A 380V 50Hz Δ-connected induction motor, its R2=2.5Ω, and I2=25A. For a rotor slip of 3% at rated conditions, Pconv equal:
Q: H.W 1: A 380 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the…
A: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and…
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A: Stator parameters R1 = 0.5ohm R2 = 0.25 ohm X1 = 0.75 ohm X2 = 0.5 ohm Xm = 100 ohm Rc =…
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A: It is given that the rotational loss is 4% of the power developed, slip is 4% . Now for the…
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Q: See attached.
A: Drawing the equivalent circuit using the given points:
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A: This problem is from electrical machine.
Q: A 460 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the…
A: As per our guide lines I solved only 3 sub parts
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Q: A 4 pole, 50 Hz, 3-phase induction machine is rated at 1480 rpm, and 240 V. A blocked rotor test…
A: “Since you have posted a question with multiple sub-parts, we will solve first three subparts for…
Q: A 460 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the…
A:
Q: A 460 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the…
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Q: A 460 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the…
A: 460V 8 pole machine run 870rpm speed Motor parameters given Determine stator current Rotor current…
Q: Q1: A: A 230 V, 50 Hz, 4 – pole single-phase induction motor has the following equivalent circuit…
A: Equivalent circuit diagram of 1 phase capacitor is Slip is 0.03
Q: Q2:A 3phase induction motor ,4pole ,star connected stator winding 380V ,5 HZ, rotor resistance and…
A: It is given that: P=4E1=380and E1Phase=E13=3803=219.39 Vf=50 HzR2=0.15 ΩX2=0.6 ΩK=0.5s=0.04
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A: Since you have asked multiple questions in a single request, we will be answering only the first…
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A: The equivalent parameters per phase is R1=R2=0.8 Ωx1=x'2=3.5 Ω Power output Pout=200…
Q: A 3phase induction motor, 6 pole, star connected stator winding 380 1, 50 H: rotor resistance and…
A: As per our policy, i am attempting first three parts. Given data, r2=0.152 Ω/phasex2=0.6 Ω/phase
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A: Given values are - f=50HzPoleP=4SpeedNr=1425rpmR2=7.8Ohm Given motor is single phase induction…
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A: Induction motor.
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Q: 380 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the following…
A: We should solve first three part if you want the remaning kindly repost them as seperate question…
Q: 4. A 3 phase,50Hz ,8 pole induction motor has a full load slip of 4%. The rotor resistance is…
A: It is given that: P=8f=50HzSfl=0.04r=0.001ΩphX=0.005Ωph
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A: “Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…
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A: As per our guide lines I solved only 3 subparts..
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A: output power or power to load = converted power - friction and windage losses-core losses- misc.…
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A 380V 50Hz Δ-connected induction motor, its R2=2.5Ω, and I2=25A. For a rotor slip of 3% at rated conditions, Pconv equal:
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- A 480V, 60Hz, six-pole, Y connected cage rotor induction motor has the followingparameters in ohms per phase referred to the stator: R1=0.5ΩX1=1.1Ω R2=0.45ΩX2=0.6Ω Xm=30Ω The total rotational losses (friction, windage, and core losses) are 1100W and are assumedto be constant. For a rotor slip of 2.2% at the rated voltage and rated frequency, calculate:d) The air-gap power (Pag) and the developed mechanical power (Pconv)e) Developed electromechanical torque, and the output mechanical torquef) EfficiencyThe stator resistance of a 1-phase induction motor is 2.5 ohm and its leakage reactance is 2.0 ohm. Onno-load, the motor takes 4 A at 96 V and at 0.25 lagging power factor. The no-load friction andwindage loss is negligible. Under the blocked-rotor condition, the input power is 130 W at 6 A and42 V. obtain the equivalent circuit parameters.A star-connected induction motor 3-phase, has the following equivalentcircuit parameters per phase:and R1 = R2 =0.8ohm; x1 = x'2 = 3.5ohmand 200-hp, 3300-V, Calculate the slip at full load if the friction andwindage loss is 3 kW. How much extra rotor resistance would be necessaryto increase the slip to three times this value with the full-load torquemaintained? How much extra stator resistance would be necessary toachieve the same object and what loss of peak torque would result?
