Speed Regulation

7.9. Speed Regulation

The speed regulation of a D.C. motor is defined as follows:

“The change in speed when the load on the motor is reduced from rated value to zero, expressed percent of the rated load speed.”

SHUNT MOTORS

Example 18. Determine the torque developed when a current of a 30 A passes through the armature of a motor with the following particulars: lap winding, 310 conductors, 4-pole, pole-shoes 16.2 cm long subtending an angle of 60° at the centre, bore radius 16.2 cm, flux density in air gap

Solution. Number of poles, p=4

Number of parallel paths, a=p=4      [Motor being lap-wound]

Number of armature conductors, Z=310

Length of pole shoe =16.2 cm

Flux density in air gap, B=0.7 tesla

Armature current, Ia=30 A

Torque developed, T :

We know that,

Hence, torque developed = 28.45 N-m. (Ans.)

Example 19. A 230 V D.C. shunt motor takes 32 A at full load. Find the back e.m.f. all full load if the resistances of motor armature and shunt field windings arc 0.2 ohm and 115ohms respectively.

Solution, Supply voltage,             V = 230 Volts

Full load current, I= 32 A

Armature resistance, Ra = 0.2 ohm

Shunt field windings’ resistance, Rsh = 115 ohms

Back e.m.f., Eb

Armature current, Ia = I – Ish

= 36.86 – 1

= 35.86 A. (Ans.)

Back e.m.f Eb = V -IaRa

= 230 – 35.86 × 0.28

= 219.96 V. (Ans.)

Fig. 62

Example 21. A six-pole lap-connected 230 V shunt motor has 410 armature conductors. It takes 41 A on full load. The flux per pole is 0.05 weber. The armature and field resistances are 0.1 ohm and 230 ohms respectively. Contact drop per brush = 1 V.

Determine the speed of motor at full load.

Solution. Number of poles, p = 6

Number of parallel paths, a = 6 [Motor being lap-connected]

Number of armature conductors, Z = 410

Full load current, I= 41 A

Flux per pole, ɸ = 0.05 weber

Armature resistance, Ra = 0.1 ohm

Shunt field resistance, Rsh = 230 ohms

Contact drop/brush = 1 V

Speed of motor on full load, N :

Hence, speed of motor on full load = 655.6 r.p.m. (Ans.)

Example 22. A 250 volt d.c. shunt motor, on no load, runs at 1000 rpm and takes 5 A. The field and armature resistances are 250 ohms and 0.25 ohm. respectively. Calculate the speed when the motor is loaded such that it takes 41 A if the armature reaction weakens the field by 3%.

[P.T.U., May, 2001]

Fig. 63

Solution. Given: V = 250 V; N0 = 1000 r.p.m. ; 10 = 5 A; Rsh = 250 Ω ;

Ra = 0.25 Ω, I = 41 A, Example 23. A 120 volt d.c. shunt motor has an armature resistance of 0.2 ohms and a field resistance of 60 ohms. The full- load line current is 60 A and full-load speed is 1800 r.p.m. If the brush contact drop is 3 V, find the speed of the motor at half-load.

(P.T.U., June, 2000)

Ia1 =I1-Ish = 60-2 = 58 A

Back e.m.f Eb1 = V – Ia1 Ra – brush contact drop

= 120 – 58 × 0.2 – 3 = 105.4 V

Example 24. A 4-pole 500 V shunt motor takes 7 A on no-load, the no-load speed being 750 r.p.m. It has a shunt field current of 2 A. Calculate the full-load spced of the motor if it takes 122 A at full-load. Armature resistance = 0.2 ohm. Contact drop/brush = 1 V. Armature reaction weakens the field by 4% on full-load.

Solution. Supply voltage, V = 500 Volts

Shunt field current, Ish=2 A

Armature resistance, Ra=0.2 ohm

Contact drop/ brush =1 V

Example 25. A 440 V, 4-pole, lap connected shunt motor has a no-load input current of 15 A and a shunt field current of 10 A At full-load it takes a current of 150 A. If armature resistance = 0.1 ohm, flux per pole on no-load = 0.05 weber, number of armature conductors = 750 and contact drop per brush = 1 V, calculate:

(iii)                      Speed regulation.

Armature reaction weakens the field by 1.5% on [ull-locui.

