductors or between the conductors and ground. Such power losses occur through leakage currents on insulators and via a corona Leakage currents are affected by: Contaminants such dirt and accumulated salt accumulated on insulators. Meteorological factors such as moisture 29

56 Conductance Corona loss occurs when the electric field at the surface of a conductor causes the air to ionize and thereby conduct. Corona loss depends on: Conductor surface irregularities Meteorological conditions such as humidity, fog, and rain Losses due to leakage currents and corona loss are often small compared to direct I 2 R losses on TLs and are typically neglected in power flow studies. 30

57 Corona loss 31

58 Corona loss 32

59 TL Parameters - Inductance 33

60 Inductance is defined by the ratio of the total magnetic flux flowing through (passing through) an area divided by the current producing that flux Φ = S B d s I, Observation L = Φ I, Definition Inductance is solely dependent on the geometry of the arrangement and the magnetic properties (permeability ) of the medium. Also, B = H where H is the mag field due to current flow only. 34

61 Magnetic Flux 35

62 Inductance When a current flows through a conductor, a magnetic flux is set up which links the conductor. The current also establishes a magnetic field proportional to the current in the wire. Due to the distributed nature of a TL we are interested in the inductance per unit length (H/m). 36

63 Inductance - Examples 37

64 Inductance of a Single Wire The inductance of a magnetic circuit that has a constant permeability (μ) can be obtained as follows: The total magnetic field B x at x via Amperes law: B x = μ o x I (Wb/m) 2πr2 The flux d through slice dx is given by: dφ = B x da = B x l dx (Wb) 38

65 The fraction of the current I in the wire that links the area da = dx l is: N f = x r 2 The total flux through the wire is consequently given by: Φ = 0 r dφ = 0 r x r 2 μ o x I 2πr 2 ldx Integrating with subsequent algebra to obtain inductance-per-m yields: L wire self = (H/m) 39

66 Inductance per unit length (H/m) for two Wires (Cables) To calculate the inductance per unit length (H/m) for two-parallel wires the flux through area S must be calculated. 40

67 The inductance of a single phase two-wire line is given by: L = ln D r (H/m) where r = r x = r y = e 1 4 r x 41

68 The inductance per phase of a 3-phase 3 wire line with equal spacing is given by: L a = ln D r (H/m) per phase 42

69 L of a Single Phase 2-conductor Line Comprised of Composite Conductors 43

70 L x = ln D xy D xx where D xy = MN N M D km and D xx = N 2 N M D km k=1 m=1 k=1 m=1 D xy is referred to as GMD = geometrical mean distance between conductors x & y D xx is referred to as GMR = geometrical mean radius of conductor x 44

71 A similar L y express exists for conductor y. L Total = L x + L y (H/m) Great NEWs!!!!! Manufacturers often calculate these inductances for us for various cable arrangements 45

72 Long 3-phase TLs are sometimes transposed for positive sequence balancing. This is cleverly called transposition The inductance (H/m) of a completely transposed three phase line may also be calculated 46

73 L a = ln D eq D s D eq = 3 D 12 D 13 D 23 L a has units of (H/m) D s is the conductor GMR for stranded conductors or r for solid conductors 47

74 In HV TLs conductors are often arranged in one of the following 3 standard configurations 48

75 L a = ln D eq D SL L a again has units of (H/m) If bundle separation is large compared to the bundle size, then D eq ~ the center-tocenter distance of the bundles 2-conductor bundle: D SL = 4 D S d 2 = D S d 3-conductor bundle: D SL = 9 D S dd 3 = 3 D S d 2 4-conductor bundle: D SL = 16 D S ddd 2 4 = D S d 3 49

76 TL Parameters - Capacitance 50

77 Q = C V C = S ε E d s l E d l RC = ε σ

78 Capacitance A capacitor results when any two conductors are separated by an insulating medium. Conductors of an overhead transmission line are separated by air which acts as an insulating medium therefore they have capacitance. Due to the distributed nature of a TL we are interested in the capacitance per unit length (F/m). Capacitance is solely dependent on the geometry of the arrangement and the electric properties (permittivity ) of the medium 52

79 Calculating Capacitance To calculate the capacitance between conductors we must calculate two things: The flux of the electric field E between the two conductors The voltage V (electric potential) between the conductors Then we take the ratio of these two quantities 53

80 Since conductors in a power system are typically cylindrical, we start with this geometry for determining the flux of the electric field and potential (reference to infinity) 54

81 Gauss s Law may be used to calculate the electric field E and the potential V for this geometry: E x = q 2πεx (V/m) The potential (voltage) of the wire is found using this electric field: V 12 = q 2πε D 2 dx x = D1 q 2πε ln D 2 D 1 Volts The capacitance of various line configurations may be found using these two equations 55

82 Capacitance - Examples 56

83 Capacitance of a single-phase 2-Wire Line The line-to-neutral capacitance is found to be: C n = 2πε ln D r (F/m) The same C n is also found for a 3-phase line-to-neutral arrangement

84 The Capacitance of Stranded Conductors e.g. a 3-phase line with 2 conductors per phase 58

85 Direct analysis yields C an = 2πε ln D eq r (F/m) where D eq = 3 D ab D ac D bc 59

86 The Capacitance of a Transposed Line 60

87 The Capacitance of a Transposed Line C an = 2πε ln D eq D SC (F/m) where D eq = 3 D ab D ac D bc 2-conductor bundle: D SC = r d 3-conductor bundle: D SC = 3 r d 2 4-conductor bundle: D SC = r d 3 61

