Power Systems With MATLAB

1Review of Chapter 2, Plus Matlab Examples 2.2 Power in single-phase circuits Let ( ) ( ) and v t i t be defined as: ( ) ( ) ( ) ( ) cos and cos m v m i v t V t i t I t ω θ ω θ = + = + then the instantaneous power is give by: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) { } ( ) ( ) ( ) ( ) ( ) { } cos cos cos 2 cos 2 cos 2 cos cos2 cos sin2 sin cos m m v i m m v i v i rms rms v v i v i v v i v v i v i p t v t i t V I t t V I t V I t V I t t ω θ ω θ ω θ θ θ θ ω θ θ θ θ θ ω θ θ θ ω θ θ θ θ θ = = + + = + + + −     = + − − + −     = + − + + − + − ( ) ( ) ( ) cos 1 cos 2 sin sin 2 v v p t V I t V I t θ ω θ θ ω θ   = + + + +   where v i θ θ θ = − . The angle θ is called the power factor angle and is the angle between the voltage and current at a particular 2-port. Note: the power factor is given by cosθ . Note that the cosine is an even function thus the sign of θ is lost! For an inductive circuit (memory helper: ELI) voltage leads the current, the carrent lags the voltage and we call this a "lagging" power factor because the current lags the voltage. For a capacitive circuit (memory helper: ICE) the current leads the voltage, and we call this a "leading" power factor. Thus, we always say clearly if the power factor is lagging (inductive) or leading (capacitive). The lag or lead is the angle by which the current lags or leads the voltage. It is also noted that θ is the impedance angle, thus: ( ) v i V V Z I I θ θ θ ∠ = = ∠ − . The instantaneous power can be expressed in two parts: ( ) ( ) ( ) R X p t p t p t = + thus ( ) ( ) cos 1 cos 2 R v p t V I t θ ω θ   = + +   and ( ) ( ) sin sin 2 X v p t V I t θ ω θ = + . It is common to define P and Q as follows: 2 ( ) ( ) ( ) ( ) cos R P avg p t avg p t V I θ = = = sin Q V I θ = Thus in terms of P and Q, the instantaneous power can be expressed as: ( ) ( ) ( ) 1 cos 2 sin 2 v v p t P t Q t ω θ ω θ   = + + + +   Thus P is the "average" value of the cosine terms (counting DC as a zero frequency cosine term) while Q is the "magnitude" of the sine term. It is noted that besides the DC term, power has twice line frequency components. Example 2.1 A supply voltage ( ) 100cos v t t ω = is applied across a load whose impedance is given by 1.25 60 Z = ∠ Ω ! . Determine ( ) i t and ( ) p t . Use Matlab to plot ( ) ( ) ( ) ( ) , , , R i t v t p t p t and ( ) X p t over an interval from 0 to 2π . The Matlab program is shown below: Vm = 100; thetav = 0; % Voltage amplitude and phase angle Z = 1.25; gama = 60; % Impedance magnitude and phase angle thetai = thetav - gama; % Current phase angle in degree theta = (thetav - thetai)*pi/180; % Degree to radian Im = Vm/Z; % Current amplitude wt=0:.05:2*pi; % wt from 0 to 2*pi v=Vm*cos(wt); % Instantaneous voltage i=Im*cos(wt + thetai*pi/180); % Instantaneous current p=v.*i; % Instantaneous power V=Vm/sqrt(2); I=Im/sqrt(2); % rms voltage and current P = V*I*cos(theta); % Average power Q = V*I*sin(theta); % Reactive power S = P + j*Q % Complex power pr = P*(1 + cos(2*(wt + thetav))); % Eq. (2.6) px = Q*sin(2*(wt + thetav)); % Eq. (2.8) PP=P*ones(1, length(wt)); % Average power with length w for plot xline = zeros(1, length(wt)); % generates a zero vector wt=180/pi*wt; % converting radian to degree subplot(2,2,1), plot(wt, v, wt, i,wt, xline), grid title(['v(t)=Vm coswt, i(t)=Im cos(wt +', num2str(thetai), ')']) xlabel('wt, degree') subplot(2,2,2), plot(wt, p, wt, xline), grid title('p(t)=v(t) i(t)'), xlabel('wt, degree') subplot(2,2,3), plot(wt, pr, wt, PP, wt,xline), grid title('pr(t) Eq. 2.6'), xlabel('wt, degree') subplot(2,2,4), plot(wt, px, wt, xline), grid title('px(t) Eq. 2.8'), xlabel('wt, degree') subplot(111) 3 S = 2.0000e+003 +3.4641e+003i 0 100 200 300 400 -100 -50 0 50 100 v(t)=Vm coswt, i(t)=Im cos(wt +-60) wt, degree 0 100 200 300 400 -2000 0 2000 4000 6000 p(t)=v(t) i(t) wt, degree 0 100 200 300 400 0 1000 2000 3000 4000 pr(t) Eq. 2.6 wt, degree 0 100 200 300 400 -4000 -2000 0 2000 4000 px(t) Eq. 2.8 wt, degree 2.3 Complex power It is observed that the term * VI gives the result: (N.B. v i θ θ θ = − ) * cos sin VI V I j V I θ θ = + which is identical to * S VI P jQ = = + where S is the complex power. OTHER FORMS for S : 2 2 2 2 * V S R I jX I Z I Z = + = = . 4 V I θ θ S P Q Phasor diagrams (V, I) and the power triangle for an inductive load V I θ θ S P Q Phasor diagrams (V, I) and the power triangle for a capacitive load, leading power factor angle For example, for an inductive load, the current lags the voltage and the phasor diagrams would be as shown below: And for a capacitive load, the current would lead the voltage , and the phasor diagrams would be as shown below: 2.4 Complex power balance The total value of S for a circuit is the SUM of S for the components of S . Example 2.2: Three impedances in parallel are suppleid by a source 1200 0 V = ∠ ! V where the impedances are given by: 1 60 0 Z j = + , 2 6 12 Z j = + and 3 30 30 Z j = − . Find the power absorbed by each load and the total complex power. As a check, compute the power supplied by the source. 5 Matlab program follows: V = 1200; Z1= 60; Z2 = 6 + j*12; Z3 = 30 - j*30; I1 = V/Z1; I2 = V/Z2; I3 = V/Z3; S1= V*conj(I1), S2= V*conj(I2), S3= V*conj(I3) S = S1 + S2 + S3 Sv = V*conj(I1+I2+I3) S1 = 24000 S2 = 4.8000e+004 +9.6000e+004i S3 = 2.4000e+004 -2.4000e+004i S = 9.6000e+004 +7.2000e+004i Sv = 9.6000e+004 +7.2000e+004i It is observed that the complex power is conserved. 6 P Q S d θ Q C Example 2.3 2.5 Power factor correction Adding a capacitor (usually in parallel) with an inductive load can improve the power factor. Example 2.3 Two loads 1 100 0 Z j = + and 2 10 20 Z j = + are connected across a 200-V 60 Hz source. (a) Find the total real and reactive power, the power factor at the source, and the total current. (b) Find the capacitance of the capacitor connected across the loads to improve the overall power factor to 0.8 lagging. The Matlab program follows: V = 200; Z1= 100; Z2 = 10 + j*20; I1 = V/Z1; I2 = V/Z2; S1= V*conj(I1), S2= V*conj(I2) I = I1 + I2 S = S1 + S2, P = real(S); Q = imag(S); PF = cos(angle(S)) thd = acos(0.8), Qd = P*tan(thd) Sc = -j*(Q - Qd) Zc = V^2/conj(Sc), C = 1/(2*pi*60*abs(Zc)) Sd = P + j*Qd; Id=conj(Sd)/conj(V) S1 = 400 S2 = 8.0000e+002 +1.6000e+003i I = 6.0000 - 8.0000i S = 1.2000e+003 +1.6000e+003i PF = 7 0.6000 thd = 0.6435 Qd = 900 Sc = 0 -7.0000e+002i Zc = 0 -57.1429i C = 4.6420e-005 Id = 6.0000 - 4.5000i Example 2.4 Three loads are connected in parallel across a 1400-V, 60 Hz single-phase supply. Given that load 1 is inductive, 125 kVA at 0.28 power factor, load 2 is capacitive, 10 kW and 40 kvar, and load 3 is resistive at 15 kW. (a) Find the total kW, kVA and the supply power factor. (b) A capacitor is connected in parallel with the loads to improve the power factor to 0.8 lagging. Find the kvar rating of the capacitor and its capacitance in µ F. The Matlab program follows: V = 1400; S1= 35000 + j*120000; S2 = 10000 - j*40000; S3 = 15000; S = S1 + S2 + S3, P = real(S); Q = imag(S); PF = cos(angle(S)) I = conj(S)/conj(V) thd = acos(0.8), Qd = P*tan(thd) Sc = -j*(Q - Qd) Zc = V^2/conj(Sc), C = 1E6/(2*pi*60*abs(Zc)) Sd = P + j*Qd; Id=conj(Sd)/conj(V) S = 6.0000e+004 +8.0000e+004i PF = 0.6000 I = 42.8571 -57.1429i thd = 0.6435 Qd = 45000 Sc = 0 -3.5000e+004i Zc = 0 -56.0000i 8 C = 47.3675 Id = 42.8571 -32.1429i 2.6 Complex Power Flow This is a very important part of chapter 2. Note the diagram below. It shows four power flows. They are: 1. 12 S is the complex power at node number 1 flowing in (i.e. from node 1 in the direction of node 2). 2. 12 S − is the complex power at node number 1 flowing out (i.e. from node 1 in the direction away from node 2). 3. 21 S is the complex power at node number 2 flowing in (i.e. from node 2 in the direction of node 1). 4. 21 S − is the complex power at node number 2 flowing out (i.e. from node 2 in the direction away from node 1). Thus the power consumed by the impedance Z can be expressed as 12 21 Z S S S = + . Important Results from section 2.6: (will be shown in detail later in the lecture) ( ) 2 1 1 2 12 1 2 V V V S Z Z γ γ δ δ = ∠ − ∠ + − Z V 1 V 2 I 12 S 12 -S 12 S 21 -S 21 + + - - 9 ( ) 2 1 1 2 12 1 2 cos cos V V V P Z Z γ γ δ δ = − + − ( ) 2 1 1 2 12 1 2 sin sin V V V Q Z Z γ γ δ δ = − + − where 1 1 1 V V δ = ∠ and 2 2 2 V V δ = ∠ and the impedance is Z Z γ = ∠ . The angle δ defined as 1 2 δ δ δ = − is often called the "power angle". N.B. This is very different from the "power factor angle" discussed earlier. If we assume that the circuit above represents two generators connected by a transmission line, then the equations above still apply. In particular the impedance of a transmission line may be assumed purely inductive, i.e. 0 Z jX = + . In this case the angle 90 γ = ! and the equations above simplify to: ( ) 1 2 12 1 2 sin V V P X δ δ = − ( ) 1 12 1 2 1 2 cos V Q V V X δ δ   = − −   Since the line is "lossless" then power consumed by the line is zero, i.e. 12 21 0 P P + = . In terms of the power angle, these equations become: 1 2 12 sin V V P X δ = 1 12 1 2 cos V Q V V X δ   = −   Observations (assuming 0 R≈ ): 1. Usually δ is very small (less than 10 degrees), thus 12 sin P δ ∝ , i.e. small changes in power angle greatly change the real power flow and not the reactive power. If 1 2 δ δ > then power flows from node one to node two. If 1 2 δ δ < , then power flows in the opposite direction (from node 2 to node 1). 2. Maximum power transfer occurs when 90 δ = ! and is given by: 1 2 max V V P X = . 10 3. Since 0 δ ≈ , 1 2 Q V V ∝ − . Thus small changes in 1 2 V V − greatly affect Q but not P . Thus, to control real power, we need to change the power angle δ . This is done be increasing prime mover power (mechanical power driving the generator). To control reactive power, we need to change the difference in voltage magnitude. This is done by changing the DC excitation of one generator. Example 2.5 Assume 1 120 5 V = ∠− and 2 100 0 V = ∠ . Let 1 7 Z j = + Ω. Determine the real and reactive power supplied or received by each source and the power loss in the line. The Matlab code follows: R = 1; X = 7; Z = R +j*X; V1 = 120*(cos(-5*pi/180) + j*sin(-5*pi/180)); V2 = 100+j*0; I12 = (V1 - V2)/Z, I21 = -I12; S12 = V1*conj(I12), S21 = V2*conj(I21) SL = S12 + S21 PL = R*abs(I12)^2, QL = X*abs(I12)^2 I12 = -1.0733 - 2.9452i S12 = -9.7508e+001 +3.6331e+002i S21 = 1.0733e+002 -2.9452e+002i SL = 9.8265 +68.7858i PL = 9.8265 QL = 68.7858 Example 2.6: Repeat example 2.5 using a Matlab program such that the angle of source 1 is changed from -30 to 30 degrees in steps of 5 degrees each. We cannot execute "chp2ex6" from within the notebook because input values are to be entered by the user. Hence we go to a Matlab window and execute this command. GO TO A MATLAB WINDOW AND EXECUTE "chp2ex6". 1 Review of Chapter 2, Plus Matlab Examples (Part II) 2.7 Balanced Three-Phase Circuits Let 1 120 a = ∠ o . Clearly then 2 * a a = and 3 1 a = . The triple (1, a , 2 a ) form a set of balanced phasors, thus 2 1 0 a a + + = . This triple represents the three cube roots of 1. Y-connected generator, positive sequence: 0 An p E E = ∠ 2 120 Bn p An E E a E = ∠− = 240 Cn p An E E aE = ∠− = Here p E is the "phase" voltage. Note that if we know An E we can find the other two voltages in terms of An E . It is noted that line-to-line voltage L V is larger than the phase voltage by a factor of 3 and is ahead by 30 o , i.e. 3 30 L L p E E − = ∠ o . However, line current and phase current are identical, i.e. L p I I = . Y-connected generator, negative sequence: 0 An p E E = ∠ 120 Bn p An E E aE = ∠ = 2 120 Cn p An E E a E = ∠− = The negative sequence connection is obtained by "interchanging" any two phases of the positive sequence connection. Power systems use only positive sequence connected generators. Also, every effort is made to maintain balanced voltages and loads. 1 a a 2 E An E Bn E Cn ω t E An E Cn E Bn ω t 2 V an n V bn V cn V ab V bc V ca 2.8 Y-Connected Loads 0 An p V V = ∠ 2 120 Bn p An V V a V = ∠− = 240 Cn p An V V aV = ∠− = The line-to-line voltages are found as follows: ( ) 1 0 1 120 3 30 ab an bn p p V V V V V = − = ∠ − ∠− = ∠ Similarly we find: 3 90 bc p V V = ∠− 3 150 ca p V V = ∠ Note: all these results apply provided the reference phasor is as shown. This is arbitrarily chosen, for example above, it is assumed that V an is the reference vector (at zero degrees). If a different reference is chosen, the diagram stays the same though it will be rotated. 2.9 Delta Connected Loads In a similar way we see that L p V V = (line voltage and phase voltage are the same for delta connected loads) and 3 30 a p I I = ∠− 3 150 b p I I = ∠− 3 90 c p I I = ∠ and in general: 3 30 L p I I = ∠− V an V bn V cn V ab V bc V ca 3 2.10 ∆-Y Transformation for Balanced Loads 3 / 3 Y Y Z Z Z Z ∆ ∆ = = 2.11 Per-Phase Analysis 1. It is assumed that everything is "balanced". 2. Change all delta connected loads/sources to Y connections, this provides a neutral point. 3. All the neutral points are at the same potential, hence all the neutral points may be connected 4. This breaks up the circuit into three entirely separate circuits, one for each phase. 5. Solve the single-phase circuit for phase "a". The other phases are the same after a phase shift of 120 ± o . 2.12 Balanced Three-Phase Power Recall that single phase instantaneous power was sinusoidal and had a frequency twice that of the line. Its average value was called the real power. Here is the single phase power equation: ( ) ( ) ( ) 1 cos2 sin 2 v v p t P t Q t ω θ ω θ   = + + + +   **Single phase** The three-phase instantaneous power equation will be derived below. Suppose the voltages of a three-phase supply are given by: ( ) 2 cos an p v v V t ω θ = + ( ) 2 cos 120 bn p v v V t ω θ = + − o ( ) 2 cos 240 cn p v v V t ω θ = + − o Assume the currents are given by: ( ) 2 cos a p i i I t ω θ = + ( ) 2 cos 120 b p i i I t ω θ = + − o ( ) 2 cos 240 c p i i I t ω θ = + − o The instantaneous power is then found by adding the powers for the three phases: 4 ( ) 3 an a bn b cn c p t v i v i v i φ = + + Thus we have: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 cos cos 2 cos 120 cos 120 2 cos 240 cos 240 p p v i p p v i p p v i p t V I t t V I t t V I t t φ ω θ ω θ ω θ ω θ ω θ ω θ = + + + + − + − + + − + − Using the identity ( ) ( ) 1 1 cos cos cos cos 2 2 A B A B A B = − + + we have: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 cos cos 2 cos cos 2 240 cos cos 2 480 p p v i v i p p v i v i p p v i v i p t V I t V I t V I t φ θ θ ω θ θ θ θ ω θ θ θ θ ω θ θ = − + + +     + − + + + −     + − + + + −     Note that the terms involving time are a balanced set hence add to zero and we have: ( ) ( ) 3 3 cos p p v i p t V I φ θ θ = − If we let v i θ θ θ = − (the impedance angle) then: ( ) 3 3 3 cos p p p t P V I φ φ θ = = It is interesting to note that the RHS is independent of time! Each phase by itself has a pulsating power, but when the three are added, the result is a constant power… compare with single phase! The concept of complex power is now extended to three phase, thus we define 3 3 sin p p Q V I φ θ = 3 3 3 S P jQ φ φ φ = + In terms of phase voltages and currents, we also have: * 3 3 p p S V I φ = 5 Since in the Y-connection we have 3 L p V V = and p L I I = , then we have: ( ) 3 3 3 cos 3 cos p p L L p t P V I V I φ φ θ θ = = = 3 3 sin L L Q V I φ θ = In case of the ∆-connection p L V V = and 3 L p I I = giving rise to the same equations above. The two equations above are valid for ∆- as well as Y-connected circuits: They involve line-to-line quantities. Very Important: The angle θ in the power equations above is the angle between the phase voltage and phase current (not line voltages nor line currents). Failure to recognize this fact can lead to serious computational errors. Example 2.7 A three phase line has an impedance of 2 4 j + Ω. The line-to-line voltage is 207.85 V. Two three-phase loads are connected to this line in parallel: one is a ∆-load, the other is a Y-load. Each leg of the ∆-load has an impedance of 60 45 j − Ω, and each leg of the Y- load has an impedance of 30 40 j + Ω. Taking the phase voltage a V as a reference, determine: (a) The current, real power, and reactive power drawn from the supply. (b) The line voltage at the combined loads. (c) The current per phase in each load. (d) The total real and reactive powers in each load and the line. The Matlab program follows: V1 = 207.85/sqrt(3); ZL = 2 + j*4; Z1 = 30 + j*40; % Note ZL is BETTER than Zl, which looks like Z1 Z2 = (60-j*45)/3; % Z2 is Z∆/3 Z = ZL + Z1*Z2/(Z1+Z2); I=V1/Z, S=3*V1*conj(I) V2 = V1- ZL*I V2ab =sqrt(3)*(cos(pi/6)+j*sin(pi/6))*V2 I1=V2/Z1, I2=V2/Z2 Iab=I2/(sqrt(3)*(cos(pi/6)-j*sin(pi/6))) S1=3*V2*conj(I1), S2=3*V2*conj(I2) SL = 3*ZL*abs(I)^2 Stotal=S1+SL+S2 V 1 V 2 Z L Z Y Z ∆ /3 I I 1 I 2 + + - - 6 I = 5.0001 S = 1.8001e+003 V2 = 1.1000e+002 -2.0000e+001i V2ab = 1.8232e+002 +6.5264e+001i I1 = 1.0000 - 2.0000i I2 = 4.0001 + 2.0000i Iab = 1.4227 + 2.1547i S1 = 4.5002e+002 +6.0002e+002i S2 = 1.2000e+003 -9.0003e+002i SL = 1.5001e+002 +3.0001e+002i Stotal = 1.8001e+003 Example 2.8 A three-phase line has impedance of 0.4 2.7 j + per phase. The line feeds two balanced three phase loads that are connected in parallel. The first load is absorbing 560.1 kVA at 0.707 power factor lagging. The second load absorbs 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 3810.5 V. Determine: (a) The magnitude of the line voltage at the source end of the line. (b) Total real and reactive power losses in the line. (c) Real and reactive power supplied at the sending end of the line. The Matlab program follows: V2 = 3810.5/sqrt(3); Zl = 0.4 +j*2.7; S1 = 560.1*(cos(pi/4) + j*sin(pi/4)); S2 = 132; SR = S1+ S2 I = conj(SR)*1000/(3*conj(V2)) V1 = V2 + Zl*I V1L = sqrt(3)*abs(V1) SL = 3*Zl*abs(I)^2/1000 SS = 3*V1*conj(I)/1000 SS = SR + SL SR = 5.2805e+002 +3.9605e+002i I = 80.0079 -60.0078i V1 = 7 2.3940e+003 +1.9202e+002i V1L = 4.1599e+003 SL = 12.0026 +81.0179i SS = 5.4005e+002 +4.7707e+002i SS = 5.4005e+002 +4.7707e+002i Note on units. In reality, P, Q, and S, all have unit of Watt, but to distinguish them apart we assign units as follows: S=VI*=P+jQ P Q |S|=|V||I| NAME Complex Power Real Power Reactive Power Apparent Power unit W+jvar or VA at power factor… W, kW, MW var, kvar, Mvar VA, kVA, MVA Chapter 3. Generator and Transformer Models; The Per-Unit System 3.1 Introduction 1. Use of the "per phase" basis to represent a three-phase balanced system. 2. Use the π model to describe transmission lines (see Chapter 5). 3. Simple models for generators and transformers to study steady state balanced operation. 4. One-line diagrams to represent a three phase system. 5. The per-unit system and the impedance diagram on a common MVA base. 3.2 Synchronous Generators • Note the coils aa', bb', and cc' are 120 o apart. Axis of a-coil is the x-axis as shown. • These are "concentrated" full pitch windings. In real machines the windings are distributed among many slots, and often are not full pitch. • We assume the windings produce a sinusoidal mmf around the rotor periphery (angle around the airgap with respect to winding axis) . • Assume the rotor is excited with DC current f I producing a flux φ which rotates with the rotor at speed ω . At time t the rotor would have moved an angle t ω . • The flux linkage for an N-turn a-winding will be maximum ( ) Nφ at 0 t ω · . The flux linkage with the a-winding will go to 0 at / 2 t ω π · . The flux linkage will go to ( ) Nφ − at t ω π · , and back to 0 at 3 / 2 t ω π · . When 2 t ω π · the flux linkage will be back to ( ) Nφ . Thus a whole cycle… This repeats every revolution of the rotor. a a' b b' c c' r r' m n F s F r F sr ψ δ r ω t ψ Rotor position shown at wt = p/2 2 • The flux linkage with winding "a" is thus given by: cos a N t λ φ ω · • Using Faraday's law, the voltage induced in phase "a" is given by: max max sin sin cos 2 a d e N t dt E t E t λ ω φ ω ω π ω · − · · ¸ _ · − ¸ , Where max 2 E N fN ω φ π φ · · , since 2 / 2 4.44 π ≈ , the rms value of the generated voltage is given by: 4.44 E fNφ · The frequency is a function of speed and the number of poles, thus: 2 60 P n f · where n is the speed in rpm (the synchronous speed) and P is the number of poles (always an even number). If the phase "a" is connected to a load, then a current a i will flow. Depending on the load, this current will have a phase angle, say ψ (see figure) lagging the generated voltage a e which is along the x- axis. Again, this is shown in the figure as the line "mn". N.B. The line "mn" is attached to and rotates with the rotor. Be careful also that the "Flux" vectors are spatial while the voltage and current vectors are phasors (in time). The same is true for phases "b" and "c" but they will "lag" the voltage in phase "a" by 120 and 240 degrees respectively. a a' b b' c c' r r' m n F s F r F sr ψ δ r ω t ψ Rotor position shown at wt = p/2 3 Since ( ) sin a e t ω ∝ , then we have: ( ) ( ) ( ) max max max sin sin 120 sin 240 a b c i I t i I t i I t ω ψ ω ψ ω ψ · − · − − · − − o o Since mmf is proportional to the current, we then have: ( ) ( ) ( ) sin sin 120 sin 240 a m b m c m F F t F F t F F t ω ψ ω ψ ω ψ · − · − − · − − o o We now take components of these phasors along the line "mn" and in quadrature with it. Along "mn" we have: ( ) ( ) ( ) ( ) ( ) ( ) 1 sin cos sin 120 cos 120 sin 240 cos 240 m m m F F t t F t t F t t ω ψ ω ψ ω ψ ω ψ ω ψ ω ψ · − − + − − − − + − − − − o o o o Using the identity ( ) 1 sin cos sin 2 2 α α α · the above equation becomes: ( ) ( ) ( ) 1 sin 2 sin 2 120 sin2 240 2 m F F t t t ω ψ ω ψ ω ψ 1 · − + − − + − − ¸ ] o o It is noted that this expression is the sum of three balanced phasors hence is equal to zero. Next we consider the components of the mmf perpendicular to "mn": ( ) ( ) ( ) ( ) ( ) ( ) 2 sin sin sin 120 sin 120 sin 240 sin 240 m m m F F t t F t t F t t ω ψ ω ψ ω ψ ω ψ ω ψ ω ψ · − − + − − − − + − − − − o o o o Using the identity ( ) 2 1 sin 1 cos2 2 α α · − we have: ( ) ( ) ( ) { } 2 3 cos2 cos2 120 cos2 240 2 m F F t t t ω ψ ω ψ ω ψ 1 · − − + − − + − − ¸ ] o o The three cosine terms add to zero (balanced phasors) thus we have: 4 3 2 S m F F · Thus the result of having three pulsating single phase fluxes produce (when applied symmetrically) a constant flux perpendicular to line "mn" and rotates at the same speed n as the rotor. This flux is called the "armature reaction" to the field of the rotor r F . the various fields are shown for one phase (say phase "a") in the diagram below: Note that F s is perpendicular to line "mn" and rotates with it at the same speed. The fact that F s (which is proportional to I a ) is perpendicular to line "mn" indicates that that we can theorize that the armature current I a is producing a reactive voltage drop parallel to line "mn" due to inductance in the machine. Run the command "rotfield" in a Matlab window to see a demo of the rotating field. θ is the angle between and a V I it is the power factor angle δ is the angle between and E V it is the power angle First the rotor field F r produces the no-load generated voltage E (at zero armature current). Note that E lags F r by 90 degrees. E is called the "excitation" voltage. It is directly proportional to the field current. The voltage/current phasors for phase "a" are shown above lagging the flux diagram by 90 deg. Note the above diagram is a "hybrid" combining spatial and temporal vectors. E F r F sr F s I a m n V R a I a jX l I a jX ar I a E sr E ar Figure 3.2 Page 53 5 Second, assume the armature carries a current to a load, now the armature reaction flux F s is produced. This is perpendicular to line "mn". The two fluxes (due to rotor and armature) combine together to form the "resultant" flux F sr . The resultant flux induces the generated on-load emf E sr . The armature mmf F s induces the voltage E ar known as the armature reaction voltage. In all cases, each mmf produces a voltage lagging the mmf by 90 degrees. Note that the voltage E ar leads F s (hence I a ) by 90 degrees. Thus we can theorize an inductor model for this relationship with reactance X ar i.e. ar ar a E jX I · . X ar is known as the reactance of armature reaction. Thus we now have the circuit equation: sr ar a E E jX I · + The terminal voltage V is found by considering the armature resistance and leakage reactance thus: ( ) a ar a E V R j X X I · + + + 1 ¸ ] l Which can be simplified to: [ ] a s a E V R jX I · + + where s ar X X X · + l is known as the synchronous reactance. The angle between V and I a is θ the power factor angle. The power angle is the angle between E and V called δ. The circuit model of the machine is as shown below: The resistance R a is much smaller than the synchronous reactance X s and is often neglected. This is shown in the simplified one line diagram below where the machine is shown connected to an infinite bus. An infinite bus has a voltage V (usually assumed at an angle zero) which does not change no matter how the generator is controlled or excited. A figure of merit of a generator is the percent voltage regulation VR defined as: 100 100 nl rated rated rated rated V V E V VR V V − − · × · × + - Load E R a jX s + - V I a jX s E V I a 6 The smaller the VR the better. In a real generator, the regulation is a few percent. 3.3 Steady-State Characteristics, Cylindrical Rotor Power factor control. Most generators are connected to a large power grid. This is an infinite bus (its voltage, angle and frequency are constant). Assuming the generator has small leakage reactance and small armature resistance, then its model is shown below: We have two ways of calculating the power (per phase): 1. 1 cos a P V I φ θ · 2. 