Question Pro ble m 1.0 A three-phase, 60 Hz induction motor runs at 890 rpm at no load and at 840 rpm at full load. a) How many poles does this motor have? b) What is the slip at full load? c) What are the rotor frequencies at full load and at no load, respectively?

Given- 3 \Phi induction motor\begin{aligned}\text { rotor speed }= & N_{r}=890 \mathrm{rpm} \text { at no-load } \\\text { rotor speed }= & N_{r}=840 \mathrm{rpm} \text { at full load } \\& f=60 \mathrm{H}_{2}\end{aligned}(a) Formula -N_{s}=\frac{120 \times f}{p}=\frac{120 \times 60}{p}since rotor speed at no load 890 \mathrm{rpm} So, syncrons speed must be near it\because pole always lie in pair so,N_{s}=\frac{7200}{p}\begin{array}{l|l|l}\text { for } P=2 & \text { at } P=4 & \text { at } P=6 \\N_{s}=\frac{7200}{2} & N_{s}=\frac{7200}{4} & N_{s}=\frac{7200}{6} \\N_{s}=3600 \mathrm{rpm} & N_{s}=1800 \mathrm{rpm} & N_{s}=1200 \mathrm{rpm}\end{array}at P=8N_{s}=\frac{7200}{8}=900 \mathrm{rpm}So, at P=8, syncronus speed near to rotor speed so,No. of Pole in this motor =8(b) Slip at full lead\begin{array}{l}S_{F L}=\frac{N_{S}-\left(N_{r}\right)_{\text {fell load }}}{N_{S}}=\frac{900-840}{900} \\S_{F L}=0.067\end{array}or \% S_{F L}=6.7 \%(c) Rotor frequenciesf_{r}=s_{s} fat full loadf_{r_{F L}}=0.666 \times 60=4 \mathrm{~Hz}at bo no- lad deip is-s=\frac{900-890}{900}=0.011so, at no ladd rotor frequencier\begin{array}{l}\text { frNL }=\operatorname{sNL} f=0.011 \times 60 \\f_{\text {INL }}=0.7 \mathrm{~Hz}\end{array} ...