Presentation on theme: "Summary of Papers 1. P. Sauer and M. Pai, “Power System Steady-State Stability and the Load Flow Jacobian,” IEEE Transactions on Power Systems, Vol. 5,"— Presentation transcript:
1 Summary of Papers1. P. Sauer and M. Pai, “Power System Steady-State Stability and the Load Flow Jacobian,” IEEETransactions on Power Systems, Vol. 5, No. 4, Nov. 19902. V. Ajjarapu and C. Christy, “The Continuation Power Flow: A Tool for Steady-State VoltageStability Analysis,” IEEE Transactions on Power Systems, Vol. 7, No. 1, Feb., 1992.3. S. Greene, I. Dobson, and F. Alvarado, “Sensitivity of the Loading Margin to Voltage Collapsewith Respect to Arbitrary Parameters,” IEEE Transactions on Power Systems, Vol. 12, No. 1,Feb. 1997, pp4. S. Greene, I. Dobson, and F. Alvarado, “Contingency Ranking for Voltage Collapse viaSensitivities from a Single Nose Curve,” IEEE Transactions on Power Systems, Vol. 14, No. 1,Feb. 1999, pp
2 Voltage SecurityVoltage security is the ability of the system to maintainadequate and controllable voltage levels at all system load buses.The main concern is that voltage levels outside of a specifiedrange can affect the operation of the customer’s loads.Voltage security may be divided into two main problems:1. Low voltage: voltage level is outside of pre-defined range.2. Voltage instability: an uncontrolled voltage decline.You should know thatlow voltage does not necessarily imply voltage instabilityno low voltage does not necessarily imply voltage stabilityvoltage instability does necessarily imply low voltage
3 ResourcesThere have been several individuals that have significantlyprogressed the field of voltage security. These include:Ajjarapu from ISUVan Cutsem: See the book by Van Cutsem and Vournas.Alvarado, Dobson, Canizares, & Greene:There are a couple other texts that provide good treatments ofthe subject:Carson Taylor: “Power System Voltage Stability”Prabha Kundur: “Power System Stability & Control”
4 Our treatment of voltage security will proceed as follows: Voltage instability in a simple systemVoltage instability in a large systemBrief treatment of bifurcation analysisContinuation power flow (path following) methodsSensitivity methods
5 Voltage instability in a simple system Consider the per-phase equivalent of a very simple threephase power system given below:V1V2Z=R+jXNode 1Node 2I++V2V1__S12SD=-S12
7 Now we can get SD=PD+jQD=-(P21+jQ21) by - exchanging the 1 and 2 subscripts in the previous equations.- negatingDefine 12 =1- 2
8 Define: is the power factor angle of the load, i.e., Then we can also express SD as:Note that phi, andtherefore beta, ispositive for lagging,negative for leading.Define β=tan. Then
9 So we have developed the following equations…. Equating the expressions for PD and for QD, we have:Square both equations and add them to get…..
10 Manipulation yields:Note that this is a quadratic in |V2|2. As such, it has the solution:
11 Let’s assume that the sending end voltage is |V1|=1.0 pu and B=2 pu. Then our previous equation becomes:% pf = 0.97 laggingbeta=0.25pdn=[ ];v2n=sqrt((1-beta.*pdn - sqrt(1-pdn.*(pdn+2*beta)))/2);pdp=[ ];v2p=sqrt((1-beta.*pdp + sqrt(1-pdp.*(pdp+2*beta)))/2);pd1=[pdn pdp];v21=[v2n v2p];% pf = 1.0beta=0pdn=[ ];pdp=[ ];pd2=[pdn pdp];v22=[v2n v2p];% pf = .97 leadingbeta=-0.25pdn=[ ];pdp=[ ];pd3=[pdn pdp];v23=[v2n v2p];plot(pd1,v21,pd2,v22,pd3,v23)You can makethe P-V plot usingthe followingmatlab code.
12 Plots of the previous equation for different power factors Real power loading, PD
13 Some comments regarding the PV curves: 1. Each curve has a maximum load. This value is typically called the maximum system load or the system loadability.2. If the load is increased beyond the loadability, the voltages willdecline uncontrollably.3. For a value of load below the loadability, there are twovoltage solutions. The upper one corresponds to one that can bereached in practice. The lower one is correct mathematically, but Ido not know of a way to reach these points in practice.4. In the lagging or unity power factor condition, it is clear that thevoltage decreases as the load power increases until the loadability.In this case, the voltage instability phenomena is detectable, i.e.,operator will be aware that voltages are declining before theloadability is exceeded.5. In the leading case, one observes that the voltage is flat, or perhapseven increasing a little, until just before the loadability. Thus, inthe leading condition, voltage instability is not very detectable.The leading condition occurs during high transfer conditions when theload is light or when the load is highly compensated.
