Question 11.11: The three-phase, 230-V, 60-Hz, 12-kW, four-pole induction mo......

As the electrical frequency f_e is varied, the motor reactances given in Example 6.4 must be varied as

X=X_{0}\left(\frac{f_{\mathrm{e}}}{60}\right)

where X_0 is the reactance value at 60 Hz. Similarly, the line-to-neutral armature voltage must be varied as

V_{\mathrm{l}}=\frac{220}{\sqrt{3}}\left(\frac{f_{\mathrm{e}}}{60}\right)=127\left(\frac{f_{\mathrm{e}}}{60}\right)  \mathrm{V}

From Eq. 4.40, the synchronous angular velocity of the motor is equal to

\omega_{\mathrm{s}}=\left(\frac{2}{\text { poles }}\right) ω_{\mathrm{e}}\quad \quad \quad (4.40)

\omega_{\mathrm{s}}=\left(\frac{4 \pi}{\text { poles }}\right) f_{\mathrm{e}}=\pi f_{\mathrm{e}}  \mathrm{rad} / \mathrm{sec}

and, at any given motor speed \omega_{\mathrm{m}}, the corresponding slip is given by

s=\frac{\omega_{\mathrm{s}}  –  \omega_{\mathrm{m}}}{\omega_{\mathrm{s}}}

Using Eqs. 11.51 through 11.53, the motor speed can be found by searching over \omega_{\mathrm{m}} for that speed at which P_{\text {load }}=\omega_{\mathrm{m}} T_{\text {mech}}. If this is done, the result is:

T_{\text {mech }}=\frac{1}{\omega_{\mathrm{s}}}\left[\frac{n_{\mathrm{ph}} V_{1, \mathrm{eq}}^{2}\left(R_{2} / s\right)}{\left(R_{1, \mathrm{eq}}+\left(R_{2} / s\right)\right)^{2}+\left(X_{1, \mathrm{eq}}+X_{2}\right)^{2}}\right]\quad \quad \quad (11.51)

R_{1, \mathrm{eq}}+j X_{1, \mathrm{eq}}=\frac{j X_{\mathrm{m}}\left(R_{1}+j X_{1}\right)}{R_{1}+j\left(X_{1}+X_{\mathrm{m}}\right)} \quad \quad \quad (11.53)

a. For f_e = 60 ~Hz:

Terminal voltage = 230 V line-to-line

Speed = 1720 r/min

Slip = 4.4%

b. For f_e = 40 ~Hz:

Terminal voltage = 153 V line-to-line

Speed = 1166 r/min

Slip = 2.8%

Here is the MATLAB script: