When the current carrying conductor is placed in the magnetic field, the conductor experiences a force which exerts turning moment or torque F x r. The current and magnetic field produce the torque and hence it is called Electromagnetic torque. In DC motor, the armature current flow in presence of air gap flux and due to an interaction of field flux and armature current, the armature conductors experiences force. Hence the torque is produced at the circumference of the rotor and rotor starts rotating.
To understand the torque equation, let us first understand the voltage equation of the DC motor.
Referring the above diagram, E is the supply voltage fed to the armature, Eb is the back emf developed across the armature. The back EMF always opposes the armature voltage and thus the it limits the armature current. Ra is the armature resistance. The IaRa drop takes place in the armature when the armature current Ia flows through the armature. The voltage equation of the DC motor is given below.
For calculating torque, first power needs to be calculated. If we multiply the back EMF(Eb) with armature current(Ia) we get power of the motor.
Derivation of Torque Equation of DC motor
Multiplying both sides of equation(1) by armature current(Ia), We get
Ia2 Ra is the power loss occurs in the armature and wasted in the form of heat energy and thus the power developed in a DC motor is;
The torque is the rotational force and it depends on the developed power of the armature at particular speed. Thus the torque of motor depends on the speed of rotation.
Putting the value of Pm from equation(3) in equation(5),
The back EMF of DC motor is mathematically expressed as;
Putting the value of back emf (Eb) from equation(8) in equation(7),we get the torque equation of DC motor.
The mechanical torque developed by DC motor can be calculated by subtracting the mechanical loss from the gross torque.
Tm is also called shaft torque(Tsh) of the DC motor.
Solved Problems on Torque Equation of DC motor
A d.c. motor takes an armature current of 100 A at 460 V. The armature circuit resistance is 0.2 Ω. The machine has 6-poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate-(i) Speed of the motor (ii) the gross torque developed by the armature.
Ia = 100 A
V = 460 V
Ra = 0.2 Ω
P = 6
A = P= 6 (Lap winding)
Z = 864
Φ = 0.05 Wb
Ta = ?
Eb = ΦZN/60 x (P/A) ——–(1)
Also, Eb = V – IaRa
= 460 – 100 x 0.2
= 460 – 20
Eb = 440 Volts
Putting value of Eb in equation (1)
Eb = ΦZN/60 x (P/A)
440 = 0.05 x 864 x N /60 x (6/6)
440 = 0.72 N
N = 440 /0.72 = 611 r.p.m.
Motor Gross Torque
T = ΦIa/2π x Z (P/A)
T = 0.159 ΦIa x Z (P/A)
T = 0.159 x 0.05 x 100 x 864 x(6/6)
T = 686.55 N-m
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