14 We consider our simple (lossless) system again, with the equations QV CurvesWe consider our simple (lossless) system again, with the equationsNow, again assume that V1=1.0, and for a given value of PDand V2, compute 12 from the first equation, and then Q from thesecond equation. Repeat for various values of V2 to obtain a QVcurve for the specified real load PD.You can make the P-V plot using the following matlab code.v1=1.0;b=1.0;pd1=0.1v2=[1.1,1.05,1.0,.95,.90,.85,.80,.75,.70,.65,.60,.55,.50,.45,.40,.35,.30,.25,.20,.15];sintheta=pd1./(b*v1.*v2);theta=asin(sintheta);qd1=-v2.^2*b+v1*b*v2.*cos(theta);plot(qd1,v2);The curve on the next page illustrates….
16 Homework1. Draw the PV-curve for the following cases, and for each, determine the loadability.a. B=2, |V1|=1.0, pf=0.97 laggingb. B=2, |V1|=1.0, pf=0.95 laggingc. B=2, |V1|=1.06, pf=0.97 laggingd. B=10, |V1|=1.0, pf=0.97 laggingIdentify the effect on loadability of power factor, sending-end voltage, and line reactance.2. Draw the QV-curves for the following cases, and for each, determine the maximum QD.a. B=1, |V1|=1.0, PD=0.1b. B=1, |V1|=1.0, PD=0.2c. B=1, |V1|=1.06, PD=0.1d. B=2, |V1|=1.0, PD=0.1Identify the effect on maximum QD of real power demand, sending-end voltage, and line reactance.
17 Some comments regarding the QV Curves In practice, these curves may be drawn with a power flow programby1. modeling at the target bus a synchronous condenser (agenerator with P=0) having very wide reactive limits2. Setting |V| to a desired value3. Solving the power flow.4. Reading the Q of the generator.5. Repeat 2-4 for a range of voltages.QV curves have one advantage over PV curves:They are easier to obtain if you only have a power flow (standardpower flows will not solve near or below the “nose” of PV curvesbut they will solve completely around the “nose” of QV curves.)
18 Voltage instability in a large system: Influential factors: Load modelingReactive power limits on generatorsLoss of a circuitAvailability of switchable shunt devicesTwo important ideas on which understanding of the above influences rest:Voltage instability occurs when the reactive power supply cannot meet the reactive power demand of the network.Transmission line loading is too highReactive sources (generators) are too far from load centersGenerator terminal voltages are too low.Insufficient load reactive compensation2. Reactive power cannot be moved very far in a network(“vars do not travel”), since I2X is large.Implication: The SYSTEM can have a var surplus but experiencevoltage instability if a local area has a var deficiency.
19 Load modelingIn analyzing voltage instability, it is necessary to consider the networkunder various voltage profiles.Voltage stability depends on the level of current drawn by the loads.The level of current drawn by the loads can depend on the voltage seenby the loads.Therefore, voltage instability analysis requires a model of how theload responds to load variations.Thus, load modeling is very influential in voltage instability analysis.
20 Exponential load model A typical load model for a load at a bus is the exponential model:where the subscript 0 indicates the initial operating conditions.The exponents and are specific to the type of load, e.g., Incandescent lampsRoom air conditionerFurnace fanBattery chargerElectronic compact florescentConventional florescent
21 Polynomial load modelThe ZIP or polynomial model is a special case of the more general exponential model, given by a sum of 3 exponential models with specified subscripts:where again the subscript 0 indicates the initial operating conditions.Usually, values p2 and q2 are the largest.So this model is composed of three components:constant impedance component (p1, q1) - lightingconstant current component (p2, q2) – motor/lightingconstant power component (p3,, q3) – loads served by LTCs
22 Effect of Load modeling Understanding the effect of each component on voltage instabilitydepends on understanding two ideas:1. Voltage instability is alleviated when the demand reduces. Thisis because I reduces and I2X reactive losses in the circuits reduce.2. Since voltage instability causes voltage decline, alleviation ofvoltage instability results if demand reduces with voltage decline.This gives the key to understanding the effect of load modeling.constant impedance load (p1) is GOOD since demandreduces with square of voltage.constant current load (p2) is OK since demand reduceswith voltage.Constant power load (p3) is BAD since demand doesnot change as voltage declines.