88 TLs Continued Dr. Greg Mowry Annie Sebastian Marian Mohamed 1

89 Maxwell s Eqns. & the Telegraph Eqns. 2

90 3

91 Maxwell s Equations & Telegraph Equations The Telegrapher's Equations (or) Telegraph Equations are a pair of coupled, linear differential equations that describe the voltage and current on a TL as functions of distance and time. The original Telegraph Equations and modern TL model was developed by Oliver Heaviside in the 1880 s. The telegraph equations lead to the conclusion that voltages and currents propagate on TLs as waves. 4

92 Maxwell s Equations & Telegraph Equations The Transmission Line equation can be derived using Maxwell s Equation. Maxwell's Equations (MEs) are a set of 4 equations that describe all we know about electromagnetics. MEs describe how electric and magnetic fields propagate, interact, and how they are influenced by other objects. 5

93 Time for History James Clerk Maxwell [ ] was an Einstein/Newtonlevel genius who took the set of known experimental laws (Faraday's Law, Ampere's Law) and unified them into a symmetric coherent set of Equations known as Maxwell's Equations. Maxwell was one of the first to determine the speed of propagation of electromagnetic (EM) waves was the same as the speed of light - and hence concluded that EM waves and visible light were really descriptions of the same phenomena. 6

94 Maxwell s Equations 4 Laws Gauss Law: The total of the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field. D = ρ ε The no name ME: The net magnetic flux through any closed surface is zero. B = 0 7

95 Faraday s law: Faraday's law of induction is a basic law of electromagnetism describing how a magnetic field interacts over a closed path (often a circuit loop) to produce an electromotive force (EMF; i.e. a voltage) via electromagnetic induction. X E = B t Ampere s law: describes how a magnetic field is related to two sources; (1) the current density J, and (2) the time-rate-of-change of the displacement vector D. X H = D t + J 8

96 The Constitutive Equations Maxwell s 4 equations are further augmented by the constitutive equations that describe how materials interact with fields to relate (B & H) and (D & E). For simple linear, isotropic, and homogeneous materials the constitutive equations are represented by: B = μ H μ = μ r μ o D = ε E ε = ε r ε o 9

97 The Telegraph Eqns. 10

98 TL Equations TL equations can be derived by two methods: Maxwell s equations applied to waveguides Infinitesimal analysis of a section of a TL operating in ACSS with LCRG parameters (i.e. use phasors) The TL circuit model is an infinite sequence of 2-port components; each being an infinitesimally short segment of the TL. 11

99 TL Equations The LCRG parameters of a TL are expressed as perunit-length quantities to account for the experimentally observed distributed characteristics of the TL R = Conductor resistance per unit length, Ω/m L = Conductor inductance per unit length, H/m C = Capacitance per unit length between TL conductors, F/m G = Conductance per unit length through the insulating medium between TL conductors, S/m 12

100 The Infinitesimal TL Circuit 13

101 The Infinitesimal TL Circuit Model 14

102 Transmission Lines Equations The following circuit techniques and laws are used to derive the lossless TL equations KVL around the abcd loop KCL at node b V L = L i t OLL for an inductor I C = C v t OLC for a capacitor Set R = G = 0 for the Lossless TL case 15

103 Apply KVL around the abcd loop: V(z) = V(z+ z) +R z I(z) + L z I(z) t Divide all the terms by z : V z z V z+ z = z + R I z + L I(z) t As z 0 in the limit and R = 0 (lossless case): V(z) z = R I z + L I(z) t V(z) z = L I(z) t (Eq.1)

104 Apply KCL at node b: I(z) = I(z+ z) +G z V(z+ z) + C z V(z+ z) t Divide all the terms by z: I z z I z+ z = z + G V z + z + C V(z+ z) t As z 0 in the limit and G = 0 (lossless case): I(z) z = G V z + C V(z) t I(z) z V(z) = C t (Eq.2)

105 Summarizing (Eqn. 1) V z,t z I z,t = L t (Eqn. 2) I z,t z V z,t = C t Equations 1 & 2 are called the Coupled First Order Telegraph Equations 18

106 Differentiating (Eq.1) wrt z yields: 2 V(z) z 2 = L 2 I(z) z t Differentiating (2) wrt t yields: 2 I(z) z t = C 2 V(z) t 2 (Eq.3) (Eq.4) Plugging equation 4 into 3: 2 V(z) z 2 = L C 2 V(z) t 2 (Eq.5) 19

107 Differentiating (Eq.1) wrt t yields: 2 V(z) z t Reversing the differentiation sequence = L 2 I(z) t 2 (Eq.6) Differentiating (2) wrt z yields: 2 I(z) z 2 = C 2 V(z) z t (Eq.7) Plugging equation 6 into 7: 2 I(z) z 2 = L C 2 I(z) t 2 (Eq.8) 20

108 Summarizing (Eqn. 5) 2 V(z) z 2 = L C 2 V(z) t 2 (Eqn. 8) 2 I(z) z 2 = L C 2 I(z) t 2 Equations 5 & 8 are the v(z,t) and i(z,t) TL wave equations 21

