Question (a) A three-phase 50-Hz four-pole star-connected cage induction motor gave the following testing results: Stator winding resistance per phase 0.75 Ω Locked rotor test at rated stator current 100 V, 28 A, 2.8 kW No load test at rated stator voltage 380 V, 5 A, 2.1 kW Friction and windage losses are negligible. Based on the approximate model, (1) calculate the rated speed and rated torque of the machine; and (i) calculate the starting torque of the machine at direct online starting. (b) Briefly explain why induction machines are preferred over de machines in most industrial application

(a) Given f=\mathrm{SOH}_{2}, P=4z_{S}=0.75 n [stator winding resistance per phase](x) By locked rotor test at rated stator curredrtV_{\text {reduced }}=100 \mathrm{~V}I_{B R}=28 \mathrm{~A} [ This is the curvent that will flow through stator windiry, which is the rated current of motor corresponding to reduced applied voltage]Full wad copper loss P_{\text {cuft }}=2.8 \mathrm{kw}(iy) No load test at rated stator voltage.\begin{array}{c}V_{\text {rated }}=380 \mathrm{~V} . \\\text { constant } \cos =2.1 \mathrm{kw}\end{array}No load cunent I_{0}=5 \mathrm{~A}., frichion & windage loss negligible (Given)(i) we have to find rated speed.\text { Power input to rotor } \begin{aligned}=P_{g} & =\sqrt{3} \times V_{\text {rated }} \times I_{\text {rated }} \\& =\sqrt{3} \times 380 \times 28 \\P_{g} & =18.43 \times \omega\end{aligned}\begin{aligned}\text { Rotorovitput power }=P_{g}-C_{u} \text { Coss } & =18.43-2.8 \\& =15.63 \mathrm{kw}\end{aligned}and. fotor ofp power =(1-s) \mathrm{Pg}\begin{aligned}15.63 k \omega & =(1-s) 18.43 \\S & =1-\frac{15.63}{18.43} \\S & =0.152\end{aligned}now, Synchronous speed N_{s}=\frac{120 f}{p}=\frac{120 \times 50}{40}N_{s}=1500 \mathrm{rm}\begin{array}{l}S=\frac{N_{S}-N_{r}}{N_{S}} \\0.152=\frac{1500-N_{r}}{1500} \Rightarrow N_{r}=1272 \mathrm{rpm}\end{array}Now, rated torque =\tau_{\text {rated }}\begin{array}{c}T_{\text {rated }}=\frac{P_{\text {output of rotor }}}{\omega_{m}}, \quad \omega_{m}=\frac{2 \pi N_{r}}{60} \\P_{\text {out put of rotor }}=15.63 \mathrm{kw} \quad \omega_{m}=\frac{2 \pi \times 1272}{60} \\T_{\text {rated }}=\frac{15.63 \times 10^{3}}{133.2 \mathrm{rad} / \mathrm{sec}} \omega \omega_{\text {rated }}=117.32 \mathrm{~N}-\mathrm{m}\end{array}(ii) [starting [starbing torque by directonline starting] GDOL method. in DOL method we have the relation:-\begin{array}{l}T_{s t}=117.32\left[\frac{292.52}{28}\right]^{2} \cdot(0.152) \\\bar{S}_{s t}=1945 \mathrm{~N}-\mathrm{m} \\\end{array}AS PER POLICY I CAN SOLVE ONLY FIRST QUESTION ,IF YOU NEED THE SOLUTION OF OTHER ONE PLEASE POST IT AGAIN ,I WILL GIVE THE ANSWER.... ...