Armature Torque of D.C. Motor

Author: arjun
Category: DC Motor, Direct Current Machines, Electrical Machines

Torque is the turning moment of a force about an axis and is measured by the
product of force (F) and radius (r) at right angle to which the force acts i.e.
D.C. Motors torque
T = F × r

In a d.c. motor, each conductor is acted upon by a circumferential force F at a distance r, the radius of the armature (Fig. 4.8). Therefore, each conductor exerts a torque, tending to rotate the armature. The sum of the torques due to all armature conductors is known as gross or armature torque (Ta).

Let in a d.c. motor
r = average radius of armature in m
l = effective length of each conductor in m

Z = total number of armature conductors

A = number of parallel paths

i = current in each conductor = Ia/A
B = average flux density in Wb/m2
Φ = flux per pole in Wb
P = number of poles

Force on each conductor, F = B i l newtons

armature torque

Torque due to one conductor = F × r newton- metre
Total armature torque, Ta = Z F r newton-metre
= Z B i l r

Now i = Ia/A, B = Φ/a where a is the x-sectional area of flux path per pole at
radius r. Clearly, a = 2πr l /P.

Ta = Z × (Ф/a)×( Ia/A)×l×r

Ta = Z × (ФP/2πr l)×( Ia/A)×l×r = Z Ф IaP/(2πA) N-m

or Ta = 0.159Z Ф Ia(P/A) N-m……………………………………(i)

so Ta

Ta α Ф Ia

Hence torque in a d.c. motor is directly proportional to flux per pole and
armature current.

(i) For a shunt motor, flux Φ is practically constant.

Ta α  Ia

(ii) For a series motor, flux Φ is directly proportional to armature current Ia
provided magnetic saturation does not take place.

Ta α Ia2

up to magnetic saturation

Alternative expression for Ta

Eb = PФZN/60A

(60×Eb) /N= PФZ/A

From Eq.(i), we get the expression of Ta as:

Ta =0.159×(60× Eb/N)× Ia

Ta =9.55×( EbIa/N)

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