A (1/2) kW, 4-pole, 50 Hz, 220 V, two-value capacitor motor has the following circuit model parameters: R1m = 4.2 Ω , X1m = 11.3 Ω R1a = 5.16 Ω , X1a = 12.1 Ω X = 250 Ω , a = 1.05 Ω R2 = 7.48 Ω , X2 = 7.2 Ω Friction, windage and core losses = 45 W (a) Calculate the starting torque and current if

Question

A (1/2) kW, 4-pole, 50 Hz, 220 V, two-value capacitor motor has the following circuit model parameters:

[latex] R_{l m}=4.2 \Omega, \quad X_{l m}=11.3 \Omega [/latex]

[latex] R_{l a}=5.16 \Omega, \quad X_{I a}=12.1 \Omega[/latex]

[latex] X=250 \Omega, \quad a=1.05 \Omega[/latex]

[latex] R_{2}=7.48 \Omega, \quad X_{2}=7.2 \Omega[/latex]

Friction, windage and core losses = 45 W

(a) Calculate the starting torque and current if the two capacitors in parallel equal [latex] 70 \mu F[/latex].

(b) Calculate the value of the run capacitor for zero backward field when the motor is running at a slip of 0.04. What is the meaning of the associated resistance value?

(c) Calculate the motor performance for the value of the run capacitor as in part (b). Assume [latex] R_{C}=0[/latex]

Accepted Answer

(a) s = 1

[latex] \bar{Z}_{f}=\bar{Z}_{b}=j 250 \|(7.48+j 7.2)[/latex]

[latex] =10.1 \angle 45.6^{\circ}=7.07+j 7.22[/latex]

[latex] \bar{Z}_{ l a}=(5.16+j 12.1)-j \frac{10^{6}}{314 imes 70}[/latex]

= 5.16 – j 33.4

[latex] \bar{Z}_{1 a} / a^{2}=4.68-j 30.29[/latex]

[latex] \bar{Z}_{12}=\frac{1}{2}\left(Z_{1 a} / a^{2}-\bar{Z}_{1 m} ight)[/latex]

[latex] =\frac{1}{2}(4.68-j 30.29-4.2-j 11.3)[/latex]

[latex] =0.24-j 20.8=20.8 \angle-89.3^{\circ}[/latex]

[latex] \bar{V}_{m f}=\frac{220}{2}\left(1-\frac{j}{1.05} ight)=151.9 \angle-43.60^{\circ}[/latex]

[latex] \bar{V}_{m b}=\frac{220}{2}\left(1+\frac{j}{1.05} ight)=151.9 \angle 43.6^{\circ}[/latex]

[latex] \bar{Z}_{1 m}+\bar{Z}_{f}+\bar{Z}_{12}=\bar{Z}_{1 m}+\bar{Z}_{b}+\bar{Z}_{12}=(4.2+j 11.3)+(7.07+j 7.22)+(0.24-j 20.8)[/latex]

[latex] =11.51-j 2.28=11.73 \angle-11.2^{\circ}[/latex]

Substituting in Eqs (10.48a) [latex] \bar{V}_{m b}\left(\bar{Z}_{1 m}+\bar{Z}_{f}+\bar{Z}_{12} ight)+\bar{V}_{m f} \bar{Z}_{12}=0[/latex] and (10.48b) [latex] \bar{Z}_{12}=\frac{-\bar{V}_{m b}\left(\bar{Z}_{1 m}+\bar{Z}_{f} ight)}{\bar{V}_{m f}+\bar{V}_{m b}}=\frac{-\bar{V}_{m b}\left(\bar{Z}_{1 m}+\bar{Z}_{f} ight)}{\bar{V}_{L}}[/latex]

[latex] \bar{I}_{m f}=\frac{151.9 \angle-43.6^{\circ} imes 11.73 \angle-11.2^{\circ}+151.9 \angle 43.6^{\circ} imes 20.8 \angle-89.3^{\circ}}{(11.73)^{2} \angle-22.4^{\circ}-(20.8)^{2} \angle-178.6^{\circ}}[/latex]