1 sin s E V P X φ δ · Equating these two equations we have sin cos s a E X I δ θ · . Assuming the power is constant, then from equation 1 we have: cos . a I const θ · This locus is shown as the vertical dashed line in the figure below. From equation 2, assuming the power is a constant we have: sin . E const δ · This locus is the horizontal dashed line shown. Note that for this load, the minimum armature current is 2 a I when the power factor of the generator is unity. Note that 2 E is directly above the voltage V . Thus 2 a I and V are in phase, producing unity power factor. If the excitation is increased, the emf of the generator is increased to some value say 1 E as shown. Clearly the voltage of the generator leads the current hence it is like an inductor consuming vars (lagging power factor). On the other hand, if the excitation is reduced below that for unity power factor, the emf of the generator is smaller, say 3 E which lags 3 a I . Now the power factor of the generator is "leading", i.e. it is like a capacitor (current leading voltage). A plot of the variation of a I as a function of field current is known as a V-curve. A family of such curves, one at each load is known as the V-curves of the generator. The minimum point on each curve corresponds to unity power factor. To the right (higher excitation) is the lagging power factor area and to the left is the leading power factor zone. jX s E V I a I a1 I a2 I a3 V E 1 E 2 E 3 δ 1 q 1 7 Power angle characteristics. Consider the machine connected to an infinite bus and neglecting armature resistance, the machine model is shown below: * 3 3 a S VI φ · But we have 0 a s E V I Z δ γ ∠ − ∠ · ∠ thus for complex power: 2 3 3 3 s s E V V S Z Z φ γ δ γ · ∠ − − ∠ thus the real and imaginary powers are given by: ( ) 2 3 3 cos 3 cos s s E V V P Z Z φ γ δ γ · − − ( ) 2 3 3 sin 3 sin s s E V V Q Z Z φ γ δ γ · − − and if we neglect the armature resistance we have: (Note that 90 γ · o and s s Z X → ) 3 3 sin s E V P X φ δ · and ( ) 3 3 cos s V Q E V X φ δ · − . Thus the real power is a function of the power angle δ and varies as sinδ . The maximum power is given by: max(3 ) 3 s E V P X φ · . The angle δ starts at zero when the machine does not deliver any power. As the machine is loaded, the power delivered increases and the angle δ increases. As the angle δ reaches 90 o , maximum power is reached. If prime mover power is increased beyond this point, the machine will speed up and loose synchronism. This means the tie between the machine and the power grid can handle only the maximum power shown above. In real operation, the power angle is very small of the order of 10 degrees and less. The power angle δ can "swing" by larger amounts under transient conditions. The equation for Q shows that ( ) 3 3 s V Q E V X φ − ; . Thus if E V > the generator delivers reactive power and the generator is said to be overexcited. If E V < , the generator is underexcited and the reactive power delivered to the bus is negative, thus the generator is receiving reactive power from the bus. Generators are usually operated in the overexcited mode since they need to supply the system with reactive power: most our domestic loads are reactive. Example 3.1 jX s E V I a 8 A 50-MVA, 30-kV, three phase, 60 Hz synchronous generator has Xs = 9 Ω per phase and a negligible resistance. The generator is delivering rated power at 0.8 power factor lagging at the rated terminal voltage to an infinite bus. (a) Determine the excitation voltage per phase E and the power angle δ. (b) With the excitation held constant at the value found in (a), the driving torque is reduced until the generator is delivering 25 MW. Determine the current and the power factor. (c) If the generator is operating at the excitation voltage of part (a), what is the steady- state maximum power the machine can deliver before losing synchronism? Also, find the armature current corresponding to this maximum power. The Matlab program follows: Ra=0; Xs = 9; Zs = Ra + j*Xs; S = 40 + j*30; V = 30/sqrt(3); % MW and kV Ia1 = conj(S)*1000/(3*conj(V)); % Amp Ia1M = abs(Ia1), Ia1ang=angle(Ia1)*180/pi E1 = V + Zs*Ia1*0.001; % kV E1M = abs(E1), delta1 = angle(E1)*180/pi % kV, deg disp('(b)') P = 25; % MW delta2 = asin(P*Xs/(3*abs(E1)*V)); delta2d = delta2*180/pi E2 = E1M*(cos(delta2) +j*sin(delta2)); % kV Ia2 = (E2 - V)*1000/Zs; Ia2M = abs(Ia2), Ia2ang=angle(Ia2)*180/pi PF = cos(angle(Ia2)) disp('(c)') Pmax = 3 *E1M*V/Xs % kW E3 = E1M*(cos(pi/2) +j*sin(pi/2)); % kV Ia3 = (E3 -V)*1000/Zs; % A Ia3M = abs(Ia3), Ia3ang=angle(Ia3)*180/pi PF = cos(angle(Ia3)) Ia1M = 962.2504 Ia1ang = -36.8699 E1M = 23.5584 delta1 = 17.1027 (b) delta2d = 10.5914 Ia2M = 807.4918 Ia2ang = -53.4284 PF = 0.5958 (c) Pmax = 136.0147 9 Ia3M = 3.2489e+003 Ia3ang = 36.3239 PF = 0.8057 Example 3.2 The generator of example 3.1 is delivering 40 MW at a terminal voltage of 30 kV. Compute the power angle, armature current, and power factor when the field current is adjusted for the following excitations: (a) The excitation voltage is decreased to 79.2 percent of the value found in example 3.1. (b) The excitation voltage is decreased to 59.27 percent of the value found in example 3.1. (c) Find the minimum excitation below which the generator will lose synchronism. The Matlab program follows: Ra=0; Xs = 9; Zs = Ra + j*Xs; V = 30/sqrt(3); P = 40; E1M = 0.792*23.558; delta1 = asin(P*Xs/(3*abs(E1M)*V)); delta1d = delta1*180/pi E1 = E1M*(cos(delta1) +j*sin(delta1)); Ia1 = (E1 - V)*1000/Zs; Ia1M = abs(Ia1), Ia1ang=angle(Ia1)*180/pi PF = cos(angle(V/Ia1)) disp('(b)') E2M = 0.5925*23.558; delta2 = asin(P*Xs/(3*E2M*V)); delta2d = delta2*180/pi E2 = E2M*(cos(delta2) +j*sin(delta2)); Ia2 = (E2 - V)*1000/Zs; Ia2M = abs(Ia2), Ia2ang=angle(Ia2)*180/pi PF = cos(angle(Ia2)) disp('(c)') E3M = P *Xs/(3*V*1) E3=E3M*(cos(pi/2)+j*sin(pi/2)); Ia3 = (E3 -V)*1000/Zs; Ia3M = abs(Ia3), Ia3ang=angle(Ia3)*180/pi PF = cos(angle(Ia3)) delta1d = 21.7975 Ia1M = 769.8005 Ia1ang = -0.0283 PF = 1.0000 (b) delta2d = 29.7592 Ia2M = 962.7210 10 Ia2ang = 36.9072 PF = 0.7996 (c) E3M = 6.9282 Ia3M = 2.0728e+003 Ia3ang = 68.1986 PF = 0.3714 Notes: in (a) the generator is operating at unity power factor. In both (b) and (c) the generator is under-erexcited and is receiving reactive power (power factor of load is leading, i.e. capacitive load and the generator is inductive.) In the real world the load is usually inductive (lagging power factor) and the generator is over-excited, delivering reactive power. Example 3.3 For the generator of example 3.1, construct the v-curve for the rated power of 40 MW with varying field excitation from 0.4 power factor leading to 0.4 powr factor lagging. Assume the open-circuit characteristic in the operating reagion is given by E = 2000I f . The Matlab program follows: P = 40; % real power, MW V = 30/sqrt(3)+ j*0; % phase voltage, kV Zs = j*9; % synchronous impedance ang = acos(0.4); theta=ang:-0.01:-ang; % Angle from 0.4 leading to 0.4 lagging pf P= P*ones(1,length(theta)); % generates P array of same size Iam = P./(3*abs(V)*cos(theta)); % current magnitude kA Ia = Iam.*(cos(theta) + j*sin(theta)); % current phasor E = V + Zs.*Ia; % excitation voltage phasor Em = abs(E); % excitation voltage magnitude, kV If = Em*1000/2000; % field current, A plot(If, Iam), grid xlabel('If, A'), ylabel('Ia, kA') text(3.4, 1, 'Leading pf'), text(13, 1, 'Lagging pf') text(9, .71, 'Upf') 11 2 4 6 8 10 12 14 16 18 0.6 0.8 1 1.2 1.4 1.6 1.8 2 If, A I a , k A Leading pf Lagging pf Upf Note that the over-excited generator (load is inductive, generator is capacitive) is the normal way to operate the generator since most loads are inductive. If the load becomes capacitive, then the generator is operated in the under-excited mode. Chapter 3. Generator and Transformer Models; The Per-Unit System (Part II) 3.4 Salient-Pole Synchronous Generators Due to the non uniform airgap of the salient poles, the flux is stronger in the direct axis than in the quadrature axis. Thus the reactance d X in the direct axis direction is larger than q X in the quadrature axis direction. The common way of dealing with this is to decompose the armature current into two components: one along the direct axis and one along the quadrature axis. Now the voltage can be found by adding the components along these directions. It is noted that an Ampere in the direct axis direction will produce more voltage than an Ampere in the quadrature direction (due to the difference in magnitude between d X and q X ). This is illustrated in the figure below. Note how the contribution to voltage in the direct axis is much larger than in the quadrature axis. Note also that the arma ture resistance is neglected. It is no longer possible to represent the machine by a simple equivalent circuit. The excitation voltage is given by: cos d d E V X I δ · + The three phase power at the generator is given by: 3 cos a P V I θ · Using the figure above we see that cos cos sin a q d I ab de I I θ δ δ · + · + Therefore the power is: ( ) 3 cos sin q d P V I I δ δ · + . From the diagram we see that sin q q V X I δ · or sin q q V I X δ · . Also from an equation above we have: cos d d E V I X δ − · , therefore the power becomes sin cos 3 cos sin q d V E V P V X X δ δ δ δ ¸ _ − · + ¸ , . Note that sin2 sin cos 2 δ δ δ · , thus: a b c d e E I d I q I a V jX q I q jX d I d δ δ θ 2 2 2 sin 3 3 si n2 3 2 d q d E V V V P X X X δ δ 1 · + − 1 1 ¸ ] Which simplifies to the equation in the book (3.31) page 64: 2 sin 3 3 sin 2 2 d q d d q X X E V P V X X X δ δ − · + Remarks: the author does not show the details shown above, but the results are well known. Also note that if the machine is cylindrical, then d s X X · and q d s X X X · · , thus the power equation becomes the normal power, namely: sin 3 s E V P X δ · . Also note that the new term known as the reluctance power is at frequency (in space) twice that of the normal power. Thus the power-angle relationship is changed from a perfect sinusoid. This is shown below: delta=0:.001:pi; Xs=1; Xd=1; Xq=0.6; E=1; V=0.7; P=3*E*V*sin(delta)/Xs; P2=3*V^2*(Xd-Xq)*sin(2*delta)/(2*Xd*Xq); Psal=P+3*V^2*(Xd-Xq)*sin(2*delta)/(2*Xd*Xq); plot(delta,P,delta,Psal,delta,P2),grid 0 0.5 1 1.5 2 2.5 3 3.5 -0.5 0 0.5 1 1.5 2 2.5 3 Note that the salient-pole machine has greater peak power (stiffer), and this peak power occurs at an angle slightly less than / 2 π radians. 3.6 Equivalent Circuit of a Transformer First consider an ideal transformer, one with no leakage and infinite permeability. Let the transformer have N 1 turns on the left side and N 2 turns on the right side. Assume the current IN on the left is 2 I′ and OUT on the right is I 2 . In such a case we have (assuming sinusoidal flux cos t φ ω · Φ ): ( ) 1 1 1 1max sin cos 90 d e N dt N t E t φ ω ω ω · · − Φ · + o where 1max 1 2 E fN π · Φ. Thus the RMS value of E 1 is 1 1 4.44 E fN · Φ . Since the same flux links both windings, we have on the right: 2 2 4.44 E fN · Φ. Also, since the core has no losses, we must have 2 1 2 2 I N I N ′ · . This leads to the relationships: 1 2 1 2 2 2 E I N E I N · · ′ I.e. in the ideal transformer, the ratio of currents and voltages is the turns ratio (though the current ratio is inversely so). The ideal transformer is shown below: In a real transformer, there would be some leakage flux, also some losses. There is also a magnetizing current. These are simulated by adding components to the ideal model, thus a real transformer would be as shown below, this is the T-equivalent model or equivalent circuit. Assuming the load on an ideal transformer has an impedance 2 Z , then we have: 2 2 2 E Z I · . I ' 2 Ε 1 Ε 2 Ι 2 The ideal tranformer I ' 2 Ε 1 Ε 2 Ι 2 Z 1 =R 1 +jX 1 Z 2 =R 2 +jX 2 V 1 + - V 2 + - R c1 jX m1 I 0 I m I c Ι 1 N 1 :N 2 Equivalent circuit of a power transformer 4 The impedance seen at the left of the ideal transformer is 1 2 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 1 N E E N E N N Z Z N I N I N I N ¸ _ ¸ _ · · · · ′ ¸ , ¸ , . Thus impedances are transformed as the turns ratio squared (or the inverse turns ratio squared going the other way). Let 1 2 N N α · . Then: 2 2 2 2 E V Z I · + . But, 1 2 E E α · and 2 2 I I α ′ · , therefore: 2 1 2 2 2 2 2 2 E V Z I V Z I α α ′ · + ′ ′ ′ · + where 2 2 2 2 2 2 2 Z R jX R X α α ′ ′ ′ · + · + . Using these transformations, the model of the transformer will be as shown in the figure below: This is then approximated with the figure below: It is easier to perform computations with the latter figure. Note that this approximate equivalent model can be either on the left side (primary) as shown above, or the right side (secondary) as shown below. V' 1 + - R c2 jX m2 I' 0 I' m I' c Ι 1 Z e2 =R e2 +jX e2 I 2 V 2 + - Ε 1 Z 1 =R 1 +jX 1 V 1 + - R c1 jX m1 I 0 I m I c Ι 1 + - Z' 2 =R' 2 +jX' 2 I' 2 =I 2 N 2 /N 1 V' 2 =V 2 N 1 /N 2 + - V 1 + - R c1 jX m1 I 0 I m I c Ι 1 Z' e1 =R' e1 +jX' e1 I' 2 =I 2 N 2 /N 1 V' 2 =V 2 N 1 /N 2 + - 5 Furthermore, if we neglect the shunt parts (which are relatively of high impedance) the result is only an RL equivalent as shown below. In most applications, only the reactance X is kept since R is so much smaller it is neglected. V 1 + - Ι 1 Z e1 =R e1 +jX e1 + - V' 2 =V 2 N 1 /N 2 V' 1 =V 1 N 2 /N 1 + - Ι 2 Z e2 =R e2 +jX e2 V 2 + - Simplified models referred to one side. 3.7 Determination of Equivalent Circuit Parameters The OC and SC tests are used to determine the equivalent circuit parameters. This is done using the approximate model of figure 3.12 (top of page 68). In the OC test, one side is excited to rated voltage 1 V (usually the low voltage side) and the other side is open circuited. The input power 0 P and current 0 I are measured. Since the current is small compared to rated value, the losses in the series arms of the model may be neglected and it is assumed that the power lost is the "core losses" (the iron core losses) in the shunt components 1 c R and 1 m X which are in parallel. The computations required are as follows: 2 1 1 0 c V R P · , 1 1 c c V I R · , 2 2 0 m c I I I · − and 1 1 m m V X I · . In the SC test a reduced voltage SC V is applied to one side while the other side is shorted. The input current SC I and input power SC P are measured. In this test the applied voltage is raised till the current on the short circuit side reaches rated value. Note that only a few percent of rated voltage is needed to produce rated currents under this SC condition. Since the currents are so high and the voltage applied is so low, the shunt branch may be neglected, and the series branch is where the losses are (approximately). Note that the series branch is the "equivalent" values of the model 1 1 1 e e e Z R jX · + (see figure 3.15, top of page 70). The equations needed to compute these parameters are: 1 SC e SC V Z I · , 1 2 SC e SC P R I · and 2 2 1 1 1 e e e X Z R · − . V 1 + - R c1 jX m1 I 0 I m I c I 0 OC Test Z e1 = R e1 +jX e1 V SC + _ I SC Short Circuit Test 6 3.8 Transformer Performance If we have the approxi mate equivalent model from the OC and SC tests, we can measure transformer performance. The efficiency of these transformers tends to be high (above 95% efficient). The efficiency η is computed as follows: ( ) 2 cu c power out n S PF power in n S PF n P P η × × · · × × + × + where n is the fraction of the full-load power, S is the rated apparent power or nominal power in volt-ampere, and cu P is the full load copper loss. For a three phase transformer these are given by: 2 2 3 S V I · and 2 2 2 3 cu e P R I · . Also, c P is the iron loss at rated voltage. As the book says, "it can be easily shown that maximum efficiency occurs when copper losses equal iron losses at n per-unit loading…" see page 71! At any rate, we do not need these details. There is also a program called "trans" which we have seen before that solves these equations for efficiency. Other important equations are those for percent regulation. These are referred to either side and have to do with voltage drop at full load compared to no load voltage. See equations on bottom of page 71. Example 3.4 Run Trans with the given data! see page 72. 3.9 Three-Phase Transformer Connections Y Y − : This is rarely used due to third harmonics. Often a third "tertiary" winding is added connected in ∆ to provide a path for the third harmonic current thus allowing the voltage to remain sinusoidal and almost distortion free. Voltage out is in phase with voltage in. ∆−∆ : This allows for the third harmonic, it lacks the neutral connection. It has the advantage that one bank (assuming three banks are used) of transformers can be removed for repairs, and it will continue to operate (at reduced power) till the repair is completed. It is then connected as a V V − or simply V-connection. Y ∆ − : Is used frequently to raise voltage. Note that there is a connection voltage gain (in addition to the inherent turns ratio gain). There is also a phase shift of thirty degrees. The neutral is often grounded. Y −∆: Is used to lower the voltage from high transmission values. Note that it also compensates for phase shifts on a Y ∆ − connection thus bringing the line back in phase. 7 3.9.1 The Per-Phase Model of a Three-Phase Transformer Usually the low voltage side is designated as the X side and the high voltage side as the H side. Thus XP V refers to the phase voltage on the low side. In the case of ∆−∆ and Y Y − connections the phase does not change going from one side to the other. Also, the voltage gain is essentially the turns ratio. This is not the same for a Y ∆ − or a Y −∆ connection. Take for example the Y ∆ − connection. First of all there is a 30 o phase shift. It is the standard (American Standards Association) to make the connection such that the line voltage on the HV side leads the corresponding line voltage on the LV side by 30 o regardless of which side the Y or the ∆ is on. Second, the Y-side has a connection gain of 3 . Assume the Y-side is the HV side, thus we have for a Y ∆ − connection: (let H HP X XP N V a N V · · ) 3 HL HP XL XP V V V V · · thus the ratio of the line-to-line voltages of a Y ∆ − assuming the Y is the high voltage side is 3 3 HL HP XL XP V V a V V · · . I.e. the total gain is the connection gain multiplied by the turns ratio. When dealing with a Y ∆ − or a Y −∆ connection, it is best to transform the delta to a Y then use a per-phase diagram. See figure 3.19 page 77. 3.10 Autotransformers When the two windings of a transformer are connected together it becomes an autotransformer. Note that the two sides are now no longer isolated from each other as in a conventional transformer. This may be undesirable. If the isolation is not an important feature, then this connection has several advantages as we show below. The two winding transformer is shown to the right. Below that we show the two sides connected into an autotransformer. Note that now the two sides are no longer isolated. The winding voltages and currents are related in the same manner for the autotransformer as for the two-winding transformer, thus: 1 1 2 2 V N a V N · · and 2 1 1 2 I N a I N · · . From the N 1 :N 2 V 1 V 2 + + _ _ X 2 X 1 H 1 H 2 I 1 I 2 V L + _ H 1 H 2 I 2 V 1 + _ X 1 I 1 I H X 2 V 2 V H I L 8 figure we have: ( ) 1 2 1 2 2 2 1 H L N V V V V V a V N · + · + · + . Thus we have 1 H L V a V · + . Since we have the ideal transformer (it is a good approximation in this case) we have 2 2 1 1 N I N I · or ( ) 2 1 1 1 L N I I N I − · or 1 2 1 2 L N N I I N + · , therefore 1 L H I a I · + . If we now compare the power rating of the two transformers we have: ( ) 1 2 1 2 2 1 1 1 1 1 1 auto w V V I S N S V I N a − + · · + · + . If 1 a < then the power rating to the autotransformer is higher than the conventional two winding transformer. This is known as the power rating advantage of the autotransformer. In addition to this, the autotransformer has the advantages of higher efficiency, lower internal impedance and smaller size (for the same rating). In three phase systems this transformer is used frequently provided the value of the voltage ratio from one side to the other does not exceed three (i.e. 3 a ≤ ). Example 3.5 A two-winding transformer is rated 60 kVA, 240/1200 V, 60 Hz. When operated as a conventional two winding transformer at rated load, 0.8 power factor, its efficiency is 0.96. This transformer is to be used as a 1440/1200 V step-down autotransformer. (a) Assuming ideal transformer find the transformer kVA rating when used as an autotransformer. (b) Find the effiecnecy with the kVA loading of part (a) and 0.8 power factor. The Matlab program follows: S_2w = 60; V1 = 240; V2 = 1200; I1= S_2w*10^3/V1; I2 = S_2w*10^3/V2; IL = I1 + I2; S_auto = V2*IL*10^-3 Ploss = S_2w*0.8*(1 - 0.96)/0.96 Eff = S_auto*.8/(S_auto*0.8 + Ploss) S_auto = 360 Ploss = 2.0000 Eff = 0.9931 We shall skip section 3.12. Below is a short program to plot the three phase voltages of amplitude one at 60 Hz. f=60; w=2*pi*f; t=0:0.0001:0.02; 9 T=1000*t; a=cos(w*t); b=cos(w*t+2*pi/3); c=cos(w*t-2*pi/3); plot(T,a,T,b,T,c),grid, xlabel('milliseconds'), ylabel('Amplitude'), title('a is blue, b is red, c is green') 0 2 4 6 8 10 12 14 16 18 20 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 milliseconds A m p l i t u d e a is blue, b is red, c is green Chapter 3. Generator and Transformer Models; The Per-Unit System (Part III) 3.13 The Per-Unit System Let us say that the measure unit of age is 80 years (this is the base unit). Then a person who is 40 years old is half of the base unit. So we say the person's age is 0.5 per unit. age in years Age per unit age base in years = Thus the age per unit is dimensionless. In power systems there are four base quantities required to define a per unit system. These are: power, voltage, current and impedance. Given two of these, all others can be defined using these two. In general for any quantity: actual quantity Quantity per unit base value of quantity = And for power systems we have: pu pu pu pu B B B B S V I Z S V I Z S V I Z = = = = N.B. The value of the numerators above may be a complex number, however, the denominator (the base value) is a positive real number. Normally we select a three-phase power base ( B S or B MVA ) and a line-to-line voltage base ( B V or B kV ). From these two the other bases can be computed using circuit laws thus: 3 3 3 B B B B B S S I V V = = and 3 B B B V Z I = Other forms for B Z follow from the above equations: ( ) ( ) 2 2 B B B B B B V Z S kV Z MVA = = Note that B V is line to line voltage base! and B I is line current base. Also, B S is 3-phase power base. Thus in dealing with per-phase quantities, the 3 and 3 need to be taken into account. 2 Per unit quantities obey the circuit laws, thus: * pu pu pu S V I = and pu pu pu V Z I = In terms of phase quantities, the three phase complex power to a load is given by: * ( 3 ) 3 L P P S V I φ = and the phase current is related to the phase voltage by the load impedance thus: P P P V I Z = Using this value of phase current in the equation above we have (solving for impedance): ( ) ( ) 2 2 * * 3 3 3 P L L P L L V V Z S S φ φ − = = Therefore the phase impedance in per-unit is given by: ( ) 2 * 3 p L L pu B B L Z V Z Z S Z φ − = = and since 2 B B B V Z S = (see last page) the above expression becomes: ( ) ( ) 2 2 * * 3 3 p L L L L B pu B B B L L Z V V S Z Z S Z V S φ φ − − = = = or ( ) 2 * pu pu L pu V Z S = 3.14 Change of Base Usually if none are specified, the pu values given are on nameplate ratings as base. For example a generator whose impedance is 0.4 pu and whose ratings are 110 MVA and 24 kV, then the base MVA is assumed 110 MVA and base line to line voltage is 24 kV. Often the base for the system is different from the base for each particualr generator or transformer, hence it is important to be able to express the pu value in terms of different bases. This is derived below. Let old pu Z be the per-unit impedance on the power base old B S and voltage base old B V . We would like to find new pu Z on the new power base new B S and voltage base new B V . Let Z Ω be the "ohmic" value of the impedance. Then we have: 3 ( ) 2 old old B pu old old B B Z S Z Z Z V Ω Ω = = and similarly ( ) 2 new new B pu new new B B Z S Z Z Z V Ω Ω = = By comparison of these two expressions it is evident that: 2 new old new old B B pu pu old new B B S V Z Z S V   =     This last expression is very useful in transforming per-unit values from one set of bases to another. Example. A transformer has a voltage rating of 110/220 V and a power rating of 1000 VA. The impedance of the transformer on the low voltage side is 5 33 j + . Find the per- unit impedance viewed from each side assuming the bases are equal to the voltages and power ratings on each side. Solution using Matlab: From the give quantities: VB1=110; VB2=220; SB1=1000; SB2=1000; Z1=5+j*33; Z2=Z1*4; ZB1=VB1^2/SB1; ZB2=VB2^2/SB2; Zpu1=Z1/ZB1 Zpu2=Z2/ZB2 Zpu1 = 0.4132 + 2.7273i Zpu2 = 0.4132 + 2.7273i Thus, with proper selection of base S and base V on each side of a transformer, the per-unit impedance of the transformer becomes the same viewed from either side! Thus the transformer may be removed and replaced with its per-unit impedance (but we have to keep in mind that now there are two different values of voltage base, one on each side of the transformer). Rules for selection of voltage bases around a transformer: Let the turns ratio of the transformer be 1 2 N N α = , and suppose there is a connection gain of k . If this is a Y Y − or ∆ − ∆ connected transformer, then 1 k = . If it is a Y −∆ connection, then 3 k = . If it is Y ∆ − then 1/ 3 k = . Select a voltage base on side 1, say 1 B V . Then the voltage base on side 2 must be: 2 1 B B V k V α = . The power base, of course is the same everywhere in the power system. 