23 Some considerations in load modeling The effects of voltage variation on loads, and thus of loads on voltage instability, cannot be fully captured using exponential orpolynomial load models because of the following three aspects.Thermostatic load recoveryInduction motor stalling/trippingLoad tap changers
24 Thermostatic load recovery Heating load is the most common type of thermostatic load, and it is one for which we are all quite familiar. Although much heating is done with natural gas as the primary fuel, some heating is done electrically, and even gas heating systems always contain some electric components as well, e.g., the fans.Other thermostatic loads include space heaters/coolers, water heaters, and refrigerators.When voltage drops, thermostatic loads initially decrease in power consumption. But after voltages remain low for a few minutes, the load regulation devices (thermostats) will start the loads or will maintain them for longer periods so that more of them are on at the same time. This is referred to as thermostatic load recovery, and it tends to exacerbate voltage problems at the high voltage level.
25 Induction motor stalling/tripping Three phase induction motors comprise a significant portion ofthe total load and so its response to voltage variation is important,especially since it has a rather unique response.Consider the steady-state induction motor per-phase equivalent model.I’2Za=R1+jX1X’2V1Zb=Rc//jXmR’2+R’2(1-s)/s =R’2 / s
26 Induction motor stalling/tripping The (referred to stator) rotor current is given by:whereandUnder normal conditions, the slip s is typically very small, less than 0.05 (5%). In this case, R’2/s >> R’2, and I’2 is small.But as voltage V1 decreases, the electromagnetic torque developed decreases as well, the motor slows down. Ultimately, the motor may stall. In this case, s=1, causing R’2/s = R’2. Thus, one sees that the current I’2 is much larger for stalled conditions than for normal conditions. Because of X1 and X’2 of the induction motor, the large “stall” current represents a large reactive load.Large motors have undervoltage tripping to guard against this, but smaller motors (refrigerators/air conditioners) may not.
27 Under low voltage conditions at the high side, the LTC will decrease t Tap changers:Load tap changers (LTC, OLTC, ULTC, TCUL) are transformers thatconnect the transmission or subtransmission systems to the distributionsystems. They are typically equipped with regulation capability that allow them to control the voltage on the low side so that voltagedeviation on the high side is not seen on the low side.t:1V1 and t aregiven in pu.HV sideV1/tLV sideV1In per unit, we say that the tap is t:1, wheret may range from pua single step may be about pu (5/8%= is very common)a change of one step typically requires about 5 seconds.there is a deadband of 2-3 times the tap step to prevent excessive tap change.Under low voltage conditions at the high side, the LTC will decrease tin order to try and increase V1/t.
28 Tap changers:Thus, as long as the LTC is regulating (not at a limit), a voltagedecline on the high side does not result in voltage decline at theload, in the steady-state, so that even if the load is constant Z,it appears to the high side as if it is constant power. So a simpleload model for voltage instability analysis, for systems using LTC,is constant power!There are 2 qualifications to using such a simple model (constant power):1. “Fast” voltage dips are seen at the low side (since LTCaction typically requires minutes), and if the dip is low enough,induction motors may trip, resulting in an immediate decrease inload power.2. Once the LTC hits its limit (minimum t), then the low sidevoltage begins to decline, and it becomes necessary to modelthe load voltage sensitivity.
29 Generator capability curve: Field current limit due to field heating,enforced by overexcitation limiter on If.QQmaxArmature current limit due toarmature heating, enforced byoperator control of P and If.Typicalapproximationused in powerflow programs.PQminLimit due to steady-state instability (small internal voltage E gives small |E||V|Bsin), and due to stator end-region heating from induced eddy currents, enforced by underexcitation limiter (UEL).
30 Effect of generator reactive power limits: 1. Voltage instability is typically preceded by generators hitting their upper reactivelimit, so modeling Qmax is very important to analysis of voltage instability.2. Most power flow programs represent generator Qmax as fixed. However, thisis an approximation, and one that should be recognized. In reality, Qmax is not fixed.The reactive capability diagram shows quite clearly that Qmax is a function of P andbecomes more restrictive as P increases. A first-order improvement to fixed Qmaxis to model Qmax as a function of P.3. Qmax is set according to the Over-eXcitation Limiter (OXL). The field circuit has arated steady-state field current If-max, set by field circuit heating limitations. Sinceheating is proportional to , we see that smaller overloads can be toleratedfor longer times.Therefore, most modern OXLs are set with a time-inverse characteristic:4. As soon as the OXL acts to limit If, then nofurther increase in reactive power is possible.When drawing PV or QV curves, theaction of a generator hitting Qmax, willmanifest itself as a sharp discontinuityin the curve.2.0OXL characteristicIfIrated1201.010Overload time (sec)
31 Effect of OXL action on PV curve: |V| o P (demand) One generator hits reactive limit|V|No reactivelimits modeledoP(demand)Note: Georgia Power Co. models its loadabilitylimit at point x, not point o.
