Lecture Notes on Power System Stability

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Lecture Notes on Power System Stability Mrinal K Pal CHAPTER 1 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS Representation of Transmission Lines A transmission line has four parameters series resistance and inductance, and shunt conductance and capacitance. These are uniformly distributed along the line. In the derivation of transmission line equations the following assumptions are usually made: The line is transposed (or symmetrical), and the three phases are balanced. In practice this is not entirely true but the unbalance and departure from symmetry are small. The line can therefore be analyzed on a per-phase basis. The line parameters, per unit length, are constant, i.e., they are independent of position, frequency, current, and voltage. This assumption, although approximate, is permissible for most power system studies. Equations of a transmission line The voltage and current on a transmission line depend, in general, upon both time and position. Consequently, a general mathematical description of the line involves partial differential equations. A transmission line section is shown schematically in Figure 1.1. The voltage and current conditions at a small section of length dx at a distance x from the receiving end of the line is as shown in the figure. The voltage and current, denoted by v and i respectively, represent instantaneous quantities. Denoting the series resistance and inductance and the shunt conductance and capacitance per unit length of the line by r, l, g and c, respectively, the series voltage for the section dx, is ∂v∂i dx= rdx i+ ldx (1.1) ∂x∂t and the shunt current is ∂i∂v dx= gdx v+ cdx (1.2) ∂x∂t Fig. 1.1 Schematic representation of a transmission line. 1-1 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS Equations (1.1) and (1.2) apply to both steady state and transient conditions. For power flow and stability analysis purposes we need to consider only the steady state phenomena. The transmission line transients are fast acting and have negligible impact on system stability except in special situations. The steady state sinusoidal time variation of voltage and current can be represented by jωt jωt v(x,t)= V (x)ε , i(x,t)= I(x)ε Therefore, in the steady state, equations (1.1) and (1.2) reduce to dV (x) =(r+ jωl) I(x)= z I(x) (1.3) dx dI(x) = (g+ jωc)V (x)= yV (x) (1.4) dx where z and y are the series impedance and shunt admittance per unit length of the line, respectively. Equations (1.3) and (1.4) can be combined to obtain equations in one unknown, yielding the following equations. In these equations V and I represent rms values. 2 d V = yzV (1.5) 2 dx 2 d I = yz I (1.6) 2 dx The solutions of equations (1.5) and (1.6) are V+ I Z V− I Z R R cγx R R c−γx V=ε+ε (1.7) 2 2 V Z+ I V Z− I R c Rγx R c R−γx I=ε−ε (1.8) 2 2 where z Z= = the characteristic impedance c y γ= yz = the propagation constant Equations (1.7) and (1.8) give the rms values of V and I, and their phase angles at any distance x from the receiving end. Both Z and γ are complex quantities. γ can be expressed as γ=α+ jβ . α is called the c attenuation constant and β the phase constant. The two terms in equation (1.7) ((1.8)) are called incident voltage (current) and reflected voltage (current), respectively. If a line is terminated in its characteristic impedance Z , V = I Z and c R R c there is no reflected voltage or current, as may be seen from equations (1.7) and (1.8). If a line is lossless, its series resistance and shunt conductance are zero and the characteristic impedance 1- 2FUNDAMENTALS OF POWER FLOW AND POWER LIMITS reduces to L / C , a pure resistance. Under these conditions, the characteristic impedance is often referred to as surge impedance. The wavelength λ is the distance along a line between two points of a wave which differ in phase of 2π radians. If β is the phase shift in radians per mile, the wavelength in miles is 2π 1 λ=≈ (1.9) β f LC where L and C are the series inductance and shunt capacitance per mile, respectively. At a frequency of 60 Hz, the wavelength is approximately 3,000 miles. The velocity of propagation of a wave in miles per second is 1 v= fλ≈ (1.10) LC Equations (1.7) and (1.8) can be rearranged using hyperbolic functions θ−θθ−θ ε−εε+ε sinh= , cosh= 2 2 to obtain V= V coshγx+ I Z sinhγx R R c and V R I= I coshγx+ sinhγx R Z c Letting x = l, where l is the length of the line, V= V coshγl+ I Z sinhγl (1.11) S R R c V R I= I coshγl+ sinhγl (1.12) S R Z