A 415V, 3-phase Induction motor has output of 30HP and operates at power factor of 0.6 lagging with an efficiency of 83%. Calculate the reading on each of the two wattmeter connected to measure the input.
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- Induction motors and welding transformers require magnetizing current, which causes a. lagging power factor. b. leading power factor. c. unity power factor. d. zero power factor.Can we connect a single-phase motor to operate from a three-phase power source? Explain your answer.A 25 kVA, 220V, 3φ alternator delivers rated kVA at a power factor of 0.84. The effective ac resistance between armature winding terminals is 0.18Ω. The field takes 9.3A at 115V. If friction and windage loss is 460W and the core loss is 610W, calculate the percent efficiency.
- DEGREE: BS ELECTRICAL ENGINEERING COURSE: AC MACHINES Note: Please solve in this way. Given>Missing>Illustration>Solution>FinalAnswer. Please include symbols, units, formulas, step by step solution, please don't cut the solution. Thank you! (#1011) A 4-pole,3-phase, y-connected induction motor has 48 stator slots 8 series conductors per slot and coil pitch of 83.3 percent. [For C Only] The rotor of has 36 slots with 4 series conductors per slot. The' winding is Y-connected with a pítch of 77.7%. (A) Determine the pitch and distribution factors. [Answer: 0.966 and 0.958] (B) Determine the maximum flux per'pole when connected to a' 220-volt,60-cycle line, neglecting local resistance and reactance drops. [Answer: 805,000 maxwells] (C) Determine the ratio of transformation of the motor [Answer: 2.74]Please solve 209V, three-phase, Six poles, Y-connected induction motor has the following parameter: R1=0.128 Ohm, R'2=0.0935 Ohm, Xeq=0.49 Ohm. the motor slip at full load is 2% and for a Fan-Type load, Calculate following below : I have the Final answers and I do need to step by step which the formula and solution. Please if you use handwriting to write clearly and big. Thanks, The final and correct answers are: Problem3 (Reducing supply voltage) a) motor speed: 1163.3 rpm, b) starting torque= 71.93 Nm, c)starting current=179.5A, efficiency= 93.04%A 230 V single-phase induction motor has the following parameters: R1 =R2= 11 ohms, X1 = X2 = 14 ohms and Xm =220 ohms. With 8% slippage, calculate:1. The impedance of the anterior branch2. Posterior branch impedance3. Total impedance4. The input current module5. Power factor6. Input power7. Power developed8. The torque developed at nominal voltage and with a speed of 1728 rpm ..
- A 500 kVA, 11 kV, 3-phase star-connected alternator has the following data : Friction and windage loss = 1500 W Open-circuit core loss = 2500 W Effective armature resistance per phase 40 ohm Field copper loss = 1000 W Find (a) alternator efficiency of half-full load and at 0.8 power factor lagging, (b) maximum efficiency of the alternator.A single-phase alternator connected to 6,600-V bus-bars has a synchronous impedance of 10Ω and a resistance of 1 Ω. If its excitation is such that on open circuit the p.d. would be 5000 V, calculate the maximum load the machine can supply to the external circuit before dropping out of step and the corresponding armature current and p.f. [2864 kW, 787 A, 0.551]Can we connect a single-phase motor to operate from a three-phase power source? Explainbriefly your answer in one(1) sentence
- A 500kW, 2400V, four-pole, 60Hz, Y connected induction machine has the following equivalent circuit parameters in ohms per phase referred to the stator:R1 = 0.122 Ω; R2 = 0.317 Ω; X1 = 1.364 Ω; X2 = 1.32 Ω; Xm = 45.8 Ω; Rc≈∞ (no core losses)When operating as a motor, supplied from a 2400V source, the machine achieves rated shaft output power (500kW) at a slip of 3.35%. The machine is to be used as a generator, driven by a wind turbine. It will be connected to a distribution system which can be represented by a 2400V infinite bus.a) From the given data calculate the total rotational losses at rated load.b) With the wind turbine driving the induction machine at a slip of -3.2 percent, calculate (i) the electric power output in kW, (ii) the efficiency (electric power output divided by the shaft input mechanical power) in percent, and (iii) the power factor measured at the machine terminals.Q/ A 4-pole, 400V, 3-phase, 50Hz induction motor runs at 1440 r.p.m. at 0.8 p.f. lag and delivers an output of 10.8KW. The stator loss is 1060W and friction, windage etc. losses are 300W. Calculate:(i) Slip.(ii) rotor copper loss.(iii) rotor frequency.(iv) line current.A three-phase, 440-V, 60-Hz, wye-connected induction motor takes a stator current of 50 A at 0.8 powerfactor while running at its rated speed of 855 rpm. The stator loss is 2500 W, and the rotational lossesare 3200 W. Calculate the efficiency of the motor.