A 250-V dc series motor with compensating winding has a total armature circuit resistance of (Ra + Rse) = 0.08 Ω. It is run at 1200 rpm by means of a primemover with armature circuit open and series field separately excited. This test yielded the following magnetisation data: Ise(A) 40 80 120 160

Question

A 250-V dc series motor with compensating winding has a total armature circuit resistance of [latex] (R_{a} + R_{se}) = 0.08 \Omega[/latex]. It is run at 1200 rpm by means of a primemover with armature circuit open and series field separately excited. This test yielded the following magnetisation data:

[latex] I_{se}(A)[/latex]	40	20	120	160	200	240	280	320	360	400
[latex] V_{OC} (V)[/latex]	62	130	180	222	250	270	280	288	290	292

Sketch the speed-torque characteristic of the series motor connected to 250 V main by calculating the speed and torque values at armature currents of 75, 100, 200, 300, 400 A.

Accepted Answer

Rather than drawing the magnetisation curve, we shall use the above data by linear interpolation between the data point given.

Sample Calculation

[latex] I_{a} = I_{se} = 75 A[/latex]

[latex] E_{a} = 250 – 0.08 imes 75 = 244 V[/latex]

Using linear interpolation, at [latex] I_{se} = 75 A[/latex]

[latex] E_{a} (1200 rpm) = 130 – \frac{130 - 62}{40} imes 5 = 121.5 V[/latex]

[latex] n = \frac{1200}{121.5} imes 244 = 2410 rpm[/latex]

[latex] T \omega = E_{a} I_{a}[/latex]

or [latex] T = \left(\frac{60}{2\pi imes 2410} ight) imes 244 imes 75=72.5 Nm[/latex]

Computations can be carried in tabular from below:

[latex] I_{a} = I_{se} (A)[/latex]	75	100	200	300	400
[latex] E_{a} (V)[/latex]	244	242	234	226	218
[latex] E_{a} (V)[/latex] at 1200 rpm	121.5	155	250	283	292
n (rpm)	2410	1874	1123	958	902
T (Nm)	72.5	123	398	676	923

The speed-torque (n – T) characteristic is drawn in Fig. 7.74.

Question 7.36: A 250-V dc series motor with compensating winding has a tota...

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