Electrical Machines Test 7

The direction of rotation of a universal motor can be changed by either:

1. Reversing the field connection with respect to those of armature

2. By using two field windings wound on the core in opposite directions so that the one connected in series with armature gives clockwise rotation, while the other in series with the armature gives a counterclockwise rotation

R₂ = 3.6 Ω

\({N_S} = \frac{{120 \times 50}}{4} = \frac{{6000}}{4}\)

= 1500 rpm

At 1440 rpm, slip, \(s = \frac{{{N_S} - N}}{{{N_S}}} = \frac{{1500 - 1440}}{{1500}}\)

= 0.04

Forward resistance \(= \frac{{{r_2}}}{s}\), backward resistance \(= \frac{{{r_2}}}{{2 - s}}\)

Where, \({r_2} = \frac{{{R_2}}}{2} = \frac{{3.6}}{2} = 1.8\ {\rm{\Omega }}\)

Forward resistance \(= \frac{{1.8}}{{0.04}} = 45\ {\rm{\Omega }}\)

Backward resistance \(= \frac{{1.8}}{{2 - 0.04}} = 0.92\ {\rm{\Omega }}\)

Concept:

The frequency of the current induced in the rotor by forward field = sf

The frequency of the current induced in the rotor by backward field = (2 – s)f

Where f is the frequency

s is the slip

Calculation:

Given that, frequency (f) = 50 Hz

Slip (s) = 4 % = 0.04

The frequency of the current induced in the rotor by forward field = sf = 0.04 × 50 = 2 Hz

Concept:

The starting torque (in synchronous watts) in a single-phase inductor motor is given by,

\({T_{st}} = 2{I_M}{I_A}aR_2^{'}\sin \theta \)

Where \(a = \frac{{{N_A}}}{{{N_M}}}\)

I_M is the RMS value of current in the main winding

I_A is the RMS value of current in the auxiliary winding

N_A is the number of turns of the auxiliary winding

N_M is the number of turns of the main winding

R₂ is the rotor resistance

θ is the phase difference between the main winding and auxiliary winding

Calculation:

Given that, I_M = 5∠-80° A and I_A = 2∠-50° A

θ = -50 – (-80) = 30°

\({T_{st}} = 2{I_M}{I_A}aR_2^{'}\sin \theta \)

Concept:

In a single-phase resistor split induction motor, the condition to get the maximum starting torque is

θ_m = 2θ_a

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right)\)

In a capacitor start induction motor,

The condition to get the maximum starting torque is

θ_m + 2θ_a = 90°

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right) = 90^\circ \)

Where θ_m is the power factor angle of the main winding

θ_a is the power factor angle of the auxiliary winding

Calculation:

Given that, power factor angle of main winding of single-phase resistor split induction motor (θ_m) = 50°

To get maximum starting torque,

θ_m = 2θ_a

Main winding, Z_m = (4.5 + j 3.7) Ω

Auxiliary winding, Z_a = (9.5 + j3.5) Ω

The condition to get the main and auxiliary winding currents in quadrature at starting is

θ_m + θ_a = 90°

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right) = 90^\circ \)

Let the starting capacitance to be added is X_C Ω

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{X_C} - {X_a}}}{{{R_a}}}} \right) = 90^\circ \)

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{3.7}}{{4.5}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{X_C} - 3.5}}{{9.5}}} \right) = 90^\circ \)

⇒ X_C = 15.05 Ω

\( \Rightarrow \frac{1}{{2\pi fC}} = 15.05\)

\( \Rightarrow C = \frac{1}{{2\pi \times 50 \times 15.05}} = 211.4\;\mu F\)

Concept:

In a single-phase resistor split induction motor, the condition to get the maximum starting torque is

θm = 2θa

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right)\)

In a capacitor start induction motor,

The condition to get the maximum starting torque is

θm + 2θa = 90°

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right) = 90^\circ \)

Where θm is the power factor angle of the main winding

θa is the power factor angle of the auxiliary winding

Calculation:

Given that, main winding impedance, Z_m = 3 + j4 Ω

Auxiliary winding impedance, Z_a = 6 + j8 Ω

Power factor angle of main winding, \({\theta _m} = {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right)\)

\( = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right) = 53.13^\circ \)

The condition to get the maximum starting torque is

θ_m + 2θ_a = 90°

⇒ 53.13° + 2θ_a = 90°

⇒ θ_a = 18.43°

Let the external capacitance to be added is X_C Ω

Now, \({\tan ^{ - 1}}\left( {\frac{{{X_C} - {X_a}}}{{{R_a}}}} \right) = 18.43^\circ \)

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_C} - 8}}{6}} \right) = 18.43^\circ \)

⇒ X_C = 10 Ω

\( \Rightarrow \frac{1}{{2\pi fC}} = 10\)

\( \Rightarrow C = \frac{1}{{2\pi \times 50 \times 10}} = 318.3\;\mu F\)

Concept:

In a single-phase resistor split induction motor, the condition to get the maximum starting torque is

θm = 2θa

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right)\)

In a capacitor start induction motor,

The condition to get the maximum starting torque is

θm + 2θa = 90°

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right) = 90^\circ \)

Where θm is the power factor angle of the main winding

θa is the power factor angle of the auxiliary winding

Calculation:

Given that, R_m = 1 Ω, X_m = √3 Ω

Power factor angle of main winding, \({\theta _m} = {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right)\)

\( = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{1}} \right) = 60^\circ \)

The condition to get the maximum starting torque is

θ_m + 2θ_a = 90°

⇒ 60° + 2θ_a = 90°