Design the 5 sections of a 6-stud starter for a 3-phase slip-ring induction motor The full-load slip is 2% and the maximum starting current is limited to twice the full-load current. Rotor resistance per phase is 0.03 Ω. (AMIE Winter 2023)
Solution
Full load slip = 2%
Slip at 2 times full load current = 2 x 2 = 4% = 0.04
\(\alpha = {(0.04)^{1/n}} = {0.04^{1/5}} = 0.76\)
\({R_1} = \frac{{0.02}}{{0.04}} = 0.5\)
Resistances of the sections are
\(\begin{array}{l}{r_1} = (1 - \alpha ){R_1} = (1 - 0.76)(0.5) = 0.12\\{r_2} = \alpha {r_1} = 0.76x0.12 = 0.0912\\{r_3} = {\alpha ^2}{r_1} = {(0.76)^2}(0.12) = 0.693\\{r_4} = {\alpha ^3}{r_1}\\{r_5} = {\alpha ^4}{r_1}\end{array}\)
In a transformer, the core loss is found to be 52 W at 40 Hz and 90 W at 60 Hz measured at same peak flux density. Compute the hysteresis and eddy current losses at 50 Hz. (AMIE Winter 2023).
Solution
\(\begin{array}{l}{W_i} = Af + B{f^2}\\\frac{{{W_i}}}{f} = A + Bf\\\frac{{52}}{{40}} = A + 40B\\and\\\frac{{90}}{{60}} = A + 60B\\\therefore \,A = 0.9;B = 0.01\end{array}\)
At 50 Hz, two losses are
\(\begin{array}{l}{W_h} = {A_f} = 0.9x50 = 45\,W\\{W_e} = B{f^2} = 0.01x{50^2} = 25\,W\end{array}\)
A short-circuit test, when performed on the h.v. side of a 10 kVA, 2000/400 V single phase transformer, gave the following data ;
60 V, 4 A, 100 IV
If the l.v. side is delivering full load current at 0.8 p.f lag and at 400 V, find the voltage applied to h.v. side. (AMIE Winter 2023)
Solution
\(\begin{array}{l}{Z_{01}} = 60/4 = 15\Omega \\{R_{01}} = 100/{4^2} = 6.25\Omega \\{X_{01}} = \sqrt {{{15}^2} - {{6.25}^2}} = 13.63\Omega \\FL,\,{I_1} = 10,000/2000 = 5A\end{array}\)
Total transformer voltage drop as referred to primary is
\({I_1}({R_{01}}\cos \phi + {X_{01}}\sin \phi ) = 5(6.25\,x\,0.8 + 13.63\,x\,0.6) = 67V\)
Hence, primary voltage has to be raised from 2000 V to 2067 V in order to compensate for the total voltage drop in the transformer. In that case secondary voltage on load would remain the same as on no-load.
A commutator with a diameter of 50 cm rotates at 1000 rpm. For a brush width of 1.5 cm. find the time of commutation. (AMIE Winter 2023)
Solution
Peripheral velocity of commutator is
\(V = \pi DN = \pi (50)\left( {\frac{{1000}}{{60}}} \right)\,cm/s\)
Now, brush width = v x time of commutation
Hence, time of commutation = brush width/v
= \(\frac{{(1.5/100)}}{{\pi (50)(1000/60)}} = 0.573\,ms\)
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