# Objective Questions on Alternator | Page – 3

1. An alternator having induced emf of 1.6 p.u. is connected to an infinite bus of 1.0 p.u. If the bus has reactance of 0.6 p.u. and alternator has reactance of 0.2 p.u. The maximum power that the can be transferred is given by

Pmax = Vf × Ef / Xs = 1.6 × 1.0 / 0.8 = 2 p.u.

2. A 500 kVA, 3 - phase star connected alternator has a rated line to line terminal voltage of 3300 V. The resistance & synchronous reactance per phase are 0.3 Ω and 4.0 Ω respectively. Calculate the voltage regulation at full load 0.8 power factor loading?

Under construction.

3. A 50 Hz, 4 pole turbo generator rated at 20 MVA, 13.2 KV has inertia constant H = 3 kW-sec / KVA. The kinetic energy stored in the rotor is

Kinetic energy stored = GH (G → rating in MVA and H → inertia constant, Kinetic energy stored = 20 × 3 = 20 × 3 = 60 MJ.

4. If two pole, 50 Hz, 50 MVA turbo generator has a moment of inertia of 8.6 × 103 Kg-m2, then inertia constant H in MJ / MVA will be

Ns = 120 × 50 / 2 = 3000 rpm (synchronous speed), so kinectic energy stored = (1/2) × Jω2 = ( 1 / 2 ) × (8.6 × 103 ) × ( 2 × π × 3000 / 60) = 424.4 MJ, H = kinetic energy stored in MJ / MVA rating = 424.4 / 50 = 8.48 MJ / MVA.

5. An alternator of 300 KW is driven by a prime mover of speed regulation 4 % and another alternator of 220 KW driven by a prime mover of speed regulation 3 %. The total load they can take is

Let's load shared by two machines are x KW and y KW. Hence, (4 / 300 )x = ( 3y ) / 200
⇒ 8x = 9y. But y = 200 KW because all shear, x = 225 KW. Total load = 225 + 200 = 425 KW

6. The p.u parameters for a 500 MVA machine on its own base are, inertia M = 20 p.u. reactance X = 2 p.u. The p.u. values of inertia and reactance on 100 MVA common bases respectively are

Per unit values inertia (Mp.u.)new =(Mp.u.)old × (MVA)old / (MVA)new = 20 × ( 500 / 100) = 100 p.u. then p.u. values of reactance( Xp.u.) = (Xp.u.)old × (MVA)new / (MVA)old × (KV)old2 / (KV)new2 or, 2 × 100 / 500 = 0.4 [(KV)old = (KV)new].

7. In a 3 phase alternator a field electric current of 40 A produced a full load current of 200 A on short circuit and 1732 V on open circuit if the resistance is 1 ohm the synchronous reactance is

Zs = synchronous impedance = OC voltage / SC electric current = 1732 / 200 = 8.66 ohm
Xs = √Z2s - R2 = √8.662 - 12 = 8.66 ohm.

8. An alternator is supplying a load of 300 kW of a p.f. of 0.6 lagging. If the power factor is raised to unity, How many more kilowatts can alternator supply for the same KVA loading

KVA = KW / cosφ = 300 / 0.6 = 500 KVA,
KW at 0.6 p.f = 300 KW,
Kw at 1 p.f. = 500 × 1 = 500 KW,
Increased power supplied by the alternator ( 500 - 300 ) = 200 KW.

9. A 3-phase, 11 KV, 5 MVA has synchronous impedance, ( 1 + j10 ) Ω per phase. Its excitation is such that the generated emf is - 14 KV. If the alternator is connected to infinite bus bar, then maximum output at the given excitation is

V = 11 / √3 KV, E = 14 / &radic3; KV, X = 10 Ω
Pm = ( V × E) / X.
∴ Pm = ( 11 × 14 ) / (10 × 3 ) = 5135 KW .

10. A 50 Hz, 4 pole turbo alternator rated at 20 MVA, 13.2 KV has an inertia constant = 4 kW sec/ KVA. The K.E. stored in rotor at synchronous speed is

K.E. stored = ( H × MVA ) = ( 4 × 20 ) = 80 MJ.

11. In a certain three-wire Y-connected generator, the phase voltages are 2 KV. The magnitudes of the line voltages are

For Y (star connection) : Line Voltage = 1.732 × Phase Voltage.
For delta connection : Line Voltage = Phase Voltage..
Accordingly, 2000 × 1.732 = 3464.10 Volts.

12. The balanced short circuit electric current of a 3 phase alternator is 50 A at a speed of 1500 rpm for the same filed electric current the balanced short circuit electric current at 1200 rpm would be

Since speed is reduced to 1200 rpm, therefore generated emf of the alternator also decreases.
Eg2 / Eg1 = 1200 / 1500 = 4 / 5
Eg2 is 80 % of Eg1. Current also reduces by 80 %, 40 amps.

