Name: Use a 100-MVA base. The base loads and generations are as follows: Bus Load (MW) Gen (MW) Min Gen (MW)

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1 Name: Eam, EE 55, Fall, Dr. McCalley Close books, closed notes, calculator permitted, 75 minute time limit. Students are permitted to brin teir W# to te eam and use it.. 4 pts Tis problem is identical to prob. in te W&W tet; if you worked it in advance and brout te solution wit you, you may turn in tat as your solution to tis part of te eam. Consider te tree-bus network sown in te fiure below were =.pu, =.4pu and =.5 pu. Te [X]-matri for tis system is also iven below:.8 [X ] Use a -MA base. Te base loads and enerations are as follows: Bus Load MW Gen MW Min Gen MW Ma Gen MW a. Find base power flows on te transmission lines. b. Calculate te eneration sift factors for line -. Calculate te sift in eneration on bus and so as to force te flow on line - to zero MW. Assume for economic reasons tat any sifts from base conditions are more epensive for sifts at te enerator on bus tat for sifts on bus, and tat te enerator on bus can be sifted witout any penalty. a

2 . pts Consider te four-bus, five-line power system sown below. All indicated values are admittances iven in per-unit. y =.- y =- y 4 =- y 4 =- y =- 4 y 4 =. a. Develop te Y-bus matri i.e., te admittance matri Y bus b. Develop te B matri used in te DC power flow calculation. Assume bus is te reference. Wit resistance nelected, and inorin sunt terms, we ave: Y bus Droppin te, and multiplyin by -, we ave: Since te reference is bus, we eliminate row and column, resultin in ' B. pts A bus system as enerators, all of wic are modeled in a power flow proram wit specified terminal voltae and reactive limits, i.e., tey are voltae-controlled buses, and tere are no oter voltae-controlled buses besides tese enerator buses. a. At te beinnin of te Newton-Rapson solution procedure initialization of iteration : i. ow many type P buses are tere in te power flow model? ii. ow many type PQ buses are tere in te power flow model?

3 iii. Wat is te number of power flow equations used? iv. ow many bus voltae manitudes are unknown? v. ow many bus voltae anles are unknown? b. After one iteration, te reactive power of two of te type P buses is computed to eceed its upper reactive limit. In initializin for iteration, relative to iteration + is increase, - is decrease: i. Wat is te cane in te number of type P buses in te power flow model? ii. Wat is te cane in te number of type PQ buses in te power flow model? iii. Wat is te cane in te number of power flow equations used? iv. Wat is te cane in te number of bus voltae manitudes tat are unknown? v. Wat is te cane in te number of bus voltae anles tat are unknown? a. i. 8 ii. 5 iii. iv. 5 v. 5 b. Lk i. - ii. + iii. + iv. + v pts A two-bus power system is interconnected by a transmission line avin series admittance of Y = pu and total line carin of Y carin =.4 pu. Bus is a load bus wit specified demand as P D pu and Q D pu. Bus is a enerator bus wit specified terminal voltae manitude of pu. We desire to solve te power flow problem for tis system. a. Form te Y-bus for tis system. b. Identify te variables in te solution vector. c. Write down te mismatc equations tat are required in te solution procedure. Epress eac equation symbolically no numbers. Denote eac equation by i, i=, d. Write down te acobian matri to be used in te solution procedure. Indicate te elements in te matri usin partial derivative notation; you do not need to differentiate anytin or provide any numerical values. e. Write down te update formula for tis system in terms of and te solution vector. f. A enerator is now brout on-line at bus so tat it becomes a type P bus. Repeat part d a. Y bus b. solution c. P Q, sc, sc P Q

4 4 d. e. Q P f pts A -bus power system is interconnected by a transmission line avin series admittance of y =- pu. Bus is a load bus. Bus is a enerator bus. A SCADA system presents te followin measurements to te state estimator:..;.; P z z z. Te standard deviations of error on te meters are.5,.5, and., respectively. a. Identify te elements for te state vector for tis problem. States are =[ θ ] T b. Epress, te measurements as functions of te states usin AC power flow epressions. sin c. Te first iteration of te state estimator assumes an initial state and solves A =b usin T R A, T z R b Epress, R, and z- for tis problem. You need not evaluate anytin, ust epress in a fasion so it is clear tat you know ow to evaluate it m R

5 . z. sin. pts Answer te followin questions: a. Consider te full Newton power flow, te fast-decoupled power flow, and te DC power flow. For eac of te followin caracteristics, rank te tree metods as best,, and worst. Metods tat are equal in one of te caracteristics sould et te same rank. Solution metod Computational Speed Accuracy Newton-Rapson Fast decoupled DC Solution metod Computational Speed Accuracy Newton-Rapson Fast decoupled DC b. An enineer modifies a full Newton power flow metod by zeroin te elements in te Pθ and Q submatrices and usin te resultin acobian matri to update te solution at eac iteration. Do you tink tis metod will be faster tan te full Newton metod? Wy or wy not? No, it would not be faster and in fact would probably not convere at all. Te reason is tat te elements of te Pθ and Q submatrices are te larest in manitude elements of te acobian, as a result of te fact tat te real power is ily sensitive to voltae anle, and reactive power is ily sensitive to voltae manitude. c. Identify te four continency cateories of te NERC disturbance-performance table, and for eac one, caracterize te event in terms of ow many elements may be lost, and indicate weter demand interruption is allowed. Cateory A: No elements lost, demand interruption is not allowed. Cateory B: Loss of a sinle element, demand interruption is not allowed. Cateory C: Loss of or more elements, demand interruption is allowed but most be controlled. CateoryD: Etreme event resultin in loss of or more elements, demand interruption is allowed. d. Wic of te below situations are in te alert state? i. Te power system is operatin wit all elements in. Tere is ust one loss of one element continency wic will result in second circuit to eceed its continuous ratin but not its emerency ratin. ii. Te power system is operatin wit all elements in. Tere is ust one loss of one element continency wic will result in a second circuit to eceed its emerency ratin. iii. Te power system is operatin wit all elements in. Tere is ust one loss of one element continency wic will result in an operatin condition suc tat a second loss of one element continency, if it occurred followin manual system adustments subsequent to te first continency, would result in a tird circuit to eceed its emerency ratin. Situation ii and iii are in te alert state. Situation i is not. 5

6 e. Optimal orderin sceme # numbered buses so tat a lower numbered node as less or equal number of actual adacent brances tan any ier numbered node. In applyin suc a sceme to a 5 bus network, wic column of te Y-bus would tend to ave te most non-zero elements, column, column, or column 4? Eplain your coice. Column 4 would tend to ave te most non-zero elements. Since column 4 corresponds to a bus numbered ier tan te oter two buses, it must be connected to as many buses, and most likely more buses, tan eiter of te buses correspondin to columns and. Since eac time a bus is connected to anoter bus, it will create anoter non-zero element in te column correspondin to tat bus, te ier numbered columns will tend to ave te most non-zero elements. f. You are iven te eneration sift factors t b,k and t b, wic are based on a reallocation policy of bus # compensates, i.e., te sift factors are computed usin bus # as a sinle slack bus. Write down an epression tat sows ow to use tese sift factors to compute te cane in flow on branc b for ΔP k = +.5 pu and ΔP = -.5 pu? Pb tb, kpk tb, P. An AC state estimator based on minimization of least square error informs tat te current condition is unobservable. i. Wat led to tis detection? Sinularity of te A matri, or, equivalently, identification of te matri as avin rank less tan n, were n is te number of columns of. ii. Wat te least epensive way of correctin te problem so tat te condition becomes observable? Add pseudo-measurements.