Question 11.4: Figure 11.8 shows the block diagram for a simple speed contr......
Figure 11.8 shows the block diagram for a simple speed control system to be applied to the dc motor of Example 11.3. In this controller, the field voltage is held constant (not shown) at its rated value of 300 V. Thus, the control is applied only to the armature voltage and takes the form
V_a = V_{a0}+ G(ω_{ref} – ω_m)
where V_{a0} is the armature voltage when ω_m = ω_{ref} and G is a multiplicative constant.
With the reference speed set to 2000 r/min (ω_{ref} =2000 × π/30), calculate V_{a0} and G so that the motor speed is 2000 r/min at no load and drops only by 25 r/min when the torque is increased to its rated full-load value.
As was found in Example 11.3, the field current under this condition will be 2.75 A. At no load, 2000 r/min,
V_{\mathrm{a}} \approx E_{\mathrm{a}}=K_{\mathrm{f}} I_{\mathrm{f}} \omega_{\mathrm{m}}=0.694 \times 2.75 \times 2000\left(\frac{\pi}{30}\right)=400 \mathrm{~V}
and thus V_{a0}=400 V.
The full load torque was found in Example 11.3 to be T_{\text {rated }}=285 \mathrm{~N} \cdot \mathrm{m} and thus the armature current required to achieve rated full-load torque can be found from Eq. 11.4
T_{\mathrm{mech}}=\frac{E_aI_a}{ω_m}=K_{\mathrm{f}}I_{\mathrm{f}}I_a \quad \quad \quad(11.4)
I_{\mathrm{a}}=\frac{T_{\text {rated }}}{K_{\mathrm{f}} I_{\mathrm{f}}}=\frac{285}{0.694 \times 2.75}=149 \mathrm{~A}
At a speed of 1975 r/min, E_{\mathrm{a}} will be given by
E_{\mathrm{a}}=K_{\mathrm{f}} I_{\mathrm{f}} \omega_{\mathrm{m}}=0.694 \times 2.75 \times 1975\left(\frac{\pi}{30}\right)=395 \mathrm{~V}
and thus
V_{\mathrm{a}}=E_{\mathrm{a}}+I_{\mathrm{a}} R_{\mathrm{a}}=395+149 \times 0.084=408 \mathrm{~V}
Solving for G gives
G=\frac{V_{\mathrm{a}}-V_{\mathrm{a} 0}}{\omega_{\mathrm{ref}}-\omega_{\mathrm{m}}}=\frac{408-400}{(2000-1975)\left(\frac{\pi}{30}\right)}=3.06 \mathrm{~V} \cdot \mathrm{sec} / \mathrm{rad}