32 Compare reactive losses with and without second circuit Loss of a circuitCompare reactive losses with and without second circuitAssume both circuits have reactance of X.I/2IXI/2XPPQloss=(I/2)2X+ (I/2)2X=I2X/2Qloss=I2XImplication: Loss of a circuit will always increase reactive lossesin the network. This effect is compounded by the factthat losing a circuit also means losing its line chargingcapacitance.
33 Kundur, on pp. 979-990, has an excellent example which illustrates many of the aforementioned effects. The illustration was done usinga long-term time domain simulation program (Eurostag).
34 Influence of switched shunt capacitors |V|WithoutcapacitorWith capacitorP(demand)
35 But, shunt compensation has some drawbacks: It produces reactive power in proportion to the square of thethe voltage, therefore when voltages drop, so does the reactivepower supplied by the capacitor.It has a maximum compensation level beyond which stableoperation is not possible (See pg. 972 of Kundur, and next slide).(A synchronous condenser and an SVC do not have these 2 drawbacks)It results in a flatter PV curve and therefore makes voltageinstability less detectable. Therefore, as the load grows in areaslacking generation, more and more shunt compensation is used tokeep voltages in normal operating ranges. By so doing, normaloperating points progressively approach loadability.
36 V1=1.0 V2 PL QL=0 Capacitive Mvars S=|V2|2B*Sbase with |V2|=1.0 |V2| Each QV curve/Capacitor characteristic intersection shows the operating point. Note that for the first three operating points, a small increase in Q-comp (indicated by arrows) results in voltage increase, but for the last operating point (950), more Q-comp (say 960) results in a voltage decrease.PLQL=0S=|V2|2B*Sbasewith |V2|=1.0|V2|675 Mvar450 Mvar300 Mvar1.2950 MvarPL=1300 mw1.0PL=1900 mw0.8QV-curves drawnusing synchronouscondensor approach.PL=1700 mwPL=1500 mw0.61600140012001000800600400200Capacitive Mvars
37 A bifurcation, for a dynamic system, is an acquisition of a new Bifurcation analysis (ref: A. Gaponov-Grekhov, “Nonlinearities in action” andalso Van Cutsem & Vournas, “Voltage stability of electric power systems.”)A bifurcation, for a dynamic system, is an acquisition of a newquality by the motion the dynamic system, caused by small changesin its parameters. A power system that has experienced a bifurcationwill generally have corresponding motion that is undesirable.Consider representing the dynamics of the power system as:Eqts. 1A differential-algebraic system (DAS):Here x represents state variables of the system (e.g., rotor angles, rotorspeed, etc), y represents the algebraic variables (bus voltage magnitudes& voltage angles), and p represents the real and reactive power injectionsat each bus. The function F represents the differential equations for thegenerators, and the function G represents the power flow equations.
38 Types of bifurcationsThere are at least two types of bifurcation:Hopf: two eigenvalues become purely imaginary:a birth of oscillatory or periodic motion.Saddle node: a disappearance of an equilibrium state.The stable operating equilibrium coalesces with an unstableequilibrium and disappears. The dynamic consequence of ageneric saddle node bifurcation is:a monotonic decline in system variables.So we think it is the saddle node bifurcation that causesvoltage instability.
39 The unreduced Jacobian: The Jacobian matrix of eqts. 1 isand it is referred to as the unreduced Jacobian of the DAS, whereEqt. 2
40 We may reduce eq. 2 by eliminating the variable y The reduced Jacobian:We may reduce eq. 2 by eliminating the variable yThis means we need to force the top right hand submatrix to 0, which we can do by multiplying the bottom row by -FYGY-1 and then adding to the top row.This results in:So that the reduced Jacobian matrix is a Schur’s complement:
41 Fact 1 : The conditions for a saddle node bifurcation are Equilibrium: Stability:Fact 1 : The conditions for a saddle node bifurcation areEquilibrium:Singularity of the unreduced Jacobian det(J)=0 (a 0 eigenvalue, J noninvertible) .Implication 1: The stability of an equilibrium point of the DAS depends on the eigenvalues of the unreduced Jacobian J. The system will experience a SNB as parameter p increases when J has a zero eigenvalue.Fact 2: The determinant of a Schur’s complement times the determinant of GY gives the determinant of the original matrix: det(J)=det(A)*det(GY)if GY is nonsingular.Implications 2:If GY is nonsingular, then singularity of A implies singularity of J so that we may analyze eigenvalues of A to ascertain stability.The fact that GY may be nonsingular, yet A singular, means that load flow convergence is not a sufficient condition for voltage stability.
