A 6 pole, 50 Hz, 3-phase, wound rotor induction motor is taking 8000 watts from the line. If core loss is 210 watts, stator copper loss is 520 watts, rotor copper loss is 430 watts, friction and windage losses are 90 watts, determine: (a) power transferred across air gap; (b) mechanical power developed by rotor; (c) mechanical power output; (d) motor efficiency; (e) slip?

What are the active components of three-phase Induction Motor? b) Derive an expression for the frequency in the rotor circuit of three-phase IM? c) Draw with details the approximate per phase equivalent circuit of three-phase IM? d) Why the iron loss in rotor of an induction motor is negligible?

7. The total rotational losses of an induction motor are a combination of ------- A) Friction losses B) Windage losses. D) Stray losses. C) Core losses. 8. The developed mechanical power can be calculated using the equation of- A) PAG - PRCL. B) PAG w sync C) (1-s) PAG. D) 3 E2Gc. 9. For an induction motor, if the synchronous speed is 1800 r/m, and the rotor speed 180rad/sec, then the slip will be: A) 5.5% C) 6% B) 4.5% D) 4% 10. The power losses in the stator part of the induction motor are B) Core losses D) Friction and windage losses A) Stator copper losses C) Air-gap losses

An 8 pole DC shunt generator with 900 wave-connected armature conductors supplies 10ohm  at 500 V when driven at 500 rpm. The armature resistance 0.4ohm and the field resistance is 250ohm . Determine:                                     . .                                   2 1.1 The flux per pole                       2.1.2 The armature current for maximum efficiency if the rotational losses are 10% of the ar- mature copper losses .                                    3.2 A series motor runs at 600 rpm while drawing 40-A from 600-V supply. The total re- sistance of the armature and the field is 0.5-ohm. Determine the value of the additional re- sistance required in series with armature to reduce the speed to 450 rpm if the load toque is:                                      3.2.1 Constant                                   3.2.2 Proportional to the square of the speed .

Explain the conditions for parallel operation of synchronous generators. A three phase, wye connected, synchronous generator is rated 150 MW, 0.85 lagging power factor, 12.6 kV, 60 Hz, and 1800 rpm. Each winding has an armature resistance of 0.05Q and a synchronous reactance of 0.6Q. Draw the phasor diagram with values, show torque angle and determine the induced voltage for the condition of rated load.

A 240 V, Short – Shunt DC Generator supplies a load current of 60 A. The armature circuit resistance is 0.10 Ohms and shunt-field circuit resistance is 120 Ohms. The series-field resistance is 0.08 Ohms.  Calculate The developed Power The percentage voltage regulation. Draw the power flow graph, showing the various losses as well the input, developed, and output power values. P rotational = 500W

machinery

8. Homework: III. IV. V. What is the effect of load on the efficiency of an induction motor? Explain the term air-gap power or power developed in the rotor and shaft power. How these terms are related to each other? Derive an expression for the rotor copper loss in terms of slip and input to the rotor? What are the various losses in an induction motor? On what factors do they depend? List the different types of stray load losses in an induction motor.

With the aid of diagrams describe the basic topology of a wound field 3 phase synchronous generator and outline the relationship between the excitation voltage and rotor current. Q2 (a) (b) Outline with the aid of a suitable diagram the necessary power conversion stages required to interface a small permanent magnet (PM) generator to a stand alone DC system. A star connected 3 phase wound field synchronous generator with a synchronous reactance of 20 N is connected to a 6.6 kV (phase voltage) grid and supplies 1.5 MW at 0.95 lagging power factor at its terminals. Calculate the phase current and resultant voltage across the synchronous reactance (Vxs), and from a scaled phasor diagram graphically determine the required excitation voltage (Eph) and load angle (8). (c) The load is now increased to 3 MW with the Excitation Voltage kept constant. Using the phasor diagram drawn in part (c) graphically determine the resultant phase current, power factor and load angle. (d)

A 25 kW, 240 V self excited dc shunt generator has a field circuit resistance of 62.5 02, and an armature resistance of 0.05 Q. When delivering rated current at rated speed and voltage, calculate % Efficiency when rotational losses are 1200W, and iron losses are 600W. Note that Copper losses must be calculated too. No Units!