A universal motor (ac-operated) has a 2-pole armature with 960 conductors. At a certain load the motor speed is 5000 rpm and the armature current is 4.6 A; the armature terminal voltage and input are respectively 100 V and 300 W.

Compute the following, assuming an armature resistance of $3.5 \Omega$ .

(a) Effective armature reactance

(b) Maximum value of useful flux/pole

Question

A universal motor (ac-operated) has a 2-pole armature with 960 conductors. At a certain load the motor speed is 5000 rpm and the armature current is 4.6 A; the armature terminal voltage and input are respectively 100 V and 300 W.

Compute the following, assuming an armature resistance of [latex] 3.5 \Omega[/latex] .

(a) Effective armature reactance

(b) Maximum value of useful flux/pole

Accepted Answer

The operating conditions in terms of voltage and current of the armature circuit are shown in Fig. 10.49.

[latex] 100 imes 4.6 \cos \phi=330 W[/latex]

Or [latex] \phi=49.3^{\circ}[/latex]

(lagging because of reactive nature of the circuit).

(a) From the circuit the following can be written

[latex] \frac{100 \angle 49.3^{\circ}-E_{a} \angle 0^{\circ}}{3.5+j X_{a}}=4.6 \angle 0^{\circ}[/latex]

[latex] E_{a}[/latex] is in-phase with [latex] \bar{I}_{a}[/latex]

Or [latex] 65.2+j 75.8-E_{a}=16.1+j 4.6 X_{a}[/latex]

Equating real and imaginary parts,

[latex] E_{a}=65.2-16.1=49.1 V[/latex]

[latex] X_{a}=\frac{75.8}{4.6}=165 \Omega[/latex]

(b) [latex] E_{a}=\frac{1}{\sqrt{2}} \cdot \frac{\Phi n Z}{60}\left(\frac{P}{A} ight)[/latex]

[latex] \Phi=\frac{\sqrt{2} imes 49.1 imes 60}{5000 imes 960}=0.868 mWb[/latex]

Question 10.7: A universal motor (ac-operated) has a 2-pole armature with 9...

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