課程名稱︰電力工程導論 課程性質︰電機系複選必修 課程教師︰劉志文 開課學院：電資學院 開課系所︰電機系 考試日期（年月日）︰2010/11/11 考試時限（分鐘）：120 mins 是否需發放獎勵金：是 謝謝 （如未明確表示，則不予發放） 試題 : Introduction to Power Engineering Midterm exam 2010/11/11 1. Please List 5 kinds of primary energy. (5%) 2. Please list 5 kinds of renewable energy. (5%) 3. State the Theorem of Conservation of Complex Power in terms of mathematical formula. (10%) 4. What is balanced three-phase system? Pleasse explain it. (10%) 5. A single phase load draws 10kW from a 416-V line at a power factor of 0.9 lagging. (a) Find S = P+ jQ (5%) (b) Find |I| (5%) 6. If a straight infinitely long wire of radius r has uniform current density in the wire and total current i, then calculate the flux linkages inside the wire. (10%) 7. Calculate the inductance per meter of each phase of a three-phase transmission line(Fig.1). Assume that 1. Conductors are equally spaced D, and have equal radii r. (5%) 2. ia+ib+ic=0 (5%) Fig.1 -> 請見課本P.63 Fig. E3.2 8. A 60-Hz 138-kV 3Φ transmission line is 200 km long.the distributed line parameters are r = 0.15 Ω/km l = 2 mH/km c = 0.012 uF/km g = 0 The transmission line delivers 40MW at 132kV with a 95% power factor lagging. Finding the sending-end voltage and current. Find the transmission line efficiency. (10%) [Hint]: Zc = √(z/y), γ = √(z*y), z = r + jωl, y = g + jωc V(x) = V2cosh(γx) + Zc*I2sinh(γx), I(X) = I1cosh(γx) + V2/Zcsinh(γx), where V2, I2 are receiving-end voltage and current. 9. In Fig. 2, assume that V1 = 1∠0度 Fig. 2 -> 請見課本P.103 Fig. 4.4 Pick QG2 so that |V2| = 1. In this case what are QG2, SG1 and ∠V2? (10%) 10.Consider the following system shown in Fig. 3. In the transmission system all the shunt elements are capacitors with an admittance Yc = 0.01j, while all the series elements are inductors with an impedence of Zc = 0.1j. Find: (a) Ybus matrix (5%) (b) power flow equations. You don't need to solve the equations. Just list equations. (5%) Fig. 3 -> 請見課本P. 347 Fig. E10.6 11.Use the Newton method to solve f1(x) = (x1)^2 + (x2)^2 - 1 = 0 f2(x) = (x1) + (x2) = 0 With an initial guess " 0" = 1 and " 0" = 0. Do two iterations. (10%) x x 1 1 //""裡 0和1分別是x的上標與下標 --

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