1/19/22, 11:29 AM (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu You’re invited to submit to Download Academia Letters, a new, experimental open access journal. Download PDF Full PDF Package Translate × LEARN MORE 1 Search 88 PREMIUM AUDIENCE Chapter 7: Induction Motors Amadou Top Top 1% 9549 Views 1 File ▾ 34 Pages Download PDF Summary Original PDF Chapter 7: 7-1. Download Full PDF Package Translate Related Induction Motors A dc test is performed on a 460-V ∆-connected 100-hp induction motor. If VDC = 21 V and I DC = 72 A, what is the stator resistance R1 ? Why is this so? SOLUTION If this motor’s armature is connected in delta, then there will be two phases in parallel with one phase between the lines tested. VDC R1 R1 R1 Therefore, the stator resistance R1 will be VDC I DC R1 = = R1 (R1 R1 + R1 ) 2 = R + (R1 + R1 ) 3 1 3 VDC 2 I DC 7-2. 3 21 V = = 0.438 Ω 2 72 A A 220-V three-phase six-pole 50-Hz induction motor is running at a slip of 3.5 percent. Find: (a) The speed of the magnetic fields in revolutions per minute (b) The speed of the rotor in revolutions per minute https://www.academia.edu/30187856/Chapter_7_Induction_Motors 1/32 1/19/22, 11:29 AM (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu (c) The slip speed of the rotor The rotor frequency in hertz Download(d) PDF Download Full PDF Package Translate SOLUTION (a) The speed of the magnetic fields is Hz ) = 1000 r/min = 120P f e = 120(50 6 nsync (b) The speed of the rotor is = (1 − s ) nsync = (1 − 0.035 )(1000 r/min ) = 965 r/min nm (c) The slip speed of the rotor is = snsync = (0.035)(1000 r/min) = 35 r/min nslip (d) The rotor frequency is fr 7-3. nslip P ( 35 r/min )( 6 ) = 120 = 120 = 1.75 Hz Answer the questions in Problem 7-2 for a 480-V three-phase four-pole 60-Hz induction motor running at a slip of 0.025. SOLUTION (a) The speed of the magnetic fields is 114 = nsync (b) 120 ( 60 Hz ) 4 = 1800 r/min The slip speed of the rotor is = snsync = ( 0.025)(1800 r/min ) = 45 r/min The rotor frequency is fr 7-4. = = (1 − s ) nsync = (1 − 0.025)(1800 r/min ) = 1755 r/min nslip (d) P The speed of the rotor is nm (c) 120 f e = nslip P 120 = ( 45 r/min )( 4 ) = 1.5 Hz 120 A three-phase 60-Hz induction motor runs at 715 r/min at no load and at 670 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 10 poles, which produces a synchronous speed of nsync (b) 120 f e P = 120 ( 60 Hz ) 10 = 720 r/min The slip at rated load is s= (c) = nsync − nm nsync × 100% = 720 − 670 720 × 100% = 6.94% The motor is operating in the linear region of its torque-speed curve, so the slip at ¼ load will be s = 0 25(0 0694) = 0 0174 https://www.academia.edu/30187856/Chapter_7_Induction_Motors 2/32 1/19/22, 11:29 AM s 0.25(0.0694) (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu 0.0174 The resulting speed is Download PDF (d) Translate The electrical frequency at ¼ load is fr 7-5. Fullr/min PDF) =Package = (1 − s ) nsync =Download 707 r/min (1 − 0.0174 )(720 nm = sf e = ( 0.0174 ) ( 60 Hz ) = 1.04 Hz A 50-kW 440-V 50-Hz two-pole induction motor has a slip of 6 percent when operating at full-load conditions. At full-load conditions, the friction and windage losses are 520 W, and the core losses are 500 W. Find the following values for full-load conditions: (a) The shaft speed nm (b) The output power in watts (c) The load torque τ load in newton-meters (d) The induced torque τ ind in newton-meters 115 (e) The rotor frequency in hertz SOLUTION (a) The synchronous speed of this machine is = nsync 120 f e = P 120 ( 50 Hz ) 2 = 3000 r/min Therefore, the shaft speed is = (1 − s ) nsync = (1 − 0.06 )( 3000 r/min ) = 2820 r/min nm (b) The output power in watts is 50 kW (stated in the problem). (c) The load torque is = τ load POUT 50 kW 2π rad 1 min = ωm (2820 r/min ) 1r (d) = POUT + PF&W + Pcore + Pmisc = 50 kW + 520 W + 500 W = 51.2 kW = τ ind Pconv ωm = 51.2 kW 2π rad 1 min (2820 r/min ) 1 r 60 s = 173.4 N ⋅ m The rotor frequency is fr 7-6. 