D.c. Motor:Comparison of Shunt and Series Motors

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Comparison of Shunt and Series Motors

(a) Shunt Motors

The different characteristics have been discussed in Art.

It is clear that

(a) speed of a shunt motor is sufficiently constant.

(b) for the same current input, its starting torque is not a high as that of series motor. Hence, it is used.

(i) When the speed has to be maintained approximately constant from N.L. to F.L. i.e. for driving a line of shafting etc.

(ii) When it is required to drive the load at various speeds, any one speed being kept constant for a relatively long period i.e. for individual driving of such machines as lathes. The shunt regulator enables the required speed control to be obtained easily and economically.

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(b) Series Motors

The operating characteristics have been discussed in Art 29.13. These motors

1. have a relatively huge starting torques.

2. have good accelerating torque.

3. have low speed at high loads and dangerously high speed at low loads. Hence, such motors are used

1. when a large starting torque is required i.e. for driving hoists, cranes, trams etc.

2. when the motor can be directly coupled to a load such as a fan whose torque increases with speed.

3. if constancy of speed is not essential, then, in fact, the decrease of speed with increase of load has the advantage that the power absorbed by the motor does not increase as rapidly as the torque. For instance, when torque is doubled, the power approximately increases by about 50 to 60% only.

4. a series motor should not be used where there is a possibility of the load decreasing to a very small value. Thus, it should not be used for driving centrifugal pumps or for a belt-drive of any kind.

Losses and Efficiency

The losses taking place in the motor are the same as in generators. These are (i) Copper losses (ii) Magnetic losses and (iii) Mechanical losses.

The condition for maximum power developed by the motor is

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The condition for maximum efficiency is that armature Cu losses are equal to constant losses.          (Art. 26.39).

Power Stages

The various stages of energy transformation in a motor and also the various losses occurring in it are shown in the flow diagram of Fig. 29.26.

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It is seen that A - B = copper losses and B - C = iron and friction losses.

Example 29.34. One of the two similar 500-V shunt machines A and B running light takes 3 A. When A is mechanically coupled to B, the input to A is 3.5 A with B unexcited and 4.5 A when B is separately-excited to generate 500 V. Calculate the friction and windage loss and core loss of each machine. (Electric Machinery-I, Madras Univ. 1985)

Solution. When running light, machine input is used to meet the following losses (i) armature Cu loss (ii) shunt Cu loss (iii) iron loss and (iv) mechanical losses i.e. friction and windage losses. Obviously, these no-load losses for each machine equal 500 ´ 3 = 1500 W.

(a) With B unexcited

In this case, only mechanical losses take place in B, there being neither Cu loss nor iron-loss because B is unexcited. Since machine A draws 0.5 A more current.

Friction and windage loss of B = 500 ´ 0.5 = 250 W

(b) With B excited

In this case, both iron losses as well as mechanical losses take place in machine B. Now, machine

A draws, 4.5 - 3 = 1.5 A more current.

Iron and mechanical losses of B = 1.5 ´ 500 = 750 W Iron losses of B = 750 - 250 = 500 W

Example 29.35. A 220 V shunt motor has an armature resistance of 0.2 ohm and field resistance of 110 ohm. The motor draws 5 A at 1500 r.p.m. at no load. Calculate the speed and shaft torque if the motor draws 52 A at rated voltage. (Elect. Machines Nagpur Univ. 1993)

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Example 29.39. A d.c. shunt machine while running as generator develops a voltage of 250 V at 1000 r.p.m. on no-load. It has armature resistance of 0.5 W and field resistance of 250 W. When the machine runs as motor, input to it at no-load is 4 A at 250 V. Calculate the speed and efficiency of the machine when it runs as a motor taking 40 A at 250 V. Armature reaction weakens the field by 4 %. (Electrical Technology, Aligarh Muslim Univ. 1989)

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Example 29.40. The armature winding of a 4-pole, 250 V d.c. shunt motor is lap connected. There are 120 slots, each slot containing 8 conductors. The flux per pole is 20 mWb and current taken by the motor is 25 A. The resistance of armature and field circuit are 0.1 and 125 W respectively. If the rotational losses amount to be 810 W find,

(i) gross torque (ii) useful torque and (iii) efficiency. (Elect. Machines Nagpur Univ. 1993)

