Question

A 375 W, 230 V, 50 Hz, capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): Z_m = (12.50 + j15.75) Ω (main winding), Z_a = (24.50 + 12.75) Ω (auxiliary winding). Neglecting the magnetizing branch, the value of the capacitance (in μF) to be added in series with the auxiliary winding to obtain maximum torque at starting is ___________.

Answer 1

Concept:

The condition to obtain maximum torque at starting is

ϕm + 2ϕa = 90°

Explanation:

Given that,

                 Z_m = (12.50 + j15.75)Ω

                  Z_a = (24.50 + j12.75)Ω

Let the reactance of capacitance added in series with auxiliary winding X_c.

Now,

            Z_a = 24.50 + j(-12.75 + X_c)

\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{15.75}}{{12.50}}} \right) = 51.56^\circ \)

\({\phi _a} = {\tan ^{ - 1}}\left( {\frac{{ - 12.75 + {X_C}}}{{24.5}}} \right)\)

using the condition to obtain maximum torque at starting is

ϕ_m + 2ϕ_a = 90°

\(\Rightarrow 51.56^\circ +2 ~{\tan ^{ - 1}}\left( {\frac{{{X_C} - 12.75}}{{24.5}}} \right) = 90^\circ \)

\(\Rightarrow 2{\tan ^{ - 1}}\left( {\frac{{{X_C} - 12.75}}{{24.5}}} \right) = 38.44^\circ\)

\(\Rightarrow \frac{{{X_C} - 12.75}}{{24.5}} = 0.3486\)

⇒ X_c = 21.29 Ω

\(\Rightarrow \frac{1}{{2\pi fC}} = 21.29\)

\(\Rightarrow C = \frac{1}{{2\pi \times 50 \times 21.29}} = 149.51\;\mu F\)