Question 7.6: A 100-kW, 250-V, 400-A, 1200-r/min dc shunt generator has th......

a. The 50 Ω field-resistance line 0a (Fig. 7.14) passes through the 250-V, 5.0-A point of the no-load magnetization curve. At I_a = 400  A

I_aR_a = 400 × 0.025 = 10  V

Thus the operating point under this condition corresponds to a condition for which the terminal voltage V_t (and hence the shunt-field voltage) is 10 V less than the generated voltage E_a.

A vertical distance of 10 V exists between the magnetization curve for I_a = 400  A and the field-resistance line at a field current of 4.1 A, corresponding to V_t = 205  V. The associated line current is

 I_L =I_a  –  I_f = 400  –  4 = 396  A

Note that a vertical distance of 10 V also exists at a field current of 1.2 A, corresponding to V_t = 60  V. The voltage-load curve is accordingly double-valued in this region. It can be shown that this operating point is unstable and that the point for which V_t = 205  V is the normal operating point.

b. For the no-load voltage to be 250 V, the shunt-field resistance must be 50 Ω and the field-resistance line is 0a (Fig. 7.14). At full load, I_f = 5.0  A  \text{because}  V_t = 250  V. Then

I_a= 400 + 5.0 = 405  A

and

E_a = V_t + I_a(R_a + R_p) = 250 + 405(0.025 + R_p)

where R_p is the parallel combination of the series-field resistance R_s = 0.005  Ω and the diverter resistance R_d

R_p= \frac{R_sR_d}{(R_s + R_d)}

The series field and the diverter resistor are in parallel, and thus the shunt-field current can be calculated as

I_s=405 \left( \frac{R_d}{R_s +R_d} \right) =405 \left( \frac{R_p}{R_s} \right)

and the equivalent shunt-field amperes can be calculated from Eq. 7.21 as

\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} I_{net} &= I_f+ \frac{4}{1000}I_s = 5.0+ \frac{4}{1000}I_s\\& = 5.0 + 1.62 \left( \frac{R_p}{R_s} \right) \end{aligned}

This equation can be solved for R_p which can be, in turn, substituted (along with R_s = 0.005  Ω) in the equation for E_a to yield

E_a = 253.9 + 1.25I_{net}

This can be plotted on Fig. 7.14 (E_a on the vertical axis and I_{net} on the horizontal axis). Its intersection with the magnetization characteristic for I_a = 400  A (strictly speaking, of course, a curve for I_a = 405  A should be used, but such a small distinction is obviously meaningless here) gives I_{net} = 6.0  A.

Thus

R_p=\frac{R_s(I_{net}  –  5.0)}{1.62}= 0.0031  Ω

and

R_d = 0.0082  Ω