hw #3

# hw #3 - /7/ﬂ EE 4420 Homework#3 Name MATIHEN EQnEﬂI—t...

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Unformatted text preview: /7//ﬂ EE 4420: Homework #3 Name: MATIHEN . .. EQnEﬂI—t- Return by 11/07/05 1. A three-phase, 5 hp, 208 V, 60 Hz induction motor runs at 1746 rpm when it delivers rated output power. Determine: a) The number of poles of the machine. b) The slip at full load. - c) The frequency of the rotor current. d) The speed of the rotor ﬁeld with respect to the: . - Stator, 0 Stator rotating ﬁeld. 2. . A three~phase, 460 V, 60 Hz induction machine produces 100 hp at the shaft at 1746 rpm. Determine the efﬁciency of the motor if rotational losses are 3500 W and stator copper losses are 3000 W. 3. A 440 V, 60 Hz, 6-pole, 3-phase induction motor is taking 50-kVA at 0.8 PF and is running at a slip of 0.025. The stator copper losses are 0.5 kW and rotational losses ‘ are 2.5 kW. Compute: a) The rotor copper losses. b) The shaft power in hp and the shaft torque. c) The efﬁciency. l. A NA naNA-L il-hwapow #5 PC: PEDBLEM #:1- GIVE/dr 3?‘ 512,; =(5’Ap)(7%)v‘ 373cm 203V EOHz n= [749 rpm 6’ EATeﬂ 0mm Paws}? Q) # or 770155 m MALHINE (7“ 05 5nd“ n: 17%rpm THEN _(}3=Jaoo rPM 120-?- ‘7 tin-:5}.~ Raﬁaﬁ ns= ‘3‘ - P‘ n; moo i 4PoLe MMHNBI b) 5m? Jar rum. LOAD _ nﬁm _ MOD—(7% _ 6‘ “I: ‘ Moo “- c3 Freemapw of: Peg-rail ﬂaw-RENT” ¥z= ~39: 3‘? 5212(425390 =/.8 A on 67355;; or Ferrari: Pram wm+ Easpec-r 7’0 —- STA-101 02+“:“5 -‘/BOOPFM x ‘nz‘ +Y\ = IBOO rpm} '" STATDZ Fla. .3 X 4/ Weaem #2 mad: 5,6 460V Gaffe IODBP= i00('7‘lI-)= 7‘4600N’Pou-r (1: E7419 ff-‘M 5PM: 350m“) 5P1: 5000M DETEKM WE Erma Em! C. 72 “5: I500 rpm S_ 95-0 _ Moo-I746 _ - ‘- '_“—-.._- .- n‘5 1900 '- '93 all: Pam: +APM : 7‘4609 4-35—00 = 78th»; N ___. - 31’s. EVZ‘ "S _ ’_'o% EFF: Pouﬂ' SINCE he IT‘f‘bpr R: an AP,“ 413p. +1; .9; Osmond ; 245w \/ H J 74! 000 74,390 + 35004 30ao+ z-{xg MA'HH Ed (‘30 NSHLL. Her/16w ow if? Womm #3 C—AVEM.’ 440V 3?; (wife (a Few ‘5: 5:3de 6’ 45,9? ‘5: .025 DR: 5'an 5%: 75000 05 E0702 Capra? LoSSEj , AP; )\" BE! '77 Ru:£)\= (50,90o3{.3>-' 1/0,00o\,J / Rid: 40,090'JJ :37.300w J Pun: Fan-AP. -‘ 140,000- 500 7.467: 39,500vd Apr 53:7: (025337.500 = N1: J ‘9 SHAFT PoME? N HP 4- éHAF-r TDZQME: SHAFr Powéiz : Bur PM (H3315: dam-331500 .— 3585:2511 'PM -‘ Pom-r +2534 :7 PM; PIN-AFN = 885125—7500 : EQDI1.§ - Pour: 390:2.5 u \/ 6HAPT" TDVDqE o n -n ﬂsz ‘1 of”), 120ml,” 5: 3—“ = n: rush-s):Izoo(1—.ozs)=ll7o h 3 [[70 (PM x be P ‘ 'T‘: .—-—M: (9.9 (quﬁ). . 2f“ 111- 01703 ‘ 2% d3 EFFIQBNC‘I J ‘PMT POU'V1 AFN +AP¢+APL "3100:2137 3W2.“ 2500+ Sou-I- 981! ; EFF: . 9 ...
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## This note was uploaded on 02/26/2012 for the course EE 4150 taught by Professor Wu during the Fall '10 term at LSU.

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hw #3 - /7/ﬂ EE 4420 Homework#3 Name MATIHEN EQnEﬂI—t...

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