4 Example 3.7 The one line diagram of a power system is shown in fig. 3.29 page 92. Select a common base of 100 MVA and 22 kV on the generator side. Draw an impedance diagram with all impedances including the load impedance marked in per unit. The manufacturer's data for each device is given on page 92. The three phase load on bus 4 absorbs 57 MVA, 0.6 power factor lagging at 10.45 kV. Line 1 and line 2 have reactances of 48.4Ω and 65.43Ω, respectively. Outline of solution: first find the voltage base in all sections of the system. In this case there are five sections. Note that in this case there is no mention of how the transformers are connected (i.e. Y or delta). Hence we consider only the turns ratio (otherwise there would be some 3 factors as well). The Matlab program follows: VLL = 10.45; VB1 = 22; VB2 = 22*220/22; VB4 = 220*11/220; VB5 = 22*110/22; Xg = 0.18*100/90 Xt1 = 0.1*100/50 Xt2 = 0.06*100/40 Xt3 = 0.064*100/40 Xt4 = 0.08*100/40 Xm = 0.185*(100/66.5)*(10.45/11)^2 ZB2 = (220)^2/100 ZB5 = (110)^2/100 Xline1 = 48.4/484 Xline2 = 65.43/121 SL = 57*(0.6 + j*sin(acos(0.6))) ZL = VLL^2/conj(SL) ZB4 = (11)^2/100 ZLpu = ZL/ZB4 Xg = 0.2000 Xt1 = 0.2000 Xt2 = 0.1500 Xt3 = 0.1600 Xt4 = 0.2000 Xm = 0.2511 ZB2 = 484 ZB5 = 121 Xline1 = 0.1000 Xline2 = 5 0.5407 SL = 34.2000 +45.6000i ZL = 1.1495 + 1.5327i ZB4 = 1.2100 ZLpu = 0.9500 + 1.2667i Note these are then incorporated into the impedance diagram on page 94. Note there are no transformers showing! Just an elementary one phase diagram with impedances. To go back to actual values, we need to know the MVA B and the V B at every point in the system. As stated earlier, the MVA B is the same everywhere, but the V B changes as we go from one side of a transformer to the other. Example 3.8 The motor of example 3.7 operates at full load 0.8 power factor leading at a terminal voltage of 10.47 kV. (a) Determine the voltage at the generator bus. (b) Determine the generator and motor emfs. The solution is found using ordinary circuit analysis. Of course, the answers will be "pu" all the way! The Matlab program follows: ZL = 0.95 + j*1.2667; V4=10.45/11 Sm = 66.5/100*(0.8 - j*.6) Im = conj(Sm)/conj(V4) IL = V4/ZL I = Im + IL X11 = 0.45*0.9/(0.45+0.9); V1 = V4+j*X11*I; V1M=abs(V1), V1ang=angle(V1)*180/pi Eg = V1 + j*0.2*I; EgM=abs(Eg), Egang=angle(Eg)*180/pi Em = V4 - j*0.25*Im; EmM=abs(Em), Emang=angle(Em)*180/pi V4 = 0.9500 Sm = 0.5320 - 0.3990i Im = 0.5600 + 0.4200i IL = 0.3600 - 0.4800i I = 0.9200 - 0.0600i V1M = 6 1.0066 V1ang = 15.9139 EgM = 1.0826 Egang = 25.1445 EmM = 1.0642 Emang = -7.5591 Chapter 5. Line Model and Performance 5.1 Introduction We shall represent lines as a "pi" section with lumped parameters. There are three cases: short, medium and long lines. In the short line model (under 50 miles), we assume the shunt capacitance (the legs of the π ) are so small that they are open circuit (i.e. neglected). This leaves only the series RL branch. In the medium line model (50 to 250 miles), we assume that the capacitance may be represented as two capacitors (legs of the π ) each equal to half the line capacitance. This is known as the nominal π model. In the long line model (over 250 miles), we assume the line has distributed parameters instead of lumped parameters. This yields exact results. After finding the exact model, an equivalent π is found. This is called the equivalent π model because it has lumped parameters which are adjusted so that they are equivalent to the exact distributed parameters model. 5.2 Short Line Model A short line with a load is shown (per phase) in the figure to the right. Using circuit theory we have: * * R R R S I V · . We also know S R R V V ZI · + . Note that S R I I · . Now we will find the transmission parameters (or ABCD parameters) of the short line: For a 2-port shown as "ABCD" the equations are as follows: S R S R V V A B I I C D 1 1 1 · 1 1 1 ¸ ] ¸ ] ¸ ] . Comparing with the equations for the short line above, we see that: 1 A· , B Z · , 0 C · and 1 D · . This completes the equations needed to represent the short line. The voltage regulation in percent is given by: ( ) ( ) ( ) 100 R NL R FL R FL V V Percent VR V · × Z = R+jX V S + _ V R + _ S R I S I R ABCD V S + _ V R + _ I S I R V S + _ V R + _ I S I R Z = R+jX Y/2 Y/2 I L 2 At no-load 0 R I · and thus ( ) S R NL V AV · , and since 1 A· , ( ) R NL S V V · . This conclusion is easily made by observation of the circuit above. Example 5.1 A 220-kV, three phase transmission line is 40 km long. The resistance per phase is 0.15Ω per km and the inductance per phase is 1.3263 mH. The shunt capacitance is negligible. Use the short line model to find the voltage and power at the sending end and the voltage regulation and efficiency when the line is supplying three phase load of (a) 381 MVA at 0.8 power factor lagging at 220 kV. (b) 381 MVA at 0.8 power factor leading at 220 kV. The Matlab program follows: VRLL=220; VR = VRLL/sqrt(3); Z = (0.15+j*2*pi*60*1.3263e-3)*40; disp('(a)') SR=304.8+j*228.6; IR = conj(SR)/(3*conj(VR)); IS = IR; VS = VR + Z*IR; VSLL = sqrt(3)*abs(VS) SS = 3*VS*conj(IS) REG = (VSLL - VRLL)/VRLL*100 Eff = real(SR)/real(SS)*100 disp('(b)') SR=304.8-j*228.6; IR = conj(SR)/(3*conj(VR)); IS = IR; VS = VR + Z*IR; VSLL = sqrt(3)*abs(VS) SS = 3*VS*conj(IS) REG = (VSLL - VRLL)/VRLL*100 Eff = real(SR)/real(SS)*100 (a) VSLL = 250.0186 SS = 3.2280e+002 +2.8858e+002i REG = 13.6448 Eff = 94.4252 (b) VSLL = 210.2884 SS = 3.2280e+002 -1.6862e+002i REG = -4.4144 Eff = 94.4252 3 5.3 Medium Line Model As the length of the line is increased to over 50 miles (but less than 150 miles) the medium line model is used. This is also known as the nominal π model as shown below. The admittance Y and impedance Z shown are shown for the whole line, per-phase (and not per mile). For the "pi" model shown we can write the following equations: 2 L R R Y I I V · + and S R L V V ZI · + . Using the first equation in the second we have: 1 2 S R R ZY V V ZI ¸ _ · + + ¸ , . We also see that 2 S L S Y I I V · + and thus we have: 1 2 2 2 1 1 4 2 S R R R R R R Y Y ZY I I V V ZI ZY ZY Y V I 1 ¸ _ · + + + + 1 ¸ , ¸ ] ¸ _ ¸ _ · + + + ¸ , ¸ , . Thus we now have the ABCD parameters: 1 2 1 1 4 2 ZY A B Z ZY ZY C Y D ¸ _ · + · ¸ , ¸ _ ¸ _ · + · + ¸ , ¸ , In general for a symmetrical 2-port A D · and 1 AD BC − · (the determinant of the ABCD matrix in this case is unity). Thus it is easy to find the inverse relationship: S R S R V V D B I I C A − 1 1 1 · 1 1 1 − ¸ ] ¸ ] ¸ ] Note the two Matlab functions for computing the ABCD and ZY parameters: [ ] ( ) , , 2 , , , , , Z Y ABCD rlc abcd r L C g f Length · [ ] ( ) , , 2 , , Z Y ABCD zy abcd z y Length · Note also that f is frequency in Hz and g is conductance to ground per unit length (zero in the specific case of the medium line model). Note also that , r g, z, and y are all per unit length of line. The series impedance is ( ) Z r j L R jX ω · + · + l and the shunt admittance is ( ) Y g j C ω · + l . Example 5.2 A 345-kV three-phase line is 130 km long. The resistance per phase is 0.036Ω per km and the inductance per phase is 0.8 mH per km. The shunt capacitance is 0.0112µ F per km. The receiving end load is 270 MVA with 0.8 power factor lagging at 325 kV. Use the medium line model to find the voltage and power at the sending end and the voltage regulation. V S + _ V R + _ I S I R Z = R+jX Y/2 Y/2 I L 4 Note that we cannot run this program from within MS-Word since it has interactive statements. A listing of the program for example 5.2 is shown below. Run example 5.2 in the Matlab window (chp5ex2). DO NOT EXECUTE IN THIS NOTEBOOK! (Reason: edit the function "rlc2abcd.m", it has interactive code!) r = .036; g = 0; f = 60; L = 0.8; % milli-Henry C = 0.0112; % micro-Farad Length = 130; VR3ph = 325; VR = VR3ph/sqrt(3) + j*0; % kV (receiving end phase voltage) [Z, Y, ABCD] = rlc2abcd(r, L, C, g, f, Length); AR = acos(0.8); SR = 270*(cos(AR) + j*sin(AR)); % MVA (receiving end power) IR = conj(SR)/(3*conj(VR)); % kA (receiving end current) VsIs = ABCD* [VR; IR]; % column vector [Vs; Is] Vs = VsIs(1); Vs3ph = sqrt(3)*abs(Vs); % kV(sending end L-L voltage) Is = VsIs(2); Ism = 1000*abs(Is); % A (sending end current) pfs= cos(angle(Vs)- angle(Is)); % (sending end power factor) Ss = 3*Vs*conj(Is); % MVA (sending end power) REG = (Vs3ph/abs(ABCD(1,1)) - VR3ph)/VR3ph *100; fprintf(' Is = %g A', Ism), fprintf(' pf = %g\n', pfs) fprintf(' Vs = %g L-L kV\n', Vs3ph) fprintf(' Ps = %g MW', real(Ss)), fprintf(' Qs = %g Mvar\n', imag(Ss)) fprintf(' Percent voltage Reg. = %g\n', REG) Example 5.3 A 345-kV, three-phase transmission line is 130 km long. The series impedance is 0.036 0.3 z j · + Ω per phase per km. The shunt admittance is 6 4.22 10 y j − · × siemens per phase per km. The sending end voltage is 345-kV and the sending end current is 400- A at 0.95 power factor lagging. Use the medium line model to find the voltage, current and power at the receiving end and the voltage regulation. Note that we cannot run this program from within MS-Word since it has interactive statements. A listing of the program for example 5.3 is shown below. Run example 5.3 in the Matlab window (chp5ex3). DO NOT EXECUTE IN THIS NOTEBOOK! The culprit is the command "zy2abcd" which has interactive code. z = .036 + j* 0.3; y = j*4.22/1000000; Length = 130; Vs3ph = 345; Ism = 0.4; %KA; As = -acos(0.95); Vs = Vs3ph/sqrt(3) + j*0; % kV (sending end phase voltage) Is = Ism*(cos(As) + j*sin(As)); [Z,Y, ABCD] = zy2abcd(z, y, Length); VrIr = inv(ABCD)* [Vs; Is]; % column vector [Vr; Ir] Vr = VrIr(1); Vr3ph = sqrt(3)*abs(Vr); % kV(receiving end L-L voltage) Ir = VrIr(2); Irm = 1000*abs(Ir);% A (receiving end current) pfr= cos(angle(Vr)- angle(Ir)); % (receiving end power factor) 5 Sr = 3*Vr*conj(Ir); % MVA (receiving end power) REG = (Vs3ph/abs(ABCD(1,1)) - Vr3ph)/Vr3ph *100; fprintf(' Ir = %g A', Irm), fprintf(' pf = %g\n', pfr) fprintf(' Vr = %g L-L kV\n', Vr3ph) fprintf(' Pr = %g MW', real(Sr)) fprintf(' Qr = %g Mvar\n', imag(Sr)) fprintf(' Percent voltage Reg. = %g\n', REG) 5.4 Long Line Model This is a distributed parameters model of the line. It is very exact and gives the voltage and current anywhere along the line. The distance x is measured from the receiving end towards the sending end. If we let the x · l (the length of the line) the equations for the long line model have the following ABCD parameters: cosh sinh 1 sinh cosh c c A B Z C D Z γ γ γ γ · · · · l l l l where c Z is the line characteristic impedance given by c z Z y · and j zy γ α β · + · . It is now possible to find an accurate π model whose ABCD parameters are equal in value to the ones shown above. Without proof the results are outlined below: sinh sinh 1 tanh / 2 tanh 2 2 / 2 c c Z Z Z Y Y Z γ γ γ γ γ γ ′ · · ′ · · l l l l l l where the primed quantities are for the exact π or equivalent π model, and the unprimed quantities are those for the nominal π . This way we can transform the nominal π to an exact or equivalent π . The Matlab functions mentioned earlier (rlc2abcd and zy2abcd) with option 2 will perform the equivalent π model for a long line. Example 5.4 A 500-kV, three phase transmission line is 250 km long. The series impedance is 0.045 0.4 z j · + Ω per phase per km and the shunt admittance is 6 4 10 y j − · × siemens per phase per km. Evaluate the equivalent π model and the transmission matrix. Below is the program for example 5.4: z = 0.045 + j*.4; y = j*4.0/1000000; Length = 250; gamma = sqrt(z*y); Zc = sqrt(z/y); A = cosh(gamma*Length); B = Zc*sinh(gamma*Length); C = 1/Zc * sinh(gamma*Length); D = A; ABCD = [A B; C D] Z = Zc * sinh(gamma*Length) Y = 2/Zc * tanh(gamma*Length/2) 6 ABCD = 0.9504 + 0.0055i 10.8778 +98.3624i -0.0000 + 0.0010i 0.9504 + 0.0055i Z = 10.8778 +98.3624i Y = 0.0000 + 0.0010i Chapter 6. Power Flow Analysis 6.1 Introduction The system is assumed 3-phase balanced operating in a steady state and stable condition. It is represented by a single phase diagram on a per-unit basis, with a system wide MVA base, and a voltage base properly chosen on each side of every transformer. The base MVA and base V are specified/known everywhere in the system. The most common way to represent such a system is to use the node-voltage method. Given the voltages of generators at all generator nodes, and knowing all impedances of machines and loads, one can solve for all the currents in the typical node voltage analysis methods using Kirchoff's current law. First the generators are replaced by equivalent current sources and the node equations are written in the form: I YV = where I is the injected current vector, Y is the admittance matrix and V is the node voltage vector. These equations are easy to write by inspection of the circuit. The problem is not so simple in real power circuits and systems. Usually in a power system the complex power may be known at load nodes, and sometimes on generator nodes only the real power and voltage are known. Thus not enough variables are known to solve an equation of the form I YV = . In fact, since the power is a nonlinear function of the current and voltage, the solution of the resulting equations (while it may exist) is not easy! In fact there is no known analytical method to find the solution. As a result iterative techniques are used to find the solution (voltages, currents, etc.). The nonlinear set of equations which are generated are called power flow equations. The solution of such equations results in a power flow study or load flow analysis. Such studies are the backbone of power system studies, for analysis, design, control, and economic operation of the power system. They are also essential for transient analysis of the system. 6.2 Bus Admittance Matrix 1. The first step is to number all the nodes of the system from 0 to n . Node 0 is the datum or reference node (or ground node). 2. replace all generators by equivalent current sources in parallel with an admittance. 3. Replace all lines, transformers, loads to equivalent admittances whenever possible. Knowing a load in MVA but not knowing the operating voltage, makes it impossible to change the load to an admittance. The rule for this is simple: 1/ y z = where and y z are generally complex numbers. 4. The bus admittance matrix Y is then formed by inspection as follows (this is similar to what we learned in circuit theory): sum of admittances connected to node ii y i = and sum of admittances connected from node to node ij ji y y i j = = − 5. The current vector is next found from the sources connected to nodes 0 to n . If no sourde is connected, the injected current would be 0. 6. The equations which result are called the node-voltage equations and are given the "bus" subscript in power studies thus: bus bus bus I Y V = The inverse of these equations results in the set 1 bus bus bus V Y I − = . It is emphasized that the matrix 1 bus Y − is not the same as the matrix Z which results from solving a circuit using mesh equations. To clearly show this difference we define 1 bus bus Z Y − = . It is noted that the matrix bus Y is the same as the Y matrix obtained from circuit theory. Thus bus Y Y = however bus Z Z ≠ . It is also noted that in general, solving a large set of linear equations is never done using the matrix inverse. Finding the inverse involves more work than is needed to find a solution and the resulting solutions are less accurate. Usually Gauss elimination or a similar factorization scheme is used in a direct solution. Iterative solutions are also very effective in larger systems. It is further noted that in a real power system with hundreds of nodes, each node is rarely connected to more than two or three other nodes, thus most of the elements of the admittance matrix are zero (the admittance matrix is sparse). Special techniques exist to solve systems with sparse matrices. In a power system the bus Z is built up directly from the power system (see section 9.5) and matrix inversion is not used. Also, we have a routine to form the admittance matrix called ybus used as follows: Y = ybus(zdata) where "zdata" is a matrix of entries having n-rows and 4 columns. Column 1 and 2 indicate the nodes between which the impedance exists (nonzero), and columns 3 and 4 give the real and imaginary parts of the impedance. The result is the bus admittance matrix bus Y . Example 6.1. The emfs shown in figure 6.1 page 190 are 1 1.1 0 E = ∠ o and 2 1.0 0 E = ∠ o . Use the function Y = ybus(zdata) to obtain the bus admittance matrix. Find the bus impedance matrix by inversion and solve for the bus voltages. Tha Matlab program is shown below: % From To R X z = [ 0 1 0 1.0 0 2 0 0.8 1 2 0 0.4 1 3 0 0.2 2 3 0 0.2 3 4 0 0.08]; [Ybus] = ybus(z) % bus admittance matrix Ibus = [-j*1.1; -j*1.25; 0; 0]; % vector of injected bus currents Zbus = inv(Ybus) % bus impedance matrix Vbus = Zbus*Ibus %Vbus = Ybus\Ibus Note: the last statement which is commented out is the better way of doing it! Why? Do not execute chp6ex1 from this workbook. Instead, go to the Matlab window and give the command to get the output. 6.3 Solution of Nonlinear Algebraic Equations The most common methods for solving nonlinear algebraic equations are Gauss-Seidel, Newtow-Rahpson, and quasi-Newton-Raphson methods. We start with one dimensional equations and then generalize to n-dimensional equations. 6.3.1 Gauss-Seidel Method This method is also known as the method of successive displacements. Consider the nonlinear equation ( ) 0 f x = . The equation is broken into two parts thus: ( ) x g x = . We assume ( ) 0 x is an initial "guess" of the solution, then "refine" the solution using: ( ) ( ) ( ) 1 0 x g x = This process is repeated thus ( ) ( ) ( ) 2 1 x g x = and on the th n iteration we have: ( ) ( ) ( ) 1 n n x g x − = . If this process is convergent, then the successive solutions approach a value which is declared as the solution. Thus if at some step 1 k + we have ( ) ( ) 1 k k x x ε + − ≤ where ε is the desired "accuracy", then we claim the solution has been found to the accuracy specified. Example 6.2 Use the Gauss-Seidel method to obtain the roots of the equation ( ) 3 2 6 9 4 0 f x x x x = − + − = First the equation is expressed in a different form thus ( ) ( ) 3 2 1 6 4 9 x x x g x = − − − = And the iteration can proceed. Take a good look at the shape of the iterations! Below is the program showing the process graphically (later showing how to do it iteratively). clear; x=0:.01:4.5; g=-1/9*x.^3+6/9*x.^2+4/9; k=1; dz=10; z=2; r(k)=z; s(k)=z; while dz >.001 k=k+2; r(k-1)=z; p=-1/9*z^3+6/9*z^2+4/9; s(k-1)=p; dz=abs(z-p); z=p; r(k)=z; s(k)=z; end plot(x,g,'-',x,x,'-', r,s,'-'),grid xlabel('x') text(0.8, 4.2,'g(x) =-1/9x^3+6/9x^2+4/9') text(3.2, 3.0,'x') 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x g(x) =-1/9x 3 +6/9x 2 +4/9 x Note: try running the program above with an initial "guess" of z = 0 and z = 0.8 Does this process converge? always? to a unique solution? Next, plot the function and find the roots using Matlab: roots([1; -6; 9; -4]) x=0:0.01:4.5; zz=zeros(size(x)); % plot the x-axis f=x.^3-6*x.^2+9.*x-4; plot(x,f,x,zz),grid ans = 4.0000 1.0000 + 0.0000i 1.0000 - 0.0000i 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 -4 -2 0 2 4 6 8 Note the double root at x = 1. Now the program to do this iteratively: dx=1; % Change in variable is set to a high value x=2; % Initial estimate iter = 0; % Iteration counter disp('Iter g dx x') % Heading for results while abs(dx) >= 0.001 & iter < 100 % Test for convergence iter = iter + 1; % No. of iterations g = -1/9*x^3+6/9*x^2+4/9 ; dx = g-x; % Change in variable x = x + dx; % Successive approximation fprintf('%g', iter), disp([g, dx, x]) end Iter g dx x 1 2.2222 0.2222 2.2222 2 2.5173 0.2951 2.5173 3 2.8966 0.3793 2.8966 4 3.3376 0.4410 3.3376 5 3.7398 0.4022 3.7398 6 3.9568 0.2170 3.9568 7 3.9988 0.0420 3.9988 8 4.0000 0.0012 4.0000 9 4.0000 0.0000 4.0000 Remarks: The roots found depend on the starting point! There may be other roots… Moreover, there is no guarantee the method is converging unless one checks on the convergence by means of some algorithm. The Gauss-Seidel method can be expressed with a parameter α known as an acceleration factor, thus: ( ) ( ) ( ) ( ) ( ) 1 k k k k x x g x x α +   = + −   Note that if 1 α = the method becomes the non-accelerated method mentioned earlier. Example 6.3. Find the roots of the equation used earlier but with an acceleration factor of 1.25. First a graphical solution using Matlab: clear; x=0:.05:4.5; alpha=1.25; g=-1/9*x.^3+6/9*x.^2+4/9; k=1; dz=10; z=2; r(k)=z; s(k)=z; while dz >.001 k=k+2; r(k-1)=z; p=-1/9*z^3+6/9*z^2+4/9; s(k-1)=p; dz=abs(z-p); z=z+alpha*(p-z); r(k)=z; s(k)=p; end plot(x,g,'-',x,x,'-', r,s,'-'),grid xlabel('x') text(0.7, 4.1,'g(x) = -1/9x^3+6/9x^2+4/9') text(3.2, 3.0,'x') 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x g(x) = -1/9x 3 +6/9x 2 +4/9 x Note how at each step we choose more than "x" trying to go "further" in the right direction. This can also produce an unstable result, especially if 1 α ? . Next the numerical solution using Matlab: %chp6fig4 % Graphical display for Example 6.3 clear; dx=1; % Change in variable is set to a high value x=2; % Initial estimate iter = 0; alpha=1.25; % Iteration counter, alpha disp('Iter g dx x') % Heading for results while abs(dx) >= 0.001 & iter < 100 % Test for convergence iter = iter + 1; % No. of iterations g = -1/9*x^3+6/9*x^2+4/9; dx = g-x; % Change in variable x = x +alpha*dx; % Successive approximation with alpha accel. factor fprintf('%g', iter), disp([g, dx, x]) end Iter g dx x 1 2.2222 0.2222 2.2778 2 2.5902 0.3124 2.6683 3 3.0801 0.4118 3.1831 4 3.6157 0.4326 3.7238 5 3.9515 0.2277 4.0084 6 4.0000 -0.0085 3.9978 7 4.0000 0.0022 4.0005 8 4.0000 -0.0005 3.9999 Next the process is generalized for n equations in n variables: ( ) ( ) ( ) 1 1 2 1 2 1 2 2 1 2 , , , , , , , , , n n n n n f x x x c f x x x c f x x x c = = = L L LLLLLLLL L Solving for one variable from each equation, the system of equations car be represented as follows: ( ) ( ) ( ) 1 1 1 1 2 2 2 2 1 2 1 2 , , , , , , , , , n n n n n n x c g x x x x c g x x x x c g x x x = + = + = + L L LLLLLLLLLL L First a starting solution is assumed to be ( ) ( ) ( ) ( ) 0 0 0 1 2 , , , n x x x L . Now there are two ways to proceed: 1. Gauss iteration: use the starting estimate ( ) ( ) ( ) ( ) 0 0 0 1 2 , , , n x x x L in the equations to compute a first iterate ( ) ( ) ( ) ( ) 1 1 1 1 2 , , , n x x x L then repeat the process using the first iterate to get a second iterate ( ) ( ) ( ) ( ) 2 2 2 1 2 , , , n x x x L . This is repeated till the solutions converge. 2. Gauss-Seidel iteration: use the starting estimate ( ) ( ) ( ) ( ) 0 0 0 1 2 , , , n x x x L in the first equation ONLY to get a new estimate for ( ) 1 1 x . Now use ( ) ( ) ( ) ( ) 1 0 0 1 2 , , , n x x x L in the second equation to get an estimate for ( ) 1 2 x . Now use ( ) ( ) ( ) ( ) 1 1 0 1 2 , , , n x x x L in the third equation to get an estimate for ( ) 1 3 x and so on till the whole set ( ) ( ) ( ) ( ) 1 1 1 1 2 , , , n x x x L is found. The process is repeated till the solutions converge. An acceleration factor α can be used as before, thus at each step upgrade the estimate as follows ( ) ( ) ( ) ( ) ( ) 1 1 k k k k i i ical i x x x x α + + = + − Clearly if 1 α = then the iteration is the same as before. Usually 1 α ≥ but not much larger than unity. Chapter 6. Power Flow Analysis (Part II) 6.3.2 Newton-Raphson Iteration In this method both the function ( ) f x and its derivative ( ) f x ′ are needed to perform the iteration. Graphically, this is shown below for ( ) 3 2 6 9 4 f x x x x · − + − . In this example, the derivative is easily found as ( ) 2 3 12 9 f x x x ′ · − + . Take a look at the program chp6fig5 using the Matlab Editor. chp6fig5 0 1 2 3 4 5 6 -10 0 10 20 30 40 50 x f(x) = x 3 -6x 2 +9x-4 Using the slope instead of the function to get the new estimate as shown above. This is a linear estimate of the root of ( ) 0 f x · . The equation used for the iterations is given by: ( ) ( ) ( ) ( ) ( ) ( ) 1 k k k k f x x x f x + · − ′ where a starting solution ( ) 0 x is needed to start the iteration. Note that there is no guarantee of convergence, and, if the iteration converges, there may be other roots as well. In general, let ( ) f x c · . Expanding ( ) f x in a Taylor series about the initial estimate ( ) 0 x results in ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 2 2 0 0 0 2 1 HOT 2! df d f f x x x c dx dx ¸ _ ¸ _ + ∆ + ∆ + + · ¸ , ¸ , L keeping only the linear terms and neglecting the higher order terms we have ( ) ( ) ( ) ( ) 0 0 0 df f x x c dx ¸ _ + ∆ ¸ , ; which can be written as ( ) ( ) ( ) ( ) 0 0 0 df c f x x dx ¸ _ − ∆ ¸ , ; . If we define ( ) ( ) ( ) 0 0 c c f x ∆ − @ then ( ) ( ) ( ) 0 0 0 df c x dx ¸ _ ∆ ∆ ¸ , ; . Thus we have ( ) ( ) ( ) 0 0 0 c x df dx ∆ ∆ · ¸ _ ¸ , and a new estimate for ( ) 1 x is now given by: ( ) ( ) ( ) ( ) ( ) ( ) 0 1 0 0 0 0 c x x x x df dx ∆ · + ∆ · + ¸ _ ¸ , . This process can be repeated several times, at the (k+1) th iteration we have: ( ) ( ) ( ) k k c c f x ∆ − @ ( ) ( ) ( ) k k k c x df dx ∆ ∆ · ¸ _ ¸ , ( ) ( ) ( ) 1 k k k x x x + · + ∆ The middle equation above can be written as: ( ) ( ) ( ) k k k df c x dx ¸ _ ∆ · ∆ ¸ , We define the derivative at step k as follows: ( ) ( ) k k df j dx ¸ _ ¸ , @ thus we have: ( ) ( ) ( ) k k k c j x ∆ · ∆ In summary: given the nonlinear equation ( ) f x c · or ( ) 0 f x c − · , the curve is approximated at ( ) k x by a linear approximation which is the tangent to the curve. The intersection of the linear approximation with the x-axis is the next estimate ( ) 1 k x + . Example 6.4. Use the N-R method to find the root of the equation 3 2 ( ) 6 9 4 f x x x x · − + − assume an initial value of ( ) 0 6 x · . This is demonstrated graphically in the plot shown above. It is next solved by iteration. dx=1; % Change in variable is set to a high value x=6; % Initial estimate iter = 0; % Iteration counter disp('iter Dc J dx x') % Heading for result while abs(dx) >= 0.001 & iter < 100 % Test for convergence iter = iter + 1; % No. of iterations Dc=4 - (x^3-6*x^2+9*x); % Residual J = 3*x^2-12*x+9; % Derivative dx= Dc/J; %Change in variable x=x+dx; % Successive solution fprintf('%g', iter), disp([Dc, J, dx, x]) end iter Dc J dx x 1 -50.0000 45.0000 -1.1111 4.8889 2 -13.4431 22.0370 -0.6100 4.2789 3 -2.9981 12.5797 -0.2383 4.0405 4 -0.3748 9.4914 -0.0395 4.0011 5 -0.0095 9.0126 -0.0011 4.0000 6 -0.0000 9.0000 -0.0000 4.0000 Note that it took only 6 iterations. This is true in almost all cases, even if the function is very complicated and of a high order. If the method is going to converge, it will converge within a few iterations (usually of the order of 10 or less). Note in the program a limit of 100 iterations is stipulated. Next the method is generalized to the n-dimensional case. The equations are: ( ) ( ) ( ) 1 1 2 1 2 1 2 2 1 2 , , , , , , , , , n n n n n f x x x c f x x x c f x x x c · · · L L LLLLLLLL L Expanding the functions in a Taylor series and keeping only the linear terms we have: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 0 1 1 1 1 1 2 1 1 2 0 0 0 0 0 0 0 2 2 2 2 1 2 2 1 2 0 0 0 0 0 0 0 1 2 1 1 2 n n n n n n n n n n f f f f x x x c x x x f f f f x x x c x x x f f f f x x x c x x x ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ + ∆ + ∆ + + ∆ · ∂ ∂ ∂ ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ + ∆ + ∆ + + ∆ · ∂ ∂ ∂ ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ + ∆ + ∆ + + ∆ · ∂ ∂ ∂ ¸ , ¸ , ¸ , L L L L which may be expressed in matrix form as follows: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 1 1 1 1 2 0 0 1 1 1 0 0 0 0 0 2 2 2 2 2 2 1 2 2 0 0 0 0 0 1 2 n n n n n n n n f f f x x x c f x f f f x c f x x x x c f f f f x x x 1 ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ 1 ∂ ∂ ∂ 1 ¸ , ¸ , ¸ , 1 − ∆ 1 1 ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ 1 1 ∆ − 1 · ∂ ∂ ∂ 1 ¸ , ¸ , ¸ , 1 1 1 1 ∆ − 1 1 ¸ ] ¸ _ ¸ _ ¸ _ 1 ∂ ∂ ∂ 1 ∂ ∂ ∂ 1 ¸ , ¸ , ¸ , ¸ ] L L M M M M O M L 1 1 1 1 1 1 ¸ ] In general, assuming we are at step k instead of 0, this may be written compactly as: ( ) ( ) ( ) k k k C J X ∆ · ∆ which can be solved for ( ) k X ∆ thus: ( ) ( ) ( ) 1 k k k X J C − 1 ∆ · ∆ ¸ ] and thus the next iterate is given by: ( ) ( ) ( ) 1 k k k X X X + · + ∆ where ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 2 1 1 1 1 2 2 2 2 1 2 2 1 2 and k k k k k k k k n n n k k k n k k k k k k k n n n n c f x x c f X C x c f f f f x x x f f f J x x x f f f x x x 1 1 − ∆ 1 1 1 ∆ − 1 ∆ · ∆ · 1 1 1 1 1 1 ∆ − ¸ ] 1 ¸ ] ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ ∂ ∂ ∂ ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ · ∂ ∂ ∂ ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ ∂ ∂ ∂ ∂ ∂ ∂ ¸ , ¸ , ¸ , M M L L M M O M L 1 1 1 1 1 1 1 1 1 1 1 1 ¸ ] ( ) k J is the generalized derivative in n-dimensions. It is called the Jacobian matrix. Elements of this matrix are partial derivatives of the nonlinear functions with respect to the variables and evaluated at ( ) k X . It is assumed that this Jacobian has an inverse at each step of the iteration. Thus, instead of dealing with nonlinear equations, the equations are first linearized, and solved for each iteration. Note that inversion of the Jacobian is not needed if we only want the solution of the linear matrix equation. In Matlab we use \ X J C ∆ · ∆ in order to solve this set of equations without obtaining the inverse of the Jacobian. Example 6.5. Use the N-R method to find the intersections of the curves: 1 2 2 1 2 2 4 1 x x x e x + · + · First a graph of the functions is shown below: chp6fig6 Current plot held -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 x x 2 +y 2 =4 e x +y=1 Next the problem is solved using the N-R iteration. The Jacobian is found from the equations by differentiation thus: 1 1 2 2 2 1 x x x J e 1 · 1 ¸ ] The Matlab program follows: iter = 0; % Iteration counter x=[0.5;-1]; % Initial estimate Dx = [1; 1]; C=[4; 1]; disp('Iter DC Jacobian matrix Dx x'); % Heading for results while max(abs(Dx)) >= .0001 & iter < 10 %Convergence test iter=iter+1; % Iteration counter f = [x(1)^2+x(2)^2; exp(x(1))+x(2)]; % Functions DC = C - f; % Residuals J = [2*x(1) 2*x(2) % Jacobian matrix exp(x(1)) 1]; Dx=J\DC; % Change in variables x=x+Dx; % Successive solutions fprintf('%g', iter), disp([DC, J, Dx, x]) % Results end Iter DC Jacobian matrix Dx x 1 2.7500 1.0000 -2.0000 0.8034 1.3034 0.3513 1.6487 1.0000 -0.9733 -1.9733 2 -1.5928 2.6068 -3.9466 -0.2561 1.0473 -0.7085 3.6818 1.0000 0.2344 -1.7389 3 -0.1205 2.0946 -3.4778 -0.0422 1.0051 -0.1111 2.8499 1.0000 0.0092 -1.7296 4 -0.0019 2.0102 -3.4593 -0.0009 1.0042 -0.0025 2.7321 1.0000 0.0000 -1.7296 5 -0.0000 2.0083 -3.4593 -0.0000 1.0042 -0.0000 2.7296 1.0000 -0.0000 -1.7296 The other intersection can be found using the initial estimate [-0.5 ; 1]. Example 6.6 Starting with an initial value [1, 1, 1] x · solve the following system of equations using the N-R method. 2 2 2 1 2 3 2 1 2 2 3 1 1 3 2 3 11 3 3 6 x x x x x x x x x x x x − + · + − · − + · First we find the Jacobian, thus: 1 3 2 1 2 3 3 1 2 2 2 2 2 3 1 x x x J x x x x x x x − 1 1 · + − 1 − − + 1 ¸ ] The Matlab program follows: Dx=[10; 10; 10]; % Change in variable is set to a high value x=[1; 1; 1]; % Initial estimate C=[11; 3; 6]; iter = 0; % Iteration counter while max(abs(Dx))>=.0001 & iter<10; % Test for convergence iter = iter + 1; % No. of iterations F = [x(1)^2-x(2)^2+x(3)^2 % Functions x(1)*x(2)+x(2)^2-3*x(3) x(1)-x(1)*x(3)+x(2)*x(3)]; DC =C - F; % Residuals J = [2*x(1) -2*x(2) 2*x(3) % Jacobian matrix x(2) x(1)+2*x(2) -3 1-x(3) x(3) -x(1)+x(2)]; Dx=J\DC; %Change in variable x=x+Dx; % Successive solution end iter, DC, J, Dx, x iter = 6 DC = 1.0e-004 * -0.3669 -0.1784 0.0957 J = 4.0000 -6.0000 8.0000 3.0000 8.0000 -3.0000 -3.0000 4.0000 1.0000 Dx = 1.0e-005 * -0.5577 -0.1130 -0.2645 x = 2.0000 3.0000 4.0000 It is noted that the convergence of the method is quadratic, and is independent of the size of the system being iterated. Usually convergence takes place in 10 iterations or less. Chapter 6. Power Flow Analysis (Part III) 6.4 Power Flow Solution Power flow studies also known as load flow are important in power system analysis. They are used in system planning, economic scheduling, control, and future plans for the power system. It is assumed the system is operating in a balanced steady state. The problem is to determine the magnitude and phase angle of voltages at every node, and the real and reactive power flow for each line. A single phase model is used for analysis (due to balanced conditions). The per-unit system is almost always used. Four quantities are associated with each bus. These are voltage magnitude V , and phase angle δ , real power P , and reactive power Q. The system buses are classified into three types as follows: Slack bus One bus, known as the slack or swing bus, is taken as reference where the magnitude and phase angle of the voltage are specified. After specifying the powers of some other buses, this bus takes the "slack" by balancing the power equation. This bus will be bus #1. Both P and Q need to be found for this bus. Load buses At these buses the real and reactive powers are known. The voltage magnitude and angle need to be found. These buses are also known as P-Q buses. Generator buses These buses are the generator buses. At these buses the voltage magnitude and real power are known. Other names for these buses: P-V buses, regulated buses and voltage controlled buses. The unknown variables on these buses are the reactive power Q and the voltage angle δ . 6.4.1 Power Flow Equation Consider a typical bus in a power system as shown in the figure on the right for bus i. Transmission lines are represented by their equivalent π models and the impedances are converted to admittances on a per-unit basis on a common MVA base and properly selected kV bases. Application of KCL we have: ( ) ( ) ( ) ( ) 0 1 1 2 2 0 1 1 1 2 2 i i i i i i i in i n i i in i i i in n I y V y V V y V V y V V y y y V y V y V y V · + − + − + + − · + + + − − − − L L L y i1 y i2 y in y i0 I i V i V 1 V 2 V n Bus i 2 or 0 1 n n i i ij ij j j j I V y y V j i · · · − ≠ ∑ ∑ . We also know the real and reactive power for bus i are given by * i i i i P jQ VI + · or * i i i i P jQ I V − · ∴ we have: * 0 1 ### n n i i i ij ij j j j i P jQ V y y V j i V · · − · − ≠ ∑ ∑ The above equation is the fundamental equation from which the power flow equations are formulated. These will result in nonlinear algebraic equations which are solved by the iterative techniques discussed earlier. 6.5 Gauss-Seidel Power Flow Solution Equation ### is solved for the voltages for iteration thus: ( ) ( ) ( ) * 1 1, 0, sch sch n k i i ij j k k j j i i i n ij j j i P jQ y V V V y + · ≠ · ≠ − + · ∑ ∑ where sch i P and sch i Q are the net power flows into bus i due to the current entering the bus. For generator buses, the value of sch i P and sch i Q is positive when power is going into the bus. For load buses, sch i P and sch i Q are positive when the power is leaving the bus. We can solve for the real and imaginary power using equation ###, thus: ( ) ( ) ( ) ( ) 1 * 0, 1, Re n n k k k k i i i ij ij j j j i j j i P V V y y V + · ≠ · ≠ ¹ ¹ 1 ¹ ¹ · − ' ; 1 ¹ ¹ ¸ ] ¹ ¹ ∑ ∑ ( ) ( ) ( ) ( ) 1 * 0, 1, Im n n k k k k i i i ij ij j j j i j j i Q V V y y V + · ≠ · ≠ ¹ ¹ 1 ¹ ¹ · − − ' ; 1 ¹ ¹ ¸ ] ¹ ¹ ∑ ∑ These three equations are the ones used for the iteration after some more manipulation of the terms involved. Usually the elements of the matrix bus Y are known, hence we will express these equations in terms of the elements of bus Y shown by upper case letters thus: off diagonal elements ij ij Y y · − and diagonal elements 0, n ii ij j j i Y y · ≠ · ∑ . The three equations above become: 3 ( ) ( ) ( ) * 1 1, * sch sch n k i i ij j k k j j i i i ii P jQ Y V V V Y + · ≠ − − · ∑ ( ) ( ) ( ) ( ) 1 * 1, Re ** n k k k k i i i ii ij j j j i P V V Y Y V + · ≠ ¹ ¹ 1 ¹ ¹ · + − ' ; 1 ¹ ¹ ¸ ] ¹ ¹ ∑ ( ) ( ) ( ) ( ) 1 * 1, Im *** n k k k k i i i ii ij j j j i Q V V Y Y V + · ≠ ¹ ¹ 1 ¹ ¹ · − + ' ; 1 ¹ ¹ ¸ ] ¹ ¹ ∑ It is noted that ii Y includes the admittance to ground. We are interested in solving fo the voltage (magnitude and angle) at all the buses. Thus there are 2n unknowns. However, at the slack bus we assume 1 1 0 V · ∠ o , thus we now have ( ) 2 1 n − unknowns which we need to fi nd by iteration. For all other buses, we start the iteration with an initial estimate of 1 0 1 0 V j · ∠ · + o . For P-Q buses we know the scheduled power components sch i P and sch i Q thus equation (*) above can be used to iterate for the voltages. For P-V buses we know sch i P and i V , thus we can use (***) to find sch i Q , then use (*) to iterate for V . However, since i V is fixed, only the imaginary part of ( ) 1 k i V + is retained, and the real part is selected so as to satisfy ( ) ( ) ( ) ( ) 2 2 2 1 1 k k i i i e f V + + + · or ( ) ( ) ( ) 2 2 1 1 k k i i i e V f + + · − where ( ) 1 k i e + and ( ) 1 k i f + are the real and imaginary components of the voltage, i.e. ( ) ( ) ( ) 1 1 1 k k k i i i V e j f + + + · + ⋅ . Often an acceleration factor is employed to increase the rate of convergence, thus ( ) ( ) ( ) ( ) ( ) 1 k k k k i i ical i V V V V α + · + − where α is in the range of 1.3 to 1.7. The updated voltages are used immediately, and the whole process is repeated till the solution converges. A good test for convergence is when ( ) ( ) 1 k k i i e e ε + − ≤ ( ) ( ) 1 k k i i f f ε + − ≤ 4 V i V j I ij I ji I l I i0 I j0 y ij y i0 y j0 where ε is in the range 0.00001 to 0.00005 pu. In practical systems it is the power mismatch vector which is used to judge convergence, thus the largest element in the column vectors and P Q ∆ ∆ is required to be smaller than a specified value (usually of the order of 0.001 pu). Once a solution is found, the real and reactive powers of the slack bus are computed from (**) and (***) above. 6.6 Line Flows and Losses Having found the voltages at all buses, the next step is to compute the line flows and losses. Consider a typical line connecting buses i and j as shown on the left. The total current from bus i to j is given by ( ) 0 0 ij l i ij i j i i I I I y V V y V · + · − + . Similarly, the total current from bus j to bus I is given by ( ) 0 0 ji l j ij j i j j I I I y V V y V · − + · − + . The complex powers ij S from bus i to bus j, and ji S from bus j to bus i are given by * * ij i ij ji j ji S V I S V I · · The power loss in line i – j is the algebraic sum of the two power expressions above, thus Li j ij ji S S S · + Example 6.7 Read the example and the longhand solution in the book page 212, then return for the Matlab solution below (chp6ex7.m): y12=10-j*20; y13=10-j*30; y23=16-j*32; V1=1.05+j*0; iter =0; S2=-2.566-j*1.102; S3=-1.386-j*.452; V2=1+j*0; V3=1+j*0; disp(' iter V2 V3') for I=1:10; iter=iter+1; V2 = (conj(S2)/conj(V2)+y12*V1+y23*V3)/(y12+y23); V3 = (conj(S3)/conj(V3)+y13*V1+y23*V2)/(y13+y23); disp([iter, V2, V3]) end V2= .98-j*.06; 5 V3= 1-j*.05; I12=y12*(V1-V2); I21=-I12; I13=y13*(V1-V3); I31=-I13; I23=y23*(V2-V3); I32=-I23; S12=V1*conj(I12); S21=V2*conj(I21); S13=V1*conj(I13); S31=V3*conj(I31); S23=V2*conj(I23); S32=V3*conj(I32); I1221=[I12,I21] I1331=[I13,I31] I2332=[I23,I32] S1221=[S12, S21 (S12+S13) S12+S21] S1331=[S13, S31 (S31+S32) S13+S31] S2332=[S23, S32 (S23+S21) S23+S32] iter V2 V3 1.0000 0.9825 - 0.0310i 1.0011 - 0.0353i 2.0000 0.9816 - 0.0520i 1.0008 - 0.0459i 3.0000 0.9808 - 0.0578i 1.0004 - 0.0488i 4.0000 0.9803 - 0.0594i 1.0002 - 0.0497i 5.0000 0.9801 - 0.0598i 1.0001 - 0.0499i 6.0000 0.9801 - 0.0599i 1.0000 - 0.0500i 7.0000 0.9800 - 0.0600i 1.0000 - 0.0500i 8.0000 0.9800 - 0.0600i 1.0000 - 0.0500i 9.0000 0.9800 - 0.0600i 1.0000 - 0.0500i 10.0000 0.9800 - 0.0600i 1.0000 - 0.0500i I1221 = 1.9000 - 0.8000i -1.9000 + 0.8000i I1331 = 2.0000 - 1.0000i -2.0000 + 1.0000i I2332 = -0.6400 + 0.4800i 0.6400 - 0.4800i S1221 = 1.9950 + 0.8400i -1.9100 - 0.6700i 4.0950 + 1.8900i 0.0850 + 0.1700i S1331 = 2.1000 + 1.0500i -2.0500 - 0.9000i -1.3860 - 0.4520i 0.0500 + 0.1500i S2332 = -0.6560 - 0.4320i 0.6640 + 0.4480i -2.5660 - 1.1020i 0.0080 + 0.0160i Since the algorithm converged in less than 10 iterations, there is no need to repeat with more iterations. Example 6.8 Read the example and the longhand solution in the book page 216, then return for the Matlab solution below (chp6ex8.m): clear; y12=10-j*20; y13=10-j*30; y23=16-j*32; y33=y13+y23; V1=1.05+j*0; format short E 6 iter =0; S2=-4.0-j*2.5; P3 = 2; V2=1+j*0; Vm3=1.04; V3=1.04+j*0; for I=1:10; iter=iter+1; %*** E2 = V2; E3=V3; V2 = (conj(S2)/conj(V2)+y12*V1+y23*V3)/(y12+y23);%*** DV2 = V2-E2;%*** Q3 = -imag(conj(V3)*(y33*V3-y13*V1-y23*V2));%*** S3 = P3 +j*Q3; Vc3 = (conj(S3)/conj(V3)+y13*V1+y23*V2)/(y13+y23);%*** Vi3 = imag(Vc3); Vr3= sqrt(Vm3^2 - Vi3^2); V3 = Vr3 + j*Vi3;%*** DV3=V3-E3;%*** if iter==9 iter V2, DV2, Q3, Vc3, V3, DV3 elseif iter==10 iter V2, DV2, Q3, Vc3, V3, DV3 end end I12=y12*(V1-V2); I21=-I12; I13=y13*(V1-V3); I31=-I13; I23=y23*(V2-V3); I32=-I23; S12=V1*conj(I12); S21=V2*conj(I21); S13=V1*conj(I13); S31=V3*conj(I31); S23=V2*conj(I23); S32=V3*conj(I32); I1221=[I12,I21] I1331=[I13,I31] I2332=[I23,I32] S1221=[S12, S21 (S12+S13) S12+S21] S1331=[S13, S31 (S31+S32) S13+S31] S2332=[S23, S32 (S23+S21) S23+S32] iter = 9 V2 = 0.9706 - 0.0457i DV2 = -8.1775e-007 -6.5578e-006i Q3 = 1.4616 Vc3 = 1.0400 - 0.0091i V3 = 1.0400 - 0.0091i DV3 = -4.5070e-008 -5.1804e-006i iter = 10 V2 = 7 0.9706 - 0.0457i DV2 = -3.3824e-007 -2.7125e-006i Q3 = 1.4617 Vc3 = 1.0400 - 0.0091i V3 = 1.0400 - 0.0091i DV3 = -1.8650e-008 -2.1428e-006i I1221 = 1.7082 - 1.1308i -1.7082 + 1.1308i I1331 = 0.3720 - 0.2107i -0.3720 + 0.2107i I2332 = -2.2828 + 1.6329i 2.2828 - 1.6329i S1221 = 1.7936 + 1.1874i -1.7096 - 1.0195i 2.1841 + 1.4085i 0.0839 + 0.1679i S1331 = 0.3906 + 0.2212i -0.3887 - 0.2157i 2.0000 + 1.4618i 0.0018 + 0.0055i S2332 = -2.2903 - 1.4805i 2.3888 + 1.6775i -3.9999 - 2.5000i 0.0985 + 0.1969i As can be seen, after 9 and 10 iterations, the accuracy of the voltages is of the order of 6 10 − . Cover the power flow diagrams for these two examples. Skip section 6.7 on tap changing transformers. Chapter 6. Power Flow Analysis (Part IV) 6.8 Power Flow Programs Read carefully! Note that if tap changing transformers are not used, instead of tap setting the author uses "1". See for example the last column on page 227. Note that these "Programs" must be run in a very specific order (see p. 222.) 6.10 Newton-Raphson Power Flow Solution Due to its quadratic convergence, the N-R method has rapid convergence independent from system size. Thus, the method usually converges in less than 10 iterations regardless of system size! However, as the system size (number of equations) increases, the function evaluations at each iteration increases. For the typical node shown on the right we can write: 1 n i ij j j I Y V = = ∑ . This equation includes bus i under the summation. Using polar form this equation is written as: ( ) 1 n i ij i ij j j I Y V θ δ = = ∠ + ∑ where ij ij ij Y Y θ = ∠ and j j j V V δ = ∠ . The conjugate of the complex power at bus i is given by: * * i i i i i S P jQ V I = − = and using the expression for current we have: ( ) ( ) 1 n i i i i ij j ij j j P jQ V Y V δ θ δ = − = ∠ − ∠ + ∑ Solving for real and imaginary parts we have: ( ) 1 cos n i i j ij ij i j j P V V Y θ δ δ = = − + ∑ and ( ) 1 sin n i i j ij ij i j j Q V V Y θ δ δ = = − − + ∑ If these equations are linearized as before around the point ( ) k i δ and ( ) k i V using the Taylor series expansion around these points we have: y i1 y i2 y in y i0 I i V i V 1 V 2 V n Bus i 2 ( ) ( ) ( ) ( ) 2 2 2 2 2 2 ( ) 2 ( ) ( ) ( ) ( ) ( ) 2 2 ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 ( ) ( ) ( ) 2 | | | | k k k k n n k k k k k n n n n k n n n k k k k n n k n k k n n n P P P P V V P P P P P V V P Q Q Q Q Q V V Q Q Q δ δ δ δ δ δ δ δ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂   ∆   ∂ ∂ ∂ ∂   ∂ ∂ ∂ ∂   ∆   = − − − − − − −     ∆ ∂ ∂ − ∂ ∂   ∂ ∂ ∂ ∂     ∆   ∂ ∂ ∂ ∂ L L M O M M O M M L L L M L M O M L ( ) 2 ( ) ( ) ( ) 2 ( ) ( ) ( ) 2 | | k k n k k k n k k n n n V V Q Q V V δ δ           ∆             ∆     −         ∆             ∆         ∂ ∂     ∂ ∂     M M M O M L Note that 1 P and 1 Q for the slack bus can be computed since for this bus we know 1 V and 1 δ , hence they are not included above. All quantities are assumed in per unit and the angles are in radians. If some bus is a generator bus (P-V bus) then we know P and V for that bus, hence the equation in Q ∆ and V ∆ for that bus is not needed. Thus we can eliminate that row and the corresponding column for the equations above. Thus for each generator bus we have one less equation (one less unknown) to solve for. If there are n buses, then we have a maximum of ( ) 2 1 n − equations. Let us assume we have also 1 m n ≤ − generator buses, then the number of equations becomes ( ) 2 2 n m − − equations. Note also that the Jacobian matrix is evaluated at every iteration step! The above equation can be put into the more compact form 1 2 3 4 J J P V J J Q δ ∆ ∆       =       ∆ ∆       where 1 J , ( ) ( ) 1 1 n n − × − is the derivatives of P with respect to δ , 2 J , ( ) ( ) 1 1 n n m − × − − is the derivative of P with respect to V , 3 J , ( ) ( ) 1 1 n m n − − × − is the derivative of Q with respect to δ , and 4 J , ( ) ( ) 1 1 n m n m − − × − − is the derivative of Q with respect to V . All these derivatives are the partial derivatives. Starting with some initial value for V and δ and solving these equations for δ ∆ and V ∆ we can proceed to the next iteration thus: ( 1) ( ) ( ) ( 1) ( ) ( ) k k k i i i k k k i i i V V V δ δ δ + + = + ∆ = + ∆ 3 We can stop the iteration process when the power residuals are smaller than a pre- specified value. These are ( ) ( ) k k sch i i i P P P ∆ = − and ( ) ( ) k k sch i i i Q Q Q ∆ = − . These are also known as the power mismatch. The general form for the Jacobian elements are as follows. For 1 J the diagonal and off diagonal elements are: ( ) ( ) sin sin i i j ij ij i j j i i i i j ij ij i j j P V V Y P V V Y j i θ δ δ δ θ δ δ δ ≠ ∂ = − + ∂ ∂ = − − + ≠ ∂ ∑ and the diagonal and off diagonal elements for 2 J are: ( ) ( ) 2 cos cos cos i i ii ii j ij ij i j j i i i i ij ij i j j P V Y V Y V P V Y j i V θ θ δ δ θ δ δ ≠ ∂ = + − + ∂ ∂ = − + ≠ ∂ ∑ and the diagonal and off diagonal elements for 3 J are: ( ) ( ) cos cos i i j ij ij i j j i i i i j ij ij i j j Q V V Y Q V V Y j i θ δ δ δ θ δ δ δ ≠ ∂ = − + ∂ ∂ = − − + ≠ ∂ ∑ and the diagonal and off diagonal elements for 4 J are: ( ) ( ) 2 sin sin sin i i ii ii j ij ij i j j i i i i ij ij i j j Q V Y V Y V Q V Y j i V θ θ δ δ θ δ δ ≠ ∂ = − − − + ∂ ∂ = − − + ≠ ∂ ∑ The procedure for the power flow solution using the Newton-Raphson method is as follows: 1. For load buses where sch i P and sch i Q are specified, the initial voltage magnitudes and angles are set equal to that of the slack bus (often unity at zero angle, i.e. (0) 1.0 i V = and (0) 0.0 i δ = ). For generator buses (voltage regulated buses) where i V and sch i P are specified, set the voltage angle i δ equal to that of the slack bus (usually zero, i.e. (0) 0.0 i δ = ). 2. For load buses ( ) k i P and ( ) k i Q are computed using the equations for i P and i Q derived earlier. Now compute the power mismatch ( ) ( ) k k sch i i i P P P ∆ = − and ( ) ( ) k k sch i i i Q Q Q ∆ = − . 4 3. For voltage controlled buses (generator buses), ( ) k i P is computed directly, hence ( ) ( ) k k sch i i i P P P ∆ = − is computed next. 4. The four elements of the Jacobian matrix are computed. 5. Now the equations involving the Jacobian are solved (in an optimal manner) for the change in voltage magnitudes and angles. 6. The new voltage magnitudes and angles are found, thus completing one iteration. 7. The process is repeated till the residuals are less than some specified value ε , i.e. ( ) k i P ε ∆ ≤ and ( ) k i Q ε ∆ ≤ for all buses. The method will converge at a small number of iterations (usually less than 10) regardless of the size of the system. In large systems the Jacobian matrix is not updated at each iteration. Instead, it is kept for a few iterations then updated. This saves a lot of time in the larger power systems. Go over example 6.10 page 235 in detail. % Example 6.10 Newton-Raphson method V = [1.05; 1.0; 1.04]; d = [0; 0; 0]; Ps=[-4; 2.0]; Qs= -2.5; YB = [ 20-j*50 -10+j*20 -10+j*30 -10+j*20 26-j*52 -16+j*32 -10+j*30 -16+j*32 26-j*62]; Y= abs(YB); t = angle(YB); iter=0; pwracur = 0.00025; % Power accuracy DC = 10; % Set the maximum power residual to a high value while max(abs(DC)) > pwracur iter = iter +1 P=[V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+V(2)^2*Y(2,2)*cos(t(2,2))+ V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); V(3)*V(1)*Y(3,1)*cos(t(3,1)-d(3)+d(1))+V(3)^2*Y(3,3)*cos(t(3,3))+ V(3)*V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2))]; Q= -V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-V(2)^2*Y(2,2)*sin(t(2,2))- V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); J(1,1)=V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))+ V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); J(1,2)=-V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); J(1,3)=V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+2*V(2)*Y(2,2)*cos(t(2,2))+ V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); J(2,1)=-V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2)); J(2,2)=V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))+ V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2)); J(2,3)=V(3)*Y(2,3)*cos(t(3,2)-d(3)+d(2)); J(3,1)=V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+ V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); J(3,2)=-V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3)); 5 J(3,3)=-V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-2*V(2)*Y(2,2)*sin(t(2,2))- V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3)); DP = Ps - P; DQ = Qs - Q; DC = [DP; DQ] J DX = J\DC d(2) =d(2)+DX(1); d(3)=d(3) +DX(2); V(2)= V(2)+DX(3); V, d, delta =180/pi*d; end P1= V(1)^2*Y(1,1)*cos(t(1,1))+V(1)*V(2)*Y(1,2)*cos(t(1,2)-d(1)+d(2))+ V(1)*V(3)*Y(1,3)*cos(t(1,3)-d(1)+d(3)) Q1=-V(1)^2*Y(1,1)*sin(t(1,1))-V(1)*V(2)*Y(1,2)*sin(t(1,2)-d(1)+d(2))- V(1)*V(3)*Y(1,3)*sin(t(1,3)-d(1)+d(3)) Q3=-V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-V(3)*V(2)*Y(3,2)* sin(t(3,2)-d(3)+d(2))-V(3)^2*Y(3,3)*sin(t(3,3)) iter = 1 DC = -2.8600 1.4384 -0.2200 J = 54.2800 -33.2800 24.8600 -33.2800 66.0400 -16.6400 -27.1400 16.6400 49.7200 DX = -0.0453 -0.0077 -0.0265 V = 1.0500 0.9735 1.0400 d = 0 -0.0453 -0.0077 iter = 2 DC = -0.0992 0.0217 -0.0509 J = 51.7247 -31.7656 21.3026 -32.9816 65.6564 -15.3791 -28.5386 17.4028 48.1036 DX = -0.0018 -0.0010 -0.0018 V = 1.0500 6 0.9717 1.0400 d = 0 -0.0471 -0.0087 iter = 3 DC = 1.0e-003 * -0.2166 0.0382 -0.1430 J = 51.5967 -31.6939 21.1474 -32.9339 65.5976 -15.3516 -28.5482 17.3969 47.9549 DX = 1.0e-005 * -0.3856 -0.2386 -0.4412 V = 1.0500 0.9717 1.0400 d = 0 -0.0471 -0.0087 P1 = 2.1842 Q1 = 1.4085 Q3 = 1.4618 Note that the N-R method took only 3 iterations. It is a good idea to practice using the "canned" routines to solve this example. See example 6.11 on page 239 for a sample on using the canned routines. 