[latex] =\frac{4928 \angle-49^{\circ}}{561.3 \angle-43^{\circ}}=8.78 \angle-44.7^{\circ}=6.24-j 6.18[/latex]

[latex] \bar{I}_{m b}=\frac{151.9 \angle 43.6^{\circ} imes 11.73 \angle-11.2^{\circ}+151.9 \angle-43.6^{\circ} imes 20.8 \angle-89.3^{\circ}}{561.3 \angle-4.3^{\circ}}[/latex]

[latex] =\frac{-1506 \angle 64.6}{561.3 \angle-4.3^{\circ}}=-2.68 \angle 68.9^{\circ}=-0.96-j 2.5[/latex]

[latex] n_{s}=1500 rpm , \omega_{s}=\frac{2 \pi imes 1500}{60}=157.1 rad / s[/latex]

[latex] T_{s}=\frac{2}{157.1} imes 7.07\left\{(8.78)^{2}-(2.68)^{2} ight\}[/latex]

= 6.31 Nm

[latex] \bar{I}_{m}=\bar{I}_{m f}+\bar{I}_{m b}=5.28-j 8.68[/latex]

[latex] \bar{I}_{a}=\frac{j}{a}\left(\bar{I}_{m f}-\bar{I}_{m b} ight)=\frac{j}{1.05}(7.2-j 3.68)=3.5+j 6.86[/latex]

[latex] \bar{I}_{L}=\bar{I}_{m}+\bar{I}_{a}=8.78-j 1.82=8.97 \angle-11.7^{\circ}[/latex]

[latex] \bar{I}_{L}( ext { start })=8.97 A[/latex]

(b) s = 0.04

[latex] \bar{Z}_{f}=j 250 \|\left(\frac{7.48}{0.04}+j 7.2 ight)[/latex]

[latex] =j 250 \|(187+j 7.2)=147 \angle 38.2^{\circ}=115.5+j 90.9[/latex]

From Eq. (10.55) Torque developed [latex] =\frac{P_{h}}{s \omega_{s}}=\frac{K_{h} s f B^{2}}{s \omega_{s}} =\frac{K_{h} f B^{2}}{\omega_{s}}= ext { constant }[/latex]

[latex] \bar{Z}_{12}=-\frac{1}{2}\left(1+\frac{j}{a} ight)\left(\bar{Z}_{1 m}+\bar{Z}_{f} ight)[/latex]

[latex] \left.=-\frac{1}{2}\left(1+\frac{j}{1.05} ight)[(4.2+j 11.3)]+(115.5+j 90.9) ight][/latex]

[latex] =-108.6 \angle 84.1^{\circ}=-11.2-j 108[/latex]

From Eq. (10.56) [latex] P_{e}=K_{e} f_{2}^{2} B^{2} =K_{e} s^{2} f^{2} B^{2}[/latex]

[latex] \bar{Z}_{1 a}=a^{2}\left(2 \bar{Z}_{12}+\bar{Z}_{1 m} ight)[/latex]

[latex] =(1.05)^{2}[2(-11.2-j 108)+(4.2+j 11.3)]=-20.1-j 225.7[/latex]

From Eq. (10.57) [latex] T=\frac{P_{e}}{s \omega_{s}}=\left(\frac{K_{e} f^{2} B^{2}}{\omega_{s}} ight) s[/latex]

[latex] R_{C}-j X_{C}=(-20.1-j 225.7)-(5.16+j 12.1)[/latex]

= – 25.3 – j238

[latex] X_{C}=\frac{1}{314 imes C}[/latex]

Or [latex] C=\frac{10^{6}}{314 imes 238}=13.4 \mu F[/latex]

[latex] R_{C}[/latex] being negative is unrealizable so that a completely balanced operation is not possible.