c Equations (1.11) and (1.12) can be solved for V and I to obtain R R V= V coshγl− I Z sinhγl (1.13) R S S c V S I= I coshγl+ sinhγl (1.14) R S Z c Equations (1.11) through (1.14) are the fundamental equations of a transmission line. For balanced three-phase lines, the current is the line current, and the voltage is the line-to-neutral voltage, i.e., the line voltage divided by 3. Equivalent circuit of a transmission line A transmission line can be represented accurately by a lumped parameter equivalent circuit (either a π or T circuit), insofar as the conditions at the ends of the line are concerned. An equivalent π circuit is shown in Figure 1.2, where the equivalent series impedance and shunt admittance are also shown. 1- 3FUNDAMENTALS OF POWER FLOW AND POWER LIMITS Fig. 1.2 Equivalent π circuit of a transmission line. Note that Z (= zl) and Y (= yl) in Figure 1.2 represent the total series impedance and shunt admittance of the line, respectively. Also note that γl = ZY . Problems 1. Derive the expressions for Z and Y /2 as shown in Fig.1.2. π π 2. Derive the equivalent T circuit. sinhγl tanh(γl / 2) and are the factors by which the total line series impedance and shunt γlγl / 2 admittance are to be multiplied in order to obtain the series impedance and shunt admittances of the equivalent π circuit. The correction factors approach unity as γl (or ZY) approaches zero, i.e., as the line becomes electrically shorter. A π circuit in which the series arm has the impedance Z and each of the shunt arms has the admittance Y/2, obtained by setting the correction factors equal to unity, is called a nominal π circuit. A T circuit with two series arms each of impedance Z/2 and one shunt arm of admittance Y is called a nominal T circuit. Nominal π and nominal T are approximations to the equivalent π and equivalent T, respectively. The approximations are valid for lines less than 100 miles long. Longer lines may be broken into two or more segments and each segment may be represented by a nominal π or T. A nominal π is more convenient for computational purposes and is therefore more widely used. Problem Derive the sending-end voltage (V ) and current (I ) in terms of the receiving-end voltage (V ) S S R and current (I ), and vice-versa, for both the nominal π and the nominal T circuits. R For short transmission lines (less than 50 miles long), the total shunt capacitance is small and can be neglected. Therefore, a short transmission line, for the purpose of power flow and stability studies, can be represented by the simple series circuit shown in Figure 1.3. 1- 4FUNDAMENTALS OF POWER FLOW AND POWER LIMITS Fig. 1.3 Equivalent circuit of a short transmission line. Y - ∇ transformations Fig. 1.4 Y- ∇ equivalent circuits. The Y circuit in Figure 1.4 can be transformed into the equivalent ∇, and the node o eliminated by a Y - ∇ transformation. The impedances of the equivalent are Z Z+ Z Z+ Z Z 1  a b b c c a Z== Z Z  ab a b Z Z  c Z Z+ Z Z+ Z Z 1  a b b c c a Z== Z Z (1.15)  bc b c Z Z  a Z Z+ Z Z+ Z Z 1  a b b c c a Z== Z Z  ca c a Z Z  b where 1 1 1 1 =++= Y+ Y+ Y a b c Z Z Z Z a b c In terms of admittances, the above equations can be written as Y Y a b Y= ab Y+ Y+ Y a b c Y Y b c Y= (1.16) bc Y Y Y ++ a b c Y Y c a Y= ca Y+ Y+ Y a b c 1- 5FUNDAMENTALS OF POWER FLOW AND POWER LIMITS Problem Derive the above expressions for Z, Z , Z ab bc ca A ∇ circuit can be transformed to an equivalent Y, where the impedances of the equivalent Y are Z Z ab ca Z= a Z+ Z+ Z ab bc ca Z Z ab bc Z= (1.17) b Z+ Z+ Z ab bc ca Z Z bc ca Z= c Z+ Z+ Z ab bc ca Problem Derive the above expressions for Z, Z , Z a b c If more than three impedances terminate on a node, the node may be eliminated by applying the general star-mesh conversion equations. However, the conversion is not reversible. Problem Convert the star circuit shown in Figure 1.5 into the equivalent mesh circuit and find the expressions for Z , etc. ab Fig. 1.5 Star-mesh equivalent circuit. Per Unit System In power system computations, great simplifications can be realized by employing a system in which the electrical quantities are expressed as per units of properly chosen base quantities. The per unit value of any quantity is defined as the ratio of the actual value to its base value. Base quantities In an electrical circuit, voltage, current, volt-ampere and impedance are so related that selection of base values of any two of them determines the base values for the remaining two. Usually, base kVA (or MVA) and base voltage in kV are the quantities selected to specify the base. For single-phase systems (or three-phase systems where the term current refers to line current, the term voltage refers to voltage to neutral, and the term kVA refers to kVA per phase), the following formulas relate the various quantities. 