13. A 3-φ 50 Hz, 6000 KVA star connected alternator having an effective resistance of 0.2Ω. A field electric current of 10 A produces 480 V on open circuit and field electric current of 5 A gives armature electric current of 105 A. The full load current and phase voltage will be

Given, rating: 6000 kVA; 6000 V
Armature resistance(Ra) = 0.2 Ω
Full load current = 6000 × 1000 / √3 × 6000 = 577.35 A
As the O.C voltage of 480 V appears at a field electric current of 10 A, hence the field electric current of 10 A under S.C. test will give an armature electric current of 105 × 2 = 210 A
Phase voltage = 6000 / √3 = 3464 V.

14. A 550 V, 55 KVA single phase alternator having effective resistance of 0.2 Ω. A field electric current is 200 A on short circuit end and an emf is 400 V on open circuit. The synchronous impedance will be

We know synchronous impedance of an alternator is given by Zs = open circuit phase voltage / short circuit armature current. Here, open circuit phase voltage = 450 V;
Short circuit armature electric current = 200 A;
∴ Zs = 450 / 200 = 2.25 Ω.

15. A 500 V, 55 KVA single phase alternator has an effective resistance of 0.5 Ω. An excitation electric current of 10 A produces 200 A armature electric current on short circuit and an emf of 450 V on per circuit. The synchronous reactance will be

We know synchronous impedance of an alternator is given by, Zs = open circuit emf / short circuit electric current = 450 / 200 = 2.25 Ω
Now the synchronous reactance of an alternator will be,
Xs = √( Zs2 - Ra2 ) = 2.241 Ω.

16. A 3-φ, 6 pole, star connected alternator revolves at 1000 r.p.m. The stator has 90 slots and 8 conductors per slot. The flux per pole is 0.05 Wb. If the winding factor is 0.96 then the voltage generated by the machine is

We know f = PNs / 120
Where P = no. of pole = 6
Ns = no. of revolution = 1000,
∴ f = 6 × 1000 / 120 = 50 Hz
Total no. of stator conductor = Conductor per slot × No. of slots = 8 × 90 = 720
Stator conductor per phase = Zp = 720 / 3 = 240
Winding factor(given) = 0.96
Generated voltage per phase = Ep = 2.22KwfφZp = 2.22 × 0.96 × 50 × 0.05 ×240 = 1278.7 V
Generated line voltage = EL = √3Ep = √3 × 1278.7 = 2214.7 V.

17. A 6 pole alternator rotating at 100 rpm has a single phase winding housed in 3 slots per pole the slots in groups of three being 20° apart. If each slot contains 10 conductors and the flux per pole is 2 × 10-2Wb, the voltage will be generated ( assuming the flux distribution to be sinusoidal )

Under construction.

18. A 5000 KVA, 3φ, 10000 V 50 Hz alternator runs at 1500 rpm connected to a constant voltage bus bars. If the moment of inertia of the rotating system is 1.5 × 104 kg m2 and the steady state short circuit is five times the normal full load current, then the natural time period of oscillation will be

We know Xs pu = Ia/Isc.
T = 0.498ns√( 15 × 104 × 0.2) / ( 5000 × 500) = 1.3638 s.

19. A 2 pole, 50 Hz, 3- φ, 100 MVA, 33 KV turbo alternator connected to infinite bus has a moment of inertia 106kg-m2 in its rotating parts. It has a synchronous reactance of 0.5 p.u. The natural frequency of oscillation will be

pns = f ; ns = f / p = 50 / 1 = 50 rps
T = 0.498 × ns × √jXs pu / (kVA) ×f
T = 0.498 × 50 × √106 × 0.5 / ( 100 × 103 ) × 50 = 7.874 s

20. A 3 - φ, star connected alternator is rated at 1600 KVA, 13500 V. The armature effective resistance and synchronous resistance are 1.5 Ω and 30 Ω respectively per phase. If the power factor is 0.8(leading) then the percentage regulation for a load of 1280 KW will be

P = √3VLILcosφ
⇒ 1280 × 103 = √3 × 13500IL × 0.8
IL = 1280 × 103 × / ( √3 × 1350 × 0.8 ) = 68.43 A = Ia
cosφ = 0.8; sinφ = 0.6;
Ra = 1.5 Ω, Xs = 30 Ω, Vp = 13500 / √3 = 7794.5 V
For leading power factor, Ep2 = ( Vpcosφ + IaRa2 - ( - Vpsinφ + IaXs )2 = ( 7794.5 × 0.8 + 68.43 × 1.5 )2 + ( - 7794.5 × 0.6 + 68.43 × 30)2 = (6338)2 + ( - 2623.8)2
⇒ Ep = 6859.6 V
Voltage regulation = ( Ep - Vp / Vp) × 100
= - 11.99%.