# Compound Generators

6.4. Compound Generators. The compound generator is used for more than any other type.

(i) It may be built and adjusted automatically to supply an approximately constant voltage at the point of use., throughout the entire range of load. This is well great advantage. It is possible to provide a constant supply voltage at the end of a long feeder by the simple expedient of over compounding the generator, because the resistance drop in the line is compensated for by the rising characteristic of the generator.

When the point of utilisation is near the generator, a flat-compounded machine may be used.

(ii) Differentially compounded generator finds an useful application as an arc welding generator where the generator is practically short circuited every time the electrode touches the metal plates to be welded.

(iii) Compound generators are used to supply power to :

• Railway circuits,
• Motors of electrified steam rail-roads,
• Industrial motors in many fields of industry,
• Incandescent lamps, and
• Elevator motors etc.

Example 15. Draw armature drop line if armature resistance is 0.2 Ω

Solution. The procedure of drawing an armature drop line is as follows (see Fig. 46):

(i) Take any value of armature current and find corresponding voltage in the
armature resistance. For example, take armature current of 200 A, voltage drop
in the armature for this value of armature current is 200 × 0.2 = 40 V.

So a point (200, 40) lying on this line is obtained.

(ii) Join the above point with origin O, the armature drop or ohmic drop line is obtained.

Fig. 46

Example 16. The open circuit characteristic (O. C. C.) of’ a separately excited genera tor, at 600 r.p.m., is as under:

Field Current, A       1.6       3.2       4.8       6.4       8.0       9.6       11.2

E.m.f., V         148     285     390     460     520     560     590

Find:

(i)                           The voltage to which the machine will excite as a shunt generator with a field circuit resistance of 60 ohm.

(ii)                        The critical resistance, at this speed.

Solution.

• Plot O.C.C. (ErjIf) as shown in Fig. 47.

Fig. 47

• Line OL represents 60 Ω line.
• The voltage to which the machine will excite a shunt generator is given by point L’ i.e. the intersection of O.C.C. and 60 Ω line. The machine will excite at 550 V.
• Draw line OM tangent to O.C.C. The slope of this line represents critical resistance. This value of critical resistance is 91 Ω.

Example 17. The open circuit characteristic of a 4-pole, 250 V shunt generator having 610 lap-connected armature conductors running at 750 r.p.m. is as follows:

Field current, A        0          0.5       1.0       2.0       3.0       4.0       5.0

E.m.f.,             V         10        50        100     175     220     245     262

Calculate:

(i) Field circuit critical resistance.

(ii) Critical speed for field circuit resistance of 80 n.

(iii) Residual flux per pole.

Solution. Number of poles of the generator, p = 4

Number of parallel paths, a = p = 4 [Generator being lap connected I

Number of armature conductors, Z = 610

• Plot O.C.C. from the given data as shown in Fig. 48.