Question 7.3: A 100-kW, 250-V, 400-A, long-shunt compound generator has an......
A 100-kW, 250-V, 400-A, long-shunt compound generator has an armature resistance (including brushes) of 0.025 Ω, a series-field resistance of 0.005 Ω, and the magnetization curve of Fig. 7.14. There are 1000 shunt-field turns per pole and three series-field turns per pole. The series field is connected in such a fashion that positive armature current produces direct-axis mmf which adds to that of the shunt field.
Compute the terminal voltage at rated terminal current when the shunt-field current is 4.7 A and the speed is 1150 r/min. Neglect the effects of armature reaction.
As is shown in Fig. 7.12, for a long-shunt connection the armature and series field-currents are equal. Thus
I_s = I_a = I_L + I_f = 400 + 4.7 = 405 A
From Eq. 7.21 the main-field gross mmf is
\text{Gross mmf} = I_f + \left( \frac{N_s}{N_f} \right) I_s \text{equivalent shunt-field amperes} \quad \quad (7.21)
\begin{aligned}\text{Gross mmf} & = I_f + \left( \frac{N_s}{N_f} \right) I_s \\ &= 4.7 + \left( \frac{3}{1000} \right) 405 = 5.9 \text{equivalent shunt-field amperes}\end{aligned}
By examining the I_a = 0 curve of Fig. 7.14 at this equivalent shunt-field current, one reads a generated voltage of 274 V. Accordingly, the actual emf at a speed of 1150 r/min can be found from Eq. 7.23
E_a = \left( \frac{n}{n_0} \right) E_{a0} = \left( \frac{1150}{1200} \right) 274 = 263 V
Then
V_t = E_a – I_a(R_a + R_s) = 263 – 405(0.025 + 0.005) = 251 V
