# Torque Equation Of A D.C. Motor

It is seen that the turning or twisting force about an axis is called torque. Consider a wheel of radius R meters acted upon by a circumferential force F newtons as shown in the Fig. 1.

Fig. 1 |

The wheel is rotating at a speed of N r.p.m. Then angular speed of the wheel is,

ω = (2?N)/60 rad/sec

So workdone in one revolution is,

W = F x distance travelled in one revolution

= F x 2 R joules

And P = Power developed = Workdone/Time

= (F x 2?R) / (Time for 1 rev) = (F x 2?R) / (60/N) = (F x R) x (2?N/60)

**. ^{.}. **P = T x ? watts

Where T = Torque in N – m

ω = Angular speed in rad/sec.

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 x ?

**. ^{.}. **E

_{b }I

_{a }= x (2N/60)

but E

_{b }in a motor is given by,

E

_{b }= (?PNZ) / (60A)

**.**(ФPNZ / 60A) x Ia = T

^{.}._{a }x (2?N/60)

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

**Example 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.

**Solution** : P = 4, A = P = 4, Z = 480

F = 20 mWb = 20 x 10^{-3 }Wb, Ia = 50 A

Now T_{a }= 0.159 x ?Ia **.** (PZ/A) = 0.159 x 20 x 10^{-3 }x 50 x (4×480/4)

= 76.394 N-m

__1.1 Types of Torque in the Motor__

Basically the torque is developed in the armature and hence gross torque produced is denoted as T_{a}.

Fig. 2 Type of torque |

The mechanical power developed in the armature is transmitted to the load through the shaft of the motor. It 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 the friction, windage and the iron loss. The torque required to overcome theses losses is called lost torque, denoted as T_{f}. 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}.

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 T_{sh }and the angular speed ? 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.

__1.2 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 be rotate at a speed say r.p.m. on no load. The motor draws an armature current of I_{a0}.

Where E_{b0} is back e.m.f. on no load, proportional to speed N_{0}.

Now armature torque T_{a} for a motor is,

T_{a}? ?Ia

As flux is present and armature current is present, hence T_{a0} i.e. armature torque exists on no load.

Now T_{a} = T_{sh + }T_{a}

but on no load, Tsh = 0

So on no load, motor keeps on rotating at a speed of N_{0 }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 E_{b0 }is also large hence (V – E_{b0}) is very small hence armature current I_{a0 }is also small. So motor draws lees 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.

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. is E_{b0 }I_{a0 }practically assumed to be constant through 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 condictions.

**Example 2** : A 4 pole, lap wound d.c. motor has 540 conductors. Its speed 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

i) Induced e.m.f. ii) Armature current iii) Stray losses iv) Lost torue

**Solution** : P = 4, A = P = 4

Running light means it is on no load.

**. ^{.}. **N

_{0 }= 1000 r.p.m.

Z = 540 and ? = 25 x 10

^{-3}Wb

**.**= (?PN

^{.}._{0 }Z)/(60A) = (25 x 10

^{-3}x 4 x 1000 x 540)/(60 x 4) = 225 V

i) Induced e.m.f., E_{b0 }= 225 V

ii) From voltage equation, V = E_{b }+ I_{a }R_{a }

**. ^{.}. **V = E

_{b0 }+ I

_{a0 }R

_{a }

**.**230 = 225 + I

^{.}._{a0 }x 0.8

**.**I

^{.}._{a0 }= 6.25 A

iii) On no load, power developed is fully the power required to overcome stray losses.

**. ^{.}. **Stray losses = E

_{b0 }I

_{a0 }= 225 x 6.25 = 1406.25 W

iv) Lost torque = (E_{b0 } I_{a0})/ ?_{a0 }= 1406.25/(2?N_{0 }/60) = (1406.25 x 60)/(2×1000) = 13.428 N-m.