The magnetization characteristic of a 4-pole de series motor may be taken as proportional to current over a part of the working range; on this basis the flux per pole is 4.5 mWb/A. The load requires a gross torque proportional to the square of the speed equal to 30 Nm at 1000 rev/min. The

Question

The magnetization characteristic of a 4-pole de series motor may be taken as proportional to current over a part of the working range; on this basis the flux per pole is 4.5 mWb/A. The load requires a gross torque proportional to the square of the speed equal to 30 Nm at 1000 rev/min. The armature is wave-wound and has 492 active conductors. Determine the speed at which the motor will run and the current it will draw when connected to a 220 V supply, the total armature resistance of the motor being [latex] 2.0 \Omega [/latex] .

Accepted Answer

Referring to Eq. (7.22)

[latex] E_{a}= \frac{\Phi nZ}{60} \left(\frac{P}{A} ight) = \frac{(4.5 imes 10^{-3} imes I_{a})n imes 492}{60} \left(\frac{4}{2} ight)[/latex]

[latex] = 0.0738 nI_{a}[/latex] (i)

Recalling Eq. (7.26) [latex] I_{a} = I_{L} + I_{f} \approx I_{L} [/latex] , the torque developed

[latex] T = \frac{1}{\pi } \Phi I_{a}Z\left(\frac{P}{A} ight) =\frac{1}{2\pi } (4.5 imes 10^{-3} imes I_{a})I_{a} imes 492\left(\frac{4}{2} ight)[/latex]

[latex] = 0.705 I_{a}^{2}[/latex] (ii)

Further [latex] E_{a} = V_{t} – I_{a} (R_{a} + R_{se}) = 220 – 2I_{a}[/latex] (iii)

Substituting Eq. (i) in (iii),

[latex] 0.0738 nI_{a} = 220 – 2I_{a}[/latex]

or [latex] I_{a} = \frac{220}{2+0.0738n}[/latex] (iv)

Substituting this expression for [latex] I_{a}[/latex] in Eq. (ii),

[latex] T = 0.705 \left[\frac{220}{2+0.0738n} ight] ^{2}[/latex]

Given: Load Torque [latex] T_{L} = K_{L}n^{2}[/latex]

From the given data [latex] K_{L}[/latex] can be evaluated as

[latex] K_{L} = \frac{30}{(1000)^{2}} =3 imes 10^{-5}Nm/rpm[/latex]

Under steady operation conditions, [latex] T_{L} = T[/latex] (developed)

or [latex] 3 imes 10^{–5} n^{2} = 0.705 \left(\frac{220}{2+0.0738n} ight) ^{2}[/latex]

Solving, [latex] n = 662.6 rpm[/latex]

Substituting for n in Eq. (iv)

[latex] I_{a}= \frac{220}{2+0.0738 imes 663.2}[/latex]

[latex] = 4.32 A[/latex]

Question 7.35: The magnetization characteristic of a 4-pole de series motor...

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