A test on the main winding of a 1 kW, 4-pole. 2 15 V, 50 Hz, single-phase induction motor gave the following results: No-load test Rotor-blocked test V0 = 215 V VSC = 85 A I0 = 3.9 A ISC = 9.80 A P0 = 185 W PSC = 390 W R1 = 1.6 Ω Given: (a) Calculate the parameters of the circuit model assuming

Question

A test on the main winding of a 1 kW, 4-pole. 2 15 V, 50 Hz, single-phase induction motor gave the following results:

Given:

(a) Calculate the parameters of the circuit model assuming that the magnetizing reactance hangs at the input terminals of the model.

(b) Determine the line current power factor, shaft torque and efficiency of the motor at a speed of 1440 rpm

Accepted Answer

(a) Parameters of the circuit model are calculated using both no-load as well as rotor-blocked tests.

(i) No-load test: Assuming the slip to be zero, the circuit model on no-load is drawn in Fig. 10.7 with magnetizing reactance at input terminals

Since the backward circuit is short-circuited for practical purposes, as X being magnetizing reactance is much larger

[latex] \frac{X}{2}=\frac{215}{3.9}=55.1 \Omega[/latex]

Rotational loss [latex] P_{0}=185 W[/latex]

(ii) Rotor-blocked test (s = 1): The circuit model on rotor-blocked test is shown in Fig. 10.8

[latex] 390=85 imes 9.8 imes \cos \phi_{S C}[/latex]

Or [latex] \phi_{S C}=62^{\circ}[/latex] lagging

With reference to Fig. 10.8

[latex] \bar{I}_{e}=\frac{V_{S C}}{j X}=\frac{85}{j 2 imes 55.1}=-j 0.77 A[/latex]

[latex] \bar{I}_{m}^{\prime}=\bar{I}_{S C}-\bar{I}_{e}=9.8 \angle-62^{\circ}-(-j 0.77)[/latex]

[latex] =4.6-j 7.88=9.13 \angle-59.7^{\circ}[/latex]

[latex] \bar{Z}_{f}^{\prime}=\left(R_{1}+R_{2} ight)+j\left(X_{1}+X_{2} ight)[/latex]

[latex] \frac{V_{S C}}{I_{m}^{\prime}}=\frac{85}{9.13 \angle-59.7^{\circ}}=9.31 \angle 59.7^{\circ}[/latex]

= 4.7 + j 8.04

[latex] R_{1}+R_{2}=4.7 \Omega[/latex]

[latex] R_{1}=1.6 \Omega[/latex] (given)

[latex] R_{2}=3.1 \Omega[/latex]

[latex] X_{1}+X_{2}=8.04 \Omega[/latex]

The circuit model with parameter values is drawn in Fig. 10.9.

(b) [latex] s=\frac{1500-1440}{1500}=0.04[/latex]

[latex] 1.55 / s=\frac{1.55}{0.04}=38.75 ; 1.55 /(2-s)=\frac{1.55}{1.96}=0.79[/latex]

[latex] \bar{Z}_{f} ext { (total) }=j 55.1 \|(0.8+38.75+j 4.02)[/latex]

[latex] =j 55.1 \|(39.55+j 4.02)=30.8 \angle 39.6^{\circ}=23.73+j 19.63[/latex]

[latex] \bar{Z}_{b} ext { (total) }=j 55.1 \|(0.8+0.79+j 4.02)[/latex]

[latex] =j 55.1 \|(1.59+j 4.02)=4.02 \angle 70^{\circ}=1.37+j 3.78[/latex]

[latex] \bar{Z}( ext { total })=(23.73+j 19.63)+(1.37+j 3.78)=25.1+j 23.41=34.3 \angle 43^{\circ}[/latex]

[latex] \bar{I}_{m}=\frac{215}{34.3 \angle 43^{\circ}}=6.27 \angle-43^{\circ}[/latex]

[latex] \bar{I}_{L}=I_{m}=6.27 A ; p f=\cos 43^{\circ}=0.731[/latex]

[latex] P_{i n}=215 imes 6.27 imes 0.731=985.4 W[/latex]

[latex] \bar{I}_{m f}=6.27 \angle-43^{\circ} imes \frac{j 55.1}{39.55+j 59.12}=4.86 \angle-9.2^{\circ}[/latex]

[latex] \bar{I}_{m b}=6.27 \angle-43^{\circ} imes \frac{j 55.1}{1.59+j 59.12}=5.84 \angle-41.4^{\circ}[/latex]

[latex] T=\frac{1}{157.1}\left[(4.84)^{2} imes 38.75-(5.84)^{2} imes 0.79 ight]=5.6 Nm[/latex]

[latex] P_{m}=157.1(1-0.04) imes 5.6=844.6 W[/latex]

Rotational loss = 185 W

[latex] P_{ ext {out }}=844.6-185=659.6 W[/latex]

[latex] \eta=\frac{659.6}{985.5}=66.9 \% [/latex]

[latex] T( ext { shaft })=659.6 / 157.1=4.12 Nm[/latex]

Question 10.2: A test on the main winding of a 1 kW, 4-pole. 2 15 V, 50 Hz,...