6.11 Fast Decoupled Power Flow Solution Much simplification can be made to the Jacobian in order to simplify the power flow equations. The Jacobian can be simplified to the point where it need not be evaluated only once instead of on each iteration step. The basic ideas are that the voltages have a magnitude very near unity, the machine angles are very small and real power is almost a function of machine angle, while imaginary power is a function of machine voltage. These statements are approximately correct, but not 100%. Recall the equations involving the Jacobian are: 7 1 2 3 4 J J P V J J Q δ ∆ ∆       =       ∆ ∆       If we assume that real power is only a function of machine angle, then 2 0 J = . If we also assume that imaginary power is a function of voltage magnitude only, then 3 0 J = . Thus the equations become: 1 4 0 0 J P V J Q δ ∆ ∆       =       ∆ ∆       or 1 4 P P J Q Q J V V V δ δ δ ∂   ∆ = ∆ = ∆   ∂     ∂ ∆ = ∆ = ∆   ∂   Thus the matrix equation is broken into two decoupled equations which are much easier to solve. Further simplification can be made by making the Jacobian 1 J and 4 J independent of the iteration process. The diagonal elements of 1 J are described as follows: ( ) ( ) 2 1 sin sin n i i j ij ij i j i ii ii i i P V V Y V Y θ δ δ θ δ = ∂ = − + − ∂ ∑ It is noted that the first term on the right is i Q − , thus we have: ( ) 2 2 sin i i i ii ii i i i ii P Q V Y Q V B θ δ ∂ = − − ∂ = − − Where ( ) sin ii ii ii B Y θ = is the imaginary part of the diagonal elements of the bus admittance matrix. ii B is the sum of the susceptance of all the elements connected to bus i . Typically, ii i B Q ? , and 2 i i V V ≈ hence we have: i i ii i P V B δ ∂ = − ∂ The off diagonal elements of 1 J are simplified as follows. Under normal operating conditions the difference j i δ δ − is so small that we can assume ij i j ij θ δ δ θ − + ≈ . Thus the off diagonal elements of 1 J become ( ) sin i i j ij ij j i j ij P V V Y V V B θ δ ∂ = − ∂ = − Further simplification results from assuming 1 j V ≈ thus i i ij j P V B δ ∂ = − ∂ 8 In a similar manner the diagonal elements of 4 J may be written as: ( ) 1 sin sin n i i ii ii i j ij ij i j j i Q V Y V V Y V θ θ δ δ = ∂ = − − − + ∂ ∑ where we have used the fact that 1 j V ≈ . The summation term may be replaced by i Q − thus we now have sin i i ii ii i i Q V Y Q V θ ∂ = − + ∂ Similar to what was done earlier, since sin ii ii ii i B Y Q θ = ? , we have: i i ii i Q V B V ∂ = − ∂ And finally, assuming ij i j ij θ δ δ θ − + ≈ , the off diagonal terms of 4 J can be written as follows, : i i ij j Q V B V δ δ = − . Now the equations can be written as i i P B V Q B V V δ ∆ ′ = − ∆ ∆ ′′ = − ∆ Where B′ and B′′ are the imaginary parts of the bus admittance matrix bus Y . Since the elements of this matrix are constant, they need to be computed only once and need not be updated at each iteration. These two matrices are triangularized once and used for the rest of the solution in all the iterations. B′ is of order 1 n − because we remove the first row and column of the bus admittance matrix (the first bus is the slack bus). Now B′′ is formed from the bus admittance matrix after deleting the rows and columns that correspond to generator buses, thus it is of size 1 n m − − where m is the number of generator buses. Clearly B′′ is a sub-matrix of B′ . Now the equations above can be solved as follows: [ ] [ ] 1 1 P B V Q V B V δ − − ∆ ′ ∆ = − ∆ ′′ ∆ = − This solution is much quicker per iteration than the N-R iteration, but it is not as good in convergence, hence more iterations are needed. This method is good when a very fast solution is needed, or as a starting solution for the N-R method. Go over example 6.12 page 243 in detail, then see the Matlab solution below. % Example 6.12 Fast decoupled method V1= 1.05; V2 = 1.0; V3 = 1.04; 9 d1 = 0; d2 = 0; d3=0; Ps2=-4; Ps3 =2.0; Qs2= -2.5; YB = [ 20-j*50 -10+j*20 -10+j*30 -10+j*20 26-j*52 -16+j*32 -10+j*30 -16+j*32 26-j*62]; Y = abs(YB); t=angle(YB); B =[-52 32; 32 -62] Binv = inv(B) iter=0; pwracur = 0.0003; % Power accuracy DC = 10; % Set the max of power mismatch to a high value while max(abs(DC)) > pwracur iter = iter +1; P2= V2*V1*Y(2,1)*cos(t(2,1)-d2+d1)+V2^2*Y(2,2)*cos(t(2,2))+ V2*V3*Y(2,3)*cos(t(2,3)-d2+d3); P3= V3*V1*Y(3,1)*cos(t(3,1)-d3+d1)+V3^2*Y(3,3)*cos(t(3,3))+ V3*V2*Y(3,2)*cos(t(3,2)-d3+d2); Q2=-V2*V1*Y(2,1)*sin(t(2,1)-d2+d1)-V2^2*Y(2,2)*sin(t(2,2))- V2*V3*Y(2,3)*sin(t(2,3)-d2+d3); DP2 = Ps2 - P2; DP2V = DP2/V2; DP3 = Ps3 - P3; DP3V = DP3/V3; DQ2 = Qs2 - Q2; DQ2V = DQ2/V2; DC =[DP2; DP3; DQ2]; Dd = -Binv*[DP2V;DP3V]; DV = -1/B(1,1)*DQ2V; d2 =d2+Dd(1); d3 =d3+Dd(2); V2= V2+DV; angle2 =180/pi*d2; angle3 =180/pi*d3; R = [iter d2 d3 V2 DP2 DP3 DQ2]; disp(R) end Q3=-V3*V1*Y(3,1)*sin(t(3,1)-d3+d1)-V3^2*Y(3,3)*sin(t(3,3))- V3*V2*Y(3,2)*sin(t(3,2)-d3+d2); P1= V1^2*Y(1,1)*cos(t(1,1))+V1*V2*Y(1,2)*cos(t(1,2)-d1+d2)+ V1*V3*Y(1,3)*cos(t(1,3)-d1+d3); Q1=-V1^2*Y(1,1)*sin(t(1,1))-V1*V2*Y(1,2)*sin(t(1,2)-d1+d2)- V1*V3*Y(1,3)*sin(t(1,3)-d1+d3); S1=P1+j*Q1 Q3 B = -52 32 32 -62 Binv = -0.0282 -0.0145 -0.0145 -0.0236 1.0000 -0.0605 -0.0089 0.9958 -2.8600 1.4384 -0.2200 2.0000 -0.0565 -0.0080 0.9653 0.1759 -0.0710 -1.5790 3.0000 -0.0442 -0.0087 0.9657 0.6403 -0.4570 0.0219 4.0000 -0.0448 -0.0090 0.9730 -0.0214 0.0012 0.3652 5.0000 -0.0477 -0.0087 0.9731 -0.1534 0.1129 0.0067 6.0000 -0.0476 -0.0086 0.9714 0.0005 0.0026 -0.0861 7.0000 -0.0469 -0.0087 0.9713 0.0360 -0.0262 -0.0041 10 8.0000 -0.0469 -0.0087 0.9717 0.0009 -0.0014 0.0201 9.0000 -0.0471 -0.0087 0.9718 -0.0084 0.0061 0.0016 10.0000 -0.0471 -0.0087 0.9717 -0.0005 0.0005 -0.0047 11.0000 -0.0471 -0.0087 0.9717 0.0020 -0.0014 -0.0005 12.0000 -0.0471 -0.0087 0.9717 0.0002 -0.0002 0.0011 13.0000 -0.0471 -0.0087 0.9717 -0.0005 0.0003 0.0002 14.0000 -0.0471 -0.0087 0.9717 -0.0001 0.0000 -0.0003 S1 = 2.1843 + 1.4085i Q3 = 1.4617 Chapter 7 Optimal Dispatch of Generation Part I 7.1 Introduction When generators are interconnected with various loads, the generation capacity is usually much larger than the loads, hence the allocation of loads on the generators can be varied. It is important to reduce the cost of electricity as much as possible, hence the "optimal" division of load (meaning the least expensive) is desired. The most expensive cost of generation is the fuel cost. Other expenses include labor, repairs and maintenance. These are "economical" factors. Other factors (not considered in this presentation) include such factors as security, stability and aesthetics. 7.2 Nonlinear Function Optimization Unconstrained Parameter Optimization The necessary condition for the function ( ) ( ) 1 2 , , , n f x f x x x · K to have a minimum is obtained by setting the derivatives 0, 1,2, , i f i n x ∂ · · ∂ K . Or using the ∇ operator, 0 f ∇ · where 1 2 , , , n f f f f x x x ¸ _ ∂ ∂ ∂ ∇ · ∂ ∂ ∂ ¸ , K is known as the gradient of f . The term H associated with second derivative: 2 i j f H x x ∂ · ∂ ∂ results in a symmetric matrix called the Hessian matrix of the function ( ) f x . Suppose ( ) ˆ 0 f x ∇ · where ( ) 1 2 ˆ ˆ ˆ ˆ , , , n x x x x · K is a local extremum. For this to be a minimum, the Hessian matrix should be positive definite. This can be checked by finding the eigenvalues of the Hessian matrix, all of which should be positive at ˆ x . In the process of doing this, two functions from Matlab will be very useful, they are shown below. If x Ab · , then the solution is found in Matlab as \ x A b · . Also, the eigenvectors of H are found using the Matlab function eig(H). Example 7.1 Find the minimum of ( ) 2 2 2 1 2 3 1 2 2 3 1 2 3 2 3 8 16 32 110 f x x x x x x x x x x x · + + + + − − − + . Finding the first derivatives and equate to zero results in three linear algebraic equations thus: 2 1 2 1 1 2 3 2 2 3 3 2 8 0 4 16 0 6 32 0 f x x x f x x x x f x x x ∂ · + − · ∂ ∂ · + + − · ∂ ∂ · + − · ∂ or: 1 2 3 2 1 0 8 1 4 1 16 0 1 6 32 x x x 1 1 1 1 1 1 · 1 1 1 1 1 1 ¸ ] ¸ ] ¸ ] Using Matlab we have: A=[2 1 0; 1 4 1; 0 1 6]; b=[8;16;32]; x=A\b eigenvalues=eig(A) x = 3.0000 2.0000 5.0000 eigenvalues = 1.5505 4.0000 6.4495 Thus the eigenvalues are all positive, thus x=(3, 2, 5) is a minimum (in this case a global minimum since there is only one extremum). 7.2.1 Constrained Parameter Optimization: Equality Constraints Here we minimize ( ) f x subject to the constraints ( ) 0, 1,2, , i g x i k · · K . Such problems can be solved using Lagrange multipliers i λ thus we let 1 k i i i f g λ · · + ∑ L . The necessary conditions for constrained local minima of L are the following: 1 0 0 k i i i i i i i i f g x x x g λ λ · ∂ ∂ ∂ · + · ∂ ∂ ∂ ∂ · · ∂ ∑ L L Note that the last equation is the same as the original constraints. Example 7.2 3 Use the Lagrange multiplier method for solving constrained parameter optimizations to determine the minimum distance from the origin of the xy-plane to a circle described by ( ) ( ) 2 2 8 6 25 x y − + − · . The distance to be minimized is given by ( ) 2 2 , f x y x y · + . First we use Matlab to show a graphical display of the two curves from which it is clear that the answer is at (4, 3). Note there is also a maximum at (12, 9). wt=0:.01:2*pi; z =8+j*6+ 5*(cos(wt) + j*sin(wt)); x=0:.01:12; y=6/8*x; plot(real(z),imag(z), x, y), grid xlabel('x'), ylabel('y') axis([0 , 14, 0, 14]) 0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14 x y Now we want to minimize ( ) , f x y subject to the constraints described by the circle equation. We first form the Lagrange function thus: ( ) ( ) 2 2 2 2 8 6 25 x y x y λ 1 · + + − + − − ¸ ] L The necessary conditions for an extremum are found as 4 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 16 0 or 2 1 16 2 2 12 0 2 1 12 8 6 25 0 x x x x y y or y y x y λ λ λ λ λ λ λ ∂ · + − · + · ∂ ∂ · + − · + · ∂ ∂ · − + − − · ∂ L L L Dividing the first two equations yields 3 4 y x · . Using this in the third equation yields 2 25 25 75 0 16 x x − + · Solving the quadratic gives 4 x · and 12 x · . Thus the extrema are at the points (4, 3) with 1 λ · , and (12, 9) with 3 λ · − . From the figure it is clear that the first point is the minimum. In many situations the equations are not only nonlinear but also cannot be solved in an analytical manner, thus the solution has to be obtained by iteration. The most common iteration method is the Newton-Raphson method. This is done below for the above problem. Solve the first two equations for x and y in terms of λ : 8 1 6 1 x y λ λ λ λ · + · + Using these values in the third equation yields: ( ) ( ) ( ) ( ) 2 2 100 200 , 75 0 1 1 f f x y λ λ λ λ λ λ λ · · − + · 1 ¸ ] + + This equation can be solved by iteration as follows using the Newton-Raphson method. Using the first order term in a Taylor series expansion we have: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 and k k k k k k f df d λ λ λ λ λ λ + −∆ ∆ · · + ∆ ¸ _ ¸ , where ( ) ( ) k f λ ∆ is the residual at ( ) k λ given by ( ) ( ) ( ) ( ) ( ) , k k k f f x y λ · . If this residual is zero, the solution is exact, otherwise we would like the residual to satisfy a very small convergence value. The iteration is started with an estimated value of λ and continued till a specified accuracy. Once λ is known, x and y can be computed from the equations above. One disadvantage to this method is the need to find the derivative of the function which in this case is found to be: ( ) ( ) ( ) ( ) 3 2 3 200 200 200 1 1 1 df d λ λ λ λ λ λ − · − · + + + 5 Having a value for (0) λ using the above equation we have (0) f λ ∂ ¸ _ ∂ ¸ , . Now we can compute (0) x and (0) y hence we have a value for use in: ( ) ( ) ( ) 2 2 (0) ( 0) (0) 8 6 25 f x y λ ∆ · − + − − and now we can find ( ) ( ) ( ) 0 (0) 0 f df d λ λ λ −∆ ∆ · ¸ _ ¸ , . Thus we have a new value for (1) λ . The process is repeated till the the error in evaluating f ∆ is less than some prescribed value. The following Matlab program will perform the iteration: clear; iter = 0; % Iteration counter Df = 10; % Error in Df is set to a high value Lambda = 0.4; % Initial estimated value of Lambda disp([' Iter Df J DLambda Lambda x y']) while abs(Df) >= 0.0001 % Test for convergence iter = iter + 1; % No. of iterations x = 8*Lambda/(Lambda + 1); y = 6*Lambda/(Lambda + 1); Df = (x- 8)^2 + (y - 6)^2 - 25; % Residual J = 200*Lambda/(Lambda + 1)^3 - 200/(Lambda + 1)^2; % Derivative Delambda =-Df/J; % Change in variable disp([iter, Df, J, Delambda, Lambda, x, y]) Lambda = Lambda + Delambda; % Successive solution end Iter Df J DLambda Lambda x y 1.0000 26.0204 -72.8863 0.3570 0.4000 2.2857 1.7143 2.0000 7.3934 -36.8735 0.2005 0.7570 3.4468 2.5851 3.0000 1.0972 -26.6637 0.0411 0.9575 3.9132 2.9349 4.0000 0.0337 -25.0505 0.0013 0.9987 3.9973 2.9980 5.0000 0.0000 -25.0001 0.0000 1.0000 4.0000 3.0000 We repeat the program with an initial value of 2 λ · − : clear; iter = 0; % Iteration counter Df = 10; % Error in Df is set to a high value Lambda = -2; % Initial estimated value of Lambda disp([' Iter Df J DLambda Lambda x y']) while abs(Df) >= 0.0001 % Test for convergence iter = iter + 1; % No. of iterations x = 8*Lambda/(Lambda + 1); y = 6*Lambda/(Lambda + 1); Df = (x- 8)^2 + (y - 6)^2 - 25; % Residual J = 200*Lambda/(Lambda + 1)^3 - 200/(Lambda + 1)^2; % Derivative Delambda =-Df/J; % Change in variable 6 disp([iter, Df, J, Delambda, Lambda, x, y]) Lambda = Lambda + Delambda; % Successive solution end Iter Df J DLambda Lambda x y 1.0000 75.0000 200.0000 -0.3750 -2.0000 16.0000 12.0000 2.0000 27.8926 76.9346 -0.3625 -2.3750 13.8182 10.3636 3.0000 8.1227 38.1258 -0.2131 -2.7375 12.6042 9.4531 4.0000 1.2823 26.9480 -0.0476 -2.9506 12.1013 9.0760 5.0000 0.0454 25.0682 -0.0018 -2.9982 12.0036 9.0027 6.0000 0.0001 25.0001 -0.0000 -3.0000 12.0000 9.0000 Now the solution converges to the "maximum" distance at (12, 9). 7.2.2 Constraint Parameter Optimization: Inequality Constraints In practice both equality and inequality constraints are needed. The problem is to minimize the cost function ( ) 1 2 , , , n f x x x K subject to the equality constraints ( ) 1 2 , , , 0 1,2, , i n g x x x i k · · K K and inequality constraints ( ) 1 2 , , , 0 1,2, , j n u x x x j m ≤ · K K The Lagrange multipliers are extended to include the inequality constraints using the multipliers µ as shown below: 1 1 k m i i j j i j f g u λ µ · · · + + ∑ ∑ L The resulting necessary conditions for constrained local minima of L are the following: 0 1, , 0 1, , 0 1, , 0 & 0 1, , i i i j j j j j i n x g i k u i m u i m λ µ µ µ ∂ · · ∂ ∂ · · · ∂ ∂ · ≤ · ∂ · > · K K K K L L L Note that the second equation above is the equality constraints and the last two equations pertain to the inequality constraints. Suppose ( ) 1 2 ˆ ˆ ˆ , , , n x x x K is a relative minimum. The inequality constraint (third equation above) is said to be inactive if strict inequality holds at ( ) 1 2 ˆ ˆ ˆ , , , n x x x K and 0 j µ · . On the other hand, when strict equality holds, the 7 constraint is active at this point (i.e. if the constraint ( ) 1 2 ˆ ˆ ˆ , , , 0 j j n u x x x µ · K and 0 j µ > ). This is known as the Kuhn-Tucker necessary condition. Example 7.3 Solve example 7.2 with an additional inequality constraint defined below. The problem is to find the minimum of the function ( ) 2 2 , f x y x y · + subject to one equality constraint ( ) ( ) ( ) 2 2 , 8 6 25 0 g x y x y · − + − − · and one inequality constraint ( ) , 2 12 u x y x y · + ≥ . The unconstrained cost function is L as shown below: ( ) ( ) ( ) 2 2 2 2 8 6 25 2 12 x y x y x y λ µ 1 · + + − + − − + + − ¸ ] L The resulting necessary conditions for constrained local minima of L are ( ) ( ) ( ) ( ) 2 2 2 2 8 2 0 2 2 6 0 8 6 25 0 2 12 0 x x x y y y x y x y λ µ λ µ λ µ ∂ · + − + · ∂ ∂ · + − + · ∂ ∂ · − + − − · ∂ ∂ · + − · ∂ L L L L Eliminating µ from the first two equations results in ( )( ) 2 4 1 8 0 x y λ λ − + + · From the fourth condition we have 12 2 y x · − Now substituting for y in the previous equation yields 4 4.8 1 x λ λ + · + And substituting for x in the fourth condition we get 4 2.4 1 y λ λ + · + Substituting for x and y in the third condition (equality constraint) results in an equation in terms of λ : 2 2 4 4.8 4 2.4 8 6 25 0 1 1 λ λ λ λ + + ¸ _ ¸ _ − + − − · + + ¸ , ¸ , from which we have the following equation 2 2 0.36 0 λ λ + + · Roots of this equation are at 0.2 λ · − and 1.8 λ · − . Substituting for these values of λ in the expressions for x and y, the corresponding extrema are 8 ( ) ( ) ( ) ( ) , 5,2 for 0.2, 5.6 , 3,6 for 1.8, 12 x y x y λ µ λ µ · · − · − · · − · − Using Matlab these are graphically depicted below. Note that the solution is valid above the line shown in the figure below, thus only two points satisfy the solution for extrema, one is a minimum, the other a maximum. These are the points where the circle and the line intersect, which are exactly the two points found above. The graphical demonstration below further illustrates this example. %Graphical Demonstration of Example 7.3 wt=0:.01:2*pi; z =8+j*6+ 5*(cos(wt) + j*sin(wt)); x=0:.01:12; y=6/8*x; y2=12-2*x; plot(real(z),imag(z), x, y, x, y2), grid xlabel('x'), ylabel('y') axis([0 , 14, 0, 14]) a =[1 2 .36]; Lambda = roots(a) X=(4*Lambda+4.8)./(1+Lambda) Y=(4*Lambda+2.4)./(1+Lambda) F=sqrt(X.^2+Y.^2) Mindist=min(F) Lambda = -1.8000 -0.2000 X = 3.0000 5.0000 Y = 6.0000 2.0000 F = 6.7082 5.3852 Mindist = 5.3852 9 0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14 x y Chapter 7 Optimal Dispatch of Generation Part II 7.3 Operating Cost of a Thermal Plant The major cost of plant operation is fuel. Other costs may be added to the fuel-cost and the sum referred to as the fuel cost. Usually the fuel cost is in $/h and is a function of the power generation in MW. The fuel cost curve is usually assumed to be parabolic of the form: 2 i i i i i i C P P α β γ = + + . The incremental fuel-cost curve is slope of the fuel-cost curve defined by 2 i i i i i dC P dP γ β = + . The incremental fuel-cost curve indicates how expensive it will be to generate the next increment of power. Units: i C is in $/h and i i dC dP is in $/MWh. Typical curves are shown below. 2 clear P1=0:1:500; C1=500+5.3*P1+0.004*P1.^2; IC1=5.3+2*0.004*P1; subplot(2,1,1),plot(P1,C1),grid title('Plot of fuel-cost'),xlabel('Power, MW'),ylabel('$/h') subplot(2,1,2),plot(P1,IC1),grid title('Plot of incremental fuel-cost'), xlabel('Power, MW'),ylabel('$/MWh') 0 100 200 300 400 500 0 1000 2000 3000 4000 5000 Plot of fuel-cost Power, MW $/h 0 100 200 300 400 500 5 6 7 8 9 10 Plot of incremental fuel-cost Power, MW $ / MWh 3 7.4 Economic Dispatch Neglecting Losses and Generator Limits The simplest economic allocation is that of power generators on the same bus (or in theory using lossless lines). Suppose we have n generators connected to a bus and supplying a load demand D P . We would like to minimize the total cost of generation ( ) 1 2 1 ng t i i n i i i i i i C C P P α β γ = = = = + + ∑ ∑ subject to the constraint 1 ng i D i P P = = ∑ This situation is shown in the figure to the right. It is noted that the number of "dispatchable" generators is ng while the total number of generators is n. This means n ng − generators are not subject to economic allocation (they may be on a flat load, such as the case of nuclear generators, or they may have reached a maximum or minimum limit of generation). As usual we now form the function L: 1 ng t D i i C P P λ =   = + −       ∑ L The necessary conditions for a minimum are: 0 0 i P λ ∂ = ∂ ∂ = ∂ L L The first condition results in ( ) 0 1 0 t i C P λ ∂ + − = ∂ Also we know that 1 2 t ng C C C C = + + + L , therefore t i i i C dC P dP λ ∂ = = ∂ Thus the condition for optimal dispatch is P 1 C 1 P D P ng C ng P 2 C 2 4 1, , i i dC i ng dP λ = = K or 2 i i i P β γ λ + = And the second condition above simply states 1 ng i D i P P = = ∑ which is the same as the constraint stated earlier. In English: Assuming no limits on generation, and no losses, then the most economical state of operation is when all the plants have the same incremental cost λ , and their total generation equals the load demand. Based on this we now can find λ for optimal generation. From the equation above we have 2 i i i P λ β γ − = The relations given by this equation are called the coordination equations. Using this equation in the constraint equation we have 1 2 ng i D i i P λ β γ = − = ∑ and solving for λ we have 1 1 2 1 2 ng i D i i ng i i P β γ λ γ = = + = ∑ ∑ Knowing the value of λ we can compute the powers using 2 i i i P λ β γ − = . In the case of lossless power systems (i.e. generators on a common bus) it was easy to analytically solve the optimal dispatch problem. In the general case where losses are significant and the generators are geographically dispersed, the solution cannot be found analytically. In such cases an iterative solution is employed. Such a solution is detailed below for the case of lossless systems (even though this can be performed analytically in this case, we do it in order to gain experience for the next case when iteration is a must). We employ the gradient method to solve the optimization problem by iteration. The constraint equation is written as follows: ( ) 1 2 ng i D i i f P λ β λ γ = − = = ∑ Taking the first order terms of the Taylor series expansion about the point ( ) k λ gives ( ) ( ) ( ) ( ) ( ) k k k D df f P d λ λ λ λ   + ∆ =     or 5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 ng k k D i k k D i k k k k k i P P P f P df df df d d d P dP d λ λ λ λ λ λ λ λ λ = − − ∆ ∆ = = =                   ∆ =       ∑ ∑ or ( ) ( ) 1 2 k k i P λ γ ∆ ∆ = ∑ thus the correction is ( ) ( ) ( ) 1 k k k λ λ λ + = + ∆ where ( ) ( ) 1 ng k k D i i P P P = ∆ = − ∑ The iteration is continued till ( ) k P ∆ is smaller than a specified accuracy. Of course, as usual, a starting value ( ) 1 λ is used to start the iteration. The steps in this process are outlined below: Step 1. Assume an initial value for lambda, say ( ) 1 λ . Step 2. Using the equation ( ) ( ) 1 1 2 i i i P λ β γ − = compute all the powers for the first iterate. Step 3. Check if the sum of powers equals the demand, and find the difference thus: ( ) ( ) 1 1 1 ng D i i P P P = ∆ = − ∑ . If this difference is smaller than the specified accuracy, a solution is reached, if not, continue. Step 4. Compute the change in lambda: ( ) ( ) 1 1 1 2 i P λ γ ∆ ∆ = ∑ and compute the second estimate for lambda, namely: ( ) ( ) ( ) 2 1 1 λ λ λ = + ∆ . The process is repeated till step 3 indicates convergence… 6 Example 7.4 The fuel cost of three thermal plants in $/h are given as: 2 1 1 1 2 2 2 2 2 3 3 3 500 5.3 0.004 400 5.5 0.006 200 5.8 0.009 C P P C P P C P P = + + = + + = + + Where the powers are in MW. The total load D P is 800 MW. Neglecting line losses and generator limits, find the optimal dispatch and the total cost in $/h (a) by the analytical method (b) by graphical demonstration (c) by iteration. (a) Lambda is computed directly, thus: 1 1 2 1 2 5.3 5.5 5.8 800 0.008 0.012 0.018 8.5 $/MWh 1 1 1 0.008 0.012 0.018 ng i D i i ng i i P β γ λ γ = = + = + + + = = + + ∑ ∑ The powers can be computed directly thus: ( ) ( ) ( ) 1 2 3 8.5 5.3 400 2 0.004 8.5 5.5 250 2 0.006 8.5 5.8 150 2 0.009 P P P − = = − = = − = = It is easy to verify that the incremental costs at the three plants are equal and the sum of the powers is equal to the power demand. (b) We can show this graphically using Matlab. %Graphical Demonstration of Example 7.4 axis([0 450 6.5 10.5]); lambda=8.5; P1=250:10:450; P2 = 150:10:350; P3=100:10:250; IC1= 5.3 + 0.008*P1; IC2= 5.5 + 0.012*P2; IC3= 5.8 + 0.018*P3; Px = 0:100:400; plot(P1, IC1, P2, IC2, P3, IC3, Px, lambda*ones(1, length(Px)),'- m'),xlabel('P MW'), ylabel(' $/MWh'), grid 7 0 50 100 150 200 250 300 350 400 450 7 7.5 8 8.5 9 9.5 10 10.5 P MW $/ MWh Choosing lambda less than 8.5 would produce less power than the demand, and larger than 8.5 would produce too much power, hence it is seen that lambda = 8.5 is the solution and can be found graphically as shown above. (c) For the iterative solution, we use Matlab thus: % Iterative solution Using Newton method alpha =[500; 400; 200]; beta = [5.3; 5.5; 5.8]; gama=[.004; .006; .009]; PD=800; DelP = 10; % Error in DelP is set to a high value lambda = 6; % estimated value of Lambda disp([' Lambda P1 P2 P3 DP grad Delambda']) iter = 0; % Iteration counter while abs(DelP) >= 0.001 % Test for convergence iter = iter + 1; % No. of iterations P = (lambda - beta)./(2*gama); DelP =PD - sum(P); % Residual J = sum( ones(length(gama), 1)./(2*gama)); % Gradient sum Delambda = DelP/J; % Change in variable disp([lambda, P(1), P(2), P(3), DelP, J, Delambda]) 8 lambda = lambda + Delambda; % Successive solution end totalcost = sum(alpha + beta.*P + gama.*P.