109 What is really cool about Equations 5 & 8 is that the solution to a wave equation was already known; hence solving these equations was straight forward. The general solution of the wave equation has the following form: f z, t = f(z ± ut) where the wave propagates with Phase Velocity u u = ω β = ω = 1 ω L C L C 22

110 The solution of the general wave equation, f(z,t), describes the shape of the wave and has the following form: f z, t f z, t = f(z ut) or = f(z + ut) The sign of ut determines the direction of wave propagation: f z ut Wave traveling in the +z direction f z + ut Wave traveling in the z direction 23

111 The +z Traveling WE Solution f z, t = f(z ut) f z ut) = f(0 z ut = 0 z = ut 24

112 The argument of the wave equation solution is called the phase. Hence the culmination of all of this is that, Waves propagate by phase change One other key point, the form of the voltage WE and current WE are identical. Hence solutions to each must be of the same form and in phase. 25

113 Wave Equation z increases as t increase for a constant phase 26

114 TL Solutions 27

115 TL Wave Equations - Recap (Eqn. 5) 2 V(z) z 2 = L C 2 V(z) t 2 (Eqn. 8) 2 I(z) z 2 = L C 2 I(z) t 2 Equations 5 & 8 are the v(z,t) and i(z,t) TL wave equations 28

116 ACSS TL Analysis Recipe 29

117 Initial Steps Typically we will know or are given the source voltage. For example, it could be: v ss t = 10 cos 10 7 t + 30 volts From the time function we form the source phasor & know the frequency that the TL operates at (in ACSS operation): ω = 10 7 s 1 V ss = 10 e j30 volts 30

118 ACSS TL Analysis Under ACSS conditions the voltage and current solutions to the WE may be expressed as: V z, t = Re V 0 + e j(ωt γz) + V 0 e j(ωt+γz) V z = V 0 + e jγz + V 0 e jγz Time Harmonic Phasor Form I z, t = Re I 0 + e j(ωt γz) + I 0 e j(ωt+γz) I z = I 0 + e jγz + I 0 e jγz = V + Z 0 e jγz + V Z 0 e jγz Time Harmonic Phasor Form

119 ACSS TL Analysis Recipe Step 1: Assemble Z 0, Z L, l, and γ Step 2: Determine Г L Step 3: Determine Z in Step 4: Determine V in Step 5: Determine V 0 + Step 6: Determine V L Step 7: Calculate I L and P L 32

120 Step 1: Assemble Z 0, Z L, l,, and γ Characteristic Impedance: Load Impedance: Z o = L C Z L = V L I L = = V + o + I = V o o I o V V 0 V + 0 V 0 Z 0 Complex Propagation Constant: ( = 0 for lossless TL) γ = α + jβ = jω L C 33

121 Additional Initial Results Phase velocity u P : u P = ω β = 1 L C Identity: L C = μ ε 34

122 Step 2: Determine Г L Reflection Coefficient: Г L = V 0 V+ = I 0 0 I+ 0 Г L = Z L Z 0 Z L +Z 0 z L = Z L Z 0 Г L = z L 1 z L + 1 Reflection Coefficient at distance z (from load to source = -l): Г z = V o e γz V o + e γz = Г Le j2βz 35

123 Additional Step-2 Results VSWR: VSWR = Γ L Γ L Range of the VSWR: 0 < VSWR < 36

124 Step 3: Determine Z in Input Impedance: Z in = V in z = l I in z = l = Z 0 Z L + Z 0 tanh(γl) Z 0 + Z L tanh(γl) For a lossless line tanh γl = j tan(βl): Z in = Z 0 Z L + jz 0 tan(βl) Z 0 + jz L tan(βl) At any distance z 1 from the load: Z(z 1 ) = Z 0 Z L + jz 0 tan(βz 1 ) Z 0 + jz L tan(βz 1 ) 37

125 Step 4: Determine V in V in = V SS Z in Z in + Z S = V in z = l 38

126 Step 5: Determine V 0 + V z = V 0 + e jβz + V 0 e jβz = V 0 + (e jβz +Г L e jβz ) V in = V TL z = l = V 0 + (e jβl +Г L e jβl ) V 0 + = V in (e jβl +Г L e jβl ) 39

127 Step 6: Determine V L Voltage at the load, V L : V L = V TL z = 0 = V o Γ L where: V 0 + = V in (e jβl +Г L e jβl ) 40

128 Step 7: Calculate I L and P L I L = V L Z L P avg = P L = 1 2 I L 2 Re(Z L ) In general, S L = V L I L = P L + j Q L 41

129 Final Steps After we have determined V L in phasor form, we might be interested in the actual time function associated with V L ; i.e. v L (t). It is straight forward to convert the V L phasor back to the v L (t) time-domain form; e.g. suppose: V L = 5 e j60 volts Then with ω = 10 7 s 1 v L t = 5 cos 10 7 t 60 volts 42

130 Surge Impedance Loading (SIL) Lossless TL with V S fixed for line-lengths < /4 43

131 2-Port Analysis 44

132 Two Port Network 45

133 A two-port network is an electrical black box with two sets of terminals. The 2-port is sometimes referred to as a four terminal network or quadrupole network. In a 2-port network, port 1 is often considered as an input port while port 2 is considered to be an output port. 46

134 A 2-port network model is used to mathematically analyze electrical circuits by isolating the larger circuits into smaller portions. The 2-port works as Black Box with its properties specified by a input/output matrix. Examples: Filters, Transmission Lines, Transformers, Matching Networks, Small Signal 47 Models

