Question 18.12: Analysis of Capacitor-Start Motor With reference to Figure 1...

Known Quantities: Motor operating characteristics; motor circuit parameters.

Find: Starting capacitance, C.

Schematics, Diagrams, Circuits, and Given Data:

Motor operating data: \frac{1}{3} \mathrm{hp} ; 120 \mathrm{~V} ; 60 \mathrm{~Hz}

Circuit parameters: R_{m}=4.5  \Omega ; C_{m}=3.7  \Omega ; R_{a}=9.5  \Omega ; X_{a}=3.5  \Omega

Analysis: The purpose of the starting capacitor is to cause the auxiliary winding current, \mathbf{I}_{\mathrm{aux}}, at standstill to lead the main winding current, \mathbf{I}_{\text {main }}, by 90^{\circ}. The 90^{\circ} phase lead will provide the maximum starting torque. Figure 18.36 shows the phasor diagram for these two currents and the voltage. The impedance angle of the main winding is:

\theta_{m}=\arctan \left(\frac{X_{m}}{R_{m}}\right)=\arctan \left(\frac{3.7 \Omega}{4.5 \Omega}\right)=39.4^{\circ}

Knowing that the desired phase shift between the main and auxiliary impedance angles is -90^{\circ} (see Figure 18.36), we compute the impedance angle of the auxiliary winding:

\theta_{a}=39.4^{\circ}-90^{\circ}=-50.6^{\circ}

The minus sign indicates that \mathbf{I}_{\text {aux }} leads the terminal voltage. The required capacitance can now be calculated from the relationship

and we can compute the desired capacitance to be:

C=\frac{1}{\omega X_{C}}=\frac{1}{377 \times 15.07}=176 \times 10^{-6} \mathrm{~F}=176  \mu \mathrm{F}