1- 6FUNDAMENTALS OF POWER FLOW AND POWER LIMITS base kVA Base current in amperes = (1.18) base voltage in kV base voltage in volts Base impedance in ohms = base current in amperes 2 (base voltage in kV)×1000 = base kVA 2 (base voltage in kV) (1.19) = base MVA Base power in kW = base kVA Base power in MW = base MVA actual quantity = (1.20) Per unit quantity base quantity Occasionally a quantity may be given in percent, which is obtained by multiplying the per unit quantity by 100. Base impedance and base current can be computed directly from three-phase values of base kV and base kVA. If we interpret base kVA and base voltage in kV to mean base kVA for the total of the three phases and base voltage from line to line, then base kVA Base current in amperes = (1.21) 3× base voltage in kV 2 (base voltage in kV/ 3)× 1000 Base impedance in ohms= base kVA/3 2 (base voltage in kV)×1000 = base kVA 2 (base voltage in kV) (1.22) = base MVA Therefore, the same equation for base impedance is valid for either single-phase or three-phase circuits. In the three-phase case line-to-line kV must be used in the equation with three-phase kVA or MVA. Line-to-neutral kV must be used with kVA or MVA per phase. Change of bases (actual impedance in ohms) × (base kVA) Per unit impedance of a circuit element = (1.23) 2 (base voltage in kV)× 1000 which shows that per unit impedance is directly proportional to the base kVA and inversely proportional to the square of the base voltage. Therefore, to change from per unit impedance on a given base to per unit impedance on a new base, the following equation applies: 1-7 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS 2 base kV  base kVA given new  Per unit Z per unit Z (1.24) = new given   base kV base kVA  new given  When the resistance and reactance of a device are given by the manufacturer in percent or per unit, the base is understood to be the rated kVA and kV of the apparatus. The ohmic values of resistance and leakage reactance of a transformer depend on whether they are measured on the high- or low-voltage side of the transformer. If they are expressed in per unit, the base kVA is understood to be the kVA rating of the transformer. The base voltage is understood to be the voltage rating of the side of the transformer where the impedance is measured. 2  kV L  Z=× Z (1.25) LT HT  kV  H where Z and Z are the impedances referred to the low-voltage and high-voltage sides of the LT HT transformer, and kV and kV are the rated low-voltage and high-voltage of the transformer, L H respectively. 2 (kV / kV )× Z× kVA L H HT ∴ Z in per unit = LT 2 (kV )×1000 L Z× kVA HT = 2 (kV )×1000 H = Z in per unit HT A great advantage in making per unit computations is realized by the proper selection of different voltage bases for circuits connected to each other through a transformer. To achieve the advantage, the voltage bases for the circuits connected through the transformer must have the same ratio as the turns ratio of the transformer windings. Per unit and percent admittance 1 1 Z= , and Y= ohm Y Z mho MVA 1 b Base admittance Y== (1.26) b 2 Z kV b b 2 Y kV Z 1 b b ∴ Y== Y= Y Z== (1.27) pu b Y MVA Z Z b b pu Z= Z×100, Y= Y×100 percent pu percent pu 4 1 10 ∴ Y=×100= (1.28) percent Z Z pu percent P= 3VI cosφ , Q= 3VI sinφ 1-8 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS 3VI cosφ ∴ P== V I cosφ pu pu pu 3V I b b (1.29) 3VI sinφ Q== V I sinφ pu pu pu 3V I b b Problem Transmission line charging is usually given in terms of total three-phase Mvar. If a transmission line has a total line charging of 50 Mvar at 485 kV, what is the per unit admittance on 100 MVA and 500 kV base? Power Limits Power System stability is dependent primarily upon the ability of the electrical system to interchange energy as required between the connected apparatus. It is therefore necessary to develop the fundamentals of power flow and power limit of electrical circuits. Consider the power flow between two points along a transmission line as shown in Figure 1.6 (the shunt admittance has been neglected). Fig. 1.6 Schematic of a two terminal transmission line, neglecting shunt admittance. At the receiving end, ˆ ˆ P+ jQ= V I (1) (1.30) R R R R or ˆ ˆ P− jQ= V I (1.31) R R R R Problem Derive equations (1.30) and (1.31). From Figure 1.6 V∠δ− V∠0 S R I= I==() V∠δ− V∠0(G+ jB) R S S R R+ jX where 1 = G+ jB , the series admittance. R+ jX Substituting the above in equation (1.31), 2 2 P− jQ=(V V∠δ−V)(G+ jB)=(V V∠δ−V)Y∠θ R R R S R R S R where (1) In this and subsequent chapters a “” is used to distinguish a phasor or complex quantity from a scalar. However, this is omitted when there is no chance of confusion. 