## 9 October 2017

### Torque and Output Power Equation of the DC Motor

• The term torque means ‘Turning movement of the force about an axis.’
T = F × r Newton – meter
Where T = Torque
F = Force in Newton

• Consider an armature of radius r meter and force F newton acts on it.
• Let us assume that the armature rotate at speed of N rpm.
• When the armature rotates one revolution, it cuts distance 2πr in time of 60 / N second. Therefore the work done per revolution
= Force × distance
= F × 2πr
But F × r = T
• So the work – done / revolution = 2πT  Newton – meter
• Now the Power developed
= Work done per unit second
= 2πT / ( 60 / N )
= 2πNT / 60
= Tω
Where ω = Angular velocity in radian / second
= 2πN / 60
• The electrical equivalent to mechanical power developed by the armature is given by
EbIa = 2πNT / 60
T = ( 60 / 2πN ) EbIa …………….( 1 )

 T = 9.55 ( EbIa / N )

If the speed is given in revolution per second ( rps )

 T = ( 9.55 / 60 ) ( EbIa / N )
• T = 0.159 ( EbIa / N )
• As the back emf Eb = ФZNP / 60A
Substitute Eb in the equation ( 1 )
T = ( 60 / 2πN ) ( ФZNP / 60A ) Ia
= ( 1 / 2π ) ( ФZNP / A ) Ia N – m
=  [ 1 / ( 2π × 9.81 ) ] ( ФZNP / A ) Ia Kg – m
• The number of conductor Z, number of poles P and number of parallel paths A is constant in the DC motor therefore
T α ФIa

Shaft Torque
• The shaft torque Tsh always less than the armature torque due to small amount of friction losses in the motor.
Shaft torque = Armature torque – Friction and windage losses
Tsh = Ta – Friction and windage losses

Output power
• Output power = Power developed in the armature
P =  T × ( 2πNT / 60 ) Watt
• The mechanical power develops at the shaft of the DC motor is always less than the armature power due to friction and windage losses.
Psh = Tsh × ( 2πNT / 60 ) Watt
• The mechanical power developed at the shaft is called as brake horse power ( BHP ).
One HP = 735.5 watt
Psh = ( Tsh ×  2πN / 60 )( 1 / 735.5 ) HP

You may also like :