^2) Lambda P1 P2 P3 DP grad Delambda 6.0000 87.5000 41.6667 11.1111 659.7222 263.8889 2.5000 8.5000 400.0000 250.0000 150.0000 0 263.8889 0 totalcost = 6.6825e+003 The Matlab program available to do this is called "dispatch". The program returns the system lambda and optimal generation dispatch. The program requires as input at least two specified reserved variables named "cost" and "Pdt" as shown in the example below. Optionally, the program accepts four other (optional) reserved variables: "mwlimits", "B", "B0", and "B00". If the optional variables are not specified, the program assumes they are zero and solves the problem as a lossless problem with no generator limits. The program called "gencost" will compute the total generation cost. This is shown below: Example 7.5 Neglecting generator limits and system losses, use "dispatch" to obtain the optimal dispatch of generation for the thermal plants in Example 7.4. We use the following Matlab commands: cost=[500 5.3 0.004 400 5.5 0.006 200 5.8 0.009]; Pdt=800; dispatch gencost Incremental cost of delivered power (system lambda) = 8.500000 $/MWh Optimal Dispatch of Generation: 400.0000 250.0000 150.0000 Total generation cost = 6682.50 $/h The use of "mwlimits" and "B B0 B00" will be shown later as we study generator limits and lossy systems. Chapter 7 Optimal Dispatch of Generation Part III 7.5 Economic Dispatch Neglecting Losses and Including Generator Limits In the real world, generators have limits on their generation, both minimum and maximum generation. Thus we have the inequality constraints ( ) ( ) min max 1,2, , i i i P P P i ng ≤ ≤ · K which are in addition to the equality constraint 1 ng i D i P P · · ∑ Now the Kuhn-Tucker conditions compliment the Lagrangian conditions to include the inequality constraints as additional terms. The necessary conditions for optimal dispatch in lossless systems but including generator limits become: ( ) ( ) ( ) ( ) min max max min for for for i i i i i i i i i i i i i dC P P P dP dC P P dP dC P P dP λ λ λ · ≤ ≤ ≤ · ≥ · The numerical solution is the same as before. Thus for an estimated value of lambda, the powers are found for each participating generator. If the power is outside the limits of a generator, that generator is "pegged" at that limit (either Pmin or Pmax), and the generator is no longer a participating generator in the optimization of dispatched power (since its power is pegged at either the high or low limit). Example 7.6 Find the optimal dispatch and the total cost in $/h for the thermal plants of Example 7.4 when the total load is 975 MW with the following generator limits (in MW): 1 2 3 200 450 150 350 100 225 P P P ≤ ≤ ≤ ≤ ≤ ≤ Assume the initial value of ( ) 1 6 λ · . From the coordination equations we compute the powers as follows: 2 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 1 3 6 5.3 87.5 2 0.004 6 5.5 41.67 2 0.006 6 5.8 11.11 2 0.009 P P P − · · − · · − · · ( ) ( ) 1 975 87.5 41.67 11.11 834.72 P ∆ · − + + · and we compute change in lambda as ( ) ( ) ( ) ( ) 1 834.72 3.1632 1 1 1 2 0.004 2 0.006 2 0.009 λ ∆ · · + + Therefore the new value of lambda is ( ) 2 6 3.1632 9.1632 λ · + · Continuing for a second iteration we have ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 2 3 9.163 5.3 482.89 2 0.004 9.163 5.5 305.26 2 0.006 9.163 5.8 186.84 2 0.009 P P P − · · − · · − · · and ( ) ( ) 2 975 482.89 305.26 186.84 0 P ∆ · − + + · Thus the equality constraint is met, but the power of generator No. 1 exceeds its upper limit. Thus we change it from 482.89 to 450, thus: ( ) ( ) 2 975 450 305.26 186.84 32.89 P ∆ · − + + · Now we continue, but we have only two generators participating in the optimal dispatch since the first generator is "pegged" at 450: ( ) ( ) ( ) 2 32.89 0.2368 1 1 2 0.006 2 0.009 λ ∆ · · + and the new value of lambda is ( ) 3 9.1632 0.2368 9.4 λ · + · And the powers are now as follows: 3 ( ) ( ) ( ) ( ) ( ) 3 1 3 2 3 3 450 9.4 5.5 325 2 0.006 9.4 5.8 200 2 0.009 P P P · − · · − · · and ( ) ( ) 3 975 450 325 200 0 P ∆ · − + + · Thus the equality constraint is met and the above is the optimal generation dispatch. The total generation cost is found using the powers found above thus: 2 2 2 500 5.3 450 0.004 450 400 5.5 325 0.006 325 200 5.8 200 0.009 200 8,236.25 $/h t C · + × + × + + × + × + + × + × · We could solve this using the following Matlab commands: cost=[500 5.3 0.004 400 5.5 0.006 200 5.8 0.009]; mwlimits=[200 450 150 350 100 225]; Pdt=975; dispatch gencost Incremental cost of delivered power (system lambda) = 9.400000 $/MWh Optimal Dispatch of Generation: 450.0000 325.0000 200.0000 Total generation cost = 8236.25 $/h 7.6 Economic Dispatch Including Losses When the loads and generators are in a small geographical region, transmission losses may be neglected and the optimal dispatch of generation is achieved with all plants operating at equal incremental productions cost. However, in a large system over long distances, transmission losses become significant and need to be considered. We shall use Kron's loss formula which is 0 00 1 1 1 ng ng ng L i ij j i i i j i P PB P B P B · · · · + + ∑∑ ∑ 4 The coefficients ij B are called loss coefficients or B-coefficients. It is assumed —with little error— that these coefficients are constant (as long as operation is near the value where these coefficients are computed). The economic dispatch problem is to minimize the overall generation cost i C , which is a function of the plant outputs ( ) 1 2 1 ng t i i n i i i i C C P P α β γ · · · · + + ∑ ∑ Note: If all generators participate in this minimization, then n ng · and either equation above may be used. If only ng n < generators are participating, then the first equation is used. This minimization is to be performed subject to the constraint that the total generation must equal the demand plus the losses thus 1 ng i D L i P P P · · + ∑ satisfying the inequality constraints: ( ) ( ) min max 1, 2, , i i i P P P i ng ≤ ≤ · K where ( ) min i P and ( ) max i P are the minimum and maximum generating limits, respectively, for plant i. Using the Lagrange multiplier and adding additional terms to include the inequality constraints, we have ( ) ( ) ( ) ( ) ( ) ( ) max max min min 1 1 1 ng ng ng t D L i i i i i i i i i i C P P P P P P P λ µ µ · · · ¸ _ · + + − + − + − ¸ , ∑ ∑ ∑ L The constraints should be understood to mean that ( ) max 0 i µ · when ( ) max i i P P < and ( ) min 0 i µ · when ( ) min i i P P > . Thus if the power is not beyond the limits both ( ) min 0 i µ · and ( ) max 0 i µ · thus L has only the equality constraint. If a generator's power is beyond one of the limits, then that limit is the power to which the generator is "pegged", and the remaining generators participate in the opti mization of the load dispatch. These statements are clear from the necessary conditions for a minimum which are: 5 ( ) ( ) ( ) ( ) max max min min 0 0 0 0 i i i i i i i P P P P P λ µ µ ∂ · ∂ ∂ · ∂ ∂ · − · ∂ ∂ · − · ∂ L L L L The last two equations imply that the power should not go beyond its limit, and when it is within its limits then ( ) ( ) min max 0 i i µ µ · · and the Kuhn-Tucker function becomes the same as the Lagrangian one. The first equation results in 0 1 0 t L i i C P P P λ ¸ _ ∂ ∂ + + − · ∂ ∂ ¸ , Since 1 2 t ng C C C C · + + + L the above equation becomes 1,2, , ** i L i i C P i ng P P λ λ ∂ ∂ + · · ∂ ∂ K The term L i P P ∂ ∂ is known as the incremental transmission loss. The second condition is given by 1 ng i D L i P P P · · + ∑ which is precisely the equality constraint imposed by the losses and load demand. The classical form of the previous equation is 1 1,2, , 1 i L i i C i ng P P P λ ¸ _ ∂ · · ∂ ∂ − ∂ ¸ , K or 1,2, , i i i C L i ng P λ ∂ · · ∂ K where i L is known as the penalty factor of plant i and is given by 1 1 L i P P ¸ _ ∂ − ∂ ¸ , . Thus the effect of transmission losses is to introduce a penalty factor with a value that depends on the location of the plant with respect to the loads. In the above equation it is clear that the 6 most economic cost of dispatch is obtained when the incremental cost multiplied by the corresponding penalty factor are equal for all the participating plants. The incremental transmission loss can be found from the loss equation and is given by 0 1 2 ng L ij j i i j P B P B P · ∂ · + ∂ ∑ and the incremental production cost is given by 2 i i i i i dC P dP γ β · + Using these two equations in the equation marked ** (reproduced below for convenience) 1,2, , ** i L i i C P i ng P P λ λ ∂ ∂ + · · ∂ ∂ K we have 0 1 2 2 ng i i i ij j i j P B P B β γ λ λ λ · + + + · ∑ or (divide by 2λ and pull out ii β from the summation) 0 1, 1 1 * 2 ng i i ii i ij j i j j i B P B P B γ β λ λ · ≠ ¸ _ ¸ _ + + · − − ¸ , ¸ , ∑ Expanding the above equation for all plants and detailing it into full matrix form we have 1 1 11 12 1 01 1 2 2 2 21 22 2 02 1 2 0 1 1 1 2 1 ng ng ng ng ng ng ng ngng ng B B B B P P B B B B P B B B B γ β λ λ γ β λ λ γ β λ λ 1 1 + − − 1 1 1 1 1 1 1 1 + − − 1 1 1 · 1 1 1 1 1 1 1 ¸ ] 1 1 + − − 1 1 ¸ ] ¸ ] L L M M M O M M L or in short matrix form EP D · To find the optimal dispatch first an estimated guess of λ , say ( ) 1 λ is made. The simultaneous linear equations above are solved using Matlab (P = E\D). Then the iterative process is continued using the gradient method. Thus, from equation * above we have i P at the kth iteration expressed as ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 2 *** 2 k k k i i ij j k j i i k i ii B B P P B λ β λ γ λ ≠ − − − · + ∑ and using this equation in the equality constraint gives 7 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 1 1 2 2 k k k ng ng i i ij j k k j i D i L k i i i ii B B P P P P B λ β λ γ λ ≠ · · − − − · · + + ∑ ∑ ∑ or ( ) ( ) ( ) k k D L f P P λ · + Expanding the left side above in a Taylor series and keeping only first order terms results in ( ) ( ) ( ) ( ) ( ) k k k D L df f P P d λ λ λ λ ¸ _ + ∆ · + ¸ , or ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 # ng k k D i L k i k k k k k i P P P df d P df d P dP d λ λ λ λ λ λ · + − ∆ · ¸ _ ¸ , ∆ · ¸ _ ¸ , ∆ · ¸ _ ¸ , ∑ ∑ where ( ) ( ) ( ) 1 ng k k k D i L i P P P P · ∆ · + − + + ∑ and ( ) ( ) ( ) ( ) ( ) 0 2 1 1 1 2 2 k k ng ng i i ii i i ij j j i i k i i i ii B B B P P B γ β γ λ γ λ ≠ · · − − − ∂ ¸ _ · + + + ∂ ¸ , + ∑ ∑ ∑ This last equation is derived by differentiation with respect to λ of *** above and summing the result 2 recall that u v u u v v v ∂ − ∂ 1 ¸ _ ∂ · 1 ¸ , ¸ ] . Now we have ( ) ( ) ( ) 1 k k k λ λ λ + · + ∆ The process is repeated till ( ) k P ∆ is less than a specified accuracy. If we simplify the loss equations and use the simpler 8 2 1 ng L ii i i P B P · · ∑ (where we assume 00 0 and 0 ij B B · · ), the solution of the simultaneous equations simplifies to ( ) ( ) ( ) ( ) ## 2 k k i i k i ii P B λ β γ λ − · + which are known as the coordination equations and equation +++ simplifies to ( ) ( ) ( ) 2 1 1 ### 2 k ng ng i i ii i k i i i ii P B B γ β λ γ λ · · ∂ − ¸ _ · ∂ ¸ , + ∑ ∑ The iteration process remains the same using these simpler functions. Example 7.7 The fuel cost in $/h of three thermal plants of a power system are 2 1 1 1 2 2 2 2 2 3 3 3 200 7.0 0.008 180 6.3 0.009 140 6.8 0.007 C P P C P P C P P · + + · + + · + + where the powers are in MW. The plant outputs are subject to the following limits 1 2 3 10 MW 85 MW 10 MW 80 MW 10 MW 70 MW P P P ≤ ≤ ≤ ≤ ≤ ≤ For this example, assume the power loss is given by the simplified expression ( ) ( ) ( ) ( ) 2 2 2 1 2 3 0.0218 0.0228 0.0179 L pu pu pu pu P P P P · + + where the loss coefficients are specified per unit on a 100-MVA base. Determine the optimal dispatch of generation when the total system load is 150 MW. First we change the power loss equation above to actual units instead of pu: 2 2 2 1 2 3 0.000218 0.000228 0.000179 L P P P P · + + where the powers are now in MW. Assume an initial value of ( ) 1 8.0 λ · . From the coordination equations we have: 9 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 1 3 8.0 7.0 51.3136 MW 2 0.008 8.0 0.000218 8.0 6.3 78.5292 MW 2 0.009 8.0 0.000228 8.0 6.8 71.1575 MW 2 0.007 8.0 0.000179 P P P − · · + × − · · + × − · · + × The real power loss is ( ) ( ) ( ) ( ) 2 2 2 1 0.000218 51.3136 0.000228 78.5292 0.000179 71.1575 2.886 L P · + + · Since 150 MW D P · , the error is (from equation ##) ( ) ( ) 1 150 2.8864 51.3136 78.5292 71.1517 48.1139 P ∆ · + − + + · − and from equation ### : ( ) ( ) ( ) ( ) 1 3 2 2 1 2 0.008 0.000218 7.0 0.009 0.000228 6.3 2 0.008 8.0 0.000218 2 0.009 8.0 0.000228 0.007 0.000179 6.8 152.4924 2 0.007 8.0 0.000179 i i P λ · ∂ + × + × ¸ _ · + + ∂ ¸ , + × + × + × · + × ∑ From equation # we have: ( ) 1 48.1139 0.31552 152.4924 λ − ∆ · · − Hence the new estimate for λ is ( ) 2 8.0 0.31552 7.6845 λ · − · Repeating the above steps it is found at the fourth step that the change in lambda becomes very small and convergence is declared at ( ) 4 7.6789 λ λ · · And the optimal dispatch for 7.6789 λ · is given by ( ) ( ) ( ) ( ) ( ) ( ) 4 1 4 2 4 3 7.6789 7.0 51.3136 MW 2 0.008 7.6789 0.000218 7.6789 6.3 64.1317 MW 2 0.009 7.6789 0.000228 7.6789 6.8 52.4767 MW 2 0.007 7.6789 0.000179 P P P − · · + × − · · + × − · · + × The real power loss is ( ) ( ) ( ) ( ) 2 2 2 4 0.000218 35.0907 0.000228 64.1317 0.000179 52.4767 1.699 L P · + + · and the total fuel cost is ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 200 7.0 35.0907 0.008 35.0907 180 6.3 64.1317 0.009 64.1317 140 6.8 52.4767 0.007 52.4767 1592.65 $/h t C · + + + + + + + + · 10 The "dispatch" program can be used to solve example 7.7. The program is designed so the loss coefficients are specified in pu. The loss coefficients are arranged in a matrix form with the reserved name "B". The base MVA must be specified in the reserved variable name "basemva". If the base MVA is not specified, it is set to 100 MVA. The following Matlab commands are then used to solve example 7.7: cost=[200 7.0 0.008 180 6.3 0.009 140 6.8 0.007]; mwlimits=[10 85 10 80 10 70]; Pdt=150; B=[0.0218 0 0 0 0.0228 0 0 0 0.0179]; basemva=100; dispatch gencost Incremental cost of delivered power (system lambda) = 7.678935 $/MWh Optimal Dispatch of Generation: 35.0907 64.1318 52.4767 Total generation cost = 1592.65 $/h It is very impressive how much less time it takes to solve example 7.7 using the Matlab code… Example 7.8 Figure 7.7 page 295 of the text shows the one-line diagram of a 5-bus 3-generator power system. Given the loss coefficients (pu on a 100 MVA base) are as follows: [ ] [ ] 0 00 0.0218 0.0093 0.0028 0.0093 0.0228 0.0017 0.0028 0.0017 0.0179 0.0003 0.0031 0.0015 0.00030523 B B B 1 1 · 1 1 ¸ ] · · Cost functions, generator limits, and total loads are as given in example 7.7. Use "dispatch" program to find the optimal dispatch of generation. 11 The following Matlab commands are used: cost=[200 7.0 0.008 180 6.3 0.009 140 6.8 0.007]; mwlimits=[10 85 10 80 10 70]; Pdt=150; B=[0.0218 0.0093 0.0028 0.0093 0.0228 0.0017 0.0028 0.0017 0.0179]; B0=[0.0003 0.0031 0.0015]; B00=0.00030523; basemva=100; dispatch gencost Incremental cost of delivered power (system lambda) = 7.767785 $/MWh Optimal Dispatch of Generation: 33.4701 64.0974 55.1012 Total generation cost = 1599.98 $/h Chapter 9 Balanced Faults 9.1 Introduction The most common types of fault are (in order) single-line-to-ground fault, line-to-line fault, and double-line-to-ground fault. All of these are unbalanced faults. The balanced (or three phase) fault is the one when all three lines are shorted to ground. It is usually rare, but can happen. When a fault occurs it is important to isolate it by opening protec- tive breakers. To properly set the breakers, the magnitude of the fault currents need to be known. This is the major point of study in this chapter. The life of a fault can be divided into the following times which occur in sequence: 1. The sub-transient period which lasts for only a few cycles 2. The transient period which lasts for a much longer period (tens of cycles) 3. The steady state period which lasts till a major change in the transmission network takes place (like a circuit breaker opening or a line failing) The magnitude of the current in each period needs to be known so as to program breakers to operate at the right time. To do this in the balanced fault case we need to know how to construct the bus impedance matrix using the building algorithm which is detailed later on in the chapter. (It is possible to invert the Ybus matrix, but it is more expedient to "build" the Zbus directly. The author presents three programs that are useful in the solution of three phase faults. These are: Zbus=zbuild(data) and Zbus=zbuildpi(linedata, gendata, yload). The lat- ter one is compatible with the power-flow input/output files. The third program symfault(zdata, Zbus,V) is developed for systematic computation of three phase bal- anced faults of a large interconnected power system. 9.2 Balanced Three-Phase Fault During a three-phase fault, the impedance of a generator is a time varying quantity. It is d X′′ in the sub-transient period (one to four cycles), d X′ in the transient period (about 30 cycles), and the synchronous reactance d X after that. Which one to use depends on the purpose of the study. Keep in mind that the sub-transient currents can be very large due to the small size of d X′′ . Due to symmetry, the three phase currents during a symmetrical fault can be solved using ordinary circuit theory. In this case we use Thevenin's theorem to analyze the fault. If the fault has zero impedance to ground, it is called a solid fault or bolted fault (all three lines shorted to ground with zero impedance). 2 Example 9.1 The one-line diagram of a three bus system is shown below. Each generator is repre- sented by an emf behind the transient reactance. All impedances are in per unit on a 100 MVA base and for simplicity, resistances are neglected. The following assumptions are made: 1. Shunt capacitances are neglected and the system is considered on no-load. 2. All generators are running at their rated voltage and rated frequency with their emfs in phase. Determine the fault current, bus voltages, and the line currents during the fault when a balanced three-phase fault with imped- ance 0.16 pu f Z = occurs on (a) Bus 3. (b) Bus 2. (c) Bus 1. Part (a): The fault on bus 3 is simulated by connecting a fault impedance Z f = j0.16 pu as shown in the figure below. According to Thevenin's theo- rem, the change due to the short circuit is equivalent to that caused by the added source V th = V 3 (0) with all other sources shorted as shown on the right, where V 3 (0) is the pre-fault voltage at bus 3 (where the fault will occur). It is now easy to solve for the changes due to the fault. These changes are then added to the pre-fault values to get the post-fault values. The fault current at bus 3 is ( ) ( ) 3 3 33 0 f V I F Z Z = + where 33 Z is the Thevenin impedance seen at the faulted bus. To solve for this, the delta is first changed to an equivalent Y then the circuit is simplified as shown next page. Note that all values are pu. j0.1 j0.1 j0.2 j0.2 j0.4 j0.4 j0.8 1 2 3 j0.2 j0.4 j0.4 j0.8 1 2 3 j0.4 Z f = j0.16 j0.2 j0.4 j0.4 j0.8 1 2 3 j0.4 Z f = j0.16 V th =V 3 (0) I 3 (F) 3 Note that the delta is not balanced, hence the Y is not either. Simplification above goes from left to right with the result that Z 33 is equal to j0.34. Now we can compute the fault current as ( ) ( ) 3 3 33 0 1 2 0.34 0.16 f V I F j Z Z j j = = = − + + (note: due to the assumptions made in the statement of the problem, ( ) ( ) ( ) 1 2 3 0 0 0 1 pu V V V = = = ). Now we go backward through the circuit to compute the currents in the original circuit. First the current divides into two parts through generators 1 and 2 thus: ( ) 1 3 0.6 1.2 pu 0.4 0.6 G j I I F j j j = = − + and ( ) 2 3 0.4 0.8 pu 0.4 0.6 G j I I F j j j = = − + . Thus the changes in bus voltages are given by: ( ) ( ) ( ) ( ) ( )( ) 1 2 3 0 0.2 1.2 0.24 pu 0 0.4 0.8 0.32 pu 0.16 2 1.0 0.68 pu V j j V j j V j j ∆ = − − = − ∆ = − − = − ∆ = − − = − Adding these changes to the pre-fault values we have the post-fault voltages thus: ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 2 3 3 3 0 1.0 0.24 0.76 pu 0 1.0 0.32 0.68 pu 0 1.0 0.68 0.32 pu V F V V V F V V V F V V = + ∆ = − = = + ∆ = − = = + ∆ = − = The short circuit currents in the lines are found as follows: j0.2 1 2 3 j0.4 Z f = j0.16 V th I 3 (F) j0.2 j0.2 j0.1 3 Z f = j0.16 V th I 3 (F) j0.1 j0.24 3 Z f = j0.16 V th I 3 (F) j0.34 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 12 12 1 3 13 13 2 3 23 23 0.76 0.68 0.1 pu 0.8 0.76 0.32 1.1pu 0.4 0.68 0.32 0.9 pu 0.4 V F V F I F j z j V F V F I F j z j V F V F I F j z j − − = = = − − − = = = − − − = = = − Part (b). The pre-fault circuit is shown below followed by the Thevenin equivalent then the reduced circuits. These are shown from left to right: The Thevenin impedance Z 22 at the fault point is given by j0.24 as shown in the figure on the right above. The fault current is given by: ( ) ( ) 2 2 22 0 1.0 2.5 pu 0.24 0.16 f V I F j Z Z j j = = = − + + This current divides between the two generators as follows: ( ) ( ) 1 2 2 2 0.4 1.0 pu 0.4 0.6 0.6 1.5 pu 0.4 0.6 G G j I I F j j j j I I F j j j = = − + = = − + Now we find the changes due to the fault as: ( )( ) ( ) ( ) ( ) 1 2 3 0 0.2 1.0 0.2 0 0.4 1.5 0.6 1.0 0.2 0.4 0.4 2 V j j pu V j j pu j V j pu ∆ = − − = − ∆ = − − = − −   ∆ = − − = −     By superposition we find the post-fault voltages thus: j0.2 j0.4 j0.4 j0.8 1 2 3 j0.4 Z f = j0.16 j0.2 j0.4 j0.4 j0.8 1 2 3 j0.4 Z f = j0.16 I 2 (F) V th j0.2 j0.4 1 2 j0.4 Z f = j0.16 I 2 (F) V th 2 j0.24 Z f = j0.16 I 2 (F) V th 5 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 2 3 3 3 0 1.0 0.2 0.8 pu 0 1.0 0.6 0.4 pu 0 1.0 0.4 0.6 pu V F V V V F V V V F V V = + ∆ = − = = + ∆ = − = = + ∆ = − = And the short circuit current in the lines are found to be: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 12 12 1 3 13 13 3 2 32 23 0.8 0.4 0.5 pu 0.8 0.8 0.6 0.5 pu 0.4 0.6 0.4 0.5 pu 0.4 V F V F I F j z j V F V F I F j z j V F V F I F j z j − − = = = − − − = = = − − − = = = − Part (c) is similar, and the three currents would be equal to one another but equal to 0.3125 j − instead of –j0.5. Please go over part (c) carefully… In this example the voltages were assumed equal to 1 pu. For more accurate calculations, a load flow study is made and the pre-fault voltages are accurately obtained. Also, the example above did not have any loads at the buses. If there were any loads they would be represented as constant impedances to ground. This is a good approximation (the loads change after a fault, but that may be neglected). The general process is summarized below: • The pre-fault bus voltages are obtained from the results of a load flow analysis. • Loads are converted to constant impedances to ground using the bus voltages. • The faulted network is reduced into a Thevenin equivalent circuit viewed from the faulted bus. Applying Thevenin's theorem, changes in bus voltages are obtained. • Bus voltages during the fault are obtained by superposition of the prefault bus volt- ages and the changes in the bus voltages computed in the previous step. • The currents during the fault may now be obtained in all branches of the faulted sys- tem. 6 9.3 Short-Circuit Capacity (SCC) The short-circuit capacity of a bus is a measure of the strength of a bus. The SCC is de- fined as the product of the magnitudes of the rated bus voltage and the fault current. It is used to determine the size of the bus bar and to size the breaker used at the bus. From the above definition we have for the short-circuit MVA at bus k: ( ) 3 3 10 MVA Lk k SCC V I F − = × where the line-to-line voltage is V Lk is in kV and I k (F) in A. The symmetrical three- phase current in per unit is given by ( ) ( ) 0 k k pu kk V I F X = where ( ) 0 k V is the per unit prefault bus voltage and kk X is the per unit reactance at the point of fault. N.B. System resistance is neglected, thus the current computed is on the pessimistic (larger) side. The base current is given by 3 10 3 B B B S I V × = where B S is the base MVA and B V is the line-to-line base voltage in kV. Thus the fault current in ampere is ( ) ( ) ( ) 3 0 10 3 k k B pu k B kk B I F I F I V S X V = × = Thus we have for the SCC: ( ) 0 k B Lk kk B V S V SCC X V = If we assume the base voltage is equal to the line voltage (i.e. Lk B V V = ) then ( ) 0 k B kk V S SCC X = The prefault bus voltage is usually assumed 1.0 pu, thus we have the following approxi- mate equation for the short-circuit capacity or the short-circuit MVA B kk S SCC X ≈ Chapter 9 Balanced Faults, Part II 9.4 Systematic Fault Analysis Using Bus Impedance Matrix In the previous analysis we employed the Thevenin model and found the Thevenin volt- age and impedance by means of circuit manipulation. That method works for small power circuits only: It is impractical for a real large power system. A systematic method is presented in this section which applies to power systems of any size. The method can also be programmed on a computer. Consider a typical bus of an n-bus power system network as shown below. Here we show buses i and k. The generator shown is a voltage source behind a reactance which may be , , or d d d X X X ′′ ′ . Transmission lines are represented by their equivalent π model and all impedances are pu on a common MVA base. A balanced three-phase fault is to be applied at bus k through an impedance f Z . The prefault voltages are obtained from a power flow solution and represented by the vector ( ) ( ) ( ) ( ) 1 0 0 0 0 k bus n V V V V         =         M M (1.1) i k S i S k Z f 2 Generally the short circuit currents are so large compared to the steady state loads, the loads may be neglected. However, the loads can be represented as impedances using the above voltages for example, the load S i may be approximated by the impedance ( ) 2 * 0 i i i V Z S = (1.2) The changes in network voltage due to the fault are found from the system with all volt- ages shorted to ground and the prefault voltage ( ) 0 k V applied as shown in the figure be- low: Note the capacitors for the pi equivalent of line ik. Also note the loads are replaced by the approximation of impedances. The voltage source at i is grounded and its impedance remains as shown above. The bus voltage changes caused by the fault in this circuit are represented by the column vector 1 k bus n V V V V ∆       ∆   ∆ =       ∆   M M (1.3) From Thevenin's theorem the voltages during the fault are found from ( ) ( ) 0 bus bus bus V F V V = + ∆ (1.4) We also know for the nodal equations that bus bus bus I Y V = (1.5) where bus I is the bus current entering the bus and bus Y is the bus admittance matrix. The diagonal elements of each bus is the sum of the admittances connected to that bus, i.e. 0 m ii ij j Y y j i = = ≠ ∑ (1.6) i k Z f I k (F) V th =V k (0) Z k Z i 3 The off-diagonal element is equal to the negative of the admittance between the nodes, i.e. ij ji ij Y Y y = = − (1.7) where ij y is the actual admittance between nodes i and j. For the Thevenin circuit above, the nodal equations are ( ) 1 11 1 1 1 1 0 0 k n k k kk kn k n nk nn n V Y Y Y V Y Y Y I F Y Y Y V ∆                   ∆   =   −                     ∆       L L M M M M M M M L L M M M M M M M L L (1.8) Note that the minus sign is due to the fact the fault current is shown leaving node k. The above matrix equation can be written as ( ) bus bus bus I F Y V = ∆ (1.9) which can be solved for the voltage change thus ( ) bus bus bus V Z I F ∆ = (1.10) where 1 bus bus Z Y − = is known as the bus impedance matrix (not to be mixed-up with the impedance matrix in mesh equations of circuit theory!) Using (1.10) in (1.4) we have: ( ) ( ) ( ) 0 bus bus bus bus V F V Z I F = + (1.11) Which can be expanded into matrix form: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 11 1 1 1 1 0 0 0 0 0 k n k k k kk kn k n nk nn n n V F V Z Z Z V F V Z Z Z I F Z Z Z V F V                             = +   −                                   L L M M M M M M M M L L M M M M M M M M L L (1.12) Using the kth equation we have: ( ) ( ) ( ) 0 k k kk k V F V Z I F = − (1.13) Also it is clear that ( ) ( ) k f k V F Z I F = (1.14) From (1.13) and (1.14) we have: ( ) ( ) 0 k k kk f V I F Z Z = + (1.15) Thus we can solve for any bus voltage using (1.12) for row i: ( ) ( ) ( ) 0 i i ik k V F V Z I F = − (1.16) And using (1.15) in (1.16) we have: ( ) ( ) ( ) 0 0 k i i ik kk f V V F V Z Z Z = − + (1.17) 4 Knowing all the bus voltages during a fault we can solve for all the fault currents, in par- ticular, for line current from bus i to bus j we have: ( ) ( ) ( ) i j ij ij V F V F I F z − = (1.18) It is noted that this method works for any bus under fault (i.e. the faulted bus k could be any one bus of the buses in the system). [Note that ij z is the impedance between bus i and bus j. It is not the i-j-th element of bus Z .] The method suggested above to find the bus