1-9 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS 1 1 2 2 Y= G+ B== 2 2 Z R+ X and B X −1−1o θ= tan=− tan=α− 90 G R where R −1 α= tan X or o α= 90+θ For normal lines with inductive reactance X, α will have a small positive value. Equating real and imaginary parts, 2 P= V V Y cos(δ+θ )− V Y cosθ (1.32) R R S R 2 Q=−V V Y sin(δ+θ )+ V Y sinθ (1.33) R R S R o Alternatively, sinceθ=α− 90 , 2 P= V V Y sin(δ+α)− V Y sinα (1.34) R R S R and 2 Q= V V Y cos(δ+α)− V Y cosα (1.35) R R S R In terms of G and B, P and Q can be written as, sincesinα= G /Y and cosα=− B /Y , R R 2 P=−V G+ V V(G cosδ− Bsinδ) (1.36) R R R S and 2 Q= V B+ V V(G sinδ− B cosδ) (1.37) R R R S When R = 0, G = 0 and B = −1/X, and equations (1.36) and (1.37) reduce to V V R S P= sinδ (1.38) R X and 2 V V V R S R Q= cosδ− (1.39) R X X Problem Derive expressions similar to equations (1.32) through (1.39) for P and Q . S S For R = 0, the maximum power that can be transferred, when V , V and X are constant, is, from R S equation (1.38), V V R S P= (1.40) R max X 1-10 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS When R is not equal to zero, the maximum power can be found by taking the partial derivative of P with respect to δ, and equating to zero. For example, from equation (1.34), R ∂P R = V V Y cos(δ+α)= 0 R S ∂δ which yields o δ+α= 90 or o δ= 90−α and 2 2 P= V V Y− V Y sinα= V V Y− V G (1.41) R max R S R R S R Problems o 1. Show that the maximum power at the sending end occurs at δ= 90+α . 2. Assuming V= V , find the value of the reactance (X) for a given value of the resistance R S (R) for maximum receiving end power. (The solution is X= 3 R .) Circle Diagram Equations (1.34) and (1.35) can be rewritten as 2 P+ V Y sinα= V V Y sin(δ+α) (1.42) R R R S 2 Q+ V Y cosα= V V Y cos(δ+α) (1.43) R R R S Squaring (1.42) and 1.43), and adding 2 2 2 2 2 ()() () P+ V Y sinα+ Q+ V Y cosα= V V Y (1.44) R R R R R S 2 2 Equation (1.44) describes a circle with centers located at −V Y sinα , −V Y cosα , if P and Q R R R R are used as the axes of coordinates. The radius of the circle is V V Y. The circle diagram is R S shown in Figure 1.7. Problems 1. Draw the sending-end circle diagram for the system shown in Figure 1.6 2. Using the circle diagram calculate δ, given P = 1.0, V = V = 1.0, R = 0.05, X = 0.3. S S R It should be noted that the maximum power as given by equation (1.41), or as obtained from the circle diagram cannot be realized in practice, since this would require an uneconomically large amount of reactive power to be supplied at the receiving end in order to hold the receiving-end voltage constant. Also, above a certain level of transmitted power, the reactive support must have some sort of continuous control, such as provided by synchronous condensers or static var systems (SVS), in order to avoid voltage instability and collapse. The subject of voltage instability will be dealt with in more detail in Chapter 10. 1-11 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS Fig. 1.7 Receiving-end circle diagram for fixed V and V . S R If there are synchronous machines at both ends of a transmission line, the power limit is set by the ability of the machines to stay in synchronism with each other. This limit is called the (synchronous) stability limit. (2) The steady-state stability limit may be defined as the maximum power that can be delivered without loss of synchronism when the load is increased gradually and the machine terminal voltages are adjusted by manual control of excitation. Similarly, the transient stability limit may be defined as the maximum power that can be delivered without loss of synchronism following a large disturbance such as a system fault and its subsequent clearing. Basic Power Flow Calculation Referring to Figure 1.6, P , Q , and the magnitude of V are given. It is desired to solve for R R S V ,δ , P , and Q . The solution can be obtained in several ways, two of which will be illustrated S R S here. Direct solution From (1.31) P− jQ R R I= I= R S V R P− jQ R R V∠δ= V+ I Z= V+ () R+ jX S R R V R  P R+ Q X P X− Q R R R R R =V++ j R  V V R R   K K 1 2  = V++ j (1.45) R  V V R R  (2) In this chapter we will follow the classical definitions of power system stability 4. A more rigorous definition of stability is given in Chapter 2, which will be followed in subsequent chapters. 