42 Singularity of load flow Jacobian: Implications 2:If GY is nonsingular, then singularity of A implies singularity of J so that we may analyze eigenvalues of A to ascertain stability.The fact that GY may be nonsingular, yet A singular, means that load flow convergence is not a sufficient condition for voltage stability.Singular (unstable)SingularSingularNonsingular (stable)NonsingularNonsingularGYAJ
43 Singularity of load flow Jacobian: So voltage instability analysis using only a load flow Jacobian may yield optimistic results when compared to results from analysis of A,that is, stable points (based on Gy) may not be really stable.=> However, I believe it is true that points identified as unstable using the load flow Jacobian will be really unstable (Schur’s complement does not support that singularity of GY implies singularity of J, however, because it is only valid if GY is nonsingular).Note: Sauer and Pai, 1990, provide an in-depth analysis of the relationbetween singularity of GY and singularity of J, and show some specialcases for which singularity of GY implies singularity of J.Singular (unstable)Singular (unstable)Singular (unstable)Nonsingular (stable)Nonsingular (stable)Nonsingular (stable)GYAJ
44 Singularity of load flow Jacobian: So, we assume that load flow Jacobian analysis provides an upper bound on stability.Fact: The bifurcation (zero eigenvalue of GY) of the load flow Jacobian corresponds to the “turn-around point” (i.e., the “nose” point) of a P-V or Q-V curve drawn using a power flow program.This can be proven using an optimization approach.See pp of the text by Van Cutsem and Vournas.We have previously denoted the power flow equations as G(x,y,p)=0, but now we denote them as G(y,p)=0, without the dependence on the state variables x (which relate to the machine modeling and include, minimally, and of each machine).
45 decreasing generation at all other generators. So we turn our effort to identifying the saddle node bifurcation(SNB) for the power flow Jacobian matrix.The Jacobian can reach a SNB in many ways. For example,increase the impedance in a key tie lineincrease the generation level at a generator with weak transmission, whiledecreasing generation at all other generators.increase the load at a single busincrease the load at all buses.In all cases, we are looking for the “nose” point of theV- curve, where is the parameter that is being increased.)Most applications focus on the last method (increase load at all buses).Key questions here are:“direction” of increase: are bus loads increased proportionally, or in some other way?dispatch policy: how do the generators pick up the load increase ?We will assume proportional load increase with “governor” load flow(generators pick up in proportion to their rating)|V|
46 Define: critical point - the operating conditions, characterized 3/31/2017Define: critical point - the operating conditions, characterizedby a certain value of , beyond which operation is notacceptable.Question 1:What can cause the critical point to differfrom the SNB point ?|V|Question 2:How can knowledge of the critical point provide a securitymeasure?Question 3:Does the P-V curve provide a forecast of the system trajectory ?
47 Approach 1: Search for * using some iterative search procedure. Solution approaches to finding *, the value of corresponding to SNB.Approach 1: Search for * using some iterative search procedure.1. i=12. Using (i), solve power flow using Newton-Raphson.Here, we iteratively solve G(y,p)=0. At each step,we must solve for y in the eqt: GY y = p3. If solved,(i+1)= (i)+ .i=i+1go to 2else if not solved,*= (i+1)endif4. EndBut big problem: as gets close to *, GY becomes ill-conditioned(close to singular). This means that at some point before the criticalpoint, step 2 will no longer be feasible.
48 Approach 2: Use the continuation power flow (CPF). Predictor stepCorrector stepPass * ?No.SelectcontinuationparameterYes.Stop
49 The predictor step:The power flow equations are functions of the bus voltages andbus angles and the bus injections:Augment the power flow equations so that they are functions of (dependence on p is carried through the dependence on ).pp0 Now recognize thatso thatIf we want to compute the change in the power flow equations dGdue to small changes in the variables , V, and ,that move us closer to the loadability pointas we move from one solution i to another “close” solution i+1, thendG= G((i),V(i),(i))- G((i+1),V(i+1),(i+1)) = 0 – 0 = 0
50 Here, each set of partial derivatives are evaluated at the operating conditions corresponding to the old solution. If the power flow equations are linear with the 3 sets of variables in the region between the old solution and the (close) new one, the following is satisfied:Eq. 3BUT, we have added one unknown, , to the power flow problem without adding acorresponding equation, i.e., in G(,V,)=0, there are are N equations but N+1 variables, so that in eq. 3, the matrix [G GV, G], has N rows (the number of eqts being differentiated) and N+1 columns (the number of variables for which each eqt is differentiated). So we need another equation in order to solve this. What to do ?
