Armature reaction and commutation:Load Sharing

Load Sharing

Because of their slightly drooping voltage characteristics, shunt generators are most suited for stable parallel operation. Their satisfactory operation is due to the fact that any tendency on the part of a generator to take more or less than its proper share of load results in certain changes of voltage in the system which immediately oppose this tendency thereby restoring the original division of load. Hence, once paralleled, they are automatically held in parallel.

Similarly, for taking a generator out of service, its field is weakened and that of the other genera- tor is increased till the ammeter of the generator to be cleared reads zero. After that, its breaker and then the switch are opened thus removing the generator out of service. This method of connecting in and removing a generator from service helps in preventing any shock or sudden disturbance to the prime-mover or to the system itself.

It is obvious that if the field of one generator is weakened too much, then power will be delivered to it and it will run in its original direction as a motor, thus driving its prime-mover.

In Fig. 27.19. and 27.20 are shown the voltage characteristics of two shunt generators. It is seen that for a common terminal voltage V, the generator No. 1 delivers I1 amperes and generator No. 2, I2 amperes. It is seen that generator No. 1, having more drooping characteristic, delivers less current. It is found that two shunt generators will divide the load properly at all points if their characteristics are similar in form and each has the same voltage drop from no-load to full-load.

If it desired that two generators of different kW ratings automatically share a load in proportion to their ratings, then their external characteristics when plotted in terms of their percentage full-load currents (not actual currents) must be identical as shown in Fig. 27.21. If, for example, a 100-kW generator is working in parallel with a 200-kW generator to supply a total of 240-kW, then first generator will supply 80 kW and the other 160 kW.

When the individual characteristics of the generators are known, their combined characteristics can be drawn by adding the separate currents at a number of equal voltage (because generators are running in parallel). From this combined characteristic, the voltage for any combined load can be read off and from there, the current supplies by each generator can be found (Fig. 27.20).

If the generators have straight line characteristics, then the above result can be obtained by simple calculations instead of graphically.

Let us discuss the load sharing of two generators which have unequal no-load voltages.

From the above equation, it is clear that bus-bar voltage can be kept constant (and load can be transferred from 1 to 2) by increasing F2 or N2 or by reducing N1 and F1.N2 and N1 are changed by changing the speed of driving engines and F1 and F2 are changed with the help of regulating shunt field resistances.

It should be kept in mind that

(i) Two parallel shunt generators having equal no-load voltages share the load in such a ratio that the load current of each machine produces the same drop in each generator.

(ii) In the case of two parallel generators having unequal no-load voltages, the load currents produce sufficient voltage drops in each so as to keep their terminal voltage the same.

(iii) The generator with the least drop assumes greater share of the change in bus load.

(iv) Paralleled generators with different power ratings but the same voltage regulation will di- vide any oncoming bus load in direct proportion to their respective power ratings (Ex. 27.14).

Procedure for Paralleling D.C. Generators

(i) Close the disconnect switch of the incoming generator

(ii) Start the prime-mover and adjust it to the rated speed of the machine

(iii) Adjust the voltage of the incoming machine a few volts higher than the bus voltage

(iv) Close the breaker of the incoming generator

(v) Turn the shunt field rheostat of the incoming machine in the raise-voltage direction and that of the other machine(s) already connected to the bus in the lower-voltage direction till the desired load distribution (as indicated by the ammeters) is achieved.

Compound Generators in Parallel

In Fig. 27.22 are shown two compound generators (designated as No. 1 and No. 2) running in parallel.

Because of the rising characteristics of the usual com- pounded generators, it is obvious that in the absence of any corrective devices, the parallel operation of such generators is unstable. Let us suppose that, to begin with, each generator is taking its proper share of load. Let us now assume that for some reason, generator No. 1 takes a slightly increased load. In that case, the cur- rent passing through its series winding increases which further strengthens its field and so raises its generated e.m.f. thus causing it to take still more load. Since the system load is assumed to be constant, generator No. 2 will drop some of its load, thereby weakening its series field which will result in its further dropping off its load. Since this effect is cumulative. Generator No. 1 will, therefore, tend to take the entire load and finally drive generator No. 2 as a motor. The circuit breaker of at least one of the two generators will open, thus stopping their parallel operation.

For making the parallel operation of over-compound and level-compound generators stable,* they are always used with an equalizer bar (Fig. 27.22) connected to the armature ends of the series coils of the generators. The equalizer bar is a conductor of low resistance and its operation is as follows :

Suppose that generator No. 1 starts taking more than its proper share of load. Its series field current is increased. But now this increased current passes partly through the series field coil of generator No. 1 and partly it flows via the equalizer bar through the series field winding of generator No. 2. Hence, the generators are affected in a similar manner with the result the generator No. 1 cannot take the entire load. For maintaining proper division of load from no-load to full-load, it is essential that

(i) the regulation of each generator is the same.

(ii) the series field resistances are inversely proportional to the generator rating.

Series Generators in Parallel

Fig 27.23 shows two identical series generators connected in parallel. Suppose E1 and E2 are initially equal, generators supply equal currents and have equal shunt resistances. Suppose E1 increases slightly so that E1 > E2. In that case, I1 becomes greater than I2. Consequently, field of machine 1 is strengthened thus increasing E1 further whilst the field of machine 2 is weakened thus decreasing E2 further. A final stage is reached when machine 1 supplies not only the whole load but also supplies power to machine 2 which starts running as a motor. Obviously, the two machines will form a short-circuited loop and the current will rise indefinitely. This condition can be prevented by using equalizing bar because of which two similar machines pass approximately equal currents to the load, the slight difference between the two currents being confined to the loop made by the armatures and the equalizer bar.

