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Torque equation of a DC Motor with solved problem

by | Last updated Oct 4, 2021 | DC Motor

The torque equation of dc motor is essential for analyzing the rotating force of a motor. Toque is nothing but the turning or twisting moment of a force about an axis.

It is measured by the product of force and the radius at which the force acts. Newton-metre(Nm) is the unit of torque.

Let us consider a pulley or a wheel with a radius of r metre. When a circumferential force of F newton acts upon the pulley, it will cause the pulley to rotate at N rpm.

Torque equation of dc motor

Torque is given by the formula, T = Force * radius.

Work done by the pulley = force * distance moved.

Work done by the pulley = F * 2πr joule

Power developed = F * 2πr * N Joules/second or Watts

Power developed = (F * r) * 2πN

where 2πN is the angular velocity ω in radians/second

Therefore, Power developed, P = T * ω watts

    \[P = T\omega = T *  \frac{2\pi N}{60} = 0.105 NT \, \, watts\]

Torque developed in motor armature

Let Ta be the torque devloped in the armature of a motor running at N rpm.

    \[Power \, developed = T_a *  2 \pi N \, \,watts  --->(1) \]

We have derived the power equation of a dc motor, where

    \[Mechanical\, power \,developed \,in\, the\, motor = E_bI_a \, \,watt --->(2)\]

Equating the equations (1) & (2), we get,

    \[ T_a *  2\pi N = E_bI_a \]

    \[ T_a = \frac{ E_bI_a }{ 2\pi N } \]

    \[T_a = 0.159 \, \frac{ E_bI_a }{N} \, Neuton-metre --->(3)\]

We know the equation of back emf of a dc motor. It is given by,

    \[E_b = \frac{\phi N Z P}{A} \, volt \]

Substituting the above back emf equation in equation (3), we get

    \[T_a = 0.159 \, \frac{\phi N Z P}{A} \, \frac{I_a }{N} \]

    \[  T_a = 0.159 \, \phi \, Z \, I_a \, \frac{P}{A} \, Neuton-metre\]

From the above torque equation of a motor armature, we can say that, the armture torque,

Ta α φIa

In case of series motor, the flux produced in the field winding(φ) is proportional to the armature current(Ia), hence, Ta α Ia2.

In case of shunt motor, the flux (φ) produced is practically constant. Hence Ta α Ia.

Shaft torque

All the torque(as calculated above) developed in the motor armature is not available for doing useful work. It is because some torque is needed for supplying the iron and friction losses in the motor.

The torque that is available for the dc motor to do the useful work is known as shaft torque(Tsh). The name indicates that the torque is available at the motor shaft.

Using the power developed equation, the motor output power is given by,

    \[Output \, power = T_{sh} * 2 \pi N \, Watts \]

    \[T_{sh} = \frac{Output \, power}{2 \pi N} \]

    \[ T_{sh} = \frac{Output \, power \, in \, watts}{2 \pi N/60} \]

    \[ T_{sh} = 9.55 \frac{Output}{N} \, Neuton-metre\]

The difference between the armature torque(Ta) and the shaft torque(Tsh) is known ad the lost torque due to iron and frictional losses of the motor.

Solved Problem

Let us take a look at some problem.

A 220 V, 4-pole series motor with 600 conductors wave-connected supplying a load of 7.5kW by taking 35 A from the motors. Calculate the developed torque and shaft torque, if the flux per pole is 15 mWb and its armature circuit resistance is 0.5 Ω.

Given Values: V = 220 v, No. of poles = 4, Z = 600, Wave-connected(A=2), Output power = 7.5 kW = 7500 W, Ia = 35 A, φ = 15 mWb = 0.015 Wb, Ra = 0.5 Ω

To Find : Ta and Tsh

Solution:

Developed torque is given by,

    \[  T_a = 0.159 \, \phi \, Z \, I_a \, \frac{P}{A} \, Neuton-metre\]

    \[ T_a = 0.159 * 0.015 * 600 * 35 * \frac{4}{2} \]

    \[ T_a = 100.17 \, Newton-metre \]

Back emf is given by,

    \[E_b = V - I_aR_a = 220 - (35 * 0.5) = 202.5 V \]

Now, We know

    \[ E_b = \frac{\phi Z N P}{60 A} \]

    \[ 202.5 = \frac{0.015 * 600 * N *4}{60 * 2} \]

    \[ N = 675 \, rpm \]

Shaft torque is given by,

    \[T_{sh} = \frac{Output \, power \, in \, watts}{2 \pi N/60} \]

    \[T_{sh} = \frac{7500}{2 * \pi * 675/60} \]

    \[ T_{sh} = 106.10 \, Neuton-metre \]

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