Solution. Supply voltage,             V = 440 Volts

Number of poles,     p = 4

Number of parallel paths, a = p = 4   [Motor being lap-connected]

Armature resistance, Ra = 0.1 ohm

Flux per pole on no-load,   ɸ0 = 0.05 weber

Number of armature conductors, Z = 750

Contact drop per brush, = 1 V

Shunt field current, Ish=10 A

Speed regulation:

Armature current,  Ia0 = I0 -Ish = 15 – 10 = 5 A

Eb0 = V – Ia0Ra – brush contact drop

= 440-5×0.1-2 × 1 = 437.5 V

Hence, no-load speed = 700 r.p.m. (Ans.)

Armature current,

Ia = I -Ish = 150 – 10 = 140 A

Eb = V -IaRa – brush contact drop

= 440-140 × 0.1-2 × 1 = 424 V

ɸ = (1- 0.015) ɸ0                [Since armature reaction weakens the field by 1.5 per cent]

= 0.985 ɸ0

Hence, full-load speed = 688.73 r.p.m. (Ans.)

(iii) Percentage speed regulation

Hence, percentage speed regulation = 1.637%. (Ans.)

Example 26. A 250 V shunt motor takes a line current of 60 A and runs at 800 r.p.m. Its armature and field resistances are 0.2 ohm and 125 ohms respectively. Contact drop/brush = 1 V. Calculate:

(ii)                        The percentage reduction in the flux per pole in order that the speed may be 1000 r.p.m. when the armature current is 40 A.

Neglect the effects of armature reaction.

Solution. Supply voltage, V=250 Volts

Armature resistance, Ra=0.2 ohm

Shunt field resistance, Rsh=125 ohms

Contact drop =2 V

Armature current, Ia2=40 A

Percentage reduction in flux/pole:

Since the motor is shunt wound and armature reaction effect is negligible so flux may be assumed to be constant and therefore speed is directly proportional to the back e.m.f developed,

Hence, no-load speed = 836.5 r.p.m, (Ans.)

At armature current, Ia2 = 40 A

Let the flux be reduced to ɸ2

Eb2 = V – Ia2Ra – brush drop

= 250 – 40 × 0.2 – 2 × 1 = 240 V

Hence, %age reduction in flux = 18.8%. (Ans.)

Example 27. A 220 V shunt motor takes 60 A when running at 800 r.p.m. It has an armature resistance of 0,1 ohm. Determine the speed and armature current if the magnetic flux is weakened by 20%, Contact drop per brush = 1 V.

Total torque developed remains constant.

Solution. Supply voltage = 220 V

Load current, = I1= I01 = 60 A [Neglecting shunt field current]

Speed at 60 A, N1 = 800 r.p.m.

Armature resistance, Ra = 0.1 ohm

Contact drop/brush = 1 V

N2 =?, Ia2 =?

Hence, Speed = 992.9 r.p.m. (Ans.)

Armature current         = 75 A. (Ans.)

Example 28. A 440 V shunt motor takes 105 A (armature current) from the supply-and runs at 1000 r.p.m. Its armature resistance is 0.15 ohm, If total torque developed is unchanged, calculate the speed and armature current if the magnetic field is reduced to 70% of the initial value.

(Punjab University)

Solution. Supply voltage, V = 440 Volts

Armature current, Ia1=105A

Speed, N1 = 1000 r.p.m.

Armature resistance, Ra = 0.15 ohm

ɸ2 = 0.7 ɸ1

(i) Armature current, Ia2 :

As the torque developed is same

SERIES MOTORS

Example 29. A 230 V series motor develops torque of 310 N-m at 800 r.p.m. The torque lost due to iron and friction loss is 10 N-m. If the efficiency is 85% determine the current taken by the motor at 800 r.p.m.

Solution. Supply voltage = 230 V

Torque developed,   Ta = 310 N-m

Torque lost due to iron and friction, Tlost = 10 N-m

Efficiency = 85%

Speed of the motor, N = 800 r.p.m.

Current taken by motor at 800 r.p.m. :

Tuseful = Ta – Tlost = 310 – 10 = 300 N-m

Watts corresponding to useful torque

Example 30. A 8-pole 240 V lap-wound, series motor has armature and series field resistances of 0.2 ohm and 0.02 ohm respectively. There are 660 armature conductors. If the flux/pole is 0.3 Wb and the total torque developed in the armature is 320 N-m, find the current taken by the motor and its speed.

