Electrical Machines (Solved Numericals from AMIE Exams)

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

= \(\frac{{(1.5/100)}}{{\pi (50)(1000/60)}} = 0.573\,ms\)

The emf per turn for a single phase, 2310/220 V, 50 Hz transformer approximately 13 volts. Calculate (i) the number of primary and secondary turns (ii) the net cross sectional area of the core, for a maximum flux density of 1.4 T.

Solution

E₁/N₁ = E₂/N₂ = 13 V

E₁ = 2310 V

E₂ = 220 V

Now N₁ = E₁/13 = 2310/13 = 178

N₂ = E₂/13 = 220/13 = 17

Now E₁ = 4.44 x N₁ x f x B x A_i

Putting values

2310 = 4.44 x 178 x 50 x 1.4 x A

From this, A = 0.042 m² =  0.042 x 100 x 100 = 420 cm²

A 230 V DC series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistances are 0.3 Q and 0.2 Ω respectively. Determine the resistance that must be added to obtain rated torque: (a) at starting (b) at 1000 rpm. (AMIE Winter 2024, 6 marks)

Solution

For series motor

φ ∝ I_a

So T_e ∝ I_a²

I_a1 = 20 A

(a) At starting E_b = 0 and

V_t = I_a(r_a + r_se + R_ext)

\(\begin{array}{l}{R_{ext}} = \frac{{{V_t} - {I_{a1}}({r_a} + {r_{se}})}}{{{I_{a1}}}}\\ = \frac{{230 - 20x0.5}}{{20}} = 11\Omega \end{array}\) 

(b) For developing rated torque at 1000 rpm

I_a2 = I_a1 = 20 amp

\(\frac{{{E_{b2}}}}{{{E_{b1}}}} = \frac{{{N_2}{\phi _2}}}{{{N_1}{\phi _1}}} = \frac{{{N_2}{I_{a2}}}}{{{N_1}{I_{a1}}}} = \frac{{{N_2}}}{{{N_1}}} = \frac{{1000}}{{1500}}\)

\(\frac{{1000}}{{1500}} = \frac{{V - {I_{a2}}({r_a} + {r_{se}} + {R_{ext}})}}{{V - {I_{a1}}({r_a} + {r_{se}})}}\) 

I_a2 = I_a1 (rated torque)

So, \({I_{a2}} = {I_{a1}} = \frac{{230 - 20x0.5}}{{230 - 20x0.5}} - \frac{{20x{R_{se}}}}{{220}}\)

Or, \(\frac{1}{{11}}{R_{se}} = \left( {1 - \frac{{10}}{{15}}} \right)\)

Giving R_se = 3.667 Ω

 

 

 

 

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