Question 6.3: For the motor of Example 6.2, determine (a) the load compone......

First reduce the circuit to its Thevenin-equivalent form. From Eq. 6.29,

\hat{V}_{1,eq} =\hat{V}_1\left(\frac{j X_m}{R_1  +  j(X_1  +  X_m)}\right) \quad \quad \quad  (6.29)

V_{1,eq} = 122.3  V and from Eq. 6.31,

Z_{1,eq}= \frac{j X_m(R_1  +  jX_1)}{R_1  +  j(X_1  +  X_m)} \quad \quad \quad (6.31)

R_{1,eq}  +  jX_{1,eq} = 0.273  +  j0.490  Ω.

a. At s = 0.03, R_2/s = 4.80. Then, from Fig. 6.13a,

I_{2}=\frac{V_{1, \text { eq }}}{\sqrt{\left(R_{1, \text { eq }}+R_{2} / s\right)^{2}+\left(X_{1, \text { eq }}+X_{2}\right)^{2}}}=\frac{122.3}{\sqrt{(5.07)^{2}+(0.699)^{2}}}=23.9 \mathrm{~A}

From Eq. 6.25

T_{\text {mech }}=\frac{P_{\text {mech }}}{\omega_{\mathrm{m}}}=\frac{P_{\text {gap }}}{\omega_{\mathrm{s}}}=\frac{n_{\text {ph }} I_{2}^{2}\left(R_{2} / s\right)}{\omega_{\mathrm{s}}}  \quad \quad \quad  (6.25)

T_{\text {mech }}=\frac{n_{\text {ph }} I_{2}^{2}\left(R_{2} / s\right)}{\omega_{\mathrm{s}}}=\frac{3 \times 23.9^{2} \times 4.80}{125.7}=65.4 \mathrm{~N} \cdot \mathrm{m}

and from Eq. 6.21

P_{\text {mech }}=n_{\mathrm{ph}} I_{2}^{2}R_{2}\left(\frac{1-s}{s}\right) \quad \quad \quad (6.21)

P_{\text {mech }}=n_{\mathrm{ph}} I_{2}^{2}\left(R_{2} / s\right)(1-s)=3 \times 23.9^{2} \times 4.80 \times 0.97=7980 \mathrm{~W}

The curves of Fig. 6.15 were computed by repeating these calculations for a number of assumed values of s.

b. At the maximum-torque point, from Eq. 6.35,

\begin{aligned}s_{\operatorname{maxT}} & =\frac{R_{2}}{\sqrt{R_{\mathrm{l}, \mathrm{eq}}^{2}+\left(X_{1, \mathrm{eq}}+X_{2}\right)^{2}}} \\& =\frac{0.144}{\sqrt{0.273^{2}+0.699^{2}}}=0.192\end{aligned}

and thus the speed at T_{max} is equal to (1  –  s_{maxT})n_s = (1  –  0.192) × 1200 = 970  r/min

From Eq. 6.36

\begin{aligned}T_{\max } & =\frac{1}{\omega_{\mathrm{s}}}\left[\frac{0.5 n_{\mathrm{ph}} V_{1, \mathrm{eq}}^{2}}{R_{1, \mathrm{eq}}+\sqrt{R_{1, \mathrm{eq}}^{2}+\left(X_{1, \mathrm{eq}}+X_{2}\right)^{2}}}\right] \\& =\frac{1}{125.7}\left[\frac{0.5 \times 3 \times 122.3^{2}}{0.273+\sqrt{0.273^{2}+0.699^{2}}}\right]=175 \mathrm{~N} \cdot \mathrm{m}\end{aligned}

c. At starting, s = 1. Therefore

\begin{aligned}I_{2, \text { start }} & =\frac{V_{1, \mathrm{eq}}}{\sqrt{\left(R_{1, \mathrm{eq}}+R_{2}\right)^{2}+\left(X_{1, \mathrm{eq}}+X_{2}\right)^{2}}} \\& =\frac{122.3}{\sqrt{0.417^{2}+0.699^{2}}}=150 \mathrm{~A}\end{aligned}

From Eq. 6.25

T_{\mathrm{start}}=\frac{n_{\mathrm{ph}} I_{2}^{2} R_{2}}{\omega_{\mathrm{s}}}=\frac{3 \times 150^{2} \times 0.144}{125.7}=77.3 \mathrm{~N} \cdot \mathrm{m}