60 s The induced torque can be found as follows: Pconv (e) = 169.3 N ⋅ m = sf e = ( 0.06 )( 50 Hz ) = 3.00 Hz A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load speed of 3440 r/min. Calculate the slip and the electrical frequency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation (4-57)]? SOLUTION The synchronous speed of this machine is 3600 r/min. The slip and electrical frequency at noload conditions is nsync s nl = f r ,nl − n nl nsync 3600 − 3580 × 100% = 3600 × 100% = 0.56% = sf e = ( 0.0056 )( 60 Hz ) = 0.33 Hz The slip and electrical frequency at full load conditions is nsync − n nl 100% 3600 − 3440 https://www.academia.edu/30187856/Chapter_7_Induction_Motors 100% 4 44% 3/32 1/19/22, 11:29 AM (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu sfl Download PDF = f r ,fl sync nl × 100% = nsync × 100% = 4.44% 3600 = sf e = ( 0.0444Download )( 60 Hz ) = 2.67 FullHzPDF Package Translate The speed regulation is SR = n nl − nfl × 100% = 3580 − 3440 × 100% = 4.1% nfl 3440 116 7-7. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components are R1 = 0.220 Ω R2 = 0.127 Ω X 1 = 0.430 Ω X 2 = 0.430 Ω Pmech = 300 W Pmisc ≈ 0 X M = 15.0 Ω Pcore = 200 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 Pconv τ ind (e) The induced torque (f) The load torque τ load (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: IA + R1 jX1 0.22 Ω j0.43 Ω j15 Vφ jX2 j0.43 Ω R2 0.127 Ω 1 − s s R2 jXM Ω 2.413 Ω - (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 jX M , and then calculate the current as the phase voltage divided by the sum of the series impedances, as shown below. IA + R1 jX1 jXF 0.22 Ω j0.43 Ω RF Vφ - The equivalent impedance of the rotor circuit in parallel with jX M is: ZF = 1 1 1 jX +Z = 1 1 j15 Ω 1 = 2.337 + j0.803 = 2.47∠19° Ω + 2 54 + j0 43 https://www.academia.edu/30187856/Chapter_7_Induction_Motors 4/32 1/19/22, 11:29 AM (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu jX M j15 Ω Z2 Download PDF 2.54 + j 0.43 Download Full117PDF Package Translate The phase voltage is 208/ 3 = 120 V, so line current I L is (b) = IA = IL R1 + jX 1 + RF + jX F = I A = 42.3∠ − 25.7° A 120∠0° V = 0.22 Ω + j 0.43 Ω + 2.337 Ω + j 0.803 Ω The stator copper losses are 2 = 3I A2 R1 = 3 ( 42.3 A ) ( 0.22 Ω ) = 1180 W PSCL (c) Vφ IL The air gap power is PAG = 3I 2 2 2 (Note that 3I A RF is equal to 3I 2 2 R2 s R2 s = 3I A 2 R F , since the only resistance in the original rotor circuit was R2 / s , and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit.) = 3I 2 2 PAG (d) 2 = 3I A2 RF = 3 ( 42.3 A ) ( 2.337 Ω ) = 12.54 kW = (1 − s ) PAG = (1 − 0.05)(12.54 kW ) = 11.92 kW The induced torque in the motor is = τ ind PAG ω (f) s The power converted from electrical to mechanical form is Pconv (e) R2 = sync 12.54 kW 1 min (1800 r/min ) 2π1 rad r 60 s = 66.5 N ⋅ m The output power of this motor is POUT = Pconv − Pmech − Pcore − Pmisc = 11.92 kW − 300 W − 200 W − 0 W = 11.42 kW The output speed is nm = (1 − s ) nsync = (1 − 0.05) (1800 r/min ) = 1710 r/min Therefore the load torque is = τ load POUT ωm 11.42 kW = (g) rad 1 min 1 r 60 s The overall efficiency is η = η= (h) = 63.8 N ⋅ m (1710 r/min ) 2π POUT PIN × 100% = POUT 3Vφ I A cos θ 11.42 kW 3 (120 V )( 42.3 A ) cos 25.7 ° × 100% × 100% = 83.2% The motor speed in revolutions per minute is 1710 r/min. The motor speed in radians per second is 118 https://www.academia.edu/30187856/Chapter_7_Induction_Motors 5/32 1/19/22, 11:29 AM Download PDF (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu Download Full PDF Package https://www.academia.edu/30187856/Chapter_7_Induction_Motors Translate 6/32 1/19/22, 11:29 AM Download PDF (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu Download Full PDF Package https://www.academia.edu/30187856/Chapter_7_Induction_Motors Translate 7/32 1/19/22, 11:29 AM Download PDF (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu Download Full PDF Package https://www.academia.edu/30187856/Chapter_7_Induction_Motors Translate 8/32 1/19/22, 11:29 AM Download PDF (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu Download Full PDF Package https://www.academia.edu/30187856/Chapter_7_Induction_Motors Translate 9/32 1/19/22, 11:29 AM Download PDF (88) (PDF) Chapter 7: Induction Motors | Amadou Top - Academia.edu Download Full PDF Package https://www.academia.edu/30187856/Chapter_7_Induction_Motors Translate 10/32 1/19/22, 11:29 AM Download PDF (88) (PDF) Chapter 7: Induction Motors | Amadou Top - 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