hw #4 - /fl/y EB 4420 Homework#4 Name MATIHEN $9...

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Unformatted text preview: /fl//y EB 4420: Homework #4 Name: MATIHEN. $9 NéfiL-L— Return by 11/21/05 1. A 3-phase, 100 kW, 4_60 V, 60 Hz, 8-pole induction machine has the following equivalent circuit parameters: R1 = 0.07 9, X1 = 0.2 Q, R 2= 0.05 D, X 2 = 0.2 Q, Xm = 6.5 Q. a) Draw the equivalent circuit of the machine ignoring the core resistance Rye. b) If the machine is connected to a 3-phase 460 V, 60 Hz supply, determine the starting torque, the maximum torque the machine can develop, and the speed at which the maximum torque is developed. 0) If the maximum torque is to occur at start, determine the external resistance required in each rotor phase. Assume a turns ratio of 1.2. 2. A 3-phase, 4—pole induction motor is supplied with 380 V and 50 Hz frequency. When the motor is fully loaded its mechanical power is 12 kW, its line current is 25 A, power factor 0.8 and it rotates at 1400 rpm. a) Find the rotor slip s, rotor frequency and load torque. b) Determine the power crossing the air-gap, power losses in the rotor winding and the motor efficiency if the mechanical power losses are negligible. c) Find the starting torque if the ratio T(a)/Tmax = 0.4, where Tm = is the torque found in question (a). 3. A 3-phase, 460 V, 60 Hz, 4-pole wound-rotor induction motor develops full-load ' torque at a slip of 0.04 when the slip rings are short-circuited. The maximum torque it can develop is 2.5 of full-load. The stator leakage impedance is negligible. The rotor resistance measured between two slip rings is 0.5 Q. Determine: a) The speed of the motor at maximum torque. b) The starting torque as the ratio to full-load torque. c) The value of the resistance to be added to each phase of the rotor circuit so that maximum torque is developed at the starting condition. d) The speed at full-load torque with the added rotor resistance of part (c). HIM #1,- L/ PG, 1&5 MATTi-ieu gaggnu G3 EOUNALEM'I' (1”?er 2! XI. 3; 1). XI: 1 d b3 QTATI‘HHC‘: TOEOuE, MM 11:20-48, 6W?!) Q MAV- TbtduE 50 AT STAinrJC-a Sr! th: 4/4? = Z65.b\/ JXH(E:‘+JX2'3 3(a.5’(.o§¢3,23 2.= 12+“. + W: '°7+J‘2.+JL.S+5.1+.95 : 117”) 3941-! = v‘lflfl'f..h.° 2 mph z @54- Hiri—= '57-:- .l-Ml-I £73.50 5,: V.— (F. was”; = 2599- Cu7+5.z>(.|33.4-36“?5 => Izeflu +3“: . 12111555“ :1: : £45.01fl'A = 133.1'1-351911 V Eu I735”: ___.——- 123.9592. +5 (0.65 a \/ J ‘ F' 5 — ‘3 4" ‘77 3 (97.6.5 '73- “; + 1/5 ‘05 + 3‘1 ,7 3 55' [_§_ = We w‘) = “6 _§<_ _. 909 '"PM Us 22;; : inita‘} 3 Q4525 Raw In (3&1 2% = p: ( 6174:?!“va 7'0 2a we 17“. J MM To? (Me 7. T - 3" CL" 1 “"501 5w . ' J ”M ”s 7.1,; ‘ 7425 am) " 3 5 NM 6865'.) @ Mm rename Pz' . Sb: ET ‘ 3g ‘25 0= mass) = 9000.255 W730”: CBMAK TO?QME Q Sven EXTHMHL. ? “VG-4.2 : _ Q ‘+e¢t+ .u + r * t/ % ah I 65* T" 2‘ x =7 l: S ‘:m‘: ELL: .6 Rfiw‘iflw Q? EKCfcg'hji JIM-{IL} Hm 4dr 4 PC: int-5 MATTHE'JO $09551. L. 05 SLIP; 3?ch Page: ,LoAn (creams. 50- 0 I500 ['15: [1%593 : l 500 r FM 520702 Feeaumw 3% LOAD «roams 51‘; b) ?AC7’) bPZ;EFF Pfl __ All. (‘1th ‘PowEi: PAC-r" I's ‘ 5 Punt? Losses ! 5-st I-s Pm: 31: 5Q: :(.ola7) go : 5.33 -l/ = 3:1 _ % (505(azoao)_ 1: ”5'1- 211'“ : 'Z'rr (M003, 11': 8|.qfl‘,‘ ‘32:. - x/ {327 (mono) =85: ‘/ 23st 862m) flabblérkfi EFFicleuo! : Vou'i': I’M-9r? EH: S‘Jlrl L056 -' 3( 2 - macaw g—E°)(2s)(.e,)= lasra w \/ EFF: 2341'" 12 000 ,._u— - EF‘ .- _.— 8L? .. 1’qu - '4 -7 Kikuhl - TMA: #7 : 204 1551M V 17* Z Z _ _ Sb .blo'? __ fix=iffifi : & °_‘;.7’ ‘4 —7 :54. sh : 5’7 557' -.3%Ssb tool-(gulp was? $1: 6b: .32 1—61- 2.” TsT' .1»- — .' — : T : TEAM-i fl +J— '7 2m! 75 1-,: +4751 7 9T (104-7536560555: Héfld-M 11,1— : ”2.? Am 1 HN £4 (/27 39:3 MATTHEW Banana. #5 0‘3 $191569 9. MM 'T'dROotE TFL 2- l 2.. __ _ .7 __ __ _ _ _ '1 TMm' it 2:; 71.5’ 35431 ‘7 5'0 -.Z‘b+.0olb\/ Sfi- + 'Hn "-"T Sb ska-hf? 11o(r,03 m= ‘71,- 1' 1800.73,.“ 55‘ “5 ' :7 n.- m (I— 5&3 = 1300(I-n‘!) : 1%ij n; b» KA-rro sznué, Tozaue -ro FMLL LOAD TD? Om; 3b en. T -- + -- .1 .04 “'21: R SSL 5h 1 4- — m —___—_— , .w ”a E} 4——'- 7h“— - .4: I 5‘: L_ 4,; | 1” m \IALLAE 0P EMMA» Izesamhmca Wasp. 51:: 11—11 :7( .J‘ir —5- =7 {3% 2.1” L X1 AT. $1327 w/rem 6b" 23‘ +25% _ I: .5? 9011'" M‘ x1. ‘7 - 2.6.5 -7 Kent : leSJ—L 0m S?E‘ED Q HALL. LOAD Urn-t Kerr M31359 " _W m+ Plumper: 55:! FOR TM“ ‘-C..\ ! E 2 '2 ,1... - q .. i T'MM 23+ 5 ‘ -‘—+ E ' ‘ “7 5."- 55 1H 35:1! 67:33. a «75 a ‘ ...
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