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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

α = (0.04)^{1 / n} = {0.04}^{1 / 5} = 0.76

$\alpha = {(0.04)^{1/n}} = {0.04^{1/5}} = 0.76$

R_{1} = \frac{0.02}{0.04} = 0.5

${R_1} = \frac{{0.02}}{{0.04}} = 0.5$ Resistances of the sections are

\begin{matrix} r_{1} = (1 - α) R_{1} = (1 - 0.76) (0.5) = 0.12 r_{2} = α r_{1} = 0.76 x 0.12 = 0.0912 r_{3} = α^{2} r_{1} = (0.76)^{2} (0.12) = 0.693 r_{4} = α^{3} r_{1} r_{5} = α^{4} r_{1} \end{matrix}

$\begin{array}{l}{r_1} = (1 - \alpha ){R_1} = (1 - 0.76)(0.5) = 0.12\\{r_2} = \alpha {r_1} = 0.76x0.12 = 0.0912\\{r_3} = {\alpha ^2}{r_1} = {(0.76)^2}(0.12) = 0.693\\{r_4} = {\alpha ^3}{r_1}\\{r_5} = {\alpha ^4}{r_1}\end{array}$ 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}...

Pulse and Digital Circuits (Short Answer Questions)

Differentiate vectored and non-vectored interrupts. Interrupts may be vectored or non-vectored, depending upon the provision of their pre-defined branching address. If the branching address of the interrupt, where its ISR to be located, is pre-defined, then the interrupt is designated as a vectored interrupt . On the other hand, if the interrupt is not assigned any such pre-defined branching address for its ISR, it would be taken as a non-vectored interrupt. List the advantage of segmented memory. An immediate advantage of having separate data and code segments is that one program can work on several sets of data. Perhaps the greatest advantage of segmented memory is that programs that reference logical addresses only can be loaded and run anywhere in memory.Explain importance of Hit ratio. Explain the locality of reference. Locality of Reference. The characteristics of a typical program that leads to a predictable access pattern is referred to as the locality of reference. What...

Mechanics of Solids (Solved Numerical Problems)

A stepped circular bar 150 mm long with diameters 20, 15, and 10 mm along length AB = 40 mm, BC - 45 mm, and CD = 65 mm, respectively is subjected to various forces as shown in Fig. P-14.12(a). Determine the change in its length if E = 200

k N / m m^{2}

$kN/m{m^2}$ . (AMIE Winter 2023) Solution Considering portion AB, a tensile force of 20 kN is to act to maintain equilibrium. For portion BC, if it is considered to be subjected to a tensile force of 5 kN, then the net force at section B will be 15 kN as shown. Similarly, for the portion CD, a compressive force of 10 kN would make a net force of 15 kN at section C. Area of cross sections

\begin{matrix} A_{1} = \frac{π}{4} (20)^{2} = 314.16 m m^{2} A_{2} = \frac{π}{4} (15)^{2} = 176.7 m m^{2} A_{3} = \frac{π}{4} (10)^{2} = 78.54 m m^{2} \end{matrix}

$\begin{array}{l}{A_1} = \frac{\pi }{4}{(20)^2} = 314.16\,m{m^2}\\{A_2} = \frac{\pi }{4}{(15)^2} = 176.7\,m{m^2}\\{A_3} = \frac{\pi }{4}{(10)^2} = 78.54\,m{m^2}\end{array}$ Extension of AB

\begin{matrix} δ L_{1} = \frac{20 x 40}{314.16 E} = \frac{20 x 40}{314.16 x 200} = 1.273 x 10^{- 2} m m \end{matrix}

$\begin{array}{l}\delta {L_1} = \frac{{20x40}}{{314.16E}}\\ = \frac{{20x40}}{{314.16x200}}\\ = 1.273x{10^{ - 2}}\,mm\end{array}$ Extension of BC \(\begin{array}{l}\delta {L_2}...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and

15^{0} C

${15^0}C$ has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density

ρ = \frac{P}{R T} = \frac{1.01325 x 10^{5}}{287} (288) = 1.226 k g / m^{3}

$\rho = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}$ Total Power

P_{t o t a l} = ρ A {V_{1}}^{3} / 2

${P_{total}} = \rho A{V_1}^3/2$ Power density

\begin{matrix} \frac{P_{t o t a l}}{A} = \frac{1}{2} ρ {V_{1}}^{3} = \frac{1}{2} (1.226) (15)^{3} = 2068.87 W / m^{2} \end{matrix}

$\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}$ Maximum power density

\begin{matrix} \frac{P_{max}}{A} = \frac{8}{27} ρ A {V^{3}}_{1} = \frac{8}{27} (1.226) (15)^{3} = 1226 W / m^{2} \end{matrix}

$\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}$ Assuming eff...

