Question 16.6: A separately excited dc generator has VF = 140, RF = 10 Ω, R...
A separately excited dc generator has V_F = 140, R_F = 10 Ω, R_{adj} = 4 Ω, R_A = 0.065 Ω, the prime mover rotates the armature at a speed of 1000 rpm, and the magnetization curve is shown in Figure 16.19 on page 802. Determine the field current, the no-load voltage, the full-load voltage, and the percentage voltage regulation for a full-load current of 200 A. Assuming that the overall efficiency (not including the power supplied to the field circuit) of the machine is 85 percent, determine the input torque, the developed torque, and the losses associated with friction, windage, eddy currents, and hysteresis.
The field current is
I_F =\frac{ V_F} {R_{adj} + R_F} =\frac{ 140}{ 4 + 10} = 10 A
Then, referring to the magnetization curve on page 802, we find that E_A = 280 V for a speed of 1200 rpm. Equation 16.38 shows that E_A is proportional to speed. So, for a speed of 1000 rpm, we have
E_A = K_Φω_m (16.38)
E_A = 280 \frac{1000}{ 1200} = 233.3 V
which is also the no-load voltage of the machine. For a load current of 200 A, we get
V_{FL} = E_A − R_AI_A = 233.3 − 200 × 0.065 = 220.3 V
Finally, we obtain
voltage regulation = \frac{V_{NL} − V_{FL}} {V_{FL}} × 100\% = \frac{233.3 − 220.3}{ 220.3} × 100% = 5.900%
The output power is
P_{out} = I_LV_{FL} = 200 × 220.3 = 44.06 kW
The developed power is the sum of the output power and the armature loss:
P_{dev} = P_{out} + R_AI^2_A = 44060 + 0.065(200)² = 46.66 kW
The angular speed is
ω_m = n_m \frac{2π} {60} = 104.7 rad/s
The input power is
P_{in} = \frac{P_{out}} {0.85} = \frac{44.06}{ 0.85} = 51.84 kW
The power losses associated with friction, windage, eddy currents, and hysteresis are
P_{losses} = P_{in} − P_{dev} = 51.84 − 46.66 = 5.18 kW
The input and developed torques are
T_{in} = \frac{P_{in}} {ω_m} = \frac{51,840}{ 104.7} = 495.1 N·m
T_{dev} = \frac{P_{dev}} {ω_m} = \frac{46,660}{ 104.7} = 445.7 N·m