135 Characterization of 2-Port Network 2-port networks are considered to be linear circuits hence the principle of superposition applies. The internal circuits connecting the 2-ports are assumed to be in their zero state and free of independent sources. 2-Port Network consist of two sets of input & output variables selected from the overall V 1, V 2, I 1, I 2 set: 2 Independent (or) Excitation Variables 2 Dependent (or) Response Variables 48

136 Characterization of 2-Port Network Uses of 2-Port Network: Analysis & Synthesis of Circuits and Networks. Used in the field of communications, control system, power systems, and electronics for the analysis of cascaded networks. 2-ports also show up in geometrical optics, wave-guides, and lasers By knowing the matric parameters describing the 2-port network, it can be considered as black-box when embedded within a larger network. 49

137 2-Port Networks & Linearity Any linear system with 2 inputs and 2 outputs can always be expressed as: y 1 y 2 = a 11 a 12 a 21 a 22 x 1 x 2 50

138 2-Port Networks 51

139 Review A careful review of the preceding TL analysis reveals that all we really determined in ACSS operation was V in & I in and V L & I L. This reminds us of the exact input and output form of any linear system with 2 inputs and 2 outputs which can always be expressed as: y 1 y 2 = a 11 a 12 a 21 a 22 x 1 x 2 52

140 Transmission Line as Two Port Network I 1 = I in I 2 = I L Z S + A B + V S V - 1 = V in V 2 = V L - C D + - Z L Source 2-Port Network Load

141 Representation of a TL by a 2-port network Nomenclature Used: V S Sending end voltage I S Sending end current V R Receiving end Voltage I R Receiving end current 54

142 Representation of a TL by a 2-port network Relation between sending end and receiving end is given as: V S = A V R + B I R I S = C V R + D I R In Matrix Form V S I S = A B C D V R I R AD BC = 1 55

143 4 Cases of Interest Case 1: Short TL Case 2: Medium distance TL Case 3: Long TL Case 4: Lossless TL 56

144 ABCD Matrix- Short Lines 57

145 Less than ~80 Km shunt Admittance is neglected Short TL Apply KVL & KCL on abcd loop: V S = V R + ZI R I S = I R ABCD Matrix for short lines: V S = 1 Z I S 0 1 V R IR ABCD Parameters: A = D = 1 per unit B = Z Ω C = 0 58

146 ABCD Matrix- Medium Lines 59

147 Medium TL Ranges from ~ 80 to 250 Km Nominal symmetrical-π circuit model used Apply KVL & KCL on abcd loop: V S = 1 + Z Y 2 V R + ZI R I s = Y 1 + YZ 4 V R YZ 2 I R 60

148 Medium TL ABCD Matrix for medium lines: V S I S = Y Y Z Y Z Z Y Z V R I R ABCD Parameters: A = D = 1 + YZ 2 per unit B = Z Ω C = Y 1 + YZ 4 Ω -1 61

149 ABCD Matrix- Long Lines 62

150 Long TL More than ~ 250 Km and beyond Full TL analysis Z C = Z 0 ABCD Matrix for long lines: V S I S = cosh(γl) 1 sinh(γl) Z C Z C sinh(γl) cosh(γl) V R I R 63

151 R = G = 0 ABCD Matrix- Lossless Lines Z 0 = L C γ = jω L C = jβ ABCD Matrix for lossless lines: cos(βl) V S = j I S sin(βl) Z C j Z C sin(βl) cos(βl) V R I R ABCD Parameters: A = D = cosh γl = cosh(jβl) = cos(βl) per unit B = Z 0 sinh γl = jz 0 sin(βl) Ω C = sinh(γl) Z 0 = j sin(βl) Z 0 S 64

152 ABCD Matrix-Summary 65

153 A Brief Recipe on How to do Power Flow Analysis (PFA) Cooking with Dr. Greg Mowry

154 Steps 1. Represent the power system by its one-line diagram 2

155 Steps Cont. 2. Convert all quantities to per unit (pu) 2-Base pu analysis: S Base kva, MVA, V Base V, kv, 3

156 Steps Cont. 3. Draw the impedance diagram 4

157 Steps Cont. 4. Determine the Y BUS 5

158 Steps Cont. 5. Classify the Buses; aka bus types 6

159 Steps Cont. 6. Guess starting values for the unknown bus parameters; e.g. 7

160 Steps Cont. 7. Using the power flow equations find approximations for the real & reactive power using the initially guessed values and the known values for voltage/angles/admittance at the various buses. Find the difference in these calculated values with the values that were actually given; i.e. form P & Q. (Δ value ) = (Given Value) (Approximated Value) 8

161 Steps Cont. 8. Write the Jacobian Matrix for the first iteration of the Newton Raphson Method. This will have the form: [Δvalues] = [Jacobian Matrix] * [ for Unknown Parameters] 9

162 10

163 Steps Cont. 9. Solve for the unknown differences by inverting the Jacobian and multiplying & form the next guess: (Unknown Value) new = (Unknown Value) old + Solved (Δ Value) 11

164 Steps Cont. 10. Repeat steps 7 9 iteratively until an accurate value for the unknown differences as they 0. Then solve for all the other unknown parameters. Once these iterations converge (e.g. via NR), we now have a fully solved power system with respect to S j & V j at all of the system buses!! 12

