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.
... Eb Ia = x (2N/60)
but Eb in a motor is given by,
Eb = (ΦPNZ) / (60A)
... (ΦPNZ / 60A) x Ia = Ta x (2πN/60)
This is the torque equation of a d.c. motor.
Φ = 20 mWb = 20 x 10-3 Wb, Ia = 50 A
Now Ta = 0.159 x ΦIa . (PZ/A) = 0.159 x 20 x 10-3 x 50 x (4x480/4)
= 76.394 N-m
The shaft torque magnitude is always less than the armature torque, (Tsh < Ta).
Where Eb0 is back e.m.f. on no load, proportional to speed N0.
Now armature torque Ta for a motor is,
Ta α ΦIa
As flux is present and armature current is present, hence Ta0 i.e. armature torque exists on no load.
Now Ta = Tsh + Ta
but on no load, Tsh = 0
Torque developed = Torque required to overcome friction, windage, iron losses.
Where Eb0 = Back e.m.f. on no load.
and Ia0 = Armature current drawn on no load.
Solution : P = 4, A = P = 4
Running light means it is on no load.
... N0 = 1000 r.p.m.
Z = 540 and Φ = 25 x 10-3 Wb
... = (ΦPN0 Z)/(60A) = (25 x 10-3 x 4 x 1000 x 540)/(60 x 4) = 225 V
i) Induced e.m.f., Eb0 = 225 V
ii) From voltage equation, V = Eb + Ia Ra
... V = Eb0 + Ia0 Ra
... 230 = 225 + Ia0 x 0.8
... Ia0 = 6.25 A
iii) On no load, power developed is fully the power required to overcome stray losses.
... Stray losses = Eb0 Ia0 = 225 x 6.25 = 1406.25 W
iv) Lost torque = (Eb0 Ia0)/ ωa0 = 1406.25/(2πN0 /60) = (1406.25 x 60)/(2x1000) = 13.428 N-m.
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| Fig. 1 |
The wheel is rotating at a speed of N r.p.m. Then angular speed of the wheel is,
ω = (2πN)/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 2πR) / (Time for 1 rev) = (F x 2πR) / (60/N) = (F x R) x (2πN/60)
... P = T x ω watts
Where T = Torque in N - m
ω = Angular speed in rad/sec.
Let Ta 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 Eb Ia, as seen from the power equation. So if speed of the motor is N r.p.m. then,
Power in armature = Armature torque x ω
but Eb in a motor is given by,
Eb = (ΦPNZ) / (60A)
... (ΦPNZ / 60A) x Ia = Ta x (2πN/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 = 480Φ = 20 mWb = 20 x 10-3 Wb, Ia = 50 A
Now Ta = 0.159 x ΦIa . (PZ/A) = 0.159 x 20 x 10-3 x 50 x (4x480/4)
= 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 Ta. |
| 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 Tf. 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 Tsh.
The speed of the motor remains same all along the shaft say N r.p.m. Then the product of shaft torque Tsh and the angular speed ω 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 Tsh = 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 Ia0.
Now armature torque Ta for a motor is,
Ta α ΦIa
As flux is present and armature current is present, hence Ta0 i.e. armature torque exists on no load.
Now Ta = Tsh + Ta
but on no load, Tsh = 0
So on no load, motor keeps on rotating at a speed of N0 r.p.m. drawing an armature current of Ia0. This is just enough to produce a torque Ta0 which satisfies the friction, windage and iron losses of the motor. On no load, speed of the motor is large hence Eb0 is also large hence (V - Eb0) is very small hence armature current Ia0 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 Eb0 = Back e.m.f. on no load.
and Ia0 = Armature current drawn on no load.
This component of stray losses i.e. is Eb0 Ia0 practically assumed to be constant through the load on the motor is changed from zero to the full capacity of the motor. So Tf 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 torueSolution : P = 4, A = P = 4
Running light means it is on no load.
... N0 = 1000 r.p.m.
Z = 540 and Φ = 25 x 10-3 Wb
... = (ΦPN0 Z)/(60A) = (25 x 10-3 x 4 x 1000 x 540)/(60 x 4) = 225 V
i) Induced e.m.f., Eb0 = 225 V
ii) From voltage equation, V = Eb + Ia Ra
... V = Eb0 + Ia0 Ra
... 230 = 225 + Ia0 x 0.8
... Ia0 = 6.25 A
iii) On no load, power developed is fully the power required to overcome stray losses.
... Stray losses = Eb0 Ia0 = 225 x 6.25 = 1406.25 W
iv) Lost torque = (Eb0 Ia0)/ ωa0 = 1406.25/(2πN0 /60) = (1406.25 x 60)/(2x1000) = 13.428 N-m.
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