Question 11.5: The permanent-magnet dc motor of Example 7.9 has an armature......
The permanent-magnet dc motor of Example 7.9 has an armature resistance of 1.03 Ω and a torque constant K_m = 0.22 V/(rad/sec). Assume the motor to be driving a constant power load of 800 W (including rotational losses), and calculate the motor speed as the armature voltage is varied from 40 to 50 V.
The motor power output (including rotational losses) is given by the product E_aI_a and thus we can write
P_{\text {load }}=E_{\mathrm{a}} I_{\mathrm{a}}=K_{\mathrm{m}} \omega_{\mathrm{m}} I_{\mathrm{a}}
Solving for \omega_{\mathrm{m}} gives
\omega_{\mathrm{m}}=\frac{P_{\text {load }}}{K_{\mathrm{m}} I_{\mathrm{a}}}
The armature current can be written as
I_{\mathrm{a}}=\frac{\left(V_{\mathrm{a}}-E_{\mathrm{a}}\right)}{R_{\mathrm{a}}}=\frac{\left(V_{\mathrm{a}}-K_{\mathrm{m}} \omega_{\mathrm{m}}\right)}{R_{\mathrm{a}}}
These two equations can be combined to give an equation for \omega_{\mathrm{m}} of the form
\omega_{\mathrm{m}}^{2}-\left(\frac{V_{\mathrm{a}}}{K_{\mathrm{m}}}\right) \omega_{\mathrm{m}}+\frac{P_{\text {load }} R_{\mathrm{a}}}{K_{\mathrm{m}}^{2}}=0
from which we can find
\omega_{\mathrm{m}}=\frac{V_{\mathrm{a}}}{2 K_{\mathrm{m}}}\left[1 \pm \sqrt{1-\frac{4 P_{\mathrm{load}} R_{\mathrm{a}}}{V_{\mathrm{a}}^{2}}}\right]
Recognizing that, if the voltage drop across the armature resistance is small, V_{\mathrm{a}} \approx E_{\mathrm{a}}=K_{\mathrm{m}}\omega_{\mathrm{m}}, we pick the positive sign and thus
\omega_{\mathrm{m}}=\frac{V_{\mathrm{a}}}{2 K_{\mathrm{m}}}\left[1+\sqrt{1-\frac{4 P_{\mathrm{load}} R_{\mathrm{a}}}{V_{\mathrm{a}}^{2}}}\right]
Substituting values, we find that for V_{\mathrm{a}}=40 \mathrm{~V}, \omega_{\mathrm{m}}=169.2 \mathrm{rad} / \mathrm{sec} (1616 \mathrm{r} / \mathrm{min}) and for V_{\mathrm{a}}=50 \mathrm{~V}, \omega_{\mathrm{m}}=217.5 \mathrm{rad} / \mathrm{sec} (2077 \mathrm{r} / \mathrm{min}).