1-12 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS where K= P R+ Q X , K= P X− Q R 1 R R 2 R R From (1.45) 2 2  K K 2 1 2  V= V++ S R  V V R R  which can be rearranged as 2 2 2 2 2 2 () V+(2K−V )V+(K+ K)= 0 R 1 S R 1 2 From the above we obtain 2 2 2 2 2 V− 2K±(V− 2K)− 4(K+ K) 2 S 1 S 1 1 2 V= R 2 V is obtained by taking the appropriate (positive) sign before the radical. Once V is known, δ R R can be calculated from equation (1.45). Iterative procedure (Newton's method) Rewriting equations (1.36) and (1.37) 2 P=−V G+ V V(G cosδ− Bsinδ)= f(V ,δ) (1.46) R R R S 1 R 2 Q= V B+ V V(G sinδ− B cosδ)= f(V ,δ) 1.47) R R R S 2 R From (1.46) and (1.47) ∂∂ () () ∆P= f V ,δ∆δ+ f V ,δ∆V (1.48) R 1 R 1 R R ∂δ∂V R ∂∂ ∆Q= f() V ,δ∆δ+ f() V ,δ∆V (1.49) R 2 R 2 R R ∂δ∂V R At the start of the iterative process ∆P= P− f(V ,δ) (1.50) R R 1 Ro o and ∆Q= Q− f(V ,δ) (1.51) R R 2 Ro o where V ,δ are the starting values of V , δ. R Ro o The partial derivatives can be easily computed from (1.46) and (1.48). The changes in δ and V R required to improve upon the values obtained from the previous iteration are obtained by solving equations (1.48) and (1.49). For example, at the start of the iteration process ∆P ,∆Q and the R R partial derivatives at V and δ are computed and ∆δ and ∆V are obtained from equations Ro o 1 R1 (1.48) and (1.49). At the end of the first iteration 1 δ=δ+∆δ (1.52) o 1 1-13 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS 1 V= V+∆V (1.53) R Ro R1 These updated values of δ and V are then used in the second iteration. The iteration is continued R until ∆Pε and ∆Qε , where ε and ε are the specified tolerances. The above iterative 1 2 R 1 R 2 procedure can be extended to systems of any size. Selection of starting values for δ and V R A good starting value for δ and V can be obtained by using the approximations R G≈ 0 , cosδ≈ 1 , and sinδ=δ since G and δ are usually small. To obtain the starting value of δ, a further approximation, V≈ V≈ 1.0 , can be made since the R S voltages are usually fairly close to unity. Using these approximations, the starting value of δ is obtained from equation (1.46) as δ=− P / B= P X (1.54) o R R In general, V will be close to V , and therefore V = V will provide a good starting value for R S Ro S V . If however, Q is appreciable, a better starting value for V can be obtained from equation R R R (1.47) using the above approximation, as V= V+ Q / B= V− Q X (1.55) Ro S R S R The above method of estimating the starting values can be extended to systems of any size. This method is advisable in order to avoid divergence in the power-flow solution. Problem 1. Referring to Figure 1.6, the following quantities are given: P = 0.85 , Q = 0.35, V = 1.15 R R S Calculate δ , V , P and Q . R S S 2. Referring to Figure 1.6, the following quantities are given: P = 0.8 , Q = 0.4, V = 1.0 R R R Calculate δ , V , P and Q . S S S Steady State Stability Consider a cylindrical rotor synchronous generator connected to a large power system (infinite bus) operating at constant speed with constant field current and no saturation, as shown in Figure 1.8. Fig. 1.8 A cylindrical rotor synchronous generator connected to infinite bus and phasor diagram 1-14 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS r and x are the armature resistance and synchronous reactance, respectively, and E is the a d internal machine voltage or the voltage corresponding to field excitation. Since E is produced by the field flux, it is associated with the rotor position. The power output of the machine is given by 2 E sinα EV P=+ sin(δ−α) (1.56) 2 2 2 2 r+ x r+ x a d a d where −1 α= tan (r / x ) a d Since r is usually small, it can be neglected. Then a EV P= sinδ (1.57) x d The maximum power output for fixed E and V is EV P= (1.58) max x d This is termed the pull-out power of a synchronous generator against an infinite bus at constant field excitation, since any attempt to increase the power output further will result in a loss of synchronism. Equation (1.56) represents the power angle characteristic of a cylindrical rotor synchronous machine connected directly to an infinite bus. Problem Calculate the power angle characteristic of a cylindrical rotor generator connected through an external reactance to an infinite bus of voltage 1.0 pu, as shown in Figure 1.9. E is such that 1.0 pu voltage is obtained at the terminal at P = 1.0 pu. Fig. 1.9 A cylindrical rotor synchronous generator connected to an infinite bus through an external reactance. Steady state stability is the stability of the system under conditions of gradual or relatively slow changes in load. The load is assumed to be applied at a rate which is slow when compared either with the natural frequency of oscillation of the major parts of the system or with the rate of change of field flux in the machine in response to the change in loading. We have already discussed the maximum power at constant field current against an infinite bus. Steady state stability limit The maximum power which a generator can deliver depends upon the field excitation of the generator. Within limit, the excitation depends upon the system operating condition. Generators 1-15 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS will usually have their excitations adjusted to hold their terminal voltages at some predetermined values. The terminal voltage is held either manually or with an automatic voltage regulator. If the load changes occur slowly or gradually so that steady state conditions may be assumed to exist, the field excitation will change slowly in order to adjust the terminal or system voltage to the prescribed value. The steady state stability limit of a generator or system can be defined as the maximum power that can be transmitted for a slow change in load, the load change occurring slowly enough to allow for a similar change in excitation to bring the terminal voltage back to normal after each successive load change. It is important to note that it has been assumed that the control of excitation is such as to correct the voltage change after each small load change has occurred. Therefore, this is a stability limit for an infinitesimal change in load with constant field current. If the change in excitation is assumed to be initiated immediately following the change in load (as happens with automatic voltage regulator action), the stability limit under such conditions may be termed dynamic stability limit. The dynamic limit will, in general, be higher than the steady state stability limit as defined above. However, the dynamic stability limit is dependent on automatic voltage regulator operation. Example Consider the system and its phasor diagram shown in Figure 1.10. Operating conditions are such that 1.0 pu voltage is maintained at both the infinite bus and the generator terminals. Saturation is neglected. At zero initial load, the generator excitation is also unity. Therefore EV 1.0×1.0 P= sinδ= sinδ= 0.5sinδ x+ x 2.0 d e Fig.1.10 A synchronous machine-infinite bus system and phasor diagram, x = x = 1.0. d e As the initial power increases, the initial excitation also increases. For example, at initial power of 0.5, the excitation voltage E can be calculated, that satisfies the given condition of 1.0 pu voltage at the generator terminal and the infinite bus, from the phasor diagram. E is found to be, E = 1.24. Problem: Verify the above value of E for P = 0.5. Therefore, the power angle relation is 1.0×1.24 P= sinδ= 0.62sinδ 2.0 Problem Calculate the power angle relations for initial power levels of 0.70, 0.75, 0.80, 0.85 and 0.90. It will be found that at P = 0.85, the initial power corresponds to an initial angular displacement o of 90 , as shown in Figure 1.11. 1-16 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS . Fig. 1.11 Power angle characteristics at constant field excitation, illustrating determination of steady state stability limit. The steady state stability limit corresponds to the point at which, if the receiving system requires a small additional load ∆P, the generator becomes just unable to deliver it without a change in the field excitation. Therefore, at the limit dP/dδ = 0. This is the condition at P = 0.85 and δ = o 90 , where dP/dδ is measured on curve c of Figure 1.11. If a higher load is taken, the initial angle o is greater than 90 , and dP/dδ becomes negative, when measured along the corresponding power angle curve for constant field excitation. Steady state stability criterion For the simple case of a single generator connected to an infinite bus, the condition for steady state stability is simply that dP/dδ be positive at the given operating point as shown above. The steady state power limit corresponds to the maximum power that can be delivered up to the critical point where dP/dδ changes from a positive to a negative value. dP/dδ is called the steady state synchronizing power coefficient since it indicates the rate at which the steady