Solution. Number of poles, p=4

Number of parallel paths, a=p=4             [Motor being lap-wound]

Armature resistance,  Ra=0.2 ohm

Series field resistance, Rse=0.02 ohm

Number of armature conductors, Z= 660

Flux/pole, ɸ = 0.03 Wb

Total torque developed, Ia = 320 N-m

(i) Current taken by motor, Ia :

Example 31. A six-pole, lap-wound 400 V series motor has the following data: Number of armature conductors = 920, flux/pole= 0.045 Wb, total motor resistance= 0.6 ohm, iron and friction losses= 2 k W. If the current taken fly the motor is 90 A, find:

(i)                           Total torque;

(ii)                        Useful torque at the shaft;

(iii)                      Power output;

(iv)                      Pull at the rim of a pulley of 40 cm diameter connected to the shaft.

Solution. Number of poles, p= 6

Supply voltage, V= 400 Volts

Number of parallel paths, a= p = 6                 [Motor being lap-wound]

Number of armature conductors, Z = 920

Flux/pole, ɸ = 0.045 Wb

Motor resistance, Rm = 0.6 ohm

Iron and friction losses = 2 kW or 2000 W

Current taken by the motor, Ia = 90 A

Radius of the pulley, = 40/2= 20 cm or 0.2 m

Using the relation; Eb= V -IaRm = 400 – 90 × 0.6

= 346V

Example 32. A 240 V series motor takes 40 A when giving its rated output at 1500 r.p.m. Its resistance is 0.3 Ω. Find what resistance must be added to obtain rated torque

(i)                           at starting and

(ii)                        at 1000 r.p.m.

Solution. Given: V = 240 volts; I (= Ia) = 40 A, N = 1500 r.p.m., R = 0.3 Ω.

(i) Resistance to be added to obtain rated torque at starting, Radd :

Since the torque remains the same in both the cases, it is obvious that the current drawn by the motor remains constant at 40 A.

Example 33. A 200 V D.C. series motor runs at 700 r.p.m. when operating at its full-load current of 20 A. The motor resistance is 0.5 Ω and the magnetic circuit can be assumed unsaturated. What will be the speed if

(i)                           the load torque is increased by 44%

(ii)                        the motor current is 10 A.

Solution. Given: V = 200 volts; N1 = 700 r.p.m, I1 = 20 A, Rm = Ra + Rsh) = 0.5 Ω

Speeds, (N2, N3) :

(i) When the load torque is increased by 44% :

T2 = 1.44T1

ɸ2I2 = 1.44 ɸ1I1

122 = 1.44I12

N3  = 1437 r.p.m. (ANS)

Example 34. A 220 V D. C. series motor draws full-load line current of 38 A at the rated speed of 600 r.p.m. The motor has armature resistance of 0.4 Ω and the series field resistance is 0.2 Ω. The brush voltage drop irrespective of load is 3.0 volts, find:

(i)                           The speed of the motor when the load current drops to 19 A.

(ii)                        The speed on removal of load when the motor takes only 1 A from supply.

(iii)                      The internal horse power developed in each of the above cases. Neglect the effect of armature reaction and saturation.

Solution. Given:

V = 220 volts; I1 = 38 A, N1 = 600 r.p.m., Ra = 0.4 Ω,

Rse = 0.2 Ω, Brush drop = 3 V; 12 = 19 A, 10 = 1 A.

(i) Speed of the motor N2 (at 19 A)

Back e.m.f.,

Eb1 = V -I1(Ra + Rse) – brush voltage drop

= 220 – 38(0.4 + 0.2) – 3 = 194.2 V.

Back e.m.f., Eb2 =220 – 19(0.4 + 0.2) – 3 = 205.6 V

No-load, back e.m.f., Eb0 = 220 – 1(0.4 + 0.2) -3= 216.4 V

(iii) Internal H.P. developed (in the above cases) :

Internal H.P. developed at a load current 38 A

Example 35. A-D.C. series motor draws a line current of 100A from the mains while running at 1000 r.p.m. Its armature resistance is 0.15 Ω and the field resistance is 0.1 Ω. Assuming that the flux corresponding to a current of 25 A is 40% of that corresponding to 100 A, find the speed of the motor when it is drawing 25 A from 230 V supply.

(GATE, 1996)

Solution. Given: V = 230 volts; I1 = 100 A ; N1 = 1000 r.p.m. ;

Ra= 0.15 Ω; Rse = 0.1 Ω, I2 = 25 A; ɸ2 = 0.4ɸ1

Speed N2 :

While drawing line current of 100 A,

Eb1 = V –I1(Ra + Rse)

= 230 -100(0.15 + 0.1) = 205 V

While drawing line current of 25 A,

Eb2 = V – I2(Ra + Rse)

= 230 – 25(0.15 + 0.1) = 223.75 V