Total Loss in a D.C. Generator

The various losses occurring in a generator can be sub-divided as follows :

where Ra = resistance of armature and interpoles and series field winding etc. This loss is about 30 to 40% of full-load losses.

Field copper loss. In the case of shunt generators, it is practically constant and I 2 R (or VIsh). In the case of series generator, it is = Ise2Rse where Rse is resistance of the series field winding.

This loss is about 20 to 30% of F.L. losses.

(iii) The loss due to brush contact resistance. It is usually included in the armature copper loss.

(b) Magnetic Losses (also known as iron or core losses),

(i) hysteresis loss, Wh µ B max f and (ii) eddy current loss, We µ Bmax f These losses are practically constant for shunt and compound-wound generators, because in their case, field current is approximately constant.

Both these losses total up to about 20 to 30% of F.L. losses.

(c) Mechanical Losses. These consist of :

(i) friction loss at bearings and commutator.

(ii) air-friction or windage loss of rotating armature.

These are about 10 to 20% of F.L. Losses.

The total losses in a d.c. generator are summarized below :

Stray Losses

Usually, magnetic and mechanical losses are collectively known as Stray Losses. These are also known as rotational losses for obvious reasons.

Constant or Standing Losses

As said above, field Cu loss is constant for shunt and compound generators. Hence, stray losses and shunt Cu loss are constant in their case. These losses are together known as standing or constant losses Wc.

Hence, for shunt and compound generators,

Example 26.24(a). A shunt generator delivers 195 A at terminal p.d. of 250 V. The armature resistance and shunt field resistance are 0.02 W and 50 W respectively. The iron and friction losses equal 950 W. FindExample 26.24(b). A 500 V, D.C. shunt motor draws a line current of 5 amps, on light load. If armature resistance is 0.15 ohm, and field resistance is 200 ohms, determine the efficiency of the machine running as a generator, delivering a load current of 40 Amp.

Solution. (i) As a motor, on Light load, out of 5 Amps of line current, 2.5 Amps are required for field circuit and 2.5 Amps are required for field circuit and 2.5 Amps are required for armature. Neglecting copper-loss in armature at no load (since it works out to be just one watt), the armature- power goes towards armature-core-loss and no-load mechanical loss at the rated speed. This amounts to (500 ´ 2.5) = 1250 watts.

Additional note :

Extension to Que : Comment on the armature output power in the two cases.

Solution. Assumption is that field side is suitably modified in the two cases.

Case (i) : Lap-wound Machine at 600 r.p.m.

Armature e.m.f.= 216 V

Let each armature-conductor be rated to carry a current of 10 amp.

In simple lap-wound machines, since a four-pole machine has four parallel paths in armature, the total armature output-current is 40 amp.

Hence, armature-output-power = 216 ´ 40 ´ 10-3 = 8.64 kW

Case (ii) : Wave-wound machine, at 500 r.p.m. Armature e.m.f. = 360 V

Due to wave-winding, number of parallel paths in armature = 2

Hence, the total armature output current = 20 amp

Thus, Armature Electrical output-power = 360 ´ 20 ´ 10-3 = 7.2 kW

Observation. With same flux per pole, the armature power outputs will be in the proportion of the speeds, as (7.2/8.64) = 5/6).

Further Conclusion. In case of common speed for comparing Electrical Outputs with same machine once lap-wound and next wave-wound, there is no difference in the two cases. Lap-wound machine has lower voltage and higher current while the wave-wound machine has higher voltage and lower current.

Example 26.34. A 4-pole, Lap-connected d.c. machine has an armature resistance of 0.15 ohm. Find the armature resistance of the machine is rewound for wave-connection.

Solution. A 4-pole lap-winding has 4 parallel paths in armature. If it is rewound for wave- connection, the resistance across the terminal becomes (4 ´ 0.15) = 0.6 ohm, as it obvious from Fig.26.67

Tutorial Problem No. 26.2

1. A 4-pole, d.c. generator has a wave-wound armature with 792 conductors. The flux per pole is 0.0121 Wb. Determine the speed at which it should be run to generate 240 V on no-load. [751.3 r.p.m.]

2. A 20 kW compound generator works on full-load with a terminal voltage of 230 V. The armature, series and shunt field resistances are 0.1, 0.05 and 115 W respectively. Calculate the generated e.m.f. when the generator is connected short-shunt. [243.25 V] (Elect. Engg. Madras Univ. April, 1978)

3. A d.c. generator generates an e.m.f. of 520 V. It has 2,000 armature conductors, flux per pole of

0.013 Wb, speed of 1200 r.p.m. and the armature winding has four parallel paths. Find the number of poles.

[4] (Elect. Technology, Aligarh Univ. 1978)

4. When driven at 1000 r.p.m. with a flux per pole of 0.02 Wb, a d.c. generator has an e.m.f. of 200 V. If the speed is increased to 1100 r.p.m. and at the same time the flux per pole is reduced to 0.019 Wb per pole, what is then the induced e.m.f. ? [209 V]

5. Calculate the flux per pole required on full-load for a 50 kW, 400 V, 8-pole, 600 r.p.m. d.c. shunt generator with 256 conductors arranged in a lap-connected winding. The armature winding resistances is 0.1 W, the shunt field resistance is 200 W and there is a brush contact voltage drop of 1 V at each brush on full- load. [0.162 Wb]

6. Calculate the flux in a 4-pole dynamo with 722 armature conductors generating 500 V when running at 1000 r.p.m. when the armature is (a) lap connected (b) wave connected.

