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IM Solutions - 4-2 4-3 Develop a table showing the speed of...

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Unformatted text preview: 4-2. 4-3. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and 400 Hz. SOLUTION The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is _ 120 fa m P n The resulting table is Number of Poles fe = 50 Hz fe = 60 Hz fg = 400 Hz 2 3000 r/min 3600 r/min 24000 r/min 4 1500 r/min 1800 r/min 12000 r/min 6 1000 r/min 1200 r/min 8000 r/min 8 750 r/min 900 r/min 6000 r/min 10 600 r/min 720 r/min 4800 r/min 12 500 r/min 600 r/min 4000 r/min 14 428.6 r/min 514.3 r/min 3429 r/min A three-phase four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot of the windings. All coils in each phase are connected in series, and the three phases are connected in A. The flux per pole in the machine is 0.060 Wb, and the speed of rotation of the magnetic field is 1800 r/min. (a) What is the frequency of the voltage produced in this winding? ( b ) What are the resulting phase and terminal voltages of this stator? SOLUTION (a) The frequency of the voltage produced in this winding is _ nmP _ (1800 r/min)(4 poles) _ 120 _ 120 (b) There are 12 slots on this stator, with 40 turns of wire per slot. Since this is a four-pole machine, there are two sets of coils (4 slots) associated with each phase. The voltage in the coils in one pair of slots is f, = 60 Hz E, = Jim”) f = @7440 t)(0.060 Wb)(60 Hz) 2 640 V There are two sets of coils per phase, since this is a four-pole machine, and they are connected in series, so the total phase voltage is V4, = 2(640 V) = 1280 V Since the machine is A~connected, VL = VI? 21280 V . A three-phase Y-connected 50~Hz two-pole synchronous machine has a stator with 2000 turns of wire per phase. What rotor flux would be required to produce a terminal (line-to—line) voltage of 6 kV? SOLUTION The phase voltage of this machine should be V¢ = VL /\/3 = 3464 V . The induced voltage per phase in this machine (which is equal to V¢ at no—load conditions) is given by the equation EA = fimwf SO EA 3464 V =—————=———-——=0.0078 Wb JinNC f Jin(2000 t)(50 Hz) ¢ 7-2. A 220-V, three-phase, two-pole, 50-Hz induction motor is running at a slip of 5 percent. Find: (a) The speed of the magnetic fields in revolutions per minute (b) The speed of the rotor in revolutions per minute (0) The slip speed of the rotor i i i z 1 (d) The rotor frequency in hertz SOLUTION (a) The speed of the magnetic fields is _ 120 f8 _120(50 Hz) sync — P _ 2 (b) The speed of the rotor is n = 3000 r/min nm = (1 — s) nsync = (1 — 0.05)(3000 r/min) = 2850 r/min i l (c) The slip speed of the rotor is l n snsync = (0.05)(3000 r/min) = 150 r/min slip : (d) The rotor frequency is n P (150 r/min)(2) _ slip __ r _ _ = 2.5 Hz 120 120 7-4. A three-phase, 60-Hz induction motor runs at 890 r/min at no load and at 840 r/min at full load. (a) How many poles does this motor have? (b) What is the slip at rated load? (c) What is the speed at one—quarter of the rated load? (d) What is the rotor’s electrical frequency at one—quarter of the rated load? SOLUTION (a) This machine has 8 poles, which produces a synchronous speed of __120fe _120(60 Hz) rtS m = 900 r/Inin y P 8 (b) The slip at rated load is s‘: MX 100% = Mx100% = 6.67% n 900 sync (c) The motor is operating in the linear region of its torque-speed curve, so the slip at 1%: load will be s = 0.25(0.0667) = 0.0167 The resulting speed is nm = (1 — s) nsync = (1 — 0.0167)(900 r/min) = 885 r/min :3 i E g1 l i l i i l ! E i E i (d) The electrical frequency at M: load is f, = sfe = (0.0167)(60 Hz) = 1.00 Hz A 208-V, two-pole, 60-Hz Y-connected wound—rotor induction motor is rated at 15 hp. Its equivalent circuit components are 121:0.2009 R2 =0.120§2 XM =15.0o X1 = 0.410 52 X2 = 0.410 9 th = 250 W PM = 0 Pm = 180 w For a slip of 0.05, find (a) The line current I L (b) The stator copper losses PSCL (c) The air-gap power PAG (d) The power converted from electrical to mechanical form PLOW (e) The induced torque ‘1:ind (f) The load torque 'cload (g) The overall machine efficiency ( h) The motor speed in revolutions per minute and radians per second SOLUTION The equivalent circuit of this induction motor is shown below: j0.4l (2 0.120 9 0.20 9 10.41 9 le g2 (a) The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F of the rotor circuit in parallel with jXM , and then calculate the current as the phase voltage divided by the sum of the series irnpedances, as shown below. 0.20 Q 1041 Q The equivalent impedance of the rotor circuit in parallel with jXM is: 1 1 ZF = 1 1 = 1 1 = 2.220+j0.745 = 2.341185" 9 + + jXM 22 leQ 2.40+j0.41 The phase voltage is 208/ \/§ = 120 V, so line current I L is A. -. .. _ H- m mmWWflw®mWWWKWWW“M~ _ V¢ _ 12040o V R1+jX1+RF +jXF 0.20 9 +j0.41§2+ 2.22 Q+j0.745 9 IA = 44.8.; — 25.5° A IL IA I L (b) The stator copper losses are PSCL = 31A2R1 = 3(44.8 A)2 (0.20 £2) = 1205 W (c) The air gap power is PAG = 3122 fi = 3IA2RF 3 (Note that 31 AZRF is equal to 3122 —Z , smce the only res1stance in the orlginal rotor c1rcu1t was R2 / s , and s the resistance in the Thevenin equivalent circuit is R F . The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit.) PAG = 3122 1:1 = 31/319, = 3(44.8 131)2 (2.220 9) = 13.4 kW (d) The power converted from electrical to mechanical form is POW = (1 — s)PAG = (l —— 0.05) (13.4 kW) 2 12.73 kW (6) The induced torque in the motor is T _PAG_ 13.4kW _355Nm ind _ _'——’—.“_‘ "‘ - ' “sync (3600 r/min) 2’: rad 16’3““ r S (f) The output power of this motor is POUT 2P —P —P —Pmc =12.73kW—250W — 180W — 0W 2 12.3 kW conv mech core The output speed is nm = (1— s) nsync = (1 ~ 0.05)(3600 r/min) = 3420 r/min Therefore the load torque is PM: 12.3kW =343N'm “’m (3420 r/min) Z’i rad 1613““ I' S Tload = (g) The overall efficiency is 77: PO“ x100% =—P°—UT><100% PIN 3V¢1Acosfl 12.3 kW 7]: ———-——X 100% = 84.5% 3(120 V)(44.8 A)cos25.5° (h) The motor speed in revolutions per minute is 3420 r/min. The motor speed in radians per second is 27r rad 1 min 1 r 60 5 com = (3420 r/min) = 358 rad/s For the motor in Problem 7-7, what is the slip at the pullout torque? What is the pullout torque of this motor? SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply, and then using that with the rotor circuit model. 1X1 R1 . _ iXM (R1 +17(1) IXM ZTH _ R1 + i(X1 + XM) z _ _jXM(Rl+jX1) w(1‘15 Q)(0.20§2 +j0.41§2) m — . . =O.1895+ j0.4016 Q=0.444464.7° Q R1+J(X1+XM) .0209 +J(0.41§2+15 Q) m V (115 9) V =————-—————-— :_ TH R1+j(X1+XM) ¢ 0.229+j(0.439+15£2) (12010° V) = 116.8107o V The slip at pullout torque is R2 2 RTH2 + (XTH + X2) 0.120 9 sum = —2———~—2 = 0.144 \/(O.1895 g2) +(0.4016 g2 +0.410 Q) S 2 max The pullout torque of the motor is 7 _ 3VT2H max 2mm RTH+JR§H+(XTH+X2)2 311 . V2 T Z < 68 > max 2(377 rad/s) 0.1895 (2+ (0.1895 (2)2 +(0.4016 Q +0410 (2)2 7-10. For the motor of Problem 7—7, how much additional resistance (referred to the stator circuit) would it be 1 necessary to add to the rotor circuit to make the maximum torque occur at starting conditions (when the 1 shaft is not moving)? Plot the torque-speed characteristic of this motor with the additional resistance 1 l inserted. : SOLUTION To get the maximum torque at starting, the smax must be 1.00. Therefore, R : SW = _.._...2___2 2 RTH +(XTH +X2) 1.00 = ——2R—2——-—-——————2— (0.1895 (2) +(0.4016 Q +0410 9) 1 R2 = 0.833 (2 i Since the existing resistance is 0.120 9, an additional 0.713 9 must be added to the rotor circuit. The resulting torque-speed characteristic is: 1 Induction Motor Torque-Speed Characteristic 60 -—-r— 5 I | I T ! — With extra rotor resistance i i 1 .i 0 500 1000 1500 2000 2500 3000 3500 4000 "m o i .4— i ...
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