ECEN 460 Exam 1 Fall 2018

Transcription

1 ECEN 460 Exam 1 Fall 2018 Name: KEY UIN: Section: Score: Part 1 / 40 Part 2 / 0 Part / 0 Total / 100 This exam is 75 minutes, closed-book, closed-notes. A standard calculator and one 8.5 x11 note sheet are allowed. Be sure to include correct units where necessary. Partial credit will be given.

2 Part I: Brief answer (4 points each) Briefly answer 9 of the following 10 questions. You may pick one to skip for full credit by circling the number. 1. What s the distinction between energy and power? Power is the time derivative of energy 2. What units for power would be appropriate for peak usage of a fast-food restaurant? kw. What is one reason AC is used for nearly all medium and large power systems today? Allows the use of transformers to change voltage levels, more efficient machines 4. What U.S. state is mostly served by its own synchronous AC interconnection? Texas 5. In Texas, what is the largest, most significant source of renewable electric energy? Wind 6. What overall effect did the Public Utilities Regulator Policies Act (PURPA) of 1978 and the National Energy Policy Act of 1992 have on the structure of the U.S. electric utility industry? Restructuring vertically integrated monopoly into regulated competitive markets 7. What major event in power engineering history happened on August 14, 200, and where? Electric power blackout in northeastern USA and eastern Canada 8. What is synchronous about a synchronous machine? The mechanical rotor speed is synchronized with the electric AC frequency 9. What two tests are used to determine an electric machine s synchronous reactance? Open circuit test and short circuit test 10. What is one phenomenon that causes power losses in transformers? Are these losses real or reactive? Possible answers: - Real losses: winding resistance, hysteresis, eddy currents - Reactive losses: leakage inductance, magnetizing current

3 Part II: Shorter calculation (6 points each) Answer each question or problem, showing your work. 1. What is the time signal corresponding to the phasor V = 2 15 kv? v(t) = 2 cos(2πft 15 ) kv 2. The current into a single-phase load is I L = A, and the voltage across the load is V L = V. What are the real power P, the reactive power Q, and the power factor pf? S L = V L I L = (100)(20 10 ) = VA = j47. VA P = W and Q = 47. var pf = cos φ = cos 10 = lagging. A three-phase delta-connected load has Z Δ = 5 + j10 Ω. What is the equivalent wyeconnected impedance? Z Y = 1 Z Δ = 5 + j10 Ω = j. Ω

4 4. What would be the steady-state mechanical speed in revolutions per minute (RPM) for an 8- pole synchronous machine connected to a 60 Hz electric grid? s mech = s elec 2 8 = 60 Hz 2 = 15 Hz = 900 RPM 8 5. The circuit below is the per-phase equivalent of a three-phase system, with the transformer turns ratio given line-to-line. Finish filling out the table below with the rest of the per-unit bases for the circuit shown. You do not need to solve the circuit. Bus 1 Bus 2 S base ( phase) 100 MVA 100 MVA V base (Line-line) 120 V 7.2 kv I base = 481 ka 100 = 8.02 ka 7.2 Z base = mω = Ω

5 Part III: Longer calculation (15 points each) Solve both problems, showing detailed steps. Box in the final answers. 1. The design diagram of a newly-installed transmission line is shown below. There are three conductors bundled together at each phase, arranged with equal spacing of 10 inches. The three phase bundles are arranged vertically and spaced 8 feet apart. Assume the line is fully transposed and operated at 60 Hz. The Finch conductor is used. Calculate the per-phase, per-distance series impedance Z, a using the data from the attached conductor table. (This includes the resistance and inductance, but not the capacitance.) R = Ω mile = 0.02 Ω mile GMR = (10 in)(10 in)(0.045 ft) = 52.2 in =.74 in GMD = (8 ft)(8 ft)(76 ft) = ft X = ωl = 2πf μ 0 GMD ln 2π GMR = 2π ln = Ω m = Ω mile Z = R + jx = j Ω mile = ft ft ft

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7 2. In the single-phase circuit below, a 120 kw load is served by a source through a transmission line with impedance Z = 5 + j18 Ω. The load has a lagging power factor of Assume the single-phase, 60 Hz system is always operated such that the load voltage is 20 kv. a. Calculate the capacitance of the capacitor that should be added in parallel with the load to correct its power factor to 0.95 lagging. b. Also calculate the real and reactive power losses in the transmission line with and without this additional capacitor. S 0.85 = P pf = 120 kw 0.85 = kva Q 0.85 = S 2 P 2 = ( kva) 2 (120 kw) 2 = 74.7 kvar S 0.95 = P pf = 120 kw 0.95 = kva Q 0.95 = S 2 P 2 = ( kva) 2 (120 kw) 2 = 9.44 kvar Hence for the capacitor: Q cap = Q 0.85 Q 0.95 = 74.7 kvar 9.44 kvar = 4.9 kvar Z cap = j V2 Q (20 kv)2 = j 4.9 kvar = j kω; C cap = 1 ωx = 1 = 21.6 nf 2π kω Without the capacitor: I = ( V ) S kva = = A 20 kv P loss = I 2 R = (7.059 A) 2 5 Ω = W Q loss = I 2 X = (7.059 A) 2 18 Ω = var With the capacitor: I = ( V ) S kva = = A 20 kv P loss = I 2 R = (6.16 A) 2 5 Ω = W Q loss = I 2 X = (6.16 A) 2 18 Ω = var

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