165 Power Flow Analysis (PFA) Dr. Greg Mowry Annie Sebastian Marian Mohamed 1

166 Introduction 2

167 Power Flow Analysis In modern power systems, bus voltages and power flow are controlled. Power flow is proportional to v 2, therefore controlling both bus voltage & power flow is a nonlinear problem. In power engineering, the power-flow study, or load-flow study, is a numerical analysis of the flow of electric power in an interconnected system. A power-flow study usually a steady-state analysis that uses simplified notation such as a one-line diagram and per-unit system, and focuses on various aspects of AC power parameters, such as voltages, voltage angles, real power and reactive power. 3

168 Power Flow Analysis - Intro PFA is very crucial during the power-system planning phases as well as during periods of expansion and change for meeting present and future load demands. PFA is very useful (and efficient) at determining power flows and voltage levels under normal operating conditions. PFA provides insight into system operation and optimization of control settings, which leads to maximum capacity at lower the operating costs. PFA is typically performed under ACSS conditions for 3-phase power systems 4

169 PFA Considerations 1. Generation Supplies the demand (load) plus losses. 2. Bus Voltage magnitudes remain close to rated values. 3. Gens Operate within specified real & reactive power limits 4. Transmission lines and Transformers are not overloaded. 5

170 Load Flow Studies (LFS) Power-Flow or Load-Flow software (e.g. CYME, Power World, CAPE) is used for investigating power system operations. The SW determines the voltage magnitude and phase at each bus in a power system under balanced three-phase steady state conditions. The SW also computes the real and reactive power flows (P and Q respectively) for all loads, buses, as well as the equipment losses. The load flow is essential to decide the best operation of existing system, for planning the future expansion of the system, and designing new power system. 6

171 LFS Requirements Representation of the system by a one-line diagram. Determining the impedance diagram using information in the oneline diagram. Formulation of network equations & PF equations. Solution of these equations. 7

172 Conventional Circuit Analysis vs. PFA Conventional nodal analysis is not suitable for power flow studies as the input data for loads are normally given in terms of power and not impedance. Generators are considered power sources and not voltage or current sources The power flow problem is therefore formulated as a set of nonlinear algebraic equations that require numerical analysis for the solution. 8

173 Bus A Bus is defined as the meeting point (or connecting point) of various components. Generators supply power to the buses and the loads draw power from buses. In a power system network, buses are considered as nodes and hence the voltage is specified at each bus 9

174 The Basic 2-Bus Analysis The Foundational Heart of PFA Quick Review 10

175 Motivation The basic 2-Bus analysis must be understood by all power engineers. The 2-Bus analysis is effectively the Thevenin Equivalent of all power systems and power flow analysis. Many variant exist, by my count, at least a dozen. We will explore the basic 2-Bus PFA that models transmission systems. 11

176 The Basic 2-Bus Power System 1 2 j X + - I V S = V S δ = V R e jδ V R = V R 0 = V R S = Send R = Receive 12

177 The Basic 2-Bus Power System 1 2 I ~ j X ~ + - V S = V S δ = V R e jδ V R = V R 0 = V R S = Send R = Receive 13

178 2-Bus Phasor Diagram V R the Reference V S j X I O I V R 90 j X I I due to j 14

179 Discussion 1) V S leads V R by 2) KVL V S j X I = V R I = V S V R jx 15

180 Discussion 3) At the receive end, S R = P R + jq R = V R I S R = V R V S δ V R jx 16

181 Discussion 4) Therefore the complex power received is (after algebra): S R = V S V R X sin δ + j V S V R X cos δ V 2 R X S R = P R + jq R Recall V S = V S δ = V S e jδ = V S cos δ + j sin δ 17

182 Discussion 5) Or: P R = V S V R X sin δ Q R = V S V R X cos δ V R X 2 18

183 Discussion 6) Note that if V S ~ V R ~ V then: P R V2 X δ Q R V X V 19

184 2-Bus, Y, and PFA Lead-in 20

185 Two Bus System arbitrary phase angles 21

186 Two Bus System The complex AC power flow at the receiving end bus can be calculated as follows: S r = V r I Similarly, the power flow at the sending end bus is: where S s = V s I V s = V s e jφ 1 is the voltage and phase angle at the sending bus. V r = V r e jφ 2 is the voltage and phase angle at the receiving bus. Z is the complex impedance of the TL. 22

187 Two Bus System I = V s V r Z In power flow calculations, it is more convenient to use admittances rather than impedances Y = Z 1 I = Y V s V r This 2-bus equation forms the basis of PF calculations. 23

188 The Admittance Matrix The use of admittances greatly simplifies PFA e.g. an open circuit between two buses (i.e. no connection) results in Y = 0 (which is computationally easy to deal with) rather than Z = (which is computationally hard to deal with). V ij is the voltage at bus i wrt bus j; bus i positive, bus j negative I ab represents the current flow from bus a to bus b Recall that a node voltage (bus voltage) is the voltage at that node (Bus) with respect to the reference. 24

189 Admittance Matrix Method (1) NV analysis results in n-1 node equations (bus equations) which is analytically very efficient for systems with large n. Given systems with n large, an efficient and automatable method is needed/required for PFA. NVM and Y-matrix methods are so. In circuit analysis, series circuit elements are typically combined to form a single impedance to simplify the analysis. This is generally not done in PFA. 25