Power Systems (Solved Numerical Problems)

Calculate the loop inductance per kilometre of a single-phase transmission line when the line conductors are spaced 1 m apart and each conductor has a diameter of 1.25 cm. Also, calculate the reactance of the line. The frequency is 50 Hz. (AMIE Winter 2023) Solution Loop inductance

\begin{matrix} L I = 2 x 10^{- 7} ln \frac{d}{r} = 2 x 10^{- 7} ln \frac{100}{0.625} = 10^{- 6} H / m = 10^{- 3} H / k m \end{matrix}

$\begin{array}{l}LI = 2\,x\,{10^{ - 7}}\ln \frac{d}{r}\\ = 2\,x\,{10^{ - 7}}\ln \frac{{100}}{{0.625}}\\ = {10^{ - 6}}\,H/m = {10^{ - 3}}\,H/km\end{array}$ Line reactance per km =

\begin{matrix} 2 π f (L I) = 2 π (50) (10^{- 3}) = 0.314 Ω \end{matrix}

$\begin{array}{l}2\pi f(LI) = 2\pi (50)({10^{ - 3}})\\ = 0.314\Omega \end{array}$ The speed of a 100 MW alternator drops by 4% from no load to full load, (a) Find the speed regulation R of the alternator. Express the results in Hz per unit MW and Hz per MW (b) If frequency drops by 0.1 Hz, find change in power output. (AMIE Winter 2023) Solution \(\begin{array}{l}R = \frac{{4\,x\,5}}{{100\,x\,1}} = 2\,Hz/pu\,MW\\R = \frac{{4\,x\,50}}{{100\,x\,100}} = 0.02\,Hz/MW\\\Delta P = - \frac{1}{R}\Delta f = - \frac{1}{{0.02}}( ...

Power Electronics (Solved Numerical Problems)

A 3 phase dual converter is operated front the secondary of a delta star connected transformer of 220 volts, 50 Hz supply. If the load resistance is 10 Ω, the circulating inductance is 7.5 mH and

α_{1} = 50^{0}

${\alpha _1} = {50^0}$ , find, (i) Peak circulating current (ii) Peak current of converter 1 (AMIE Summer 2023) Solution Secondary phase voltage

V_{p h} = \frac{220}{\sqrt{3}} = 127 V

${V_{ph}} = \frac{{220}}{{\sqrt 3 }} = 127\,V$ Peak value of phase voltage

\begin{matrix} V_{m} = \sqrt{2} V_{p h} = \sqrt{2} x 127 = 179.63 V ω = 2 π f = 2 π x 50 = 314.16 r a d / s \end{matrix}

$\begin{array}{l}{V_m} = \sqrt 2 {V_{ph}} = \sqrt 2 \,x\,127 = 179.63\,V\\\omega = 2\pi f = 2\pi \,x\,50 = 314.16\,rad/s\end{array}$ Peak circulating current

i_{c i r} (t) = \frac{3 V_{m}}{ω L_{r}} [sin (ω t - \frac{π}{6}) - sin α_{1}]

${i_{cir}}(t) = \frac{{3{V_m}}}{{\omega {L_r}}}\left[ {\sin \left( {\omega t - \frac{\pi }{6}} \right) - \sin {\alpha _1}} \right]$ This current will be maximum when

sin (ω t - \frac{π}{6}) = 1 \Rightarrow ω t = \frac{2 π}{3}

$\sin \left( {\omega t - \frac{\pi }{6}} \right) = 1 \Rightarrow \omega t = \frac{{2\pi }}{3}$ \({i_{cir(peak)}} = \frac{{3x179.63\,V}}{{314.16(7.5\,x\,{{10}^{ - 3}})}}\left[ {\sin \left( {\frac{{2\pi }}{3} - \frac{{2\pi }}{6}} \right)} \right] =...

Analysis and Design of Structures (Solved Numerical Problems)

Structural Analysis