EECE 471 Final Solution, Fall 2001

ELEG 071 Power Systems Analysis Final Exam – Fall 2001/2002 Prof. S. KARAKI Page 1 of 5 ELEG 071E POWER SYSTEMS FUNDAMENTALS FINAL EXAM CORRECTION – Feb 4, 2002 PROBLEM 1 a) km r / .0 105 2 .1 605 .0 3372 Ω = × = m D with R D l b .7 114 6 10 6 : ln 10 2 3 7 = × × = × = − m Rb .0 0458 3.0 .0 305 .0 0229 2 = × × = km wl x and H km l Then / .0 317 10 50 10.09 10 2 : / .1 01 10 .0 0458 .7 114 ln 2 10 : 3 7 3 7 Ω = × × × × = = × = × = + − − − π b) km F R D C bc / 10 10 ln .8 854 10 2 3 6 12 µ π × × ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × = − S km wC y and F km C Then m R with c bc / .3 55 .0 0113 50 2 : / .0 0113 10 10 .0 052 .7 114 ln .8 854 10 2 : .0 051 3.0 .0 009 : 3 6 12 2 µ π µ π = × × = = = × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × = = × = − c) A I 608 2 2 .0 5067 300 = × × × = MVA Z V P MVA S C SIL 177 242 608 10 230 3 2 3 = = = × × × = − d) From the curve, for a short line of about 75 km, its power handling capability can be about × PSIL 5.2 which is much higher than 242 MVA MVA Z V P C SIL 177 299 2302 2 = = = .1 37 177 242 : = = PSIL P Then From the line capability curve the distance is about about 200 miles. ELEG 071 Power Systems Analysis Final Exam – Fall 2001/2002 Prof. S. KARAKI Page 2 of 5 PROBLEM 2 a) S B10MVA KV VB 1 = 12 line to line (l-l) Ω = = Ω = = = Ω = = − = − = = 12 1. 10 11 435 6. 10 66 14 4. 10 12 ) ( 11 ) ( 66 2 4 3 2 2 2 1 4 3 2 B B B B B B B Z Z Z Z l KV l V l KV l V V b) 1 . . 2 j p u PD + = . . 2.0 12 10 50 12 0.1 .0 01 . . 66 10 36 .4 .0 05 . . 12 10 40 12 2.0 .0 08 . . 11 10 25 11 2.0 2 2 2 2 2 1 2 2 2 p u X p u X p u X p u X S l t t = × × = = × = = × × = = × × = ∆-Y G Y-∆ 12:66kV 40 MVA XT1= 20% 66:11kV 25 MVA XT2= 20% XL= 4.36Ω 3 4 2 1 20+j10 MVA 50 MVA at 0.9 pf Xs = 100% ELEG 071 Power Systems Analysis Final Exam – Fall 2001/2002 Prof. S. KARAKI Page 3 of 5 c) . . 2.0 4.0 1 2 1 * 2 4 * 4 2 4 4 p u j j S V I V V I V jX R Z D D D D D D + = − = = = = + = Total impedance seen by the generator is: ( ) .0 34 4.0 2.0 .0 08 .0 01 .0 05 4.0 j j ZT + = + + + + = Generator current in p.u.: .1 23 . . .1 45 .0 34 4.0 1 1 p u j j Z V I T G − = + = = Total active and reactive power supplied by generator: ( )( ) . . .1 23 .1 45 0 1 1 * p u j j V I jQ P G G G + + = = + In real quantities: MVA j jQ P G G 12 3. 14 5. + = + d) ( )( ) . . 13.74 .0 8503 .0 202 .0 826 2.0 4.0 .1 23 .1 45 4 p u j j j Z I V D G o ∠− = − = + − = × = ( ) ( )( ) . . 