impedance matrix by inversion of the bus admittance matrix is not feasible for very large power systems. An alternate method for the direct formation (or building) of the matrix bus Z will be discussed in the next section. Example 9.2 A three-phase fault with a fault impedance of 0.16 pu f Z j = occurs at bus 3 in the network for Example 9.1. Using the bus impedance matrix method, compute the fault current, the bus voltages, and the line currents during the fault. First the network in Example 9.1 is redrawn to the right us- ing impedances. By inspection we can find the bus Y thus: 8.75 1.25 2.5 1.25 6.25 2.5 2.5 2.5 5 bus j j j Y j j j j j j −     = −   −     Note that this is the bus Y before the fault occurs. By inver- sion using Matlab we have: Ybus = j*[-8.75 1.25 2.5 1.25 -6.25 2.5 2.5 2.50 -5.0]; Zbus = inv(Ybus) Zbus = 0 + 0.1600i 0 + 0.0800i 0 + 0.1200i 0 + 0.0800i 0 + 0.2400i 0 + 0.1600i 0 + 0.1200i 0 + 0.1600i 0 + 0.3400i From equation (1.15) we have: Zf = j*.16; V0=[1; 1; 1]; I3F = V0(1)/(Zbus(3,3)+Zf) I3F = -j5 -j2.5 -j2.5 -j1.25 1 2 3 -j2.5 Z f = j0.16 V th =V 3 (0) I 3 (F) 5 0 - 2.0000i and from (1.16) we have: VF = V0-I3F*Zbus(:,3) VF = 0.7600 0.6800 0.3200 Finally from (1.17) we have: z12 = j*.8; z13 = j*.4; z23 = j*.4; % I12 = (VF(1) - VF(2))/z12 I13 = (VF(1) - VF(3))/z13 I23 = (VF(2) - VF(3))/z23 I12 = 0 - 0.1000i I13 = 0 - 1.1000i I23 = 0 - 0.9000i These results are identical to those found earlier in Example 1. Note that the need for the repeated simplification to find the Thevenin impedance is removed, but a matrix inver- sion is introduced. This inversion will also be eliminated by the direct building method of the bus impedance matrix to be discussed in the next section. Chapter 9 Balanced Faults, Part III 9.5 Algorithm for Formation of the Bus Impedance Matrix Using this algorithm we shall build the Impedance Matrix bus Z starting with a single element and adding elements one by one till all the elements of the system are accounted for in the Impedance Matrix. Some definitions from graph theory are introduced below. The system shown in (a) above has a graph shown in (b). Each bus becomes a node or a corner in the graph. The datum, or ground bus is the node number 0 in the graph. Elements in the power system become edges or elements of the graph. For example, the line from bus 1 to bus 2 whose impedance is j0.8 Ohm becomes the element 4 in the graph. A tree of a connected graph is a connected subgraph connecting all the nodes without forming a loop. The elements of a tree are called branches. A graph contains multiple trees. The number of branches in a tree is always one less than the number of nodes, i.e. 1 b n = − (1.1) where n is the number of nodes and b the number of branches in the graph. Once a tree of a graph is determined, the remaining branches are called links. In figure (c) above one such tree is shown using solid lines and the links are shown with dashed lines. The collection of links is called a cotree. If e is the total number of elements in a graph, then the number of links l in a cotree is 1 l e b e n = − = − + (1.2) A loop that contains one link is called a basic loop. The number of basic loops is unique: it is equal to the number of links which is also equal to the number of independent loop equations the system has. A cut set is a minimal set of elements which when cut will divide the graph into two connected subgraphs. A fundamental cut set is a cut set which contains only one branch. The number of fundamental cut sets is unique: it is equal to the number of branches which is also equal to the number of independent node equations the system has. j0.2 j0.4 j0.4 j0.8 1 2 3 j0.4 0 1 2 3 2 1 3 4 5 0 1 2 3 2 1 3 4 5 (a) (b) (c) 2 The bus impedance matrix can be built up starting with a single element and the process is to add an element at a time till all the elements of the system are included. Suppose we have a partial network as shown on the right with a total of m buses (in addition to the ground or reference bus). The corresponding equation for this partial network is m m m bus bus bus V Z I = . Here the size of m bus Z is m m × . It is assumed that the network has n buses where n m ≥ . Next we add branches or elements, one by one, till all the elements of the network are accounted for. The added element may be a branch or a link described as follows. Addition of a Branch A branch can be added in one of two ways: either as a link between two existing nodes, or as a new branch with a new node. We consider the latter case now (the first case is considered next). The addition of a new node and branch is shown in the figures below. First consider (a) above where the new branch is not connected to the reference. The existing bus impedance matrix is of size m m × and after the addition of the branch p q − its size will be ( ) ( ) 1 1 m m + × + . The impedance of the added branch is pq z . The network equations become: 11 1 12 1 1 1 1 21 22 2 2 2 2 1 2 2 1 2 1 2 | | | | | | | q q pq p m p m p p p p pp pm m m m m m mq p m q q q qp q q m m q Z Z Z Z V Z Z Z Z Z Z Z Z Z V I Z Z Z Z V I V I Z Z Z Z V I Z Z Z Z                         =                 − − − − − − − − − − − − − − − − − − −               L L L L M M M M M M M M L L M M M M M L M M L M L L q I                     −       (1.3) Partial Network Z m bus 1 2 i m 0 . . . Reference . . . Partial Network Z m bus 1 2 p m 0 . . . Reference . . . Partial Network Z m bus 1 2 p m 0 . . . Reference . . . (a) (b) q q 3 Note: black is the old equation, red is due to the added branch. The new elements are computed one by one as follows: first compute the elements qi Z by connecting a current source i I whose current is 1 (unity) and let all other currents be 0 (zero). Expanding the equations in the matrix above we have: 1 1 2 2 i i p pi m mi q qi V Z V Z V Z V Z V Z = = = = = M M (1.4) From the figure above q p pq V V v = − (1.5) where pq v is the voltage across the added branch whose impedance is pq z , therefore pq pq pq v z i = (1.6) Since there is no current in the added branch, 0 pq i = , hence 0 pq v = and (1.5) becomes q p V V = (1.7) And from (1.4) we conclude that qi pi Z Z = (1.8) This means that all the off-diagonal terms in row q are the same as those in row p. Thus we have found all the red terms except for the term qq Z . Now we inject a current at the qth bus, i.e. 1 q I = and all other currents set to zero, we get from (1.3) q qq V Z = (1.9) Note that since the current is injected into the node q, we have 1 pq q i I = − = − thus from (1.6) we have pq pq v z = − (1.10) And using this in (1.5) we have q p pq V V z = + (1.11) In this case the injected current is at node q, thus setting i q = in (1.4) we have q qq p pq V Z and V Z = = hence (1.11) becomes qq pq pq Z Z z = + (1.12) Next we consider the case of figure (b) where the new node and added branch are connected to the reference node. Clearly this addition is a simple one whose equation is (by inspection of figure (b) above) q pq q V z I = (1.13) 4 Therefore we conclude that all the new elements would be zero except the diagonal element whose value is qq pq Z z = . Addition of a Link A link can be added in two ways to two existing nodes as shown below The impedance is connected either between two existing buses as in (c), or between an existing bus and ground as in (d). In either case the number of nodes is not changed, hence the size of the bus impedance matrix must remain the same. First consider (c) above. Let I l be the current in the link pq in the direction shown in the figure. Then pq p q z I V V = − l (1.14) or 0 q p pq V V z I − + = l (1.15) As seen in the figure, the addition of the link modifies the old currents by changing the old current p I to ( ) p I I − l and the old current ( ) to q q I I I + l . This changes the old equations as follows ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 11 1 1 1 1 1 1 1 1 1 1 p p q q m m p p pp p pq q pm m q q qp p qq q qm m m m mp p mq q mm m V Z I Z I I Z I I Z I V Z I Z I I Z I I Z I V Z I Z I I Z I I Z I V Z I Z I I Z I I Z I = + + − + + + + = + + − + + + + = + + − + + + + = + + − + + + + l l l l l l l l L L M L L L L M L L (1.16) And using the values of and p q V V from (1.16) in (1.15) we have one more equation: ( ) ( ) ( ) ( ) ( ) 1 1 2 0 q p qp pp qq pq qm pm pq pp qq pq Z Z I Z Z I Z Z I Z Z I z Z Z Z I − + + − + + − + + − + + + − = l L L L (1.17) And now combining the above equation into matrix form we have Partial Network Z m bus 1 p m 0 . . . Reference . . . Partial Network Z m bus 1 m 0 . . . Reference . . . (c) (d) q p q 5 11 1 1 1 1 1 1 1 1 1 1 0 p q p q m p pp pq pm p p q qp qq qm q q m p m mp m q m m q m m m Z Z Z Z I V Z Z Z Z Z I Z Z Z Z I V Z Z Z Z I V Z Z Z Z Z Z Z Z I V                               =                                           l l l l l l l l l ll L L M M M M M M M M M L L L L M M M M M M M M M L L L L (1.18) where i i iq ip Z Z Z Z = = − l l (1.19) and 2 pq pp qq pq Z z Z Z Z = + + − ll (1.20) Now we partition the matrix equation (1.18) above using submatrices: 0 old bus bus bus T Z Z I V I Z Z   ∆     =         ∆       l ll (1.21) Where each part is clear from the colors in (1.18). Expanding the second equation in (1.21) we have 0 T bus Z I Z I = ∆ + ll l (1.22) From which we have 0 T bus Z I Z I = ∆ + ll l (1.23) and using (1.23) in the first equation in (1.21) we have T old bus bus bus Z Z V Z I Z   ∆ ∆ = −     ll (1.24) or new bus bus bus V Z I = (1.25) where T new old bus bus Z Z Z Z Z ∆ ∆ = − ll (1.26) It is clear that the new bus impedance matrix is of the same size as the old bus impedance matrix as stated earlier. This is due to the fact that we did not add a node, we added a link which does not change the number of equations. In general the bus impedance matrix can be constructed by adding links, nodes and branches in any order. However, it is best to start by selecting a tree that contains the elements connected to the reference node. If more than one element connect a node p to the reference node, only one element can be selected as a branch leaving the other elements in the cotree. The step by step procedure for building the bus impedance matrix by adding elements is described below. 6 The case where the added link is between node p and ground will be discussed under rule 3 below. Rule 1: Addition of a Tree Branch to the Reference Start with the branches connected to the reference node. Addition of a branch 0 q z between node q and the reference node 0 changes the old bus impedance matrix by adding a row and column as shown below: 0 0 0 old new bus bus q Z Z z   =       (1.27) Note that the red zeros are vectors of size m (same size as the old bus impedance matrix). Rule 2: Addition of a Tree Branch from a New Bus to an Old Bus Addition of a branch pq z between a new node q and an existing node p will increase the size of the old bus impedance matrix by one, from ( ) m m × to ( ) ( ) 1 1 m m + × + . The change is shown below: old bus ip new bus pi pp pq Z Z Z Z Z z   =   +     (1.28) where pi Z is identical to row p of old bus Z and ip Z is identical to column p of old bus Z . Note that by symmetry T pi ip Z Z = . Also, pp Z is the diagonal element of the old bus impedance matrix in the pth position. Rule 3: Addition of a Cotree Link Between Two Existing Nodes When a link is added between existing nodes p and q, the size of the old bus impedance matrix is increased by one as follows iq ip old bus new b q i u i p s Z Z Z Z Z Z Z   =      − −  ll (1.29) where T qi pi Z Z Z − = ∆ is the difference between row q and row p of the old bus impedance matrix and iq ip Z Z Z − = ∆ is the difference between column q and column p of the old bus impedance matrix. The element Z ll is given by 2 pq pp qq pq Z z Z Z Z = + + − ll . The new row and column are eliminated using the equation 7 T new old bus bus Z Z Z Z Z ∆ ∆ = − ll . (1.30) If the added link is between bus p and the reference bus 0, the above results are changed as follows: i old bus n p pi ew bus Z Z Z Z Z   =     −  −  ll (1.31) where T pi ip Z Z − = − is the negative of the pth row (column) of the old bus impedance matrix and pq pp Z z Z = + ll . Example 9.3 Construct the bus impedance matrix for the network of example 9.1 shown again below: Note the proper tree is chosen in (c) above. We start with the branches connected to the reference node. First we include branch 1, 10 0.2 z j = between nodes 1 and 0 (ref. node). According to rule 1 we have (1) 11 10 0.2 bus Z Z z j = = = . Next add branch 2, 20 0.4 z j = between nodes 2 and 0, we now have 11 (2) 22 0 0.2 0 0 0 0.4 bus Z j Z Z j     = =         Note the zeros added and we only have diagonal terms so far. There are no more elements to ground, so we go to the next element, branch 3. We are adding a new node (3) and a new branch (3) to an existing node (1). We use rule 2: 11 12 11 (3) 21 22 21 11 12 11 13 0.2 0 0.2 0 0.4 0 0.2 0 0.6 bus Z Z Z j j Z Z Z Z j Z Z Z z l j         = =     +         This completes all the tree branches. Now we go to the cotree and add the links. We add link (4), 12 0.8 z j = between existing nodes 1 and 2. We have j0.2 j0.4 j0.4 j0.8 1 2 3 j0.4 0 1 2 3 2 1 3 4 5 0 1 2 3 2 1 3 4 5 (a) (b) (c) 8 11 12 13 12 11 21 22 21 22 21 (4) 11 12 11 13 32 31 44 21 11 22 21 32 31 44 0.2 0 0.2 0.2 0 0.4 0 0.4 0.2 0 0.6 0.2 0.2 0.4 0.2 bus Z Z Z Z Z j j j Z Z Z Z Z j j Z Z Z Z z Z Z l j j j j j Z Z Z Z Z Z Z Z − −         −     = = + −   −       − − − − −     where ( ) 44 12 11 22 12 2 0.8 0.2 0.4 2 0 1.4 Z z Z Z Z j j j j j = + + − = + + − = and [ ] 44 0.2 1 0.4 0.2 0.4 0.2 1.4 0.2 0.0286 0.0571 0.0286 0.0571 0.1143 0.0571 0.0286 0.0571 0.0286 T j Z Z j j j j Z j j j j j j j j j j j −   ∆ ∆   = − −   −     −     = − −   −     and from (1.30) we have (4) 0.2 0 0.2 0.0286 0.0571 0.0286 0 0.4 0 0.0571 0.1143 0.0571 0.2 0 0.6 0.0286 0.0571 0.0286 0.1714 0.0571 0.1714 0.0571 0.2857 0.0571 0.1714 0.0571 0.0571 bus j j j j j Z j j j j l j j j j j j j j j j j j j −         = − − −     −             =       Repeating the same process by adding link 5, 23 0.4 z j = between nodes 2 and 3 we finally have: 0.16 0.08 0.12 0.08 0.24 0.16 0.12 0.16 0.34 bus j j j Z j j j j j j     =       which is the same as that found by inversion of the bus Y matrix obtained earlier in example 9.2. Below we repeat the solution using Matlab starting where the links are added: Z1=[j*0.2 0 j*0.2 0 j*.4 0 j*0.2 0 j*0.6] DZ1 = [-j*0.2; j*0.4; -j*0.2] zl1 = j*0.8 + j*0.2 + j*.4 ZX1=1/zl1*DZ1*(-DZ1)' Z2 = Z1 - ZX1 DZ2 = [Z2(1,3)-Z2(1,2); Z2(2,3)-Z2(2,2); Z2(3,3)-Z2(3,2)] zl2=j*0.4+Z2(2,2)+Z2(3,3)-2*Z2(2,3) ZX2=1/zl2*DZ2*(-DZ2)' Z3=Z2-ZX2 9 Z1 = 0 + 0.2000i 0 0 + 0.2000i 0 0 + 0.4000i 0 0 + 0.2000i 0 0 + 0.6000i DZ1 = 0 - 0.2000i 0 + 0.4000i 0 - 0.2000i zl1 = 0 + 1.4000i ZX1 = 0 + 0.0286i 0 - 0.0571i 0 + 0.0286i 0 - 0.0571i 0 + 0.1143i 0 - 0.0571i 0 + 0.0286i 0 - 0.0571i 0 + 0.0286i Z2 = 0 + 0.1714i 0 + 0.0571i 0 + 0.1714i 0 + 0.0571i 0 + 0.2857i 0 + 0.0571i 0 + 0.1714i 0 + 0.0571i 0 + 0.5714i DZ2 = 0 + 0.1143i 0 - 0.2286i 0 + 0.5143i zl2 = 0 + 1.1429i ZX2 = 0 + 0.0114i 0 - 0.0229i 0 + 0.0514i 0 - 0.0229i 0 + 0.0457i 0 - 0.1029i 0 + 0.0514i 0 - 0.1029i 0 + 0.2314i Z3 = 0 + 0.1600i 0 + 0.0800i 0 + 0.1200i 0 + 0.0800i 0 + 0.2400i 0 + 0.1600i 0 + 0.1200i 0 + 0.1600i 0 + 0.3400i Example 9.4 The bus impedance matrix for the network shown below is found to be 0.183 0.078 0.141 0.078 0.148 0.106 0.141 0.106 0.267 bus j j j Z j j j j j j     =       The line between buses 1 and 3 with impedance 13 0.56 Z j = is removed by the simultaneous opening of breakers at both ends of the line. Determine the new bus impedance matrix. One way of solving this problem is to add a line between the buses 1 and 3 whose impedance is the negative of 13 Z to be removed, i.e. we add 0.56 j − between nodes 1 and 3. We proceed using the equation for adding a bus: j0.35 1 3 2 j0.56 j0.4 j0.5 j0.3 j0.2 10 11 12 13 13 11 21 22 21 23 21 11 12 11 13 33 31 31 11 32 21 33 31 44 bus Z Z Z Z Z Z Z Z Z Z Z Z Z Z z Z Z Z Z Z Z Z Z Z −     −   = + −     − − −   where 44 13 11 33 13 2 Z z Z Z Z = + + − . Also note that Z ∆ is the last column of the matrix less the element 44 Z . Now it is easy to use Matlab to solve the problem: Z = [ j*0.183 j*0.078 j*0.141 j*0.078 j*0.148 j*0.106 j*0.141 j*0.106 j*0.267] DZ = Z(:, 3)-Z(:,1) Z44 = -j*0.56+Z(1,1)+Z(3,3)-2*Z(1,3) DZZ =DZ*DZ.'/Z44 Zbus = Z - DZZ Z = 0 + 0.1830i 0 + 0.0780i 0 + 0.1410i 0 + 0.0780i 0 + 0.1480i 0 + 0.1060i 0 + 0.1410i 0 + 0.1060i 0 + 0.2670i DZ = 0 - 0.0420i 0 + 0.0280i 0 + 0.1260i Z44 = 0 - 0.3920i DZZ = 0 - 0.0045i 0 + 0.0030i 0 + 0.0135i 0 + 0.0030i 0 - 0.0020i 0 - 0.0090i 0 + 0.0135i 0 - 0.0090i 0 - 0.0405i Zbus = 0 + 0.1875i 0 + 0.0750i 0 + 0.1275i 0 + 0.0750i 0 + 0.1500i 0 + 0.1150i 0 + 0.1275i 0 + 0.1150i 0 + 0.3075i 9.6 Zbuild and Symfault Programs These two functions are developed for the formation of the bus impedance matrix. The first zbuild is Zbus = zbuild(zdata) zdata is an 4 e × matrix containing the impedance data of an e-element network. Columns 1 and 2 contain the element bus numbers and columns 3 and 4 contain the element resistance and reactance respectively, in pu. Bus number 0 to generator buses contain generator impedances. These may be the subtransient, transient, or synchronous reactances. Also any other loads and capacitors to ground may be included in this matrix. The second sbuild function for the formation of the bus impedance matrix is zbus = zbuildpi(linedata, gendata, yload) 11 which is compatible with the power flow programs. The arguments are linedata this is the data required for the power flow solution. Columns 1 and 2 contain the bus numbers, columns 3 through five contain the resistance, reactance and half the susceptance for the line all in pu on the same MVA base. The last column number 6 is for tap settings of transformers, if a line, the enter 1 (one). gendata is where the generator data is entered. This is an 4 g n × matrix, where each row contains bus 0, generator bus number, resistance and reactance for each generator. yload is optional. It contains the results of a load flow solution (bus number and load impedance) as generated by one of the programs lfgauss, lfnewton, or decouple. yload is automatically generated following application of one of these programs. These two programs (zbuild and zbuildpi) obtain the bus impedance matrix using the algorithms developed in the chapter. First a tree is selected which contains elements to the reference node. Then the remaining branches of the tree are connected, and finally the cotree links are added. symfault(zdata, Zbus, V) was developed for three phase balanced fault studies. The arguments zdata and Zbus are required while V is optional. If V is not included, it is assumed that all pre-fault voltages are 1.0 per unit. If V is included, then it is an array describing the bus numbers and the complex bus voltage. Again, this data is automatically generated following a load flow study. symfault will prompt the user to supply the faulted bus number and the fault impedance. The program computes the fault current and the bus voltages and line currents during the fault. Go through examples 9.5 though 9.9 using the Matlab window. Chapter 10: Symmetrical Components and Unbalanced Faults 10.1 Introduction When an unbalanced three-phase fault occurs, we can solve the three-phase circuit using ordinary circuit theory. This is much more numerically complicated than the single- phase circuit normally used in balanced three phase circuits. The degree of difficulty increases with the third power of the system size. For this reason, it is apparent that if we were to solve three different single-phase circuits, it would be numerically simpler than solving the one three-phase circuit in one set of equations. The purpose of this chapter is to break up the large three-phase circuit into three circuits, each one third the size of the whole system. Next, we solve the three components individually, and then combine the results to obtain the total system response. The two programs abc2sc and sc2abc will transform the system representation from the normal "abc" form to the "sc" or symmetrical components form (and the other way around.) Three other programs are developed, these are: Line to ground fault studies: lgfault(zdata0, zbus0, zdata1, zbus1, zdata2, zbus2, V) Line to line fault studies: llfault(zdata1, zbus1, zdata2, zbus2, V) Double line to ground fault studies: dlgfault(zdata0, zbus0, zdata1, zbus1, zdata2, zbus2, V) As can be seen above, these programs use the three components which are developed first, namely the "0", "1" and "2" components. These are often referred to as the zero- sequence, positive-sequence and negative-sequence components, respectively. Some other books use "0", " + " and " − " for these same components. 10.2 Fundamentals of Symmetrical Components It was Fortescue in 1918 who developed the idea of breaking up asymmetrical three-phase voltages and currents into three sets of symmetrical components. These three basic components are: (1) The positive sequence currents and voltages (known also as the "abc" and often denoted by the superscript "1" or " + ",) shown on the right. Note this is the regular three-phase balanced phasors we have been dealing with from the start of this course. Ι a 1 Ι b 1 Ι c 1 2 (2) The negative sequence currents and voltages (known also as the "acb" and often denoted by the superscript "2" or " − ".) These are the regular balanced negative sequence components we have seen previously. Note the sequence of the phasors is the opposite direction from the positive sequence (acb instead of abc.) (3) The zero sequence components of currents and voltages (often denoted by the superscript "0".) These components are shown on the right. Note that these zero sequence phasors are all in-phase and equal in magnitude. We will need the vector 1 120 a · ∠ o , which is a unit vector at an angle of 120 degrees. It is easy to see that 2 * 1 240 1 120 a a · · ∠ · ∠ − o o and 3 1 a · . It is also clear that 2 1 0 a a + + · . Positive sequence: 1 1 1 1 1 2 1 1 1 1 0 240 120 a a a b a a c a a I I I I I a I I I aI · ∠ · · ∠ · · ∠ · o o o Negative sequence: 2 2 2 2 2 2 2 2 2 2 0 120 240 a a a b a a c a a I I I I I aI I I a I · ∠ · · ∠ · · ∠ · o o o Zero sequence: 0 0 0 a b c I I I · · Now consider a three-phase unbalanced set of currents a I , b I , and c I as shown in the figure to the right. Note that the vectors are arbitrary in length and direction. The same argument would hold if these were voltages instead of currents. These "abc" currents can be expressed in terms of the "sc" currents as follows: 0 1 2 0 1 2 0 1 2 a a a a b b b b c c c c I I I I I I I I I I I I · + + · + + · + + And we know that all the currents can be expressed in terms of the "a" components, thus: 0 1 2 0 2 1 2 0 1 2 2 a a a a b a a a c a a a I I I I I I a I aI I I aI a I · + + · + + · + + Ι a 2 Ι c 2 Ι b 2 I a 0 I b 0 I c 0 I a I b I c 3 Or in matrix form: 0 2 1 2 2 1 1 1 1 1 a a b a c a I I I a a I I a a I ¸ _ ¸ _ ¸ _ · ¸ , ¸ , ¸ , Which can be written as a single matrix equation as: 012 abc a I AI · , where A is known as the symmetrical components transformation matrix, which transforms the currents abc I into component currents 012 a I and is given by: 2 2 1 1 1 1 1 A a a a a ¸ _ · ¸ , . Note that T A A · , thus the A-matrix is symmetric. Using matrix inversion we have: 012 1 abc a I A I − · and the inverse matrix is given by: 1 2 2 1 1 1 1 1 3 1 A a a a a − ¸ _ · ¸ , And since 2 * a a · , we see that 1 * 1 3 A A − · . Using the above equation it is seen that: 0 1 2 2 2 1 1 1 1 1 3 1 a a a b a c I I I a a I I a a I ¸ _ ¸ _¸ _ · ¸ ,¸ , ¸ , Or in detail: ( ) ( ) ( ) 0 1 2 2 2 1 3 1 3 1 3 a a b c a a b c a a b c I I I I I I aI a I I I a I aI · + + · + + · + + It is noted that 0 a I is the sum of the three "abc" components, thus if the abc components are balanced, their sum is zero, hence the zero-sequence component is also zero. The exact similar expressions are valid for voltages abc V and 012 a V . These equations need not be repeated here. Often it is understood that the quantities with superscript "0", "1", or "2" will have subscript "a" and the "a" is omitted. Thus instead of 0 a I we use 0 I and instead of 1 a I we use 1 I and so on… Next we investigate the three phase complex power. We know that ( ) * 3 T abc abc S V I φ · 4 The equation for complex power may be expressed in terms of symmetrical components thus: ( ) ( ) ( ) * 012 012 3 T a a S AV AI φ · And dropping the subscript "a" these may be written as: ( ) ( ) ( ) * 012 012 3 * 012 * 012 T T T S AV AI V A A I φ · · Keeping in mind that "A" is symmetric and * 3 T A A · , we have: ( ) ( ) * 012 012 3 * * * 0 0 1 1 2 2 3 3 3 3 T S V I V I V I V I φ · · + + The last equation shows that the total unbalanced three-phase power is equal to the sum of the three symmetrical component powers. Note that in the above equations the subscript "a" was dropped from all those terms having superscript "012," "0," "1," or "2." Please read about the functions defined on page 404: sctm, phasor, abc2sc, sc2abc, zabc2sc, rec2pol, and pol2rec. Example 10.1 Obtain the symmetrical components of a set of unbalanced currents 1.6 25 a I · ∠ o , 1.0 180 b I · ∠ o , and 0.9 132 c I · ∠ o . The Matlab program follows: Iabc = [1.6 25 1.0 180 0.9 132]; I012 = abc2sc(Iabc); % Symmetrical components of phase a I012p= rec2pol(I012) % Converts rectangular phasors into polar form I012p = 0.4512 96.4529 0.9435 -0.0550 0.6024 22.3157 5 -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 a-b-c set -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 Zero-sequence set -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 Positive-sequence set -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 Negative-sequence set Example 10.2 The symmetrical components of a set of unbalanced three-phase voltages are 0 0.6 90 V · ∠ o , 1 1.0 30 V · ∠ o , and 2 0.8 30 V · ∠ − o . Obtain the original unbalanced phasors. The Matlab program follows: V012 = [.6 90 1 30 0.8 -30]; Vabc = sc2abc(V012); % Unbalanced phasors from symmetrical components Vabcp= rec2pol(Vabc) % Converts rectangular phasors into polar form Vabcp = 1.7088 24.1825 0.4000 90.0000 1.7088 155.8175 6 -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 a-b-c set -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 Zero-sequence set -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 Positive-sequence set -1 0 1 -1.5 -1 -0.5 0 0.5 1 1.5 Negative-sequence set 10.3 Sequence Impedances Each of the sequence currents (or voltages) will encounter a different impedance in the network. In other words, each network component will present a different impedance as seen be each of the sequences. The network components are: loads, transmission lines, generators and transformers. We shall consider the impedances each of these components presents. 