Question 9.3: The equivalent-circuit parameters of a 5-hp 220-V 60-Hz four......

We will solve this example using MATLAB.^†

a. Let \hat{V}_α  and  \hat{V}_β denote the voltages applied to the two phases, respectively. Then

\hat{V}_α = 230∠0° = 230 + j0  V\\ \hat{V}_β = 210∠80° = 36.4 + j207  V

From Eqs. 9.21 and 9.22 the forward and backward components of voltages are, respectively,

\hat{V}_f=\frac{1}{2}(\hat{V}_α  –  j \hat{V}_β) \quad \quad \quad (9.21)

\hat{V}_b=\frac{1}{2}(\hat{V}_α + j \hat{V}_β) \quad \quad \quad (9.22)

\hat{V}_f = 218.4  –  j18.2= 219.2∠-4.8°  V\\ \hat{V}_b = 11.6 + j 18.2 = 21.6∠57.5°  V

b. Because of the ease with which MATLAB handles complex numbers, there is no need to use approximations such as are derived in Example 9.2. Rather, the forward- and backward-field input impedances of the motor can be calculated from the equivalent circuits of Figs. 9.12a and b. Dividing the forward-field voltage by the forward-field input impedance gives

\hat{I}_f =\frac{\hat{V}_f}{R_1+jX_1+Z_f}= 9.3  –  j6.3 = 11.2∠ -34.2°  A

Similarly, dividing the backward-field voltage by the backward-field input impedance gives

\hat{I}_b =\frac{\hat{V}_b}{R_1+jX_1+Z_b} =3.7  –  j 1.5 = 4.0 ∠-21.9°  A

c. The winding currents can be calculated from Eqs. 9.19 and 9.20

\hat{I}_α= \hat{I}_f + \hat{I}_b = 13.0  –  j7.8 = 15.2∠-31.0°  A\\ \hat{I}_β= j \hat{I}_f +j \hat{I}_b = 4.8 + j5.6 = 7.4∠49.1°  A

Note that the winding currents are much more unbalanced than the applied voltages. Even though the motor is not overloaded insofar as shaft load is concerned, the losses are appreciably increased by the current unbalance, and the stator winding with the greatest current may overheat.

d. The power delivered across the air gap by the forward field is equal to the forward-field equivalent-circuit input power minus the corresponding stator loss

P_{\mathrm{gap,f}} = 2\left(Re[\hat{V}_f \hat{I}_f^*]  –  I^2_f R_1 \right) = 4149  W

where the factor of 2 accounts for the fact that this is a two-phase motor. Similarly, the power delivered to the backward field is

P_{\mathrm{gap,b}}= 2\left(Re[\hat{V}_b \hat{I}_b^*]  –  I^2_b R_1 \right) = 14.5  W

Here, the symbol Re[ ] indicates the real part of a complex number, and the superscript * indicates the complex conjugate.

Finally, from Eq. 9.14, the internal mechanical power developed is equal to (1 – s) times the total air-gap power or

P_{mech}= (1  –  s)ω_sT_{mech}=(1  –  s)(P_{\mathrm{gap,f}}  –  P_{\mathrm{gap,b}}) \quad \quad \quad (9.14)

P_{mech}= (1  –  s)(P_{\mathrm{gap,f}}  –  P_{\mathrm{gap,b}}) = 3927  W

If the core losses, friction and windage, and stray load losses are known, the shaft output can be found by subtracting them from the internal power. The friction and windage losses depend solely on the speed and are the same as they would be for balanced operation at the same speed. The core and stray load losses, however, are somewhat greater than they would be for balanced operation with the same positive-sequence voltage and current. The increase is caused principally by the (2 – s)-frequency core and stray losses in the rotor caused by the backward field.

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