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

                    r = Radius in Meter

• 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

        E_bI_a = 2πNT / 60

            T = ( 60 / 2πN ) E_bI_a …………….( 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 ( E_bI_a / N )
• As the back emf E_b = ФZNP / 60A

        Substitute E_b in the equation ( 1 )

       T = ( 60 / 2πN ) ( ФZNP / 60A ) I_a

          = ( 1 / 2π ) ( ФZNP / A ) I_a N – m

          =  [ 1 / ( 2π × 9.81 ) ] ( ФZNP / A ) I_a 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 ФI_a

Shaft Torque

• The shaft torque T_sh always less than the armature torque due to small amount of friction losses in the motor.

        Shaft torque = Armature torque – Friction and windage losses

        T_sh = T_a – 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.

        P_sh = T_sh × ( 2πNT / 60 ) Watt

• The mechanical power developed at the shaft is called as brake horse power ( BHP ).

        One HP = 735.5 watt

        P_sh = ( T_sh ×  2πN / 60 )( 1 / 735.5 ) HP

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