- A 480V, 60Hz, six-pole, Y connected cage rotor induction motor has the followingparameters in ohms per phase referred to the stator: R1=0.5Ω X1=1.1Ω R2=0.45ΩX2=0.6Ω Xm=30Ω The total rotational losses (friction, windage, and core losses) are 1100W and are assumedto be constant. For a rotor slip of 2.2% at the rated voltage and rated frequency, calculate:a) Rotor speedb) Stator currentc) Complex, real, and reactive input power and the power factor of the machined) The air-gap power (Pag) and the developed mechanical power (Pconv)e) Developed electromechanical torque, and the output mechanical torquef) EfficiencyAn approximate per phase equivalent circuit of a 3-phase, 220-V, wye-connected, 6 pole, 60 Hz induction motor is shown in figure below for which R1 = 0.3 ohm; X1 = 0.4 ohm; R’2 = 0.4 ohm; X’2 = 0.6 ohm; Xm = 15 ohms. For the given numerical values calculate the total developed(electromagnetic) torque in (kg-cm) .The motor speed is 1080 rpm.A 460V, 25hp, 60Hz, 4 pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1 = 0.641 Ω R2 = 0.332 Ω X1 = 1.106 Ω X2 = 0.436 Ω Xm = 26.3 Ω The total rotational losses are 1100W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 3.5% at the rated voltage and rated frequency, find the motor’s Speed Stator current Power factor Pconv and Pout Ʈǐnd and Ʈload Efficiency
- The stator resistance of a 1-phase induction motor is 2.5 ohms and its leakage reactance is 2.0 ohms. Onno-load, the motor takes 4 A at 96 V and at 0.25 lagging power factor. The no-load friction andwindage loss is negligible. Under the blocked-rotor condition, the input power is 130 W at 6 A and42 V. Obtain the equivalent circuit parameters.A 440-V, 60-Hz, 2-pole, Y-connected, three-phase induction motor has the following parameters in ohms/phase: R1 = 0.3, X1 = 0.9, R2 = 0.6, X2 = 0.9, Rc = 150, Xm = 60. If the rotational loss is 4% of the powerdeveloped, determine the efficiency of the motor when it runs at 4% slip.A 3-phase, 6 pole, 440 V, 60 Hz, Y-connected induction motor has the following impedances per phase referred to the stator circuit: R1=0.43 Ω, R2=0.54 Ω, X1=1.263 Ω, X2=0.608 Ω, XM=43Ω The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2 percent at the rated voltage and rated frequency, Find the motor speed, stator current, output power
- A 200V, 3, star connected induction motor takes 30A at a line voltage of 35V with rotor locked. With this line voltage, power input to motor is 450W and core loss is 60W.The dc resistance between a pair of stator terminals is 0.15Ω. If the ratio of ac to dc resistance is 1.6, find the equivalent leakage reactance/phase of the motor and the stator and rotor resistance per phase.a)the figure below shows the per-phase equivalent circuit of a two - pole three- phase induction motor operating at 50 Hz . The voltage (E1) across the magnetizing inductance is 210 v , and the slip (s) is 0.05 . What is the developed torque produced by the motor ? b) A 4 poles , 50Hz , Y-connected three phase induction motor has the per phase equivalent circuit parameters of (R1 =R2=0.12 , X1=X2=0.6 , Xm =12 ) ohm and I2 =20 (0 . Use exact equivalent circuit to determine the following at rated slip of 0.03 : I. Stator current II. Magnetizing current III. Power factor IV. Developed torque C) What is the physical meaning for each of the following induction motor parameters : R1 , R2 , Rc , X1 , X2 , Xm d) What is synchronous wattA 3 phase 50Hz 12 pole 420V delta connected induction motor has the following equivalent circuit parameters: Stator impedance = (2.95+j6.82) Ω per phase; Standstill rotor impedance referred to stator = (2.08+ j 4.11) Ω per phase. Neglect exciting branch admittance. Determine efficiency and useful torque if rotational losses are 750W.