PS12_Solution[2] - Name Section 3 w V EE331 Homework PSl2...

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Unformatted text preview: Name: Section: 3 w/ V EE331 Homework PSl2 — spring 2011 Assigned Problems: CH 12 - 12.40(Ans: 1553 W) c 12.47 0 12.81 (Hint: first compute the total complex power of load, then add in total complex power of line to get the total complex power of the source) Additional assigned problems: 0 Next page of this PS Additional Problems (Instructor Option): 0 Any as assigned by instructor Page 1 of 3 Chapter 12, Problem 40. For the three-phase circuit in Fig. 12.59, find the average power absorbed by the delta-connected load with ZA = 21 + j24 Q. 14mg}; \ mu .131 ,,,, W .............. (D COMch DeH“ Low} qwfe 55:19 \i :9?) I, x K 2 film/1211“ \' rum '0 Figure 12.59 04 m 00“) . ark @ \Mfit EGANAQA’ J7 \ 9 E7 | ?_ +gd ’ ft. : IOOQVM 2 g-L<’qL-7’A-s:ff (6‘33?) " ( “3.53) Afifimr -3 :- [f‘filafz ‘: ($«L4m)2.?fl 2 517.3103 Chapter 12, Problem 47. The following three parallel-connected three-phase loads are fed by a balanced three-phase source. Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading Load 3: 450 kVA, unity pf If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the line impedance is zero. g i’ S) 4" SB : TOT DSonLf + 30°WA<-t‘l9° + HSowtco“ : finguAM? Chapter 12, Problem 81. A professional center is supplied by a balanced three-phase source. The center has four plants, each a balanced three-phase load as follows: Load 1: 150 kVA at 0.8 pf leading Load 2: 100 kW at unity pf Load 3: 200 kVA at 0.6 pf lagging Load 4: 80 kW and 95 kVAR (inductive) If the line impedance is 0.02 + j0.05 £2 per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source. LCoQ'Ofi) J, /S\ : I50 kw; <~3w¥o A A /\ A . 0 SW): sl+sa+53+§ : uaomsd: HSLX’JWS‘Wo/A TDTAL ~ /\ o wavy? do“? “’7 C0“?"‘\€ SLuMz o (ELI: (3%" : : Star-LA ' WU fl NJ '5 L ATLMAS . A r: ’3 Q + K ) SLW: 3.11! - 2L:3-(SHQ.4) (0.01 dons Tbhd : 0,101+ wigs \<\/A Name: Section: EE331 Homework P812 - spring 2011 Problem 1: In one ship electric drive alternative, the generated voltage is produced at 240Hz in order to reduce the size and weight of the required transformers. If the prime mover gas turbine must operate at 3600rpm a. How many poles must the stator and rotor be wound for? b. What is the required frequency of the field (rotor) winding? 0. At what speed (rpm) does the stator produced magnetic field rotate? \(I a) N P : \30 ‘g (fi “ Nudeaw {30$wa DO +‘\M€$ 1“” “CW/L L? P“- ‘go'm 1 ‘6 PURE shoe BB EA) \DNIXMS oQQa‘foxr TSOQ—W (Imzo HZ C) 5‘000 \er‘ —. 13¢.\Mq madam/Fo¥o¢ Rpm (Sbwchvowws> Problem 2 a): Given a 4-pole generator must produce SOHz, what is the required speed (rpm) of the prime mover? L47 N : 9%0 1 \SOO VPM b): For a 6-9019 generator a. TRUE FALSE\ The @rotor winding current will produce 6 poles b. TRUE ( FALSE ) The slip rings and brushes convert AC voltage into DC voltage c. TRUE FALSE If the prime mover direction is reversed, then the induced stator rotating field will also reverse . TRUE FALSE The phase sequence from the generator can be reversed by reversing the direction of rotation of the prime mover e. TRUE The phase sequence from the generator can be reversed by changing the direction of the field current TRUE FALSE From the perspective of the system, the phase sequence from the generator can be reversed by swapping connections to the b and c phases Page 2 of 3 Name: Section: EE331 Homework P812 — spring 2011 Problem 3: A 4-pole, 60Hz, Y-connected, 3-phase generator has a regulated line—to-neutral voltage of Vas = 260V 40° , a synchronous reactance of 0.069, a stator resistance of 0.0039, and Lsf = 0.02H. rms The balanced 3-phase load draws 2MW at a lagging 0.8 power factor. a Compute the 3-phase complex power of the load (magnitude and angle). b. Compute the generator phase current (magnitude and angle; hint: the angle of the current should be negative for the load to have a lagging power factor). 0. Find the required excitation voltage (Ea ). (1. Calculate the required field current. e. State the required speed of the prime mover. Problem 4: A 3-phase, Y-connected, 6-pole, 50Hz generator has X5 = IQ, negligible RS (stator resistance), and Lsf= 0.2H. The machine operates at a power angle of 31.48° with a field current of 100A. The line-to-neutral terminal voltage is 2400Vm15. Find the speed of the prime mover in rad/sec. Determine the excitation voltage. Find the line current. Find the power consumed by the load and its power factor. Find the developed torque. 99.0w» Page 3 of 3 92.: co§‘o.¥2 397° (gr: MU A<3L7 D m ,A_>\< 4n Q.SMA<3L-7°_ m” :30 J, A) : _3905<3 . A \:)> ST 45 45 ‘Las 3.;Eov <00 /L 00, fm: 3905<~3lofi°A / * /l A C) E‘ 21‘“ (QOOBSHaogDOJr DEM/<00 g” ‘— IO. VON) 5' Li 1 SfiPonAfik /- 9 AF W5 Z l A) Le: Ital 9 : LHDZMI - LSF-Qe 0.03H '(DW (10%) C) M?: DO? L E ‘ ‘3 ~ I; < . (+3 ( (wok: wad“ A): /§—: 'IA’V ( as :B-quo<o°.94oqA (307° ...
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PS12_Solution[2] - Name Section 3 w V EE331 Homework PSl2...

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