Parallel Operation of Two Alternators

Parallel Operation of Two Alternators  AU : Oct.-99, May-04, 07, April-04, Dec.-lO

It has already been seen that in case of interconnected systems at power stations number of alternators are operating in parallel. In fact the electricity demand of a country is fed by many alternators operating in parallel. There are various advantages of operating the alternators in parallel. 

1. Necessity of Parallel Operation of Alternators

i) For a given capacity of a generating station, either a single large unit or many small units may be installed. If there are many small units operating in parallel instead of single large unit then number of alternators operating at a time can be changed depending upon electricity requirements or load demands. This will help in operating the alternator near its full load capacity so that efficiency will also be better.

ii) The operating cost will be significantly reduced compared to single large unit.

iii) A particular unit may be shut down for certain period during the maintenance and inspection at the power stations. For that period the load can be transferred to other units if number of units are operating in parallel.

iv) Several power stations are interconnected by a grid which is economical and advantageous. This will make sure the optimum utility of the alternators.

v) The continuity and reliability of the supply can be maintained at better level due to interconnections.

vi) The future load demand can be satisfied by adding more alternators in parallel without disturbing the original setup.

The interconnection of alternators i.e. the process of synchronization is already discussed in previous section. Now we will consider how the two alternators will operate in parallel. What will happen if the driving torque or the excitation for any of the alternator is changed. In practice it is rare to have two alternators operating in parallel. But the concept of parallel operation of alternators can be well understood by considering two alternators in parallel.

2. Load Sharing between Two Alternators

Consider two identical alternators connected in parallel as shown in the Fig. 3.13.1.

Here cylindrical rotor alternators are assumed for simplicity, but the results obtained are equally applicable to both salient and non-salient machines.

The terminal voltage V is given by,

If no load is connected to the alternators only circulating current ISY will flow in the circuit. This current is given by,

Example 3.13.1 Two single phase alternators operating in parallel have induced e.m.f.s on open circuit of 220 ∠ 0° and 220 ∠ 10° V and respective reactances of j 3 Ω and j4 Ω . Calculate i) Terminal voltage ii) Currents and iii) Power delivered by each of the alternators to a load of resistance 6 Ω .

Solution :

Example 3.13.2 Two identical round rotor alternators supply inductive load of 1.414 p.u. at 0.707 p.f. lagging and at rated voltage p.u. synchronous reactance of each is 0.6. Find e.m.f. and power angle. State assumptions made.

Solution :

Following assumptions are made.

1. The terminal or bus bar voltage at the point of connection is constant.

2. The alternators are identical and are initially equally excited.

3. The power supplied by the prime movers is adjusted so that each machine carries half the load represented by external impedance Z = R + j 2π fL, where R and L are constant.

4. The stator resistance is negligible.

Let the rated line voltage be 1 p.u. VL =1 p.U. 

Example 3.13.3 Two alternators working in parallel supply following loads i) lighting load of 500 kW ii) 1000 kW at 0.9 p.f. lag iii) 500 kW at 0.9 p.f. lead iv) 800 kW at 0.8 lag. One alternator is supplying 1500 kW at 0.95 p.f. lagging. Calculate the load on the other machine.

Solution :

The kW and kVAR components of each load are as follows. For lagging loads kVAR is considered positive whereas for leading loads kVAR is considered negative.

Total kW = 500 + 1000 + 500 + 800 = 2800

Total kVAR = 0 + 484.30 - 242.15 + 600 = 842.15

If ϕc is total p.f. angle of combined load.

Example 3.13.4 Two 3 phase 3.3 kV star connected alternators supply a load of 1500 kW at 0.8 pf lagging. The synchronous impedance per phase of machine A is (0.5+jl0) Ω and of machine B is (0.4 + jl2) Ω. These excitaion of machine A is adjusted so that it delivers 150 A at a lagging p.f. and governers are so set that machines share load equally. Determine for each machine the current, power factor, induced e.m.f. and load angle.

Solution :

The load current can be calculated as,

Now as the load is shared equally, each alternator will supply a load of 3000 kW/2 = 1500 kW. From this we can find out operating p.f. of alternator 1.

IL1 = 150 A is given

Examples for Practice

Example 3.13.5 Two similar single-phase alternators are running in parallel. Their emfs are 100 V and 105 V respectively and the impedance of each is (0.2 + j 1.0) Ω. Find the terminal voltage, current and power supplied by each machine to a load impedance of (2 + j3) Ω.

[Ans.: 90.5438 ∠-2.726° V, 10.2804 ∠- 54.44° A, 541.327 Vf, 14.8876 ∠- 62.214“ A, 628.388 W]

Example 3.13.6 Two 3 ϕ synchronous generators operate in parallel on the same load. Determine the kW output and power factor of each machine under the following conditions. Synchronous impedance of each generator :

0.2 + j Ω / phase

Equivalent impedance of load : 3 + j4 Ω /phase

Induced e.m.f. per phase 2000 + j0 volts for machine 1 and 2200 + jl00 for machine 2.

[Ans.: Pout1 = 399.15 kW, 0.6054 lag, P out2 = 749.86 kW, 0.597 lag]

Example 3.13.7 Two 6600 volt, star connected alternators operating in parallel supply the following loads.

i) 400 kW at UPF ii) 1000 kW at 0.71 pf lag.

iii) 400 kW at 0.8 pf lag. iv) 300 kW at 0.9 pf lag.

The armature current of one machine is 110 A, at a pf. of 0.9 lag. Find the output, armature current and pf. of the other machine.

[Ans.: 1131.722 kW, 0.7365 lag, 968.27 kW]

Example 3.13.8 Two identical, 3-phase alternator operating in parallel, share equally a load of 1000 kW at 6600 V and 0.8 lagging power factor. The field excitation of the first machine is adjusted so that the armature current is 50 A at lagging pf.

Determine : i) The armature current of the second alternator

ii) The power factor at which each machine operates.

[Ans.: i) 60.216 A, ii) 0.8747 lag, 0.7263 lag]

Example 3.13.9 A 5 MVA, 10 kV, 1500 r.p.m. 3-phase, 50 Hz, 4 pole alternator is operating on infinite bus bar. Find the synchronizing power per mechanical degree of angular displacement under no load condition Xs - 20 %. AU : April-04

[Ans.: 872.66 kW]

Review Questions

1. Discuss about the parallel operation of two alternators with identical speed/load characteristics.

AU : May-04, 07, Dec.-lO, Marks 8

2. State the requirements of alternators to be connected in parallel.