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)

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°  \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 440 V rms. The phase current is

I_s = \frac{V_s}{Z_s} = \frac{440 \angle 0°}{20.28 \angle 27.59°}=21.70 \angle -27.59° A rms

Thus, the magnitude of the line current is
I_{line} = I_s \sqrt{3} = 21.70 \sqrt{3} = 37.59 A rms
The input power is

P_{in} = 3I_sV_s \cos θ
= 3(21.70)440 \cos(27.59°)
= 25.38 kW

Next, we compute V_x and I_r^{'}:

V_x = I_s \frac{j50 (0.6+19,4+j0.8)}{j50 +0.6+19,4+j0.8}
= 21.70 ∠−27.59° × 18.33 ∠23.78°
= 397.8 ∠−3.807° V rms

I_r^{'} = \frac{V_x}{j0.8 + 0.6 + 19.4}
=\frac{397.8 \angle -3.807°}{20.01 \angle 1.718°}
= 19.88 \angle -5.52° A rms

The copper losses in the stator and rotor are

P_s = 3R_sI_s^2
= 3(1.2)(21.70)^2
= 1695 W
and
P_r = 3R_r^{'}(I_r^{'})^2
= 3(0.6)(19.88)^2
= 711.4 W
Finally, the developed power is

P_{dev} = 3 \times \frac{1-s}{s}R_r^{'}(I_r^{'})^2
=3(19.4)(19.88)^2
=23.00 kW

As a check, we note that
P_{in} = P_{dev} + P_s + P_r
to within rounding error.
The output power is the developed power minus the rotational loss, given by
P_{out} = P_{dev} − P_{rot}
= 23.00 − 0.900
= 22.1 kW
This corresponds to 29.62 hp, so the motor is operating at nearly its rated load. The output torque is

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

The efficiency is

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