10.6 A 3-phase star-connected Alternator is rated at 11 kV (line). The effective resistance and Synchronous reactance per phase are 2.0 and 20 2 respectively. Calculate the percentage voltage regulation for a load current of 100 A at pf of (i) unity and (ii) 0.8 lag [Ans: (i) 7.85% : (ii) 23.634%]
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- A three – phase 11 kV wye connected synchronous alternator has a synchronous reactance of 8 ohms per phase but negligible resistance. If the excitation is such that the open circuit voltage is 14kV, determine the power factor at the maximum output. a. 0.786 b. 0.772 c. 0.793 d. 0.708The step down DC chopper uses DC source voltage V(dc) = 200 V and resistive load impedance of 10 ohm. The duty cycle is 0.4 the voltage drop across the switch may be taken as zero, Calculate : (a) Average and RMS value of output voltage. (b) Average and RMS value of output current.Generate a conclusion on this following statements: Determine and analyze the voltage regulation characteristics of the alternator with resistive, capacitive, and inductive loading.
- v. With reference to an 50 Hz3 phase alternator explain how P,Q,V and F are controlled.Two alternators A and B operate in parallel and supply a load of 15MW at 0.5 power factor lagging. a) By adjusting steam supply of Alternator A, its power output is adjusted to 6MW and by changing its excitation, its power factor is adjusted to 85% lagging. Find the power factor of Alternator B. b) If steam supply of both machines is left unchanged, but excitation of Alternator B is reduced so that its power factor becomes 85% leading. Find the new power factor of Alternator A.A three-phase star-connected alternator delivers at rated load a current I = 200 A under a voltage of 5 kV-50 Hz (between phases). The load is inductive of power factor equal to 0.87, the resistance of each stator winding is 0.292 and the speed of the rotor is n = 750 r.p.m. The sum of the constants losses and the stator Joule's losses and the rotor Joule's losses is 55 kW. The no load test, at rated frequency, gave: E-4200 V (between phases) for an excitation current lexc = 40 A. The short-circuit test, for an excitation current lexc = 40 A, gave a short-circuit current in the stator windings Ise =2.5 kA. 1- Calculate the numbers of poles of the alternator. 2- Calculate the synchronous reactance (X). 3- Calculate the electromotive force of the alternator (E) between phases for the same excitation current. 4- Calculate the power absorbed by the alternator (Pa).
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- A single-phase AC voltage controller has a resistive load R = 10 Ohm, and the RMS input voltage is Vs = 230 V, 60 Hz. The delay angle is 120 degree. Then power factor of the circuit is: Select one: O a. 1 O b. 0.75 O c. 0.44 O d. None of the above O e. 0.9A 3 phase 10 pole 600 rpm delta connected alternator has 12 slots/pole with 8 conductors per slot. The windings are short chorded by 2 slots. The flux per pole contains a fundamental of 100 mwb, the third harmonic having an amplitude of 33% and fifth harmonic of 20% of the fundamental. Determine the rms value of the phase ,line voltages and Circulating Current Please answer in typing format pleaseQ5: A 3-phase, 50 Hz, 8-pole alternator has a star-connected winding with 120 slots and 8 conductors per slot.The flux per pole is 0.05 Wb, sinusoidally distributed. Determine the phase and line voltages.