51 If we define to be the “step size,” then we can rewrite this as The answer to this can be found by identifying how we will be using using the solution to eqt. 3. Note the solution corresponding to the “new” point is:Here the “p” indicatesthat this is the“predicted” point.If we define to be the “step size,” then we can rewrite this aswhere
52 We call the update vector (with the differentials) the “tangent” vector, denoted by t.This vector provides the direction to move in orderto find a new solution (i+1,p) from the old one (i).We can think of this in terms of the following picture…..
54 For example, consider a 2-dimensional vector…. Note: In specifying a direction using an n-dimensional vector, only n-1 of the elements are constrained - one element can be chosen to be any value we like.For example, consider a 2-dimensional vector….x2=x1tan(30) so:- the direction is specified byselecting x1=1, x2=0.5774,selecting x1=0.5, x2=x2Direction= 30ox1So we can set one of the tangent vector elements toany value we like, then compute the other elements.This provides us with our other equation….
55 Suppose that we set the k-th parameter in the tangent vector to be 1 Suppose that we set the k-th parameter in the tangent vector to be 1.0. Then our equation given as eq. 3 can be augmented to become:wheree=[...1...]kkTo select , we would have:e=[......1]Which wouldforce d=1.k
56 The parameter for which we select k is called the continuation parameter, and it can be any load level (or group of load levels),or it can be a voltage magnitude. Initially, when the solution isfar from the nose, the continuation parameter is typically .The parameter is called the step size, and it can be selectedusing various techniques. The simplest of these is to justset it to a constant. Let’s try this on our simple problemformulated at the beginning of these slides.
57 HOMEWORK #2, Due Monday, Jan 26. 1. Using the equations at the bottom of slide 7, with the left-hand side (PDand QD) and also V1 given by the problem statement, we know everythingexcept V2 and theta.2. Now, just bring the right hand side of these 2 equations over to theleft-hand side, and you have the 2 equations that correspond to G(y,p)=0.3. Solve these equations to get the corresponding power flow solution (butyou do not need Newton-Raphson to do this – you can just use the equationat the bottom of slide 10).4. Now you need to replace the value specified in the equations for PD(which is 0.4 according to the problem statement) with 0.4*lambda. Thisgives you the equations in the form of slide 49: 0=G(theta,V,lambda).Note, however, that G is really two equations: G1 and G2.5. Now you need to formulate the equations on the slide 55. This is a matterof taking derivatives and then evaluating those derivatives at thesolution that you obtained above. Note, however, the each element in thematrix of slide 55 actually represents 2 elements. That is:| dG1/dtheta dG1/dV dG1/dlambda|| dG2/dtheta dG2/dV dG2/dlambda|| |6. Evaluate each of the above matrix elements at the solution obtained instep 3.7. Then solve these equations for the tangent vector. You can do this byinverting the above matrix (use matlab or a calculator to do this) andthen multiply the right-hand-side by this inverted matrix.8. Then take a “step” using an appropriately chosen step size per theequation on slide 56.9. Beginning from your predicted point that you identified in step 8 of #2a, developequations for approach a, solve them, and identify the resulting corrected point in terms of voltage and power.10. Repeat #9 except implement approach b.#9 and #10 will be explained in next few slides.
58 Corrector stepNote, however, that the predicted point will satisfy thepower flow equations only if the power flow equations arelinear, which they are not.So our point needs correction. This leads to the corrector step.There are two different approaches for performing thecorrector step.Approach a: Perpendicular intersection method.Approach b: Parameterization method
59 Approach a: perpendicular intersection Here, we find the intersection between the power flowequations (the PV curve) and a plane that is perpendicular tothe tangent vector.|V| ty(i)Solve simultaneously,for y(i+1)y(i+1,p)y(i+1)The last equation says the inner(dot) product of 2 vectors is zero.Use Newton-Raphson to solve the above (requires only 1-3 iterations since we havegood starting point). If no convergence, cut step size () by half and repeat.