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Example 29.41. A 20-hp (14.92 kW); 230-V, 1150-r.p.m. 4-pole, d.c. shunt motor has a total of 620 conductors arranged in two parallel paths and yielding an armature circuit resistance of 0.2 W. When it delivers rated power at rated speed, it draws a line current of 74.8 A and a field current of 3 A. Calculate (i) the flux per pole (ii) the torque developed (iii) the rotational losses (iv) total losses expressed as a percentage of power. (Electrical Machinery-I, Banglore Univ. 1987)

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Example 29.43. A d.c. series motor drives a load, the torque of which varies as the square of the speed. Assuming the magnetic circuit to be remain unsaturated and the motor resistance to be negligible, estimate the percentage reduction in the motor terminal voltage which will reduce the motor speed to half the value it has on full voltage. What is then the percentage fall in the motor current and efficiency ? Stray losses of the motor may be ignored.

(Electrical Engineering-III, Pune Univ. 1987)

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Example 29.44. A 6-pole, 500-V wave-connected shunt motor has 1200 armature conductors and useful flux/pole of 20 mWb. The armature and field resistance are 0.5 W and 250 W respectively. What will be the speed and torque developed by the motor when it draws 20 A from the supply mains ? Neglect armature reaction. If magnetic and mechanical losses amount to 900 W, find

(i) useful torque (ii) output in kW and (iii) efficiency at this load.

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Tutorial Problems 29.3

1. A 4-pole 250-V, d.c. series motor has a wave-wound armature with 496 conductors. Calculate

(a) the gross torque (b) the speed

(b) the output torque and (d) the efficiency, if the motor current is 50 A The value of flux per pole under these conditions is 22 mWb and the corresponding iron, friction and windage losses total 810 W. Armature resistance = 0.19 W, field resistance = 0.14 W.

[(a) 173.5 N-m (b) 642 r.p.m. (c) 161.4 N-m (d) 86.9%]

2. On no-load, a shunt motor takes 5 A at 250 V, the resistances of the field and armature circuits are 250 W and 0.1 W respectively. Calculate the output power and efficiency of the motor when the total supply current is 81 A at the same supply voltage. State any assumptions made.

[18.5 kW; 91%. It is assumed that windage, friction and eddy current losses are independent of the current and speed]

3. A 230 V series motor is taking 50 A. Resistance of armature and series field windings is 0.2 W and

0.1 W respectively. Calculate :

(a) brush voltage

(b) back e.m.f.

(c) power wasted in armature

(d) mechanical power developed

[(a) 215 V (b) 205 V (c) 500 W (d) 13.74 h.p.] (10.25 kW)

4. Calculate the shaft power of a series motor having the following data; overall efficiency 83.5%, speed 550 r.p.m. when taking 65 A; motor resistance 0.2 W, flux per pole 25 mWb, armature winding lap with 1200 conductor.

(15.66 kW)

5. A shunt motor running on no-load takes 5 A at 200 V. The resistance of the field circuit is 150 W and of the armature 0.1 W. Determine the output and efficiency of motor when the input current is 120 A at 200 V. State any conditions assumed. (89.8%)

6. A d.c. shunt motor with interpoles has the following particulars :

Output power ; 8,952 kW, 440-V, armature resistance 1.1 W, brush contact drop 2 V, interpole winding resistance 0.4 W shunt resistance 650 W, resistance in the shunt regulator 50 W. Iron and friction losses on full-load 450 W. Calculate the efficiency when taking the full rated current of 24 A.

(85%)

7. A d.c. series motor on full-load takes 50 A from 230 V d.c. mains. The total resistance of the motor is 0.22 W. If the iron and friction losses together amount to 5% of the input, calculate the power delivered by the motor shaft. Total voltage drop due to the brush contact is 2 A. (10.275 kW)

8. A 2-pole d.c shunt motor operating from a 200 V supply takes a full-load current of 35 A, the no- load current being 2 A. The field resistance is 500 W and the armature has a resistance of 0.6 W. Calculate the efficiency of the motor on full-load. Take the brush drop as being equal to 1.5 V per brush arm. Neglect temperature rise.

[Rajiv Gandhi Tech. Univ. Bhopal,2000] (82.63%)

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