Question 9.5: A two-pole, single-phase induction motor has the following p......

MATLAB, with its ease of handling complex numbers, is ideal for the solution of this problem.

a. The main winding of this motor is directly connected to the single-phase source. Thus we directly set \hat{V}_{main} = \hat{V}_ s. However, the auxiliary winding is connected to the single-phase source through a capacitor and its polarity is reversed. Thus we must write

\hat{V}_{aux} + \hat{V}_C =- \hat{V}_s

where the capacitor voltage is given by

\hat{V}_C=j \hat{I}_{aux} X_C

Here the capacitor impedance X_C is equal to

X_{\mathrm{C}}=-\frac{1}{\left(\omega_{\mathrm{e}} C\right)}=-\frac{1}{\left(120 \pi \times 35 \times 10^{-6}\right)}=-75.8  \Omega

Setting \hat{V}_{\mathrm{s}}=V_{0}=230  \mathrm{~V} and substituting these expressions into Eqs. 9.43 and 9.44 and using Eqs. 9.39 and 9.40 then gives the following matrix equation for the main- and auxiliary-winding currents.

where

\begin{array}{c}\hat{A}_{1}=L_{\text {main }}-j L_{\text {main }, r}^{2}\left(\hat{K}^{+}+\hat{K}^{-}\right) \\ \hat{A}_{2}=L_{\text {main }, \mathrm{r}} L_{\text {aux }, \mathrm{r}}\left(\hat{K}^{+}-\hat{K}^{-}\right)\end{array}

and

\hat{A}_{3}=L_{\text {aux }}-j L_{\text {aux.r}}^{2}\left(\hat{K}^{+}+\hat{K}^{-}\right)

The parameters \hat{K}^{+} and \hat{K}^{-} can be found from Eqs. 9.41 and 9.42 once the slip is found using Eq. 6.1

\hat{K}^+ =\frac{sω_e}{2(R_r+j s ω_e L_r)} \quad \quad \quad (9.41)

\hat{K}^+ =\frac{(2  –  s)ω_e}{2(R_r+j (2  –  s) ω_e L_r)} \quad \quad \quad (9.42)

s=\frac{n_{\mathrm{s}}-n}{n_{\mathrm{s}}}==\frac{3600  –  3500}{3600}=0.278

This matrix equation can be readily solved using MATLAB with the result

\begin{array}{l}\hat{I}_{\text {main }}=15.9 \angle-37.6^{\circ} \mathrm{A} \\\hat{I}_{\text {aux }}=5.20 \angle-150.7^{\circ} \mathrm{A}\end{array}

and

\hat{I}_{\mathrm{s}}=18.5 \angle-22.7^{\circ}  \mathrm{A}

The magnitude of the capacitor voltage is

\left|\hat{V}_{\mathrm{C}}\right|=\left|\hat{I}_{\mathrm{aux}} X_{\mathrm{C}}\right|=374 \mathrm{~V}

b. Using MATLAB the time-averaged electromagnetic torque can be found from Eq. 9.49 to be

The shaft power can then be found by subtracting the rotational losses P_{rot} from the air-gap power

\begin{aligned}P_{\text {shaft }} & =\omega_{\mathrm{m}}<T_{\text {mech }}>-P_{\mathrm{rot}} \\& =\left(\frac{2}{\text { poles }}\right)(1-s) \omega_{\mathrm{e}}\left(<T_{\text {mech }}>\right)-P_{\mathrm{rot}} \\& =3532  \mathrm{~W}\end{aligned}

c. The power input to the main winding can be found as

P_{\text {main }}=\operatorname{Re}\left[V_{0} \hat{I}_{\text {main }}^{*}\right]=2893 \mathrm{~W}

and that into the auxiliary winding, including the capacitor (which dissipates no power)

P_{\text {aux }}=\operatorname{Re}\left[-V_{0} \hat{I}_{\text {aux }}^{*}\right]=1043 \mathrm{~W}

The total input power, including the core loss power P_{core} is found as

P_{\text {in }}=P_{\text {main }}+P_{\text {aux }}+P_{\text {core }}=4041 \mathrm{~W}

Finally, the efficiency can be determined

\eta=\frac{P_{\text {shaft }}}{P_{\text {in }}}==0.874=87.4 \%

d. The plot of < T_{mech} > versus speed generated by MATLAB is found in Fig. 9.18.

Here is the MATLAB script: