Armature Torque of D.C. Motor
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
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)
Written by arjun on September 13th, 2009 with
no comments.
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