4. A 2008 -kVA, 3-phase, 8-pole alternator runs at 750 rpm in parallel with other machines on 6,008 V bus-bars. Find the (a) No-load phase voltage in kV; (b) the synchronizing power on full- load 0.8 pf lagging per mechanical degree of displacement and (c) the corresponding synchronizing torque. The synchronous reactance is 6 ohms per phase.

The resistance and reactance per phase of the rotor of a 3-phase induction motor are 0.6 ohm and 5 ohms respectively. The induction motor has a star-connected rotor and when the stator is connected to a supply of normal voltage, the induced e.m.f. between the slip rings at standstill is 80 V. Calculate the current in each phase and the power factor at starting when (i) the slip-rings are shorted, (ii) slip-rings are connected to a star-connected resistance of 4 ohm per phase.

A 4-pole, 3-phase alternator drives at 1500 r.p.m. is supplying to an induction motor which has full load speed of 4320 r.p.m. Find the percentage slip and number of poles.

6. A 2000KVA, 3300V, 3-phase star -connected alternator delivers rated load at a P.F. of 0.85(lagging). The effective resistance between armature winding terminals is 0.12. The field takes 72A at 125V. Friction and windage loss is 18kW and the core losses are 35 kW. Determine the efficiency of the alternator. [Ans. 94.5%]

See attached.

32. An alternator requires an alternating current of 2.1 A to produce on short circuit its full-load current of 50 A. The same on open-circuit an emf of 320 V. The armature resistance is 1Ohms . Find the full-load percentage regulation with a lagging p.f. of 0.6 when terminal p.d. on load is 2000 V. A. 7.2% B. 28.8% C. 3.6% D. 14.4%

there may be more than one answer

750kW alternators P₁ = 750kW, P₂=750kW are operated in parallel. Speed regulation of one machine is 100% to 103%. Speed regulation of second machine is 100% to 104%. Shared load, (P₁)' + (P₂)' arrow_forward Step 2 = (P₂) 1000kW....... (A) So the shared load equation for alternator-1 will be, 103-100-103- 103- (B)' = 103- (Pi)...... (₁) The shared ad equation for alternator-2 will be, 104 100 104 (P₂) 750 = 104-1 = 104 (P₂)'...... (²) Now from equation (1) and (2), 103 (P₁) 104- (P₂)' 103-104 (P)' - (P₂)' (P₁)-(P₂)' = -1 3(P₁)-4(P₂)-750kW....... On solving equation (A) and (B), (P₁) 464. 28kW (P₂) 535. 714kW How did he get the value inside the circle? Can you explain step by step and clear line please?

A three-phase 25-kVA, 480-V, 60-Hz alternator, operating under balanced steady state conditions, supplies a line current of 20 A per phase at a 0.8 lagging power factor and at rated voltage. Determine the power triangle for this operating condition.

There are two alternators, 100kW, 3phase in parallel are driven by shunt motors whose speed load characteristics are as follows: Alternator No. 1, no-load speed 600rpm and the full-load speed 530 rpm, while alternator No. 2, no-load speed 590 rpm and full-load speed 550rpm while alternator 2, no load speed 590 rpm and full load speed 550 rpm. What is the greatest load that can be delivered without overloading either alternator?