Torque Equation of a D.C. Motor

Torque Equation of a D.C. Motor

AU: May-07,08,09,14,16, April-04, Dec.-07.09,14

• It is seen that the turning or twisting force about an axis is called torque. Consider a wheel of radius R metres acted upon by a circumferential force F newtons as shown in the Fig. 4.6.1.

The wheel is rotating at a speed of N r.p.m.

The angular speed of the wheel is,

• Let T_a be the gross torque developed by the armature of the motor. It is also called armature torque. The gross mechanical power developed in the armature is E_b I_a, as seen from the power equation. So if speed of the motor is N r.p.m. then,

Power in armature = Armature torque × ω

This is the torque equation of a d.c. motor.

Ex. 4.6.1 A 4 pole d.c. motor takes a 50 A armature current. The armature has lap connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the armature of the motor.

AU: Dec.-07, Marks 8

Sol. :

1. Types of Torque in the Motor

• Basically the torque is developed in the armature and hence gross torque produced is denoted as T_a.

• The mechanical power developed in the armature is transmitted to the load through the shaft of the motor. is impossible to transmit the entire power developed by the armature to the load. This is because while transmitting the power through the shaft, there is a power loss due to the friction, windage and the iron loss. The torque required to overcome these losses is called lost torque, denoted as T. These losses are also called stray losses.

• The torque which is available at the shaft for doing the useful work is known as load torque or shaft torque denoted as T_sh.

 T_a = T_f + T_sh

• The shaft torque magnitude is always less than the armature torque, (T_sh < T_a).

• The speed of the motor remains same all along the shaft say N r.p.m. Then the product of shaft torque Te and the angular speed o rad/sec is called power available at the shaft i.e. net output of the motor. The maximum power a motor can deliver to the load safely is called output rating of a motor. Generally it is expressed in H.P. It is called H.P. rating of a motor.

Net output of motor = P_out =T_sh × ω

2. No Load Condition of a Motor

• On no load, the load requirement is absent. So T_sh = 0. This does not mean that motor is at hault. The motor can rotate at a speed say N₀ r.p.m. on no load. The motor draws an armature current of I_a0.

I_a0 =V-E_b0 / R_a

where E_b0 is back e.m.f. on no load, proportional to speed N₀.

Now armature torque T for a motor is,

T_a  α ϕ I_a

As flux is present and armature current is present, hence T_a0 i.e. armature torque exists on no load.

Now T_a = T_f + T_sh

But on no load,T_sh = 0

T_a0 = T_f

• So on no load, motor keeps on rotating at a speed of N₀ r.p.m. drawing an armature current of I_a0 This is just enough to produce a torque T_a0 which satisfies the friction, windage and iron losses of the motor. On no load, speed of the motor is large hence Ebo is also large hence (V – E_b0) is very small hence armature current I_a0 is also small. So motor draws less current on no load and takes more and more current as motor load increases.

So on no load,

Torque developed = Torque required to overcome friction, windage, iron losses.

Power developed (E_b0× I_a0) = Friction, windage and, iron losses.

where E_{b0 =} Back e.m.f. on no load

and  I_a0 = Armature current drawn on no load

• This component of stray losses i.e. E_b0 L_a0 is practically assumed to be constant though the load on the motor is changed from zero to the full capacity of the motor. So T_f is practically assumed constant for all load conditions.

Ex. 4.6.2 A 4 pole, 220 V d.c. shunt motor, lap wound has 960 conductors. The flux per pole is 20 mWb. Determine the torque developed by the armature and the useful torque in Nm when the current drawn by the motor is 28 A. The armature resistance is 0.1 Ω and the shunt field resistance is 125 Ω. The rotational losses of the machine amounts to 800 watts. AU: Dec.-07, Marks 8

Sol.

Ex 4.6.3 Determine developed torque and shaft torque of 220 V, 4 pole series motor with 800 conductors wave connected supplying a load of 8.2 kW by taking 45 A from the mains. The flux per pole is 25 mWb and its armature circuit resistance is 0.6 Ω. AU: May-08, Marks 8

Sol. :

Ex. 4.6.4 A 60 kW, 400 V d.c. shunt motor has 4 poles and a wave connected armature of 450 conductors. The flux per pole is 45 mWb. R_a = 0.1Ω and R_sh = 200 Ω.If the full load efficiency is 90.5 %, find the 1) Speed 2) Armature torque 3) Useful torque. AU: May-09, Marks 8

Sol. :

Ex. 4.6.5 A 4 pole dc motor is lap wound with 400 conductors. The pole-shoe is 20 cm long and the average flux density over one- pole- pitch is 0.4 T, the armature diameter being 30 cm. Find the torque and gross mechanical power developed when the motor is drawing 25 A and running at 1500 rpm. AU: May-16, Marks 8

Sol. :

Review Questions

1. Derive from the first principle an expression for the torque developed in a d.c. motor. AU: May-07, 08, 09, 14, Dec.-09, 14, Marks 8

 2. A 200 V, 4 pole, lap wound, d.c. shunt motor has 800 conductors on its armature. The resistance of the armature winding is 0.5 Ω and that of shunt field winding is 200 Ω. The motor takes a current of 21 A, the flux per pole 30 mWb. Find the speed and the gross torque developed in the motor. 

(Ans. 475 r.p.m., 76.38 N-m)

3. A d.c. motor takes an armature current of 74 A when flux is 10 mWb. It develops a torque of 120 N-m. If armature current is changed to 45 A keeping the flux constant, determine the new torque developed. With the same flux, if entire load on the motor is removed, armature current is observed to be 6.8 A. Determine the torque required to overcome stray losses.

(Ans. 72.97 N-m, 11.02 N-m)

 4. A 4 pole d.c. motor takes an armature current of 50 Amps. The armature has lap connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the motor. AU: April-04 (Ans.: 76.3943 N-m)

5. A 4 pole, lap wound d. C. motor has 540 conductors. Its speed is found to be 1000 r.p.m. when it is made to run light. The flux per pole is 25 mWb. It is connected to 230 V d.c. supply. The armature resistance is 0.8Ω. Calculate i) Induced e.m.f. ii) Armature current iii) Stray losses iv) Lost torque.

W 8810.0 [Ans.: 225 V, 6.25 A, 1406.25 W, 13.428 Nm]

6. Determine the torque developed when a current of 30 A passes through the armature of a motor with lap winding of 310 conductors, 4 pole with pole shoes 16.2 cm long and subtending angle of 60° at the centre. The bore radius is 16.2 cm and flux density in the air gap is 0.7 T. 

[Ans.: 28.44 Nm]