Question 16.1: Induction-Motor Performance A certain 30-hp four-pole 440-V-...

From Table 16.1, we find that synchronous speed for a four-pole motor is n_s = 1800 \text{ rpm}. Then, we utilize Equation 16.16 to compute the slip:

s=\frac{\omega_s-\omega_m}{\omega_s}=\frac{n_s-n_m}{n_s} \quad \quad \quad \quad \quad (16.16) \\ 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 16.14 for one phase of the motor. The impedance seen by the source is

\begin{matrix} 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 \underline{/27.59^\circ} \ \Omega \end{matrix}

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

\text{power factor}=\cos(27.59^\circ)=88.63 \% \text{ 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

\textbf{I}_s=\frac{\textbf{V}_s}{Z_s}=\frac{440\underline{/0^\circ } }{20.28\underline{/27.59^\circ }}=21.70\underline{/-27.59^\circ }\text{ A rms}

Thus, the magnitude of the line current is

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

In ac machine calculations, we take the rms values of currents and voltages for the phasor magnitudes (instead of peak values as we have done previously).

The input power is

\begin{matrix} P_{\text{in}}=3I_sV_s\cos \theta \\ =3(21.70)440 \cos(27.59^\circ) \\=25.38 \text{ kW} \end{matrix}

Next, we compute \pmb{\text{V}}_x \text{ and }\pmb{\text{I}}_r^\prime:

\begin{matrix} \pmb{\text{V}}_x=\pmb{\text{I}}_s\frac{j50(0.6+19.4+j0.8)}{j50+0.6+19.4+j0.8} \\ =21.70 \underline{/-27.59^\circ} \times 18.33 \underline{/23.78^\circ} \\ =397.8 \underline{/-3.807^\circ} \text{ V rms} \\ \pmb{\text{I}}^\prime_r =\frac{\pmb{\text{V}}_x}{j0.8+0.6+19.4} \\ =\frac{397.8 \underline{/-3.807^\circ}}{20.01\underline{/1.718^\circ}} \\ =19.88 \underline{/-5.52 ^\circ} \text{ A rms} \end{matrix}

The copper losses in the stator and rotor are

\begin{matrix} P_s=3R_sI_s^2 \\=3(1.2)(21.70)^2 \\ =1695 \text{ W} \end{matrix}

and

\begin{matrix} P_r=3R_r^\prime(I^\prime_r)^2 \\ =3(0.6)(19.88)^2 \\ =711.4 \text{ W} \end{matrix}

Finally, the developed power is

\begin{matrix} P_{\text{dev}}=3 \times \frac{1-s}{s}R^\prime_r(I^\prime_r)^2 \\ =3(19.4)(19.88)^2 \\ =23.00 \text{ kW} \end{matrix}

As a check, we note that

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

to within rounding error.

The output power is the developed power minus the rotational loss, given by

\begin{matrix} P_{\text{out}}=P_{\text{dev}}-P_{\text{rot}} \\= 23.00-0.900 \\ =22.1 \text{ kW} \end{matrix}

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

\begin{matrix} T_{\text{out}}=\frac{P_{\text{out}}}{\omega_m} \\ = \frac{22,100}{1746(2\pi/60)} \\ =120.9 \text{ Nm} \end{matrix}

The efficiency is

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

Table 16.1 Synchronous Speed Versus Number of Poles for f = 60 Hz