[(a) 41.56 mWb (b) 20.78 mWb] (City & Guilds, London)

7. A 4-pole machine running at 1500 r.p.m. has an armature with 90 slots and 6 conductors per slot. The flux per pole is 10 mWb. Determine the terminal e.m.f. as d.c. Generator if the coils are lap-connected. If the current per conductor is 100 A, determine the electrical power.

[810 V, 324 kW] (London Univ.)

8. An 8-pole lap-wound d.c. generator has 120 slots having 4 conductors per slot. If each conductor can carry 250 A and if flux/pole is 0.05 Wb, calculate the speed of the generator for giving 240 V on open circuit. If the voltage drops to 220 V on full load, find the rated output of the machine.

[600 V, 440 kW]

9. A 110-V shunt generator has a full-load current of 100 A, shunt field resistance of 55 W and constant losses of 500 W. If F.L. efficiency is 88%, find armature resistance. Assuming voltage to be constant at 110 V, calculate the efficiency at half F.L. And at 50% overload. Find the load current.

[0.078 W ; 85.8% ; 96.2 A]

10. A short-shunt compound d.c. Generator supplies a current of 100 A at a voltage of 220 V. If the resistance of the shunt field is 50 W, of the series field 0.025 W, of the armature 0.05 W, the total brush drop is 2 V and the iron and friction losses amount to 1 kW, find

(a) the generated e.m.f. (b) the copper losses (c) the output power of the prime-mover driving the generator and (d) the generator efficiency.

[(a) 229.7 V (b) 1.995 kW (c) 24.99 kW (d) 88%]

11. A 20 kW, 440-V, short-shunt, compound d.c. generator has a full-load efficiency of 87%. If the resistance of the armature and interpoles is 0.4 W and that of the series and shunt fields 0.25 W and 240 W respectively, calculate the combined bearing friction, windage and core-loss of the machine.

[725 W]

12. A long-shunt, compound generator delivers a load current of 50 A at 500 V and the resistances of armature, series field and shunt field are 0.05 ohm and 250 ohm respectively. Calculate the generated electro- motive force and the armature current. Allow 1.0 V per brush for contact drop.

[506.2 V ; 52 A] (Elect. Engg. Banaras Hindu Univ. 1977)

13. In a 110-V compound generator, the resistances of the armature, shunt and the series windings are 0.06 W, 25 W and 0.04 W respectively. The load consists of 200 lamps each rated at 55 W, 110 V.

Find the total electromotive force and armature current when the machine is connected

(i) long shunt

(ii) short shunt. Ignore armature reaction and brush drop.

[(a) 1200.4, 104.4 A (b) 120.3 V, 104.6 A] (Electrical Machines-I, Bombay Univ. 1979)

14. Armature of a 2-pole, 200-V generator has 400 conductors and runs at 300 r.p.m. Calculate the useful flux per pole. If the number of turns in each field coil is 1200, what is the average value of e.m.f induced in each coil on breaking the field if the flux dies away completely in 0.1 sec ?

(JNTU, Hyderabad, 2000)

Hint: Calculate the flux per pole generating 200 V at 300 rpm. Calculate the e.m.f. induced in 1200-turn field coil due to this flux reducing to zero in 0.1 sec, from the rate of change of flux-linkage.

[f = 0.1 Wb, e = 1200 V]

15. A 1500 kW, 550-V, 16 pole generator runs at 150 rev. per min. What must be the useful flux if there are 2500 conductors lap-connected and the full-load copper losses are 25 kW? Calculate the area of the pole shoe if the gap density has a uniform value of 0.9 wb/m2 and find the no-load terminal voltage, neglecting armature reaction and change in speed.

(Rajiv Gandhi Techn. Univ., Bhopal, 2000) [0.09944 m2, 559.17 V]

SUMMARY of inducation machiens

Electrical DC Servo Motors:Speed Torque Characteristic of DC Servo Motors

Lighting Circuits:Dry Lining Boxes

Microphone Technology:Microphone Connectors, Cables, and Phantom Power

Cooling System part1

Devices, symbols, and circuits:Reading and understanding electrical drawings

AUDIO AND VIDEO SYSTEMS - RECORDER/AMPLIFIER CIRCUITS

Wye-Delta Starting:Dual Voltage Connections

Feedback Control Theory:Second Order Transfer Function

Tape Recording:Bias

REVERSIBLE SPEED CONTROL