190 Admittance Matrix Method (2) In PFA, combining series elements is not typically done as we seek to keep track of the impedances of various assets and their power flows; e.g. in: i. Generators ii. TLs iii. XFs iv. FACTs Voltages sources (e.g. generators) have an associated impedance (the Thevenin impedance) that is typically inductive in ACSS 26

191 Y-Matrix Example 27

192 Example Circuit 28

193 Step 1: Identify nodes (buses), label them, and select one of them as the reference or ground. 29

194 Step 2 & 3: (2) Replace voltage sources and their Thevenin impedance with their Norton equivalent. (3) Replace impedance with their associated admittances. 30

195 Step 4: Form the system matrix Y Bus V = I (Eqn-1) where, Y bus = Bus Admittance matrix of order (n x n); known V = Bus (node) voltage matrix of order (n x 1); unknown I = Source current matrix of order (n x 1); known 31

196 Step 5: Solve the system matrix for bus voltages 1 V = Y Bus I (Eqn-2) Note: at this point the various known bus voltages (amplitude and phase) may be used to calculate currents and power flow. 32

197 Stuffing the Y-Matrix Y-matrix elements are designated as Y kn where k refers to the k th bus and n refers to the n th bus. Diagonal elements Y kk : Y kk = sum of all of the admittances connected to the k th bus Off-diagonal elements Y kn : Y kn = Y nk = -(sum of the admittances in the branch connecting bus k to bus n) 33

198 Y-Matrix Terminology Y kk, the main diagonal elements of Y, are referred to as the: Self-admittance ; or Driving point admittance of bus k Y kn, the off diagonal elements of Y, are referred to as the: Mutual-admittance ; or Transfer admittance between bus k and bus n 34

199 Y-Matrix Notes If there is no connection between bus k and bus n, then Y kn = 0 in the Y-matrix; hence there are 2 zeros in the Y-matrix for this situation since Y kn = Y nk. If a new connection is made between 2 buses k & n in a network, then only 4 elements of the Y matrix change, all of the others remain the same. The elements that change are: Y kk Y nn Y kn = Y nk 35

200 The stuffed Y-Matrix I 1 I 2 I 3 I 4 = Y 11 Y 12 Y 13 0 Y 21 Y 22 Y 23 0 Y 31 Y Y 33 Y 43 Y 34 Y 44 V 1 V 2 V 3 V 4 Note that since no direct connections exist between Bus 1 & 4 or Bus 2 & 4, the corresponding entries in the Y matrix are zero 36

201 Y-Matrix Math 37

202 PSC Reminder For a passive network element(s), the PSC yields: V 1 + V I I = V 1 V 2 Z - I = Y V 1 V 2 V 2 38

203 Y-Matrix Equations Consider the following bus in a network: I a Bus 3 Z 13 V 3 V 1 Z C1 I b V 2 I c Z 23 Z C2 I d I 3 39

204 Y-Matrix Equations KCL at bus 3 yields: I 3 = I b + I d + I a + I c I 3 = V 3 Z C1 + V 3 Z C2 + V 3 V 1 Z 13 + V 3 V 2 Z 23 I 3 = V 3 Y 3,total to gnd + m 3 V 3 V m Z m3 40

205 Y-Matrix Equations Algebra yields: I 3 = V 3 Y 3,total to gnd + m 3 V 3 V m Z m3 I 3 = V 3 Y 3,total to gnd + m 3 Y m3 V 3 V m I 3 = V 3 Y 3,total to gnd + m 3 Y m3 m 3 Y m3 V m 41

206 Y-Matrix Equations Finally: I 3 = V 3 Y 3,total to gnd + m 3 Y m3 m 3 Y m3 V m Or in general I k = V k Y k,total to gnd + m k Y mk m k Y mk V m 42

207 Y-Matrix Equations Hence: I k = V k Y k,total to gnd + m k Y mk m k Y mk V m Admittance connected to the k th bus Admittance between the k th and m th bus (i) (ii) 43

208 Y-Matrix Equations In terms of Y-matrix elements: Y kk = Y k,total to gnd + Y mk (i) m k Y km = Y mk = 1 Z km (ii) 44

209 Y-Matrix Equations Using the short-hand notation: I k = V k Y kk + m k V m Y mk Y km = Y mk 45

210 Consider a 3-Bus System Bus 1 Bus 3 Z 13 I 1 I 3 Bus 2 I 2 46

211 Let s focus on Bus 2 I 2 = V 2 Y 22 + V 1 Y 21 + V 3 Y 23 I 2 = V 1 Y 21 + V 2 Y 22 + V 3 Y 23 I 2 = Y 21 Y 22 Y 23 V 2 V 3 V 1 47

212 The General Case For a n-bus system we have: I 1 I n = Y 11 Y 1n Y n1 Y nn V 1 V n Known sources The Y-Matrix To be calculated 48

213 Power Flow Equations 49

214 Power Flow Equation The current at bus k in an n-bus system is: I k = Y k1 V 1 + Y k2 V Y kn V n I k = n m=1 Y km V m where V k = V k e jθ k V m = V m e jθ m Y km = Y km e jδ km 50

215 Power Flow Equation The complex power S at bus k is given by: S k = V k I k = P k + jq k n S k = V k m=1 Y km V m 51

216 Power Flow Equation Substituting the voltage and admittance phasors into the complex power equations yields: n S k = P k + jq k = V k e jθ k m = 1 Y km e jδ km V m e jθ m By Euler s Theorem V k V m = Vk V m cos θ km + j sin θ km where θ km = θ k θ m 52

