A dc shunt motor runs at 1200 rpm on no-load drawing 5 A from 220 V mains. Its armature and field resistances are 0.25Ω and 110Ω respectively. When loaded (at motor shaft), the motor draws 62 A from the mains. What would be its speed? Assume that the armature reaction demagnetizes the field to

Question

A dc shunt motor runs at 1200 rpm on no-load drawing 5 A from 220 V mains. Its armature and field resistances are $0.25 \Omega$ and $110 \Omega$ respectively. When loaded (at motor shaft), the motor draws 62 A from the mains. What would be its speed? Assume that the armature reaction demagnetizes the field to the extent of 5%.

Also calculate the internal torque developed at no-load and on load. What is the motor shaft torque at load (this torque drives the mechanical load).

Accepted Answer

In a dc shunt motor

[latex] I_{L} = I_{a} + I_{f}[/latex]

Shunt field current, [latex] I_{f} = \frac{220}{110} = 2 A[/latex] (constant)

At no load

[latex] I_{a0}[/latex] = 5 – 2 = 3 A

[latex] E_{a0}= 220 – 0.25 imes 3[/latex] = 219.25 V

[latex] n_{0}[/latex] = 1200 rpm

[latex] E_{a0} = \acute{K_{a}} \Phi n_{0}[/latex]

or 219.25 = [latex] \acute{K_{a}} \Phi imes 1200[/latex]

[latex] T_{0} \omega _{0}=E_{a0}I_{a0}[/latex]

[latex] \left(\frac{2\pi imes 1200}{60} ight) T_{0} = 219.25 imes 3[/latex]

or [latex] T_{0} = 5.23 Nm[/latex]

This torque is absorbed in iron loss, and windage and friction losses of the motor (the shaft being at no load).

On load

[latex] I_{a1}[/latex] = 62 – 2 = 60 A

[latex] E_{a1} = 220 – 0.25 imes 60 = 205 V[/latex]

flux/pole [latex] \Phi _{1} = 0.95 \Phi [/latex]

[latex] E_{a1} = \acute{K_{a}} \Phi_{1} n_{1}[/latex]

[latex] 205 = \acute{K_{a}} imes 0.95 \Phi n_{1}[/latex]

Dividing Eq. (ii) by Eq. (i)

[latex] \frac{0.95 n_{1}}{1200}=\frac{205}{219.25}[/latex]

or [latex] n_{1} = 1181 rpm[/latex]

Drop in motor speed is only 1.6% on being loaded. This because the reduction in flux/pole due to armature reaction counters the drop in speed caused by armature resistance drop.

[latex] T_{1} \omega_{1} = E_{a1} I_{a1}[/latex]

or [latex] T_{1} = 99.45 Nm[/latex]

[latex] T_{1} (shaft) = 99.45 – 5.23[/latex]

[latex] = 94.22 Nm[/latex]

Question 7.34: A dc shunt motor runs at 1200 rpm on no-load drawing 5 A fro...

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