Question

A 240 V DC shunt motor draws 40 A while driving a constant torque load. The armature resistance is Ra = 0.2 Ω, the field resistance is Rf = 240 Ω and machine constant is Ka = 120. The part of magnetization characteristic (field flux versus field current) related to the question is:

If (A) :      0.70 0.80 0.86 0.90 1.00 1.10 1.20 1.30 1.40

ϕ (mWb) : 5.80 6.30 6.59 6.70 7.20 7.60 7.96 8.20 8.45

The constant rotational loss is 240 W and the armature reaction is neglected. Find

a. the internal torque developed by the motor

b. speed    

c. efficiency    

Answer 1

As per question, following values are given:

Terminal voltage V = 240 V

Motor current I = 40 A

Armature resistance Ra = 0.2 ohm

Field resistance Rf = 240 ohm

Machine constant Ka = 120

Rotational losses Pr = 240 W

(a)

Current drawn by motor is equal to armature current and field current. So

I = Ia + If..........(1)

where If is the field current

and If = V/Rf

If = 240/240 = 1 A

Using values of I and If in equation 1 we get: 40 = Ia + 1

Ia = 40 - 1 = 39 A

In DC machine, back emf Eb = V - Ia.Ra

Eb = 240 - (39*0.2)

Eb = 232.2 V

Also Eb = Ka............(2)

where Ka is machine constant

is flux of motor in Weber

is speed in radians per second

As per question, motor given is DC shunt. In shunt motor, flux remains constant.

So, equation 2 becomes Eb = Ka.

= Eb/Ka

= 232.2/120 = 1.935 rad/s

So, internal torque developed by motor is given by T = (Eb.Ia)/

T = (232.2*39)/1.935

T = 4680 N-m

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(b)

Speed as calculated above = 1.935 rad/s

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(c)

Output power of motor Po = Internal power developed- rotational losses

Internal power developed = Eb.Ia

= 232.2*39 = 9055.8 W

Given rotational losses = 240 W

So, output power Po = 9055.8 - 240

Po = 8815.8 W

And, input power of motor Pi = V.Ia

Pi = 240*39 = 9360 W

So, efficiency = Po/Pi

= 8815.8/9360

= 0.9418 = 94.18 %

Efficiency = 94.18 %