Question 16.3: Induction-Motor Performance A 220-V-rms 60-Hz three-phase wy...
Induction-Motor Performance
A 220-V-rms 60-Hz three-phase wye-connected induction motor draws 31.87 A at a power factor of 75 percent lagging. For all three phases, the total stator copper losses are 400 W, and the total rotor copper losses are 150 W. The rotational losses are 500 W. Find the power crossing the air gap P_{\text{ag}}, the developed power P_{\text{dev}} , the output power P_{\text{out}}, and the efficiency.
The phase voltage is V_s = V_{\text{line}} /\sqrt{3} = 127.0 \text{ V rms}. Next, we find the input power:
\begin{matrix}P_{\text{in}}&=&3V_sI_s\cos(\theta) \\ &=&3(127)(31.87)(0.75) \\ &=&9107 \text{ W} \end{matrix}
The air-gap power is the input power minus the stator copper loss:
\begin{matrix} P_{\text{ag}}&=&P_{\text{in}}-P_s \\ &=& 9107-400 \\ &=&8707 \text{ W} \end{matrix}
The developed power is the input power minus the copper losses:
P_{\text{dev}}=9107-400-150=8557 \text{ W}
Next, by subtracting the rotational losses, we find that the output power is
\begin{matrix} P_{\text{out}}&=&P_{\text{dev}}-P_{\text{rot}} \\ &=&8557-500 \\ &=&8057 \text{ W} \end{matrix}
Finally, the efficiency is
\begin{matrix} \eta&=&\frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \\ &=&94.0 \% \end{matrix}