or sustained power changes with changes in electrical angular displacement. It should be noted that the synchronizing power coefficient can increase considerably by the action of automatic voltage regulators. In a simplified analysis, the effect of automatic voltage regulators can be approximately accounted for by suitably adjusting the equivalent machine reactance. However, as will be seen later, the usual form of instability in the steady state, under automatic voltage control, is due to lack of damping rather than synchronizing power. Nevertheless, the concept of steady state stability limit as discussed above is useful in developing a proper understanding of the subject of power system stability. Steady state stability of a two-machine system Many steady state stability problems resolve themselves into a two-machine problem. For example, a large remotely located generating plant delivering power to the main system can often be treated as a two-machine problem with the system represented as an equivalent machine in effect as a large equivalent synchronous motor. A value of reactance corresponding to the total three-phase short circuit current contributed by the receiving system may be used as the 1-17 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS equivalent reactance of the receiving-end system. A graphical method may be utilized to determine the stability limit. For the simplified two-machine case shown in Figure 1.12, the steady-state stability limit occurs when maximum power is obtained at the receiving end, i.e., when the total angle between the equivalent synchronous machine groups corresponds to the total impedance angle. Fig. 1.12 Equivalent two-machine system At the limit −1 δ= x r , where x= x+ x+ x tan ( / ) 12 12 12 g m (Note that dP/dδ becomes negative first at the receiving end, which determines the steady state stability limit.) A further condition is imposed by the operating conditions of the system, i.e., the magnitudes of the terminal voltages of the equivalent synchronous generator and motor at the sending and receiving ends, respectively. Using the graphical method, the stability limit can be determined directly without resorting to a trial and error procedure. The method is as follows: Select a reference direction for current I. Draw phasors jIx , I(r + jx), and jIx to any convenient m g scale as shown in Figure 1.13. In this figure, 1 and 2 refer to the points in the generator and motor at which voltages are maintained when the final increment of load is added; a and b are the terminals of the generator and the motor, respectively. Fig.1.13 Illustration of the graphical method for determination of stability limit. To locate the origin of the phasor diagram: Connect points 1 and 2 by a straight line and draw a perpendicular at the mid-point M of this line. Extend the line 2b until it intersects this perpendicular at point P. With P as center draw a circle passing through points 1 and 2 (i.e., with 1-18 FUNDAMENTALS OF POWER FLOW AND POWER LIMITS a radius P2). Any point O on the larger of the two arcs (as shown on Fig.1.13) satisfies the −1 condition that δ=θ= tan (x / r) (verify this from the geometry of Fig. 1.13). 12 12 In order to satisfy the voltage condition, the point O must be so located on the arc that the distances Oa and Ob are in the ratio of the terminal voltages V and V . This may be done by a b determining a series of points with a and b as centers and radii in the ratio V /V . The intersection a b of the locus of such points with the arc will completely define the location of the origin O of the phasor diagram. The scale of the diagram will be determined by the magnitudes of the specified voltages V and V at the generator and motor terminals. a b If a π circuit is used to represent the transmission line as shown in Figure 1.14, the system can be reduced to the form of Fig. 1.12 by using the Thevenin equivalent shown in Figure 1.15. Fig. 1.14 Equivalent two-machine system including the effect of shunt capacitance. Fig. 1.15 Thevenin equivalent of the system of Fig. 1.14. The same procedure as that used for the system of Figure 1.12 can therefore be used. Problem ′′′′ Verify the values of E , E , x and x . 1 2 g m If the line resistance is neglected and equal sending- and receiving-end machine terminal voltages are assumed, Figure 1.13 simplifies to Figure 1.16. At the steady state stability limit δ 12 o will be equal to 90 . Fig. 1.16 Simplification of Fig. 1.13 neglecting r. 1-19

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