Graphical Solution.

In Fig. 27.24, total load current of 60 A has been plotted along X-axis and the terminal voltage along Y–axis. The linear characteristics of the two generators are drawn from the given data. The common bus-bar voltage is given by the point of intersection of the two graphs. From the graph, it is seen that V = 236.3 V ; I1 = 23.6 A ; I2 = 36.4 A.

Example 27.16. Two shunt generators each with an armature resistance of 0.01 W and field resistance of 20 W run in parallel and supply a total load of 4000 A. The e.m.f.s are respectively 210 V and 220 V. Calculate the bus-bar voltage and output of each machine. (Electrical Machines-1, South Gujarat Univ. 1988)

It is represented by the point C, in graph, as an intersection, satisfying the condition that two currents (IA and IB) add up to 300 amp.

Example 27.20. In a certain sub-station, there are 5 d.c. shunt generators in parallel, each having an armature resistance of 0.1 W, running at the same speed and excited to give equal induced Each generator supplies an equal share of a total load of 250 kW at a terminal voltage of 500 V into a load of fixed resistance. If the field current of one generator is raised by 4%, the others remaining unchanged, calculate the power output of each machine and their terminal voltages under these conditions. Assume that the speeds remain constant and flux is proportional to field current.

Example 27.22. Two shunt generators and a battery are working in parallel. The open circuit voltage, armature and field resistances of generators are 250 V, 0.24 W, 100 W are 248 V, 0.12 W and 100 W respectively. If the generators supply the same current when the load on the bus-bars is 40 A, calculate the e.m.f. of the battery if its internal resistance is 0.172 W.

Solution. Parallel combination is shown in Fig. 27.29.

Values of currents and induced e.m.fs. are shown in the diagram.

Example 27.23. Two d.c. generators A and B are connected to a common load. A had a constant e.m.f. of 400 V and internal resistance of 0.25 W while B has a constant e.m.f. of 410 V and an internal resistance of 0.4 W . Calculate the current and power output from each generator if the load voltage is 390 V. What would be the current and power from each and the terminal voltage if the load was open-circuited ? (Elect. Engg; I, Bangalore Univ. 1987)

Solution. The generator connections are shown in Fig. 27.30 (a).

Tutorial Problem No. 27.2

1. Two separately-excited d.c. generators are connected in parallel supply a load of 200 A. The machines have armature circuit resistances of 0.05 W and 0.1 W and induced e.m.fs. 425 V and 440 V respectively. Determine the terminal voltage, current and power output of each machine. The effect of armature reaction is to be neglected. (423.3 V ; 33.3 A ; 14.1 kW ; 166.7 A ; 70.6 kW)

2. Two shunt generators operating in parallel given a total output of 600 A. One machine has an armature resistance of 0.02 W and a generated voltage of 455 V and the other an armature resistance of

0.025 W and a generated voltage of 460 V. Calculate the terminal voltage and the kilowatt output of each machine. Neglect field currents. (450.56 V ; 100 kW ; 170.2 kW)

3. The external characteristics of two d.c. shunt generators A and B are straight lines over the working range between no-load and full-load.

4. Two shunt generators operating in parallel have each an armature resistance of 0.02 W. The combined external load current is 2500 A. If the generated e.m.fs. of the machines are 560 V and 550 V respectively, calculate the bus-bar voltage and output in kW of each machine. (530 V; 795 kW; 530 kW)

5. Two shunt generators A and B operate in parallel and their load characteristics may be taken as straight lines. The voltage of A falls from 240 V at no-load to 220 V at 200 A, while that of B falls from 245 V at no-load to 220 V at 150 A. Determine the current which each machine supplies to a common load of 300 A and the bus-bar voltage at this load.

(169 A ; 131 A ; 223.1 V)

6. Two shunt-wound d.c. generators are connected in parallel to supply a load of 5,000 A. Each machine has an armature resistane of 0.03 W and a field resistance of 60 W, but the e.m.f. of one machine is 600 V and that of the other is 640 V. What power does each machine supply ?

(1,004 kW ; 1,730 kW including the fields)

7. Two shunt generators running in parallel share a load of 100 kW equally at a terminal voltage of 230 V. On no-load, their voltages rise to 240 V and 245 V respectively. Assuming that their volt-ampere characteristics are rectilinear, find how would they share the load when the total current is reduced to half its original value ? Also, find the new terminal voltage. (20 kW ; 30 kW, 236 V)

8. Two generators, each having no-load voltage of 500 V, are connected in parallel to a constant resistance load consuming 400 kW. The terminal p.d. of one machine falls linearly to 470 V as the load is increased to 850 A while that of the other falls linearly to 460 V when the load is 600 A. Find the load current and voltage of each generator.

If the induced e.m.f. of one machine is increased to share load equally, find the new current and voltage. (I1 = 626 A ; I2 = 313 A ; V = 479 V ; I = 469.5 A ; V = 484.4 V)

9. Estimate the number of turns needed on each interpole of a 6-pole generator delivering 200 kW at 200 V ; given : number of lap-connected armature conductors = 540 ; interpole air gap = 1.0 cm ; flux- density in interpole air-gap = 0.3 Wb/m2. Ignore the effect of iron parts of the circuit and of leakage.

[10] (Electrical Machines, B.H.U. 1980)

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