60 Approach b: Parameterization The corrector step is performed by identifying a continuation parameter (see slide 62) – can be λfixing it at the value found in the predictor step;then solving the power flow equations.|V|y(i) tSolve simultaneously,for y(i+1)y(i+1,p)Vertical correctionscorrespond to a fixedload-continuationparameter, horizontalcorrections to a fixedvoltage-continuationparameter.y(i+1)Here, yk(i+1) is the continuation parameter; it is the variable yk(i+1) that corresponds to the k-thelement dyk(i+1) in the tangent vector and is usually λ at first but often becomes something elseas the nose point is neared. The parameter is the value to which yk is set, which would be thevalue found in the predictor step. As in approach a, we can solve this using Newton-Raphson.If no convergence, cut step size () by half and repeat.
61 Detection of critical point: We will know that we have surpassed the critical point when the sign of d in the tangent vector becomes negative, because it is at this point where the loading reaches a maximum point and begins to decrease. increasing|V|x decreasing
62 Selection of continuation parameter: The continuation parameter is selected from among and the state variables in y according to the one that ischanging the most with . This will be the parameter thathas the largest element in the tangent vector.relatively unstressed conditions (far from nose): generally relatively stressed conditions (close to nose): generally thevoltage magnitude of the weakest bus, as it changes a greatdeal as is changed, when we are close to *.The one changingthe most with λ ismost sensitive andrepresents avariable that wewant to be carefulwith as we look foranother solution,so it makes sense tokeep it constant.Typically, ykis going to beone of these.
63 Selection of continuation parameter (unstressed condition): The continuation parameter is selected from among and the state variables in y according to the one that ischanging the most with . This will be the parameter thathas the largest element in the tangent vector.relatively unstressed conditions (far from nose): generally .=> This looks like below.|V|y(i)y(i+1,p)y(i+1)Here, is fixed.
64 Selection of continuation parameter (stressed condition): relatively stressed conditions (close to nose): generally thevoltage magnitude of the weakest bus. Here, the voltage beingplotted is chosen as the continuation parameter.|V|y(i)y(i+1,p)y(i+1)Here, |V| is fixed.“Essentially, a variable is fixed as a parameter (the voltage), andthe parameter () is treated as a variable. This process of selectinga variable to fix is sometimes called the parameterization step.”-Scott Greene, Ph.D. dissertation, 1998.
65 How does the continuation technique alleviate the ill- A central question:How does the continuation technique alleviate the ill-conditioning problem experienced by a regular power flow ?Refer to the solutions procedures for the two corrector approaches.Perpendicular interesectionSolve simultaneously,for y(i+1)ParameterizationIn both cases, we use Newton-Raphson to solve, so we need to obtain theJacobian. But the Jacobian is slightly different than in normal power flow.
66 The Jacobian of the power flow equations is just Gy, but the Jacobian of the equations in the two corrector approaches will have an extra row and column.Here, C is the additional equation, and xk is the selectedcontinuation parameter.This addition of a row and column to the Jacobian has theeffect of improving the conditioning so that the previouslysingular points can in fact be obtained. In other words, theadditional row and column provides that this Jacobian isnonsingular at * where the standard Jacobian is singular.
67 Known codes for continuation methods: Claudio Canizarres at University of Waterloo: C-codeSeeUWPFLOW is a research tool that has been designed to calculate local bifurcations related to systemlimits or singularities in the system Jacobian. The program also generates a series of output files thatallow further analyses, such as tangent vectors, left and right eigenvectors at a singular bifurcationpoint, Jacobians, power flow solutions at different loading levels, voltage stability indices, etcI have Matlab code that does it – from Scott Greene.Venkataramana Ajjarapu (ISU): Fortran codePowertech has a program
68 Calculation of sensitivities for voltage instability analysis What is a sensitivity ?It is the derivative of an equation with respect to a variable.It shows how parameter 1 changes with parameter 2.It is: exact when parameter 2 depends linearly on parameter 1.It is approximate when parameter 2 depends nonlinearly on parameter 1,but it is quite accurate if it is onlyused close to where it is calculated.
69 Consider the system characterized by G(y). Then is the sensitivity of the equation G with respect to y,evaluated at y*.G(y)Slope is G/yevaluated at y*.yyy*yIt’s usefulness is that once it is calculated, it can be used to QUICKLY evaluate f(y) from G(y)G(y*)+ (G/y|y*)y,BUT ONLY AS LONG AS y IS CLOSE TO y*.
70 Consider parameter p: we desire to obtain the sensitivity of G(y,p) to p. Typical parameters p would be a bus load, a buspower factor, or a generation level.Very important to distinguish betweenvoltage sensitivitiesvoltage instability sensitivitiesWhat is the difference between them in terms ofwhat they mean ?how to compute them ?
