Question 9.2: A 1/4 -hP, 110-V, 60-Hz, four-pole, capacitor-start motor ha......

The first step is to determine the values of the forward- and backward-field impedances at the assigned value of slip. The following relations, derived from Eq. 9.4, simplify the computations of the forward-field impedance Z_f:

Z_f ≡ R_f + jX_f ≡\left( \frac{R_{2,main}}{s}+jX_{2,main}\right)  \text{in parallel with}  jX_{m,main} \quad \quad \quad (9.4)

R_f=\left( \frac{X_{m,main}^2}{X_{22}}\right) \frac{1}{sQ_{2,main} +1/(sQ_{2,main})} \quad X_f=\frac{X_{2,main}X_{m,main}}{X_{22}}+\frac{R_f}{sQ_{2,main}}

where

X_{22}=X_{2,main}+X_{m,main} \quad and \quad Q_{2,main}=\frac{X_{22}}{R_{2,main}}

Substitution of numerical values gives, for s = 0.05,

Z_f = R_f + jX_f = 31.9 + j40.3 ~Ω

Corresponding relations for the backward-field impedance Z_b are obtained by substituting 2 – s for s in these equations. When (2  –  s) Q_{2,main} is greater than 10, as is usually the case, less than 1 percent error results from use of the following approximate forms:

R_b=\frac{R_{2,main}}{2  –  s} \left( \frac{X_{m,main}}{X_{22}}\right)^2 \quad X_b=\frac{X_{2,main}X_{m,main}}{X_{22}}+\frac{R_b}{(2  –  s)Q_{2,main}}

Substitution of numerical values gives, for s = 0.05,

Z_b = R_b + jX_b = 1.98 + j2.12 ~Ω

Addition of the series elements in the equivalent circuit of Fig. 9.11c gives

\begin{array}{l} R_{1, \text { main }}+j X_{1, \text { main }}=2.02+j 2.79 \\ \\ 0.5\left(R_{\mathrm{f}}+j X_{\mathrm{f}}\right)=15.95+j 20.15 \\ \\ \underline{{0.5\left(R_{\mathrm{b}}+j X_{\mathrm{b}}\right)}}= \underline{0.99+j1.06} \\ \\{\text { Total Input } Z}=18.96+j 24.00=30.6 \angle 51.7^{\circ}\\ \\ \text { Stator current } I=\frac{V}{Z}=\frac{110}{30.6}=3.59  \mathrm{~A} \\ \\ \text { Power factor }=\cos \left(51.7^{\circ}\right)=0.620 \end{array}

\text{Power input}=P_{\text {in }}=V I \times \text{power factor} =110 \times 3.59 \times 0.620=244 \mathrm{~W}

The power absorbed by the forward field (Eq. 9.7) is

P_{\mathrm{gap}, f}=I^{2}\left(0.5 R_{\mathrm{f}}\right)=3.59^{2} \times 15.95=206 \mathrm{~W}

The power absorbed by the backward field (Eq. 9.9) is

P_{\mathrm{gap}, \mathrm{b}}=I^{2}\left(0.5 R_{\mathrm{b}}\right)=3.59^{2} \times 0.99=12.8 \mathrm{~W}

The internal mechanical power (Eq. 9.14) is

P_{\text {mech }}=(1  –  s)ω_s T_{mech}=(1-s)\left(P_{\text {gap }, \mathrm{f}}-P_{\mathrm{gap}, \mathrm{b}}\right) \quad \quad \quad (9.14)

P_{\text {mech }}=(1-s)\left(P_{\text {gap }, \mathrm{f}}-P_{\mathrm{gap}, \mathrm{b}}\right)=0.95(206-13)=184 \mathrm{~W}

Assuming that the core loss can be combined with the friction and windage loss, the rotational loss becomes 24 + 13 = 37 W and the shaft output power is the difference. Thus

P_{\text {shaft }}=184-37=147 \mathrm{~W}=0.197  \mathrm{hp}

From Eq. 4.40, the synchronous speed in rad/sec is given by

\omega_{\mathrm{s}}=\left(\frac{2}{\text { poles }}\right) \omega_{\mathrm{e}}=\left(\frac{2}{4}\right) 120 \pi=188.5  \mathrm{rad} / \mathrm{sec}

or in terms of r/min from Eq. 4.41

n_{s}=\left(\frac{120}{\text { poles }}\right) f_{\mathrm{e}} \quad r/min \quad \quad \quad (4.41)

\begin{array}{l}n_{s}=\left(\frac{120}{\text { poles }}\right) f_{\mathrm{e}}=\left(\frac{120}{4}\right) 60=1800  \mathrm{r} / \mathrm{min} \\ \\ \begin{aligned} \text { Rotor speed } &=(1-s)(\text { synchronous speed }) \\ \\ &=0.95 \times 1800=1710  \mathrm{r} / \mathrm{min}\end{aligned}\end{array}

and

\omega_{\mathrm{m}}=0.95 \times 188.5=179  \mathrm{rad} / \mathrm{sec}

The torque can be found from Eq. 9.14.

T_{\text {shaft }}=\frac{P_{\text {shaft }}}{\omega_{\mathrm{m}}}=\frac{147}{179}=0.821 \mathrm{~N} \cdot \mathrm{m}

and the efficiency is

\eta=\frac{P_{\text {shaft }}}{P_{\text {in }}}=\frac{147}{244}=0.602=60.2 \%

As a check on the power bookkeeping, compute the losses:

\begin{array}{l} \begin{aligned}I^{2} R_{1, \text { main }}=(3.59)^{2}(2.02)=26.0 \\ \\ \text { Forward-field rotor } I^{2} R(\text { Eq. } 9.11)=0.05 \times 206=10.3 \\ \\ \text { Backward-field rotor } I^{2} R(\text { Eq. } 9.12)=1.95 \times 12.8=25.0 \\ \\ \text { Rotational losses }=\underline{37.0} \\ \quad \quad 98.3 \mathrm{~W}\end{aligned}\end{array}

\text{Forward-field rotor}  I^2 R = s  P_{\mathrm{gap,f}} \quad \quad \quad (9.11)

\text{Backward-field rotor}  I^2 R = (2  –  s)  P_{\mathrm{gap,b}} \quad \quad \quad (9.12)

From P_{\text {in }}-P_{\text {shaft }}, the total losses = 97 W which checks within accuracy of computations.