Question

The fuel-cost curves for a three-generator system are given as follows: G(Pcı) = 2000 + 45. Pai + 0.05. (Pa)2 C2 (PG2) = 500 + 50. PG2 + 0.025 (P62)2 Cz(PC3) = 400 + 60 P3 + 0.025. (Pc)2 Generator limits are: 50 <P61 < 300 200 < PG2 < 800 100 < P03 < 400 For a load of 1000 MW, use the lambda iteration method to: 1) test to ensure the solution falls within the provided 1 and 1" range (5 points), 2) determine initial values for 2M, PG12M), PG2(2), Pc3(M) (5 points), and 3) calculate new 1 and " values after evaluating the initial 2M (5 points). Use starting values of 1 = 60 and 1 = 80. Be sure to consider the generator limits. Please ignore transmission loss penalty factors.

Answer 1

XI Sotahiem :-_- ___The fuel.cest curves fow-three generator system_03. grian G. (24.) = 2000 445. Pean +0.05.(P4092 C2 (PG2) = 500+ 50. PGL 4 0.025, CPG 212 C3 (94) = 400 460. Pas 4 0.025. (Pag) 2 Generator limits ase; 50 < pa s 300 200 € VGL 2 800 100 = las 2400 Pla) a 2-45 at a-50 + 2-60 - 1000 0.1 0.05 0.05 but limits need to be considered PC60) - 150 + 200 - 1000 = -650 (Pal of low Limit) P (BO) = 3rot 600 4400 - 1000 = 350 CPa, at high limit) Initial values bound Ser Am a 60+80 the solutions - 50 PCHD) 250 + 400 + 200 - 1000 -150 Less than gero so pa pm new am 70 180_215_ 100 IP675 - 300 + 500+ 300 - 1000 cllScaGreaforth than seve se 21 am CamScanner

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