217 Power Flow Equation Expanding the complex power summation into real and imaginary parts and associating these with P & Q respectively yields our fundamental PFA equations P k = n m = 1 V k V m Y km cos θ k θ m + δ km Q k = n m = 1 V k V m Y km sin θ k θ m + δ km 53

218 Power Flow Equation Key SUMMARY points: The bus voltage equations, I=YV, are linear The power flow equations are quadratic, V 2 appears in all P & Q terms Consequently simultaneously controlling bus voltage and power flow is a nonlinear problem In general this forces us to use numerical solution methods for solving the power flow equations while controlling bus voltages 54

219 Network Terminology 55

220 Simple 3-Bus Power System Z 12 PV Bus ~ 1 2 ~ Slack Bus 3 PQ Bus 56

221 Bus Types Bus Type Load Bus (or) PQ Bus Generator Bus (or) Voltage Controlled Bus (or) Total number of buses Quantities Specified Quantities to be obtained n-m P, Q V, δ m-1 P, V Q, δ PV Bus Slack Bus (or) Swing Bus (or) 1 V, δ P, Q Reference Bus 57

222 Bus Terminology P = P G P D Q = Q G Q D P G & Q G are the Real and Reactive Power supplied by the power system generators connected P D & Q D are respectively the Real and Reactive Power drawn by bus loads V = Magnitude of Bus Voltage δ = Phase angle of Bus Voltage 58

223 The Need for a Slack Bus TL losses can be estimated only if the real and reactive power at all buses are known. The powers in the buses will be known only after solving the load flow equations. For these reasons, the real and reactive power of one of the generator bus is not specified. This special bus is called Slack Bus and serves as the system reference and power balancing agent. 59

224 The Need for a Slack Bus It is assumed that the Slack Bus Generates the real and reactive power of the Transmission Line losses. Hence for a Slack Bus, the magnitude and phase of bus voltage are specified and real and reactive powers are obtained through the load flow equation. Sum ofcomplex Power of Generator = Sum of Complex Power of Loads + Total Complex Power loss in Transmission Lines or Total Complex power loss in Transmission Lines = Sum of Complex Power of Loads - Sum of Complex Power ofgenerator 60

225 Solving the PF equations 61

226 Power Flow Analysis - Intro There are several methods available for solving the power flow equations: Direct solutions to linear algebraic equations Gauss Elimination Iterative solutions to linear algebraic equations Jacobi method Gauss-Siedel method Iterative solutions to nonlinear algebraic equations Newton-Raphson method 62

227 Gauss-Seidel Method In numerical linear algebra, the Gauss Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a linear system of equations. It is named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von Seidel, and is similar to the Jacobi method. Though it can be applied to any matrix with non-zero elements on the diagonals, convergence is only guaranteed if the matrix is either diagonally dominant, or symmetric and positive definite. It was only mentioned in a private letter from Gauss to his student Gerling in

228 The Jacobi Method Carl Gustav Jacob Jacobi was a German mathematician, who made fundamental contributions to elliptic functions, dynamics, differential equations, and number theory. 64

229 Newton-Raphson Method 65

230 Newton-Raphson Method In numerical analysis, Newton's method (also known as the Newton Raphson method), is named after Isaac Newton and Joseph Raphson. It is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. x; f x = 0 The Newton Raphson method in one variable is implemented as follows: The method starts with a function f defined over the real numbers x The function's derivative f', An initial guess x 0 for a root of the function f. If the function is well behaved then a better approximation to the actual root is x 1 : x 1 = x 0 f(x 0) f (x 0 ) 66

231 Newton-Raphson Method The Newton-Raphson method is a very popular nonlinear equation solver as: NR is very powerful and typically converges very quickly Almost all PFA SW uses Newton-Raphson Very forgiving when errors are made in one iterations; the final solution will be very close to the actual root 67

232 Newton-Raphson Method Procedure: Step 1: Assume a vector of initial guess x(0) and set iteration counter k = 0 Step 2: Compute f 1 (x (k) ), f 2 (x (k) ), f n (x (k) ) Step 3: Compute m 1, m 2, Δm n. Step 4: Compute error = max[ m 1, m 2, mn ] Step 5: If error (pre-specified tolerance), then the final solution vector is x(k) and print the results. Otherwise go to step 6 Step 6: Form the Jacobin matrix analytically and evaluate it at x = x (k) Step 7: Calculate the correction vector x = [ m 1, m 2, Δm n ] T by using equation d Step 8: Update the solution vector x (k+1) = x (k) + x and update k=k+1 and go back to step 2. 68

233 Newton-Raphson Method Consider a 2-D example where in general there are n equations are given for the n unknown quantities x 1, x 2, x 3 x n. f 1 (x 1, x 2 ) = k 1 k 1 and k 2 are constants f 2 (x 1, x 2 ) = k 2 (Equation set a) Let s assume that x 1 ~ [x 1 (0) + x 1 (0) ] x 2 ~ [x 2 (0) + x 2 (0) ] Then k 1 = f 1 (x 1 (0) + x 1 (0), x 2 (0) + x 2 (0) ) k 2 = f 1 (x 1 (0) + x 1 (0), x 2 (0) + x 2 (0) ) 69

