Question 7.7: A 100-hp, 250-V dc shunt motor has the magnetization curves ......
A 100-hp, 250-V dc shunt motor has the magnetization curves (including armature-reaction effects) of Fig. 7.14. The armature circuit resistance, including brushes, is 0.025 Ω. No-load rotational losses are 2000 W and the stray-load losses equal 1.0% of the output. The field rheostat is adjusted for a no-load speed of 1100 r/min.
a. As an example of computing points on the speed-load characteristic, determine the speed in r/min and output in horsepower (1 hp = 746 W) corresponding to an armature current of 400 A.
b. Because the speed-load characteristic observed to in part (a) is considered undesirable, a stabilizing winding consisting of 1-1/2 cumulative series turns per pole is to be added.
The resistance of this winding is assumed negligible. There are 1000 turns per pole in the shunt field. Compute the speed corresponding to an armature current of 400 A.
a. At no load, E_a = 250 V. The corresponding point on the 1200-r/min no-load saturation curve is
E_{a0}=250 \left( \frac{1200}{1100} \right)= 273 V
for which I_f = 5.90 A. The field current remains constant at this value.
At I_a = 400 A, the actual counter emf is
E_a = 250 – 400 × 0.025 = 240 V
From Fig. 7.14 with I_a = 400 \text{and} I_f = 5.90, the value of E_a would be 261 V if the speed were 1200 r/min. The actual speed is then found from Eq. 7.23
E_a = \left( \frac{n}{n_0} \right) E_{\mathrm{a0}} \quad \quad \quad (7.23)
n = 1200 \left( \frac{240}{261} \right) = 1100 r/min
The electromagnetic power is
E_a I_a = 240 × 400 = 96 kW
Deduction of the rotational losses leaves 94 kW. With stray load losses accounted for, the power output P_0 is given by
94 ~kW – 0.01 P_0 = P_0
or
P_0 = 93.1 kW = 124.8 hp
Note that the speed at this load is the same as at no load, indicating that armature- reaction effects have caused an essentially fiat speed-load curve.
b. With I_f = 5.90 A \text{and} I_s = I_a = 400 A, the main-field mmf in equivalent shunt-field amperes is
5.90 + \left( \frac{1.5}{1000} \right) 400 = 6.50 A
From Fig. 7.14 the corresponding value of E_a at 1200 r/min would be 271 V. Accordingly, the speed is now
n=1200 \left( \frac{240}{271} \right)=1063 r/min
The power output is the same as in part (a). The speed-load curve is now drooping, due to the effect of the stabilizing winding.