1 Z is the usual impedance for positive sequence currents, 2 Z is impedance for negative sequence currents, and 0 Z is the impedance for zero sequence currents. 10.3.1 Sequence Impedances of Y-Connected Loads The balanced three-phase Y-load is shown in the figure to the right. There is a mutual coupling between the three loads making up the Y-load. The return current through the neutral goes through an impedance n Z . The "abc" voltages to ground are: a s a m b m c n n b m a s b m c n n c m a m b s c n n V Z I Z I Z I Z I V Z I Z I Z I Z I V Z I Z I Z I Z I · + + + · + + + · + + + Z m Z m Z m I n Z n V a V b V c - - - 7 From Kirchhoff's current law, we have n a b c I I I I · + + . Using this equation above, we have: a s n m n m n a b m n s n m n b c m n m n s n c V Z Z Z Z Z Z I V Z Z Z Z Z Z I V Z Z Z Z Z Z I + + + ¸ _ ¸ _ ¸ _ · + + + + + + ¸ , ¸ , ¸ , Or in compact form: abc abc abc V Z I · . Using symmetrical components (recall that 012 a I is denoted by 012 I for simplicity,) this equation is written as: 012 012 abc AV Z AI · and solving for the voltage we have: 012 1 012 012 012 abc V A Z AI Z I − · · Where: 012 1 abc Z A Z A − · Using previously derived matrices we have: 012 2 2 2 2 1 1 1 1 1 1 1 1 1 3 1 1 s n m n m n m n s n m n m n m n s n Z Z Z Z Z Z Z a a Z Z Z Z Z Z a a a a Z Z Z Z Z Z a a + + + ¸ _ ¸ _ ¸ _ · + + + + + + ¸ , ¸ , ¸ , Performing the above matrix multiplications, we have: 012 3 2 0 0 0 0 0 0 s n m s m s m Z Z Z Z Z Z Z Z + + ¸ _ · − − ¸ , And if there is no mutual coupling between the legs of the Y-load, we have: 012 3 0 0 0 0 0 0 s n s s Z Z Z Z Z + ¸ _ · ¸ , N.B. The impedance matrix has non-zero elements only on the diagonal, therefore for balanced load the three sequences are independent. This means currents in one sequence will produce voltages in that sequence only! This is a very important property since it allows us to analyze each sequence network on a per-phase basis. 10.3.2 Sequence Impedances of Transmission Lines The transmission lines present the same geometry to the positive as well as the negative sequence currents, thus for lines we have 1 2 Z Z · . However, the zero-sequence impedance is much larger (in general) and is given by: 0 1 3 n X X X · + where 2 0.2ln m /km n n D X f D π ¸ _ · Ω ¸ , . Here n D is the distance between the line and the 8 neutral (usually earth or ground,) and D is the distance between the three lines (assuming they are equidistant.) 10.3.3 Sequence Impedances of Synchronous Machine Since the machine has rotation, the inductances seen will differ depending on the direction of rotation of the armature currents. Thus the negative sequence currents will appear to have twice line frequency. However, zero sequence currents will rotate with the windings, thus here we would see only the leakages. As an approximation, the sequence impedances of synchronous machines are approximated as follows: Positive Sequence: 1 or or d d d X X X X ′′ ′ · , depending on analysis (sub-transient, transient or steady state.) Negative Sequence: 2 d X X′′ ; , this is approximately correct. Zero Sequence: 0 X X l ; , where X l is the leakage reactance. 10.3.4 Sequence Impedances of Transformers We will study the per-phase parameters of the equivalent circuit of the transformer. The shunt (magnetization and core losses representation,) are usually neglected since they have a high impedance and they represent less than 1% of the rated transformer power. Also, since the transformer has no rotating fields or components, all the sequence impedances are the same (though sometimes the zero sequence current does not have a path to flow.) Hence we have: 0 1 2 Z Z Z Z · · · l . In a balanced delta connection and also in a 3-wire Y-connection, there will be no zero sequence currents outside the transformer; hence, these impedances will be represented with an open circuit (infinite zero-sequence impedance.) Recall that it is the convention to connect Y − ∆ and Y ∆ − transformers so that the high voltage side will lead the low voltage side by 30 o (this is for the normal, or positive-sequence voltages.) For the negative-sequence voltages, this phase shift would be 30 − o . The zero-sequence impedance will depend on the connection of the two sides of the transformer. These are shown schematically below, for detailed diagrams, see page 413 of the book. Four-wire Ys: g a a' 9 Four-wire Y-Y: Four-wire Y-Delta: Y-Delta: Delta-Delta: N.B. The neutral (fourth wire) impedance to ground is very important since it appears with a factor of "3" times n Z . (This is due to the fact that if 0 I flows, then the neutral current will be 0 3 n I I · .) Example 10.3 A balanced three phase voltage of 100-V line to neutral is applied to a balanced Y-connected load with ungrounded neutral. The three- phase load consists of three mutually coupled reactances. Each phase has a series reactance of 12 s Z j · Ω, and the mutual coupling between phases is 4 m Z j · Ω. (a) Determine the line currents by mesh analysis without using symmetrical components. (b) Determine the line currents using symmetrical components. g a a' g a a' g a a' g a a' Z m Z m Z m V a V b V c - - - 10 (a) Using KVL to the two meshes we have: / 6 / 2 s a m b s b m a a b L s a m c s c m b b c L Z I Z I Z I Z I V V V Z I Z I Z I Z I V V V π π + − − · − · ∠ + − − · − · ∠ − And from KCL we have: 0 a b c I I I + + · Writing these three equations in matrix form: ( ) ( ) ( ) ( ) 0 / 6 0 / 2 1 1 1 0 s m s m a L s m s m b L c Z Z Z Z I V Z Z Z Z I V I π π − − − ∠ ¸ _ ¸ _ ¸ _ − − − · ∠ − ¸ , ¸ , ¸ , Or in compact form: abc mesh mesh Z I V · . Solving this equation using matrix inversion: 1 abc mesh mesh I Z V − · The following Matlab program performs these operations: disp('(a) Solution by mesh analysis:'); Zs=j*12; Zm=j*4; Va=100; VL=sqrt(3)*Va; Z=[(Zs-Zm) -(Zs-Zm) 0 0 (Zs-Zm) -(Zs-Zm) 1 1 1]; V=[VL*cos(pi/6)+j*VL*sin(pi/6) VL*cos(-pi/2)+j*VL*sin(-pi/2) 0] ; Y=inv(Z); Iabc=Y*V; Iabcp=[abs(Iabc), angle(Iabc)*180/pi] 12.5000 (a) Solution by mesh analysis: 12.5001 Iabcp = 12.5002 12.5000 -90.0000 12.5003 12.5000 150.0000 12.5004 12.5000 30.0000 (b) Using symmetrical components we have: 012 012 012 V Z I · where: 012 0 0 a V V ¸ _ · ¸ , and from past equations (page 7 with no n Z ,) we have: 012 2 0 0 0 0 0 0 s m s m s m Z Z Z Z Z Z Z + ¸ _ · − − ¸ , And now we have: 1 012 012 012 I Z V − 1 · ¸ ] . This is done using the Matlab program below: 11 disp('(b) Solution by symmetrical components method:'); Z012=[Zs+2*Zm 0 0 0 Zs-Zm 0 0 0 Zs-Zm]; V012=[0; Va; 0]; I012=inv(Z012)*V012; a=cos(2*pi/3)+ j*sin(2*pi/3); A=[1 1 1; 1 a^2 a; 1 a a^2]; Iabc=A*I012; Iabcp=[abs(Iabc), angle(Iabc)*180/pi] (b) Solution by symmetrical components method: Iabcp = 12.5000 -90.0000 12.5000 150.0000 12.5000 30.0000 Which is identical with the answer obtained in part (a). Example 10.4: A three-phase unbalanced source with the following phase to neutral voltages 200 25 100 155 80 100 abc V ¸ _ ∠ · ∠ − ∠ ¸ , o o o is applied to the circuit shown. The load series impedance per phase is 8 24 s Z j · + and the mutual impedance between phases is 4 m Z j · . The load and source neutrals are solidly grounded. Determine: (a) The load sequence impedance matrix 012 1 abc Z A Z A − · . (b) The symmetrical components of voltage. (c) The symmetrical components of current. (d) The load phase currents. (e) The complex power delivered to the load in terms of symmetrical components. (f) The complex power delivered to the load in terms of the power per phase. The following Matlab program is the solution: clear Vabc= [200 25 100 -155 80 100]; Zabc=[8+j*24 j*4 j*4 j*4 8+j*24 j*4 j*4 j*4 8+j*24]; Z012=zabc2sc(Zabc) V012=abc2sc(Vabc); V012p=rec2pol(V012) I012=inv(Z012)*V012; I012p=rec2pol(I012) Iabc=sc2abc(I012); 12 Iabcp=rec2pol(Iabc) S3ph_sc=3*(V012.')*conj(I012) Vabcr=Vabc(:, 1).*(cos(pi/180*Vabc(:, 2))+j*sin(pi/180*Vabc(:,2))); S3ph=(Vabcr.')*conj(Iabc) Z012 = 8.0000 +32.0000i 0 + 0.0000i 0 + 0.0000i 0.0000 + 0.0000i 8.0000 +20.0000i -0.0000 + 0.0000i 0 + 0.0000i 0.0000 - 0.0000i 8.0000 +20.0000i V012p = 47.7739 57.6268 112.7841 -0.0331 61.6231 45.8825 I012p = 1.4484 -18.3369 5.2359 -68.2317 2.8608 -22.3161 Iabcp = 8.7507 -47.0439 5.2292 143.2451 3.0280 39.0675 S3ph_sc = 9.0471e+002 +2.3373e+003i S3ph = 9.0471e+002 +2.3373e+003i -5 0 5 -6 -4 -2 0 2 4 6 a-b-c set -5 0 5 -6 -4 -2 0 2 4 6 Zero-sequence set -5 0 5 -6 -4 -2 0 2 4 6 Positive-sequence set -5 0 5 -6 -4 -2 0 2 4 6 Negative-sequence set Chapter 10: Symmetrical Components and Unbalanced Faults, Part II 10.4 Sequence Networks of a Loaded Generator In the figure to the right is a generator supplying a three-phase load with neutral connected through impedance n Z to ground. The generator generates three-phase balanced voltages defined as: 2 1 abc a E a E a ¸ _ · ¸ , The terminal voltage of the generator is found using Kirchoff's voltage law: a a s a n n b b s b n n c c s c n n V E Z I Z I V E Z I Z I V E Z I Z I · − − · − − · − − And since n a b c I I I I · + + , we have: a a s n n n a b b n s n n b c c n n s n c V E Z Z Z Z I V E Z Z Z Z I V E Z Z Z Z I + ¸ _ ¸ _ ¸ _ ¸ _ · − + + ¸ , ¸ , ¸ , ¸ , Or in compact form: abc abc abc abc V E Z I · − . This equation can be transformed to the "012" or symmetrical component form thus: 012 012 012 abc a a a AV AE Z AI · − Multiplying by 1 A − we have: 012 012 1 012 012 012 012 abc a a a a a V E A Z AI E Z I − · − · − Where 012 1 2 2 2 2 1 1 1 1 1 1 1 1 1 3 1 1 abc s n n n n s n n n n s n Z A Z A Z Z Z Z a a Z Z Z Z a a a a Z Z Z Z a a − · + ¸ _ ¸ _ ¸ _ · + + ¸ , ¸ , ¸ , Performing the above matrix multiplication we have: 0 012 1 2 3 0 0 0 0 0 0 0 0 0 0 0 0 s n s s Z Z Z Z Z Z Z Z ¸ _ + ¸ _ · · ¸ , ¸ , E a E b E c Z s Z s Z s Z n I a I b I c I n + + + - - - V a V c V b 2 It is important to note that the matrices above are diagonal, and the equations are completely decoupled. Expanding into three voltage equations we have: 0 0 0 1 1 1 2 2 2 0 ( ) 0 a a a a a a a V Z I V E Z I V Z I · − · − ++ · − These equations represent three separate circuits as shown to the right, one for the positive-sequence, one for the negative- sequence, and one for the zero-sequence networks. The following important observations are made: • The three sequences are independent. • The positive-sequence network is the same one used in the one-line diagram for studying balanced three-phase currents and voltages. • Only the positive sequence-network has a voltage source. Thus the positive- sequence voltage will generate only positive-sequence currents. • There is no voltage source in the negative- and zero-sequence networks. • Negative- and zero-sequence currents cause only negative- and zero-sequence voltages. • The neutral of the system is the reference point for the positive- and negative- sequence networks. The ground is the reference point for the zero-sequence; hence the zero-sequence current cannot flow unless there is a connection to ground. • The grounding impedance is reflected in the zero sequence network as 3 n Z . • Each of the three sequence networks can be solved separately on a per phase basis. The phase currents and voltages can then be found by superposition, by adding their symmetrical components of current and voltage respectively. 10.5 Single Line-To-Ground Fault This is the most common fault on a three-phase system. It is illustrated using the following simple network: E a + - V 1 a + - V 2 a + - V 0 a + - I 1 a Z 1 I 2 a Z 2 I 0 a Z 0 Positive-sequence Negative-sequence Zero-sequence 3 Here the fault occurs between line "a" and ground through a fault impedance f Z . Assuming the generator was initially at no load, the boundary conditions at the fault are: 0 a f a b c V Z I I I · · · Using t-he second equation in the symmetrical components of currents we have: 0 1 2 2 2 1 1 1 1 1 0 3 1 0 a a a a I I I a a I a a ¸ _ ¸ _¸ _ · ¸ ,¸ , ¸ , Therefore 0 1 2 1 3 a a a a I I I I · · · . Thus the three symmetrical components are equal and each is equal to one-third phase "a" current. Phase "a" voltage in terms of symmetrical components is: 0 1 2 a a a a V V V V · + + Using equation ++ last page, and the fact the three symmetrical components of current are 0 1 2 1 3 a a a a I I I I · · · we have: ( ) 0 1 2 0 a a a V E Z Z Z I · − + + Where 0 3 s n Z Z Z · + and 1 2 s Z Z Z · · . Recall that a f a V Z I · and that 0 3 a a I I · , the above equation becomes: ( ) 0 1 2 0 0 3 f a a a Z I E Z Z Z I · − + + Or: 0 1 2 0 ( ) 3 a a f E I Z Z Z Z · + + + + And the fault current is: 0 1 2 0 3 3 3 a a a f E I I Z Z Z Z · · + + + Using the symmetrical components of current in equation ++ the symmetrical components of voltage are found, hence by transformation, the voltages during the fault. Observing the equations + and ++ we notice that the sequence networks derived earlier may be connected as shown here which satisfies these two equations. E a E b E c Z s Z s Z s Z n I a I b =0 I c =0 I n + + + - - - V a V c V b Z f E a + - V 1 a + - V 2 a + - V 0 a + - I 1 a Z 1 I 2 a Z 2 I 0 a Z 0 3Z f 4 If the generator neutral is grounded, then 0 n Z · and 0 s Z Z · , and for a bolted fault 0 f Z · . 10.6 Line-To-Line Fault This fault is shown in the figure to the right. Note the neutral is not grounded. The fault is shown between phases b and c. It is assumed the generator is initially on no-load. The boundary conditions at the fault point are: 0 0 b c f b b c a V V Z I I I I − · + · · Using 0 a I · and c b I I · − , the symmetrical components of currents become: 0 1 2 2 2 1 1 1 0 1 1 3 1 a a b a b I I a a I I a a I ¸ _ ¸ _¸ _ · − ¸ ,¸ , ¸ , Expanding the matrix equation we have: ( ) ( ) ( ) 0 1 2 2 2 0 1 3 1 3 a a b a b I I a a I x I a a I · · − · − Note that 1 2 a a I I · − from (x) above. From the equations for phase voltages (reproduced below) we have: 0 1 2 0 2 1 2 0 1 2 2 a a a a b a a a c a a a V V V V V V a V aV V V aV a V · + + · + + · + + Hence ( )( ) 2 1 2 b c a a f b V V a a V V Z I − · − − · Using the values of 1 a V and 2 a V from ++, and note that 2 1 a a I I · − , we have: ( ) ( ) 2 1 2 1 a a f b a a E Z Z I Z I 1 − − + · ¸ ] And substituting for b I from (x) above we have: E a E b E c Z s Z s Z s I a =0 I b I c + + + - - - V a V c V b Z f 5 ( ) ( ) ( ) 1 1 2 1 2 2 3 a a a f I E Z Z I Z a a a a − + · − − Since ( ) ( ) 2 2 3 a a a a − − · , solving for 1 a I we have: 1 1 2 a a f E I Z Z Z · + + The phase currents are: 2 1 2 1 1 1 1 0 1 1 a b a c a I I a a I I a a I ¸ _ ¸ _ ¸ _ · − ¸ , ¸ , ¸ , The fault current is: ( ) 2 1 1 3 b c a a I I a a I j I · − · − · − Using the symmetrical components of currents in equation ++, the symmetrical components of voltage (hence phase voltages by transformation,) at the fault point are obtained. Note that 0 0 a I · and 2 1 a a I I · − , hence from xx above we see that the sequence network for the line-to-line fault is as shown on the right. 10.7 Double Line-To-Ground Fault This fault is not as common as the previous two, but more likely than the symmetrical fault (all lines bolted, or connected through a fault to the neutral.) The double line-to-ground fault is illustrated on the right. The fault is a short between lines b and c, which are grounded through fault impedance f Z . Again, the generator is assumed initially at no-load. The boundary conditions at the fault point are: (0.1) ( ) 0 1 2 0 b c f b c a a a a V V Z I I I I I I · · + · + + · We know that the voltages b V and c V are given by: E a + - V 1 a + - V 2 a + - I 1 a Z 1 I 2 a Z 2 Z f E a E b E c Z s Z s Z s I a =0 I b I c + + + - - - V a V c V b Z f Z n 6 (0.2) 0 2 1 2 0 1 2 2 b a a a c a a a V V a V aV V V aV a V · + + · + + And since b c V V · , we have: (0.3) 1 2 a a V V · Substituting "012" for the "abc" currents in (0.1) we have: (0.4) ( ) ( ) 0 2 1 2 0 1 2 2 0 1 2 0 2 3 b f a a a a a a f a a a f a V Z I a I aI I aI a I Z I I I Z I · + + + + + · − − · Using (0.4) and (0.3) in the top equation of (0.2) we have: (0.5) ( ) 0 0 2 1 0 1 3 f a a a a a Z I V a a V V V · + + · − Using (++) for the symmetrical components of voltage in the equation above, and solving for 0 a I we have: (0.6) 1 1 0 0 3 a a a f E Z I I Z Z − · − + Using the equation (++) in (0.3) we have: (0.7) 1 1 2 2 a a a E Z I I Z − · − Now using the above two equations in the second equation of (0.1) we have (after some algebra:) (0.8) ( ) 1 2 0 1 2 0 3 3 a a f f E I Z Z Z Z Z Z Z · + + + + Note that a E drives the current 1 a I , hence the impedance seen by a E is: ( ) 2 0 1 2 0 3 3 f f Z Z Z Z Z Z Z + + + + . Clearly, this is the impendence 1 Z in series with the parallel combination of 2 Z and 0 3 f Z Z + . Thus the sequence networks can be connected as shown in the circuit on the right. Once the sequence currents are found from (0.8), (0.7) and (0.6) [in that order,] the phase currents are found using the transformation matrix A. Finally the fault current is found from: (0.9) 0 3 f b c a I I I I · + · E a + - V 1 a + - V 2 a + - V 0 a + - I 1 a Z 1 I 2 a Z 2 I 0 a Z 0 3Z f 7 [This is seen from the equation ( ) 0 1 3 a a b c I I I I · + + and knowing in this case 0 a I · .] Example 10.5 The one-line diagram of a simple power system is shown to the right. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit n a 100-MVA base. The system data expressed in per unit on a common 100- MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. Determine the fault current for the following faults: (a) A balanced three-phase fault at bus 3 through fault impedance 0.1 f Z j · per unit. (b) A single line-to-ground fault at bus 3 through fault impedance 0.1 f Z j · per unit. (c) A line-to-line fault at bus 3 through fault impedance 0.1 f Z j · per unit. (d) A double line-to-ground fault at bus 3 through fault impedance 0.1 f Z j · per unit. Item Base MVA Voltage rating 1 X 2 X 0 X 1 G 100 20 kV 0.15 0.15 0.05 2 G 100 20 kV 0.15 0.15 0.05 1 T 100 20/220 kV 0.1 0.1 0.1 2 T 100 20/220 kV 0.1 0.1 0.1 12 L 100 220 kV 0.125 0.125 0.3 13 L 100 220 kV 0.15 0.15 0.35 23 L 100 220 kV 0.25 0.25 0.7125 Preliminary calculations: First we find the Thevenin impedance viewed from bus 3, the faulted bus. First the delta is changed to a Y as can be seen in the figure below on the right. Now the Y-impedances are calculated thus: 1 2 3 G 1 G 2 T 1 T 2 8 ( ) ( ) ( ) ( ) ( )( ) 1 2 3 0.125 0.15 0.0357143 0.525 0.125 0.25 0.0595238 0.525 0.15 0.25 0.0714286 0.525 S S S j j Z j j j j Z j j j j Z j j · · · · · · Combining parallel branches, and adding the series branch, the positive- sequence Thevenin impedance is: ( ) ( ) 1 33 0.2857143 0.3095238 0.0714286 0.5952381 0.22 j j Z j j j · + · The positive- and negative-sequence networks are shown to the right. The only difference between them in this case is that the source is missing in the negative-sequence network. Hence we have: 2 1 33 33 0.22 Z Z j · · Now, the zero-sequence network is constructed based on the transformer connections and is shown in the figures to the right. Note that the delta is reduced to its equivalent Y first, and then the impedances are combined to find the equivalent zero- sequence circuit. The result is the simple circuit shown below. Zero-sequence j0.35 j0.25 j0.15 j0.25 j0.125 1 2 3 j0.25 1 2 3 j0.25 j0.25 j0.035714 j0.059524 j0.071428 S j0.25 j0.35 j0.7125 j0.3 1 2 3 j0.25 j0.05 j0.05 j0.1 j0.1 j0.25 j0.25 j0.05 j0.05 j0.1 j0.1 1 2 3 S j0.183026 j0.156881 j0.077064 E a + - j0.22 Positive-sequence Negative-sequence j0.22 9 Now we are ready to solve the problem! (a) Balanced three-phase fault at bus 3. We assume the generators at no-load have a voltage of 1.0 per unit, hence the fault current is: ( ) 3 3 1 33 0 1.0 3.125 pu 0.22 0.1 =820.1 90 V a a f V I j Z Z j j · · · − + + ∠ o (b) Single line-to-groud fault at bus 3. We know the sequence currents are the same and given by: ( ) 0 1 2 3 3 3 3 1 2 0 33 33 33 0 3 1.0 0.22 0.22 0.35 3 0.1 0.9174 pu a f V I I I Z Z Z Z j j j j j · · · + + + · + + + × · − And using the matrix A we have: 0 0 3 3 3 2 0 3 3 2 0 3 3 1 1 1 3 2.7523 1 0 0 pu 1 0 0 a b c I I I j I a a I I a a I ¸ _ ¸ _ ¸ _ − ¸ _ ¸ _ · · · ¸ , ¸ , ¸ , ¸ , ¸ , (c) Line-to-line fault at bus 3. The zero-sequence component of current is zero, i.e. 0 3 0 I · . Also from the analysis of the line-to-line done previously we have: ( ) 1 2 3 3 3 1 2 33 33 0 1 1.8519 pu 0.22 0.22 0.1 a f V I I j Z Z Z j j j · − · · · − + + + + The fault current is: 3 2 3 2 3 1 1 1 0 0 1 1.8519 3.2075 pu 1 1.8519 3.2075 a b c I I a a j I a a j ¸ _ ¸ _¸ _ ¸ _ · − · − ¸ ,¸ , ¸ , ¸ , (d) Double line-to-ground fault at bus 3. From the equations developed earlier for the sequence currents of this fault we have: 10 ( ) ( ) ( ) 1 3 3 2 0 33 33 1 33 2 0 33 33 0 1 2.6017 pu 0.22 0.35 0.3 3 0.22 0.22 0.35 0.3 3 a f f V I j j j j Z Z Z j Z j j j Z Z Z · · · − + + + + + + + + ( ) ( ) ( ) 1 1 2 3 33 3 3 2 33 0 1 0.22 2.6017 1.9438 0.22 a V Z I j j I j Z j − − − · − · − · ( ) ( )( ) 1 1 0 3 33 3 3 0 33 0 1 0.22 2.6017 0.6579 3 0.35 0.3 a f V Z I j j I j Z Z j j − − − · − · − · + + And using the A matrix we have the phase currents: 3 2 3 2 3 1 1 1 0.6579 0 1 2.6017 4.058 165.93 pu 1 1.9438 4.058 14.07 a b c I j I a a j I a a j ¸ _ ¸ _¸ _ ¸ _ · − · ∠ ∠ ¸ ,¸ , ¸ , ¸ , o o The fault current is: ( ) 3 3 3 1.9732 90 b c I F I I · + · ∠ o . Chapter 10: Symmetrical Components and Unbalanced Faults, Part III 10.8 Unbalanced Fault Analysis Using Bus Impedance Matrix We showed that the symmetrical components are independent, and for each component there is an impedance matrix bus Z . Thus we can find 0 bus Z , 1 bus Z and 2 bus Z in order to deal with each component separately. If the fault is at bus k , then we know that the k-th diagonal element of the bus is the Thevenin impedance of the network viewed from that bus. This then allows us to find the sequence networks, which we connect according to the fault, and then go on to solve for the fault currents and voltages. The impedances obtained from the bus Z matrices are called 0 kk Z , 1 kk Z and 2 kk Z which are connected according to figures 10.11, 10.13, and 10.15 in the book. In the material to follow, the subscript "a" is implied, but it is left out. It is understood that the symmetrical components refer to phase a. 10.8.1 Single Line-To-Ground Fault Using Z bus Consider the single line to ground fault on line "a" through impedance f Z to ground on bus k as shown in the figure. This fault requires the zero- positive- and negative-sequence networks for phase "a" be placed in series in addition to 3 f Z in order to compute the sequence currents for phase "a", thus: ( ) 0 1 2 0 1 2 0 3 k k k k kk kk kk f V I I I Z Z Z Z = = = + + + Where 0 1 2 , ,and kk kk kk Z Z Z are the diagonal elements in the k axis of the corresponding impedance matrix, and ( ) 0 k V is the pre-fault voltage at bus k. The fault phase current is: 012 abc k k I AI = 10.8.2 Line-To-Line Fault Using Z bus Consider a fault between phases b and c through an impedance f Z at bus k as shown in the figure. Connecting the sequence networks for phase "a" as found earlier (positive- and negative-sequence networks in opposition,) the symmetrical components of the fault current are: Z f a b c Z f a b c 2 ( ) 0 1 2 1 2 0 0 k k k k kk kk f I V I I Z Z Z = = − = + + Where 1 kk Z , and 2 kk Z are on the diagonal elements in the k-axis of the corresponding bus impedance matrix. The fault phase current is then obtained using the A matrix as usual. 10.8.3 Double Line-To-Line Fault Using Z bus Here phases b and c are shorted and connected to ground through a fault impedance f Z . From before, the equations for the sequence currents of phase a at bus k are given by: ( ) ( ) ( ) ( ) 1 2 0 1 2 0 1 1 2 2 1 1 0 0 0 3 3 0 0 3 k k kk kk f kk kk kk f k kk k k kk k kk k k kk f V I Z Z Z Z Z Z Z V Z I I Z V Z I I Z Z = + + + + − = − − = − + As usual, the 0 1 2 , ,and kk kk kk Z Z Z are the diagonal elements in the k axis of the corresponding impedance matrix, and ( ) 0 k V is the pre-fault voltage at bus k. The phase currents are obtained using the A matrix, and the fault current is ( ) b c k k k I F I I = + . 10.8.4 Bus Voltages and Line Currents During Fault Knowing the fault currents at bus k, the symmetrical components of the i-th bus voltage during fault are obtained using the equation (++): ( ) ( ) ( ) ( ) 0 0 0 1 1 1 1 2 2 2 0 0 0 i ik k i i ik k i ik k V F Z I V F V Z I V F Z I = − = − = − Where ( ) ( ) 1 0 0 i i V V = is the pre-fault phase voltage at bus i and ik Z is the ik-th element of the proper Z matrix. The phase voltages during the fault are found using the A matrix thus: 012 abc i i V AV = The symmetrical components of fault current in line i to j are given by: a b c Z f 3 ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 1 1 1 1 2 2 2 2 i j ij ij i j ij ij i j ij ij V F V F I z V F V F I z V F V F I z − = − = − = Where 0 1 2 , ,and ij ij ij z z z are the zero-, positive-, and negative-sequence components of the actual line impedance between buses i and j. Using the A matrix the phase fault currents in line i to j are: 012 abc ij ij I AI = Example 10.6 Solve Example 10.5 using the bus impedance matrix. In addition, for each type of fault determine the bus voltages and line currents during the fault. Go over the example in detail, page 435. 10.9 Unbalanced Fault Programs Go over the various programs for unbalanced fault analysis.