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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} ...