Question 17.1: Induction-Motor Performance A certain 30-hp four-pole 440-V-...

From Table 17.1, we find that synchronous speed for a four-pole motor is n_s = 1800 rpm. Then, we utilize Equation 17.16 to compute the slip:

s=\frac{\omega_s-\omega_m }{\omega_s}=\frac{n_s-n_m}{n_s}                (17.16)

s= \frac{n_s-n_m}{n_s} =\frac{1800-1746}{1800}=0.03

We can use the data given to draw the equivalent circuit shown in Figure 17.14 for one phase of the motor. The impedance seen by the source is

Z_s=1.2+j2+\frac{j50(0.6+19.4+j0.8)}{j50+0.6+19.4+j0.8}=1.2+j2+16.77+j7.392 =17.97+j9.392=20.28\angle 27.59^\circ \Omega

The power factor is the cosine of the impedance angle. Because the impedance is inductive, we know that the power factor is lagging:

power factor = cos(27.59°) = 88.63% lagging

For a delta-connected machine, the phase voltage is equal to the line voltage, which is specified to be 440V rms. The phase current is

\mathrm{I}_s =\frac{\mathrm{V}_s}{Z_s}=\frac{440\angle 0^\circ}{20.28\angle 27.59^\circ}=21.70\angle -27.59^\circ \mathrm{~A~rms}

Thus, the magnitude of the line current is

I_{\mathrm{line}}=I_s\sqrt{3}=21.70\sqrt{3}=37.59\mathrm{~A~rms}

The input power is

P_{\mathrm{in}}=3I_sV_s\cos{\theta}=3(21.70)440 \cos{(27.59^\circ)}=25.38\mathrm{~kW}

Next, we compute V_x and I‘_r:

\mathrm{V}_x=\mathrm{I}_s\frac{j50(0.6+19.4+j0.8)}{j50+0.6+19.4+j0.8}=21.70\angle -27.59^\circ \times 18.33\angle 23.78^\circ =397.8\angle -3.807^\circ \mathrm{~V~rms}

\mathrm{I}'_r=\frac{\mathrm{V}_x}{j0.8+0.6+19.4}=\frac{397.8\angle -3.807^\circ}{20.01\angle 1.718^\circ}=19.88\angle -5.52^\circ\mathrm{~A~rms}

The copper losses in the stator and rotor are

P_s=3R_sI_s^2=3(1.2)(21.70)^2=1695\mathrm{~W}

and

P_r=3R'_r(I'_r)^2=711.4\mathrm{~W}

Finally, the developed power is

P_{\mathrm{dev}}=3 \times \frac{1-s}{s}R'_r(I'_r)^2=3(19.4)(19.88)^2=23.00\mathrm{~kW}

As a check, we note that

P_{\mathrm{in}}=P_{\mathrm{dev}}+P_s+P_r

to within rounding error.
The output power is the developed power minus the rotational loss, given by

P_{\mathrm{out}}=P_{\mathrm{dev}}-P_{\mathrm{rot}}=23.00-0.900=22.1 \mathrm{~kW}

This corresponds to 29.62 hp, so the motor is operating at nearly its rated load. The output torque is

T_{\mathrm{out}}=\frac{P_{\mathrm{out}}}{\omega_{m}}=\frac{22,100}{1746(2\pi/60)}=120.9\mathrm{Nm}

The efficiency is

\eta =\frac{P_\mathrm{out}}{P_{\mathrm{in}}}\times 100%=\frac{22,100}{25,380}\times 100%=87.0%