15 6. .0 963 .0 26 .0 927 5.0 .0 52 85 .1 2.5 5. 18 .0 52 .1 85 .0 14 5.0 1 .0 14 5.0 .0 08 .0 01 .0 05 5.0 . . 5.0 4 1 p u j j V MVA j jQ P j j Z V I j j Z p u Z G G T G T D o ∠ − = − = − = + = + − = + = = + = + + + = = e) . . 5.4 45 9.0 50 p u MW Prated = = × = ( ) ( ) ( ) MVAr Q X V P Q MVAr Q G S a G G G 5 1 45 tan 5.0 2.0 1 1 5.4 tan 21 8. .0 436 50 9.0 sin cos 50 min 2 max min 1 max − = = = − − = − = = × = × = − o δ Ea j0.2 j0.05 j0.01 j0.08 XD RD V1 V2 V3 V4 ID ZT ELEG 071 Power Systems Analysis Final Exam – Fall 2001/2002 Prof. S. KARAKI Page 4 of 5 PROBLEM 3 a) ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − + + + − + + + − = 9 5 4 5 15 10 4 10 14 j YBUS b) ( ) ( ) 2.0 5sin 10sin 3 2 3 2 1 2 2 1 2 − − + − = ∆ θ θ θ θ V V VV P ( ) ( ) ( ) ( ) ( ) ( ) 2.0 9 5cos cos 10 1.0 5cos 15 cos 10 4.0 5sin sin 4 2 3 2 3 3 2 1 3 1 3 2 3 2 3 2 2 2 1 2 1 2 2 2 3 2 3 1 3 1 3 3 + + − − − = − ∆ − − − + − = − ∆ + − + − = ∆ V V V V V Q V V V V V Q V V V V P θ θ θ θ θ θ θ θ θ θ θ θ c) ⎥⎦ ⎤ ⎢⎣ ⎡ = 3 2 θ θ θ and ⎥⎦ ⎤ ⎢⎣ ⎡ = 3 2 V V V ⎥⎦ ⎤ ⎢⎣ = ⎡ ⎥⎦ ⎤ ⎢⎣ ⎡ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∆ ∆ ⎥⎦ ⎤ ⎢⎣ ⎡ − − ⎥ − ⎦ ⎤ ⎢ ⎣ ⎥ = ⎡ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∆ ∆ ⎥⎦ ⎤ ⎢⎣ ⎡ − − ⎥ − ⎦ ⎤ ⎢ ⎣ ⎥ = ⎡ ⎦ ⎤ ⎢ ⎣ ⎡ − − + + − + + .0 136 .0 045 .0 045 .0 082 15 5 5 9 10 1 9 5 5 15 9 5 5 15 1 3 2 1 3 2 1 3 1 2 3 2 1 3 2 1 3 1 2 J Q Q V V V V P P r r r r r r r r r r r r θ θ θ θ 2.0 2 = − ∆ Po 1.0 2 = − ∆ Qo 4.0 3 = − ∆ Po 2.0 3 = ∆ Qo .0 045 9773 .0 .0 0016 9992 .0 .0 0227 .0 0008 2.0 1.0 045 .0 0016 .0 4.0 2.0 3 2 1 1 3 1 2 1 1 3 1 2 ∠ − = ∠ − = ⎥⎦ ⎤ ⎢⎣ = ⎡ ⎥⎦ ⎤ ⎢⎣ ⎡− ⎥ = ⎦ ⎤ ⎢ ⎣ ⎡ ∆ ∆ ⎥⎦ ⎤ ⎢⎣ = ⎡ ⎥⎦ ⎤ ⎢⎣ ⎡− ⎥ = ⎦ ⎤ ⎢ ⎣ ⎡ ∆ ∆ − − V V J V V J θ θ G 3 2 1 G SD3=40+j20 SD2=20+j10 SG2=40+j20 j0.2 j0.25 j0.1 ELEG 071 Power Systems Analysis Final Exam – Fall 2001/2002 Prof. S. KARAKI Page 5 of 5 PROBLEM 4 a) ( ) 0 .0 012 2 7 1 = − × + λ PG (1) ( ) 0 .0 018 2 8 2 = − × + λ PG (2) 700 2 1 = + G G P P .0 012 2 8 1 × − = λ PG .0 018 2 7 2 × − = λ PG ( ) MW P MW P MWh G G 263 3. 2) ( 436 7. )1( 17.48$/ 69.44 513 9. 700 513 9. 69.44 .0 018 2 8 .0 012 2 7 .0 018 2 1 .0 012 2 1 700 2 1 = ⇒ = ⇒ = + = − ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ × + × ⎟ − ⎠ ⎞ ⎜ ⎝ ⎛ × + × = ⇒ λ λ λ b) 21 8. $/ MWh 69.44 513 9. 1000 = + λ = MW P MW P G G 3. 383 7. 616 2 1 = = PG1 is above limit, so set it to limit. ( ) MWh MW P MW P G G 22 4. $/ 400 .0 018 2 8 400 600 2 1 = × × + = = = λ