71 Sensitivities for bus voltage These we compute at the current operating condition.For a given continuation parameter, they can be obtainedfrom the first predictor step in the continuation power flow.|V|Recall that this provides us withthe tangent vector, given by:The tangent vector is the vector ofsensitivities with respect to a smallchange in , so the portion of the vectordesignated as dV is exactly the voltagesensitivities.Currentoperating point
72 Sensitivities for voltage instability Here, it is important to realize that the measure of voltage instability,the loading margin, depends on an operating conditiondifferent from the present operating condition.The implication is that we must look at sensitivitiesof the loading margin, not of the voltage.|V|So we want the sensitivitiesevaluated at this point, i.e.,the SNB point.Loading marginCurrentoperating point
73 Derivation of loading margin sensitivities at SNB point. Let S be the vector of real and reactive load powers,and k be the direction of load increase.Also, define L as the loading margin (a scalar), so thatthe load powers resulting in the SNB point are given by:We desire to find the sensitivity of the loading margin L to achange in the parameter p. We denote this sensitivity by Lp.
74 Consider the system characterized by G(y,S, p) = 0 Assumption: the system has a SNB at (y*,S*, p*), i.e., :1. G(y*,S*, p*) = 0 (an equilibrium point)2. Gy(y*,S*, p*) is singular (zero eigenvalue), andw is a left eigenvector of Gy(y*,S*, p*), correspondingto the zero eigenvalue so that (by definition of the lefteigenvector)wT Gy(y*,S*, p*) =0 wT=0Note that Gy(y*,S*, p*), being singular, cannot be inverted, butwe can compute it (that is, Gy (y*,S*, p*)), and its eigenvectors.3. wT GS(y*,S*, p*) 0We want the sensitivity ofthe loading margin to p.
75 and we can obtain a curve of such points by The points (y,S, p) satisfying numbers 1 and 2 correspond to SNB points,and we can obtain a curve of such points byvarying p about its nominal value p*.Linearization of this curve about the SNB point results inwhere the notation |* indicates the derivatives are evaluated at the SNB point.Pre-multiplication by the left eigenvector w results in:By #2 on the previous slide, the first term in the above is zero. So...
76 Now recall the relation of the load powers to the loading margin…. 3/31/2017Eqt. *Now recall the relation of the load powers to the loading margin….Substituting this expression for the load powers into eqt. *,And the loading margin sensitivity to parameter p is:So p may be, for example,real power load at a bus (todetect the most effective loadshedding) or reactive powerat a bus (to determine whereto site a shunt cap).
77 Some comments about computing Lp The left eigenvector w must be computed for theJacobian Gy evaluated at the SNB point.You only need to compute w and GS once, independent ofhow many sensitivities you need. Methods to compute the left eigenvectorw include QR or inverse iteration.The vector of derivatives with respect to the parameter p, which is Gp, istypically sparse. For example, if you want to compute the sensitivity to abus power, then there would be only 1 non-zero entry in Gp.The matrix of derivatives with respect to the load powers, GS, using constantpower load models, is a diagonal matrix with ones in the rows correspondingto load buses. This is because a particular load variable would ONLY occurin the equation corresponding to the bus where it is located, and for theseequations, these variables appear linearly with 1 as coefficient.
78 Some comments about extensions Multiple sensitivities may be computed using Gp (a matrix) instead of Gp (a vector).In this case, the result is a vector.Getting multiple sensitivities can be especially attractive when we want to findthe sensitivity to several simultaneous changes. One good example is to find thesensitivity to changes in multiple loads.A special case of this is to find the sensitivity to changes at ALL loads, which isvery typical, given a particular loading direction k . ThenA sensitivity to a line outage may be obtained by letting p contain elementscorresponding to the outaged line parameters.
79 3 Some comments about extensions Zpq=R + jX p q jB A sensitivity to a line outage may be obtained by letting p contain elementscorresponding to the outaged line parameters: R (series conductance), X (seriesreactance), and B (line charging). Then use the multiple parameter approach.3Zpq=R + jXpqjBHere, p = [R X B]T.Note that p is NOT SMALL ! Therefore L may have considerable error.For that reason, this one needs to be careful about using this approach tocompute the actual loading margins following contingencies.However, it certainly can be used for RANKING contingencies. One mightconsider having a “quick approximation” and a “long exact” risk calculation.
80 Some comments about alternatives Greene, et al., also propose a quadratic sensitivity which requires calculation of a second order term Lpp . This is used together with the linear sensitivity according toIt requires significantly more computation but can provide greateraccuracy over a larger range of p.Invariant Subspace Parametic Sensitivity (ISPS) by Ajjarapu.Advantages:based on differential-algebraic modelprovides sensitivities at ANY point on the P-V curve