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- A single-phase, 120V(rms),60Hz source supplies power to a series R-L circuit consisting of R=10 and L=40mH. (a) Determine the power factor of the circuit and state whether it is lagging or leading. (b) Determine the real and reactive power absorbed by the load. (c) Calculate the peak magnetic energy Wint stored in the inductor by using the expression Wint=L(Irms)2 and check whether the reactive power Q=Wint is satisfied. (Note: The instantaneous magnetic energy storage fluctuates between zero and the peak energy. This energy must be sent twice each cycle to the load from the source by means of reactive power flows.)Consider two interconnected voltage sources connected by a line of impedance Z=jX, as shown in Figure 2.27. (a) Obtain expressions for P12 and Q12. (b) Determine the maximum power transfer and the condition for it toAlternator A(100 kVA,3-phase, 240V,60HZ,1800 rpm) is operating with alternator B(125 kVa,3- phase,240 V,60 Hz,1800 rpm). The load of alternator A is 60kW at 90% pf lagging and the load of alternator B is 80kW at 70% lagging. Determine pf of load.
- Q3/A) A 1500 kVA, 6600V, three phase star connected alternator with Ra=0.502/phase and Xs-52/phase delivers a full load current at 0.8 lagging power factor load. Find the induced emf per phase. If this induced emf is maintained constant but load is changed to full load unity power factor estimate terminal voltage per phase. B) Answer the following:Q2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Ztoad 500 236.87° ohm , and the transmission line's impedance is Zine = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.The reactance of a generator is given as 0.1 pu based on the generator of 17 KV, 300 MVA. Determine the pu reactance on a base of 17 KV, 250 MVA. New pu reactance is.......
- The p.u. impedance value of an alternator corresponding to base values of 13.2 kV and 30 MVA is 0.2 p.u. Then the p.u. impedance value of an alternator for the new base values of 13.8 kV and 50 MVA is......For the same alternator, the ratio of rated KVA, when connected as 3-phase winding to when connected as single phase alternator (by utilizing the coils) is, a.1.33 b.0.50 c.0.67 d.1.50Two identical Y‐connected alternators are running in parallel. The generated phase voltages are: E1 = 6.7 kV and E2 = 6.5 kV with E2 leading E1 by 10°. Calculate the voltage across the load per phase and the power factor of the load if the total current is 500A lagging E1 by 37°. The synchronous reactance of each machine is 10 Ω/ phase and resistance is negligible.
- In a Three-phase alternator, a given field current produces an armature current of 250 A on short-circuit and a generated e.m.f. of 1500 V on open-circuit. Calculate the terminal p.d. when a load of 250 A at 6.6 kV and 0.8 p.f. lagging is switched off. Calculate also the regulation of the alternator at the given load. Subject: Electrical engineeringi. A 2300 V, With 2500 kVA three phase delta-connected alternator has a resistance of 0.1ohm/phase and a synchronous reactance of 1.5 ohms/phase. The alternator is adjustedto rated voltage at no load. Calculate its terminal voltage when carrying ratedcurrent at no load power factor of 0.6 lagging.An alternator with internal vallage of 125, p.u. and synchronous reactance of 0.4 p.u. is connected by a transmission line of reactance 0.1 p.u. c a synchronous meto having synchronous reactance 2.35 p u. and internal voltage of 0.85/6, p.u. If the real power supplied by the alternator is D 806 p... than (8, & is. degrees. (Round off to 2 decimal places) (Machines are of non-salient typa. Naglact resistances)