(c) s = 0.04

[latex] \bar{Z}_{f}=147 \angle 38.2^{\circ}=115.5+j 90.9[/latex]

(as calculated in part (b))

[latex] \bar{Z}_{b}=j 250 \|\left(\frac{7.48}{1.96}+j 7.2 ight)[/latex]

[latex] =j 250 \|(3.82+j 7.2)[/latex]

[latex] =7.92 \angle 63^{\circ}=3.6+j 7.06[/latex]

[latex] \bar{Z}_{1 a}=(5.16+j 12.1)-j \frac{10^{6}}{314 imes 13.4}[/latex]

= 5.16 – j 225.6

[latex] \bar{Z}_{1 a} / a^{2}=4.68-j 204.6[/latex]

[latex] \bar{Z}_{12}=\frac{1}{2}(4.68-j 204.6-4.2-j 11.3)[/latex]

[latex] =0.24-j 108=108 \angle-89.9^{\circ}[/latex]

[latex] \bar{Z}_{1 m}+\bar{Z}_{f}+\bar{Z}_{12}=4.2+j 11.3+115.5+j 90.9+0.24-j 108[/latex]

[latex] =119.9-j 5.8=120 \angle-2.8^{\circ}[/latex]

[latex] \bar{Z}_{1 m}+\bar{Z}_{b}+\bar{Z}_{12}=4.2+j 11.3[/latex]

3.6 + j7.06

0.24 – j108

[latex] =8.04-j 89.64=90 \angle-84.9^{\circ}[/latex]

Substituting in Eqs (10.48a) and (10.48b)

[latex] \bar{I}_{m f}=\frac{151.9 \angle-43.6^{\circ} imes 90 \angle-84.9^{\circ}+151.9 \angle 43.6^{\circ} imes 108 \angle-89.9^{\circ}}{120 \angle-2.8^{\circ} imes 90 \angle-84.9^{\circ}-(108)^{2} \angle-179.8^{\circ}}[/latex]

[latex] =\frac{22735 \angle-82.9^{\circ}}{16177 \angle-41.6^{\circ}}=1.405 \angle-41.3^{\circ}=1.06-j 0.927[/latex]

[latex] \bar{I}_{m b}=\frac{151.9 \angle 43.6^{\circ} imes 120 \angle-2.8^{\circ}+151.9 \angle-43.6^{\circ} imes 108 \angle-89.9^{\circ}}{120 \angle-2.8^{\circ} imes 90 \angle-84.9^{\circ}-(108)^{2} \angle-179.8^{\circ}}[/latex]

[latex] =\frac{2506 \angle 0.24^{\circ}}{16177 \angle-41.6^{\circ}}=0.155 \angle 41.8^{\circ}=0.116+j 0.103[/latex]

[latex] T=\frac{2}{157.1} imes\left[(1.405)^{2} imes 115.5-(0.155)^{2} imes 3.6 ight]=2.9 Nm[/latex]

[latex] \bar{I}_{m}=\bar{I}_{m f}+\bar{I}_{m b}=1.06-j 0.927+0.12-j 0.103=1.18-j 0.824[/latex]

[latex] \bar{I}_{a}=\frac{j}{a}\left(\bar{I}_{m f}-\bar{I}_{m b} ight)=\frac{j}{1.05}(1.06-j 0927-0.12-j 0.103)[/latex]

[latex] =\frac{j}{1.05}(0.94-j 1.13)=1.107+j 0.895[/latex]

[latex] \bar{I}_{L}=\bar{I}_{m}+\bar{I}_{a}=2.25-j 0.205[/latex]

[latex] I_{L}=2.26 ; \quad p f=1[/latex] (almost unity)

[latex] P_{m}=2\left[(1.405)^{2} imes 115.5-(0.155)^{2} imes 3.6 ight](1-0.04)=437.8 W[/latex]

[latex] P_{ ext {out }}=437.8-45=392.8 W[/latex]

[latex] P_{ ext {in }}=2.26 imes 220=497.2 W[/latex]

[latex] \eta=\frac{392.8}{497.2}=79 \% [/latex]

Question 10.5: A (1/2) kW, 4-pole, 50 Hz, 220 V, two-value capacitor motor ...

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