234 Newton-Raphson Method Applying the first order Taylor Expansion to f 1 and f 2 k 1 f 1 (x 1 (0), x 2 (0) ) + x 1 (0) f 1 x 1 + x 2 (0) f 1 x 2 k 2 f 2 (x 1 (0), x 2 (0) ) + x 1 (0) f 2 x 1 + x 2 (0) f 2 x 2 (Equation set b) Equation set b expressed in matrix form, k 1 f 1 (x 1 (0), x 2 (0) ) k 2 f 2 (x 1 (0), x 2 (0) ) = f 1 x 1 f 2 f 1 x 2 f 2 x 1 (0) x 2 (0) (Equation c) x 1 x 2 The Baby Jacobian Matrix 70

235 Newton-Raphson Method Then k 1 (0) k 2 (0) = J (0) x 1 (0) x 2 (0) Adjusting the guess according to x 1 (1) = [x 1 (0) + x 1 (0) ] x 2 (1) = [x 2 (0) + x 2 (0) ] Iterate until x i (N) < 71

236 Newton-Raphson Method For n equations in n unknown quantities x 1, x 2, x 3 x n. f 1 (x 1, x 2 x n ) = b 1 f 2 (x 1, x 2 x n ) = b 2 f n (x 1, x 2, x n ) = b n (Equation set a) 72

237 Newton-Raphson Method In equation set a the quantities b 1, b 2, b n as well as the functions f 1, f 2 f n are known. To solve the equation set an initial guess is made; the initial guesses be denoted as x 1 (0), x 2 (0), x n (0). First order Taylor s series expansions (neglecting the higher order terms) are carried out for these equation about the initial guess The vector of initial guesses is denoted as x (0) = [x 1 (0), x 2 (0), x n (0) ] T. Applying Taylor s expansion to the equation set a will give: 73

238 f 1 (x 1 (0), x 2 (0) x n (0) ) + f 1 x 1 x 1 + f 1 x 2 x f 1 xn xn = b 1 f 2 (x 1 (0), x 2 (0) x n (0) ) + f 2 x 1 f n (x 1 (0), x 2 (0) x n (0) ) + fn x 1 x 1 + f 2 x 2 x f 2 xn xn = b 2 x 1 + fn x 2 x f 2 xn xn = b n (Equation set b) Equation set b can be written in matrix form as: f 1 (x 0 ) f 2 (x 0 ) f n (x 0 ) = f 1 x 1 f 2 x 1 f n x 1 f 1 x 2 f 2 x 2 f n x 2 f 1 xn f 2 xn f n xn x 1 x 2 xn = b 1 b 2 b n (Equation c) 74

239 f 1 x 1 f 1 x 2 f 1 xn f 2 x 1 fn x 1 f 2 x 2 f 2 xn fn fn xn xn Jacobin Matrix (J) Hence equation c can be written as x 1 x 2 xn = [J -1 ] b 1 f 1 (x 0 ) b 2 f 2 (x 0 ) b n fn(x 0 ) = [J -1 ] m 1 m 2 m n (Equation d) Equation d is the basic equation for solving the n algebraic equations given in equation set a. 75

240 NR Method Flowchart 76

241 Power Flow Example (using Power World) 77

242 Power Flow Example 78

243 Step 1: Calculate Y-Bus: Power Flow Example, cont. j10 j10 0 Y Bus = j10 j30 j20 0 j20 j20 79

244 Power Flow Example, cont. Step 2: Define Bus types and their quantities: Bus No Bus Type Quantities Specified Quantities to be obtained One Slack Bus V 1, δ Two Generator Bus PV P 2, V 2 δ 2, Q 2 Three Load Bus PQ P 3, Q 3 V 3, δ 3 80

245 Power Flow Example, cont. Step 3: User Power Equations to solve for unknowns: P 2 = P 3 = 3 m = 1 3 m = 1 V 2 V m Y 2m cos θ 2k δ 2 + δ m V 3 V m Y 3m cos θ 3k δ 3 + δ m 3 Q 3 = V 3 V m Y 3m sin θ 3k δ 3 + δ m m = 1 81

246 Power Flow Example, cont. 1 = 10 sinδ 2 20 V 3 sin( δ 2 +δ 3 ) Eq.1 2 = 20 V 3 sin( δ 3 +δ 2 ) Eq = 20 V 3 cos δ 3 +δ 2 20 V 3 2 Eq.3 82

247 Power Flow Example, cont. Step 4: Apply Newton-Raphson on Eq.2 and Eq.3 to find V 3 and ( δ 3 +δ 2 ) ; then on Eq.1 to find δ 2 followed by δ 3 V 3 = pu δ 2 = Deg δ 3 = Deg 83

248 Power Flow Example, cont. Step 5: Use V 3, δ 2, and δ 3 to find Q 2 and Q 1 : 3 Q 2 = V 2 V m Y 2m sin θ 2k δ 2 + δ m Q 1 = m = 1 3 m = 1 V 1 V m Y 1m sin θ 1k δ 1 + δ m Q 2 = pu Q 2 = pu 84

249 Final Soliton Power Flow Example, cont. Bus No V δ P Q One 1 0 o Two o Three o

250 Power World Example 86

251 Y-Bus Power World Example, cont. 87

252 Power World Example, cont. Jacobin Matrix 88

253 Power World Example, cont. Power Flow Soliton 89