Pb1 y13 =-j10 Pb5. Pb4. y34 =-j10

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1 EE 55, Exam, ake-home. Due Monday, April, 06, 5:00pm. You may use class notes or any reference materials (e.g., books, etc.) that you like; however, you must work alone, i.e., you should not be communicating with anyone else about how to work the problems on this exam.. Generation planning (5 pts): Develop the complete set of equations for generation expansion plan formulation -b (GEP-b), based on the following data and assumptions. a. Plan for 30 years, at 5 year intervals (thus you should have 6 investment periods (at years 5, 0, 5, 0, 5, and 30). b. Use annual discount rate of 9%. c. Maintain a reserve of 5% above peak load. d. Assume a CO cost of $5/M (you will need to add a term to the objective function to account for this). e. ransmission topology and data: wo comments follow Note the definitions of positive branch flow for each P bk, k=,,5. he admittances y ij are given on a 00 MVA power base. Pg Pg Pb3 y =-j0 Pb3 y4 =-j0 Pb y3 =-j0 Pb5 Pd y3 =-j0 4 Pb4 y34 =-j0 3 Pg4 Pd3 f. Line flow constraints: -500MW P bk 5000MW, k=,5. he per-unit expressions of these constraints, on a 00 MVA power base, are -500 P bk 500. Given the level of loads (see item (j) below), this implies the transmission has infinite capacity. g. Existing generation data: Bus Bus Bus 4 echnology PC (j=) NGCC (j=3) C (j=4) Capacity (MW) h. echnology characteristics PC (j=) PC (j=) NGCC(j=3) C (j=4) Wind (j=5) Capacity credit Investment cost (000$/MW) Fixed O&M costs (000$/MW/yr) Var O&M ($/MWhr) Full-load heat rate (MBU/MWhr) Fuel cost ($/MBU) CO emissions (lbs/mwhr) Capacity factor 0.35 Salvage value (000$/MW)

2 i. Load duration curve and corresponding blocks. Assume a load growth rate of %/year, applied to each value d, d, d 3, for each of the two loads in the system, with the durations h, h, and h 3 remaining the same. herefore, for example, for the load at bus, d would be.0,.084,.95,.395,.4568,.6084,.7758 pu for years, 5, 0, 5, 0, 5, and 30, respectively. You will need to compute the values of d and d 3 for these years as well. hen you will need to do the same for the load at bus 3. d d L (MW) d3 h h+h h+h+h3 Number of hours that Load > L Initial load at bus (Pg) Block Block Block 3 Load (pu) d = d = d 3 =0.6 Duration (hrs) h =000 h =5000 h 3 =760 Initial load at bus 3 (Pg3) Block Block Block 3 Load (pu) d =. d =0.9 d 3 =0.7 Duration (hrs) h =900 h =500 h 3 =660. Production costing ( pts): a. (4 pts) A load duration curve characteries the next year for a power system is given below. he power system is supplied by one 3 MW unit. Identify the minimum load, the loss of load probability, the loss of load expectation, and the expected unserved energy, assuming the 3 MW unit is perfectly reliable. 0.6 F Dr ( d e ) 0. d e (MW) Solution: Minimum Load= MW, LOLP=P(d>3)=0.333, LOLE=0.333*8760=97 hrs EUE=Area*=0.5()(.333)(8760)=0.666(8760)=460MWhrs

3 Solution: b. (5 pts) Now consider that the power system is supplied with one 4 MW unit having a forced outage rate of 0.5. he load is still characteried by the load duration curve above. Compute loss of load probability, loss of load expectation, and expected unserved energy. f Dj (d j ) 0.6 F Dr ( d e ) * 0. d e (MW) 0, (b) C = (c) (d) d e d e = F r ( () d D e ) (e) d e So LOLP=P(d>4)=0.5, LOLE=0.5*8760=90hrs EUE=8760{()(0.5)+0.5(3)(0.5)}=8760{ }=5475MWhrs c. (3 pts) Consider the two situations you have assessed in parts (a) and (b) of this problem. Complete the following sentence: Although situation has the larger LOLP; situation has the larger EUE because. Solution: Although situation (a) has a larger LOLP, situation (b) has a larger EUE because when load is interrupted in situation (a), it will only be for 0 to MW, whereas when load is interrupted in situation (b), it will be for 0 to 4 MW.

4 d. (4 pts) Compute the energy expected to be produced by the 4 MW unit of the year. Solution: E=A**Area where Area is computed from the previous load duration curve, i.e., the one before the unit was convolved in, according to Area x x j j F ( j) D e ( ) d his is the load duration curve given in the problem statement. In this case, x j- =0 and x j =4, and so the area is A=()()+0.5(3)()=.5. he availability is 0.75 so that E=0.75*8760*.5=645 MWhrs. e. (3 pts) Assume the 4 MW unit is a thermal unit. What is the minimum additional information you need to know to compute the cost incurred by unit 4 in producing the amount of energy developed in part (d)? Solution: We need the heat rate and the fuel price. f. (3 pts) After reviewing a production cost program which computes annual production costs in a way that is similar to the procedure suggested in parts (a)-(c) of this problem, an engineer decides to write a production cost simulation program that will allow representation of the power system with much greater operational fidelity. What is the single most important change that the engineer needs to make, relative to that suggested in parts (a)-(e)? Solution: he engineer needs to represent the power system chronologically, hour by hour. 3. Benders (7 pts): he relation between primal and dual optimiations can be stated as: Primal max s. t. x 0 c A x b subproblem A w * Consider the following problem: max subject to : x 4x 5w w 0, : x w 0 x 3w 8 w, x 0, w Dual subproblem : min s. t. A integer 0 c b A w* a. (4 pts) Express this problem in the form of Benders, i.e., express an appropriate master problem and an appropriate (primal) subproblem.

5 Solution: Master problem: max s.t 5w Q : w 0, Subproblem: x 0 * * max 4x subject to x 0 w x 8 3w M, b. (3 pts) Identify an initial solution to the master. Solution: * M, w 0 w 0, w integer c. (4 pts) Formulate the subproblem dual in terms of w*, and based on the dual formulation, and the solution to the master problem found in part (b), explain why you know that the primal subproblem is infeasible. Solution: We first express the subproblem dual. c 4, c 5, A, A, 3 b min s. t. 0 A c ( b A w*), 0 Substituting w*=0, the subproblem dual becomes: min s. t. 4, min s. t * 8 3 w

6 Because the all coefficients in the objective are negative, and the variables are constrained to be non-negative, the subproblem dual is unbounded; an unbounded dual necessarily implies that the corresponding subproblem primal must be infeasible. d. (3 pts) Given the sub-problem primal is infeasible, explain why it is possible that we may still find a solution to this problem (and thus, that this solution will be feasible in the subproblem primal). Solution: he reason that we may still find a solution to this problem, in spite of the fact that the subproblem primal is infeasible in this step, is that the subproblem primal is expressed in terms of w*. If we further constrain w*, it will affect the subproblem primal s feasibility region. e. (3 pts) Find a constraint on w which will make the subproblem primal feasible. Solution: he feasibility constraint can be found by expressing (b-a w) λ 0, or 0 w 8 3 which results in from which we observe that 0 (0 w) (8 3 w) 0 0 w 0 83w 0 or 8 0 w w 3 And we conclude that w 8/3. 4. Planning principles (0 pts): A good paper on planning (you should read in its entirety - it can be found in IEEE XPlore) is: M. Awad, S. Broad, K. Casey, J. Chen, A. Geevarghese, J. Miller, A. Pere, A. Sheffrin, M. Zhang, E. oolson, G. Drayton, A. Rahimi, and B. Hobbs, he California ISO transmission economic assessment methodology (EAM): principles and application to Path 6, Power Engineering Society General Meeting, 006. his paper lists five principles to economic evaluation of proposed transmission upgrades. I have chosen from the paper an excerpt for each principle. For each one, identify how you would implement this principle. Where appropriate, base your answer on information, concepts, and/or methods we have covered in EE 55. I am expecting from you at minimum a -3 sentence paragraph for each principle. First key principle: Benefit Framework: Second key principle: Network Representation:

7 hird key principle: Market Prices: Fourth key principle: Uncertainty:

8 Fifth key principle: Resource alternatives to transmission expansion: Fifth key principle (continued):

9 5. ransmission line parameters (6 pts): A 765 kv transmission line has phase separation of 36 feet. Each phase is a 4-conductor bundle with 4-inch diameter. he conductor type is ern (795 kcmil). he following data is obtained from tables: Ind reactance ft spacing for 4-conductor bundle at 4 : X a =0.08Ω/mile Ind reactance spacing factor for 36 phase separation: X d =348Ω/mile Cap reactance at ft spacing for 4 conductor bundle at 4 : X a =0.005MΩ-mile Cap reactance spacing factor for 36 phase separation: X d =0.06MΩ-mile a. (6 pts) Compute the SIL for this line configuration with 4 conductors per bundle. Solution: Get per-unit length inductive reactance: So X L =X a +X d = =68 ohms/mile. Now get per-unit length capacitive reactance. X C =X a +X d = =0.3E6ohms-mile So =jx L =j376 Ohms/mile, and y=/-jx C =/-j( )=j mhos/mile j.468 Z C 6.96ohms y -6 j VLL 7650 PSIL.5785e Z 6.96 C he SIL with 4 conductors is 578 MW. b. (4 pts) A designer considers reducing the number of conductors per bundle from 4 to 3, while maintaining conductor spacing and phase spacing at 4 inches and 36 feet, respectively. Would you expect this change to increase or decrease the surge impedance loading? Support your answer. Solution: he 3 conductor bundle is an equilateral triangle with a GMR of (r *4*4) /3, whereas the 4 conductor bundle is a square with a GMR of (r *4*4*33.94) //4 where 33.94=sqrt(4 +4 ). herefore, the GMR R b of the 3 conductor bundle is slightly smaller than the GMR of the 4 conductor bundle. his causes the 3 conductor bundle to have larger inductive and capacitive reactances, per the below equations, X X L C f ln.0 0 f ln Dm R b X X a d ln.7790 ln f c m R f b X X ' a d /mile D - mile and therefore a larger surge impedance and therefore a smaller SIL. c. (6 pts) he designer desires to maximie the power transfer capability of the line. Relative to the design of part (a) If the line is 30 miles long, what changes would you consider?

10 Solution: Use a conductor with a larger diameter and therefore higher ampacity. If the line is 300 miles long, what changes would you consider? Solution: Increase SIL by decreasing phase spacing, increasing the number of subconductors in a bundle, or increasing the bundle diameter.