E.M.F. Equation of D.C. Generator

E.M.F. Equation of D.C. Generator

AU : Dec-03, 06, 07, 08, 10, 14, 19 May-10, 15 16,19

Let P = Number of poles of the generator

Φ = Flux produced by each pole in webers ( Wb )

N = Speed of armature in r.p.m.

Z = Total number of armature conductors

A = Number of parallel paths in which the 'Z'  number of conductors are divided

So A = P for lap type of winding

A = 2 for wave type of winding

• Now e.m.f. gets induced in the conductor according to Faraday's law of electromagnetic induction. Hence average value of e.m.f. induced in each armature conductor is,

e = Rate of cutting the flux = dϕ / dt

• Now consider one revolution of conductor. In one revolution, conductor will cut total flux produced by all the poles i.e. ϕ × P. While time required to complete one revolution is 60 / N seconds as speed is N r.p.m.

• This is the e.m.f. induced in one conductor. Now the conductors in one parallel path are always in series. There are total Z conductors with A parallel paths, hence Z / A number of conductors are always in series and e.m.f. remains same across all the parallel paths.

ஃ  Total e.m.f. can be expressed as,

This is nothing but the e.m.f. equation of a d.c. generator.

Ex. 3.6.1 A 4 pole, lap wound, d.c. generator has a useful flux of 0.07 Wb per pole. Calculate the generated e.m.f. when it is rotated at a speed of 900 r.p.m. with the help of prime mover. Armature consists of 440 number of conductors. Also calculate the generated e.m.f. if lap wound armature is replaced by wave wound armature.

Sol :

Ex. 3.6.2 A wave wound, 6 pole, long shunt compound d.c. generator has 600 armature conductors. The generator is driven at 300 r.p.m. Calculate the e.m.f. generated if the flux/pole is 0.060 Wb. If now, the generator is required to produce e.m.f. of 550 V at a reduced value of flux/pole of 0.055 Wb, calculate the speed at which the armature of the generator must be driven.

Sol.

• It can be observed that to increase the generated e.m.f. with reduced flux, the speed must be increased.

Ex. 3.6.3 A 4 pole, lap wound, d.c. generators has 42 coils with 8 turns per coils. It is driven at 1120 r.p.m. If useful flux per pole is 21 mWb, calculate the generated e.m.f. Find the speed at which it is to be driven to generate the same e.m.f. as calculated above, with wave wound armature.

Sol.:

Ex. 3.6.4 A 12 pole D.C. generator has simple wave wound armature containing 144 coils of 10 turns each. The resistance of each turn is 0.011 Q. Its flux per pole is 0.05 Wb and it is running at a speed of 200 r.p.m. Obtain the induced armature voltage and the effective armature resistance.

AU: Dec.-10, Marks 8

Sol.

For A = 2 parallel paths, there are 2880/2 = 1440 conductors per parallel path i.e. 720 turns per parallel path.

Resistance/parallel path = 720 × 0.011 = 7.92 Ω

Such two paths in parallel hence Ra = 7.92 || 7.92 Ω

Ra = 3.96 Ω

… Effective armature resistance

Review Questions

1. Derive the e.m.f. equation of a d.c. generator.

AU: Dec.-03,06,07,08,10,14,19, May-15,16,19, Marks 8

2. A d.c. machine has 8 poles, lap connected armature with 960 conductors and flux per pole is patby 40 mWb. It is driven at 400 r.p.m. Calculate the generated e.m.f. If now lap connected armature is 989 replaced by wave connected, calculate the speed at which it should be driven to generate 400 V.

(Ans.: 250 V, 156 r.p.m.)

3. A four pole lap wound d.c. armature has a bore diameter of 0.7 m. It has 560 conductors and ratio of pole arc/pole pitch is 0.63. If the armature is running at 600 r.p.m. and flux density in the air gap is 1.2 Wb/sq.m., determine the generated e.m.f. in the armature if effective length of armature conductor is 20 cm.

 [Ans.: 465.508 V]

4. A d.c. generator is connected with lap winding. The number of poles in machine are 6. If the resistance of each conductor is 2 milliohm and if there are 60 slots with 4 conductors per slot, find the resistance of armature. Repeat the calculation if the winding is wave winding. [Ans. : 0.013332, 0.24 92, 0.12]

5. A 4-pole generator has an induced emf of 262 V when driven at a speed of 400 r.p.m. The armature is lap wound and has 652 conductors, its resistance being 0.15 Ω. The bore of the pole shoe is 42 cm diameter; the pole subtends an angle of 60° and is 20 cm long. Calculate the flux density in the air gap. 

[Ans. : 1.37 Wb/m2]