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
\(\begin{array}{l}LI = 2\,x\,{10^{ - 7}}\ln \frac{d}{r}\\ = 2\,x\,{10^{ - 7}}\ln \frac{{100}}{{0.625}}\\ = {10^{ - 6}}\,H/m = {10^{ - 3}}\,H/km\end{array}\)
Line reactance per km
= \(\begin{array}{l}2\pi f(LI) = 2\pi (50)({10^{ - 3}})\\ = 0.314\Omega \end{array}\)
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
\(\begin{array}{l}R = \frac{{4\,x\,5}}{{100\,x\,1}} = 2\,Hz/pu\,MW\\R = \frac{{4\,x\,50}}{{100\,x\,100}} = 0.02\,Hz/MW\\\Delta P = - \frac{1}{R}\Delta f = - \frac{1}{{0.02}}( - 0.1) = 5\,MW\end{array}\)
Thus, a decrease in frequency by 0.1 Hz means an increase in output by 5 MW.
A steam power station of 100 MW capacity uses coal of calorific value 6400 kcal/kg. The thermal efficiency of the station is 30% and electrical generation efficiency is 92%. Find the coal require per hour when plant is working at full load. (AMIE Winter 2023)
Solution
Energy output/hr
= \(100\,x\,{10^6}\,x\,3600 = 36\,x\,{10^{10}}\,Watt\)
Energy input/hr
= \(\frac{{36\,x\,{{10}^{10}}}}{{0.3\,x\,0.92}} = 130.43\,x\,{10^{10}}\,Watt\,\sec \)
1 watt sec = 1 Joule
Energy input
= \(\begin{array}{l}130.43\,x\,{10^{10}}J\\ = 130\,x\,{10^{10}}\,x\,0.239\,Calories\\ = 31.17\,x\,{10^{10}}\,Calories\\ = 31.17\,x\,{10^7}\,kval\end{array}\)
Coal required
= \(\frac{{31.17\,x\,{{10}^7}}}{{6400}} = 48703.1\,kg/hr\)
A 2 pole, 50 Hz, 11 kV turbogenerator has a rating of 1000 MW at 0.85 p.f lagging. The rotor has a moment of inertia of, 100000 kg-m². Find H and M. (AMIE Winter 2023)
Solution
\(\begin{array}{l}\omega = 2\pi f = 2\pi x50 = 314\,rad/s\\MJ = \frac{1}{2}I{\omega ^2}\\ = \frac{1}{2}(100,000){(314)^2} = 4929.8\,MH\end{array}\)
MVA rating
\(\begin{array}{l}G = \frac{{1000}}{{0.85}} = 1176.5\\H = \frac{{MJ}}{{MVA}} = \frac{{4929.8}}{{1176.5}} = 4.2\,MJ/MVA\\M = \frac{{GH}}{{180f}} = \frac{{1176.5\,x\,4.2}}{{180\,x\,50}}\\ = 0.55\,MLS/elec - \deg ree\end{array}\)
What is the percentage saving in feeder copper if the line voltage in a 2-wire DC systems is raised from 100 V to 200 V for the same power transmitted over the same power distance and having the same power loss? (AMIE Winter 2021)
Solution
\(\begin{array}{l}{P_1} = {V_1}{I_1} = 200{I_1}\\{P_2} = {V_2}{I_2} = 400{I_2}\\200{I_1} = 400{I_2}\\\therefore \,{I_2} = \frac{{200}}{{400}} = 0.5{I_1}\end{array}\)
Power loss in 200 V system
\({W_1} = 2{I_1}^2{R_1}\)
Power loss in 400 V system
\(\begin{array}{l}{W_2} = 2{I_2}^2{R_2}\\ = 2{(0.5{I_1})^2}{R_2}\\ = 0.5{I_1}^2{R_2}\end{array}\)
As power loss is same,
\(\begin{array}{l}{W_1} = {W_2}\\2{I_1}^2{R_1} = 0.5{I_1}^2{R_2}\\\frac{{{R_2}}}{{{R_1}}} = 4\end{array}\)
Now, resistance is inversely proportional to area.
\(\begin{array}{l}\frac{{{A_1}}}{{{A_2}}} = 4\\\frac{{{V_1}}}{{{V_2}}} = 4\\\% \,saving\,in\,copper\, = \,\frac{{{V_1} - {V_2}}}{{{V_1}}}x100\\ = \frac{{4{V_2} - {V_2}}}{{4{V_2}}}x100\\ = \frac{{3{V_2}}}{{4{V_2}}}x100\, = 75\% \end{array}\)
A transmission line conductor having a dia of 19.5 mm weights 0.85 kg/m. The span is 275 meters. The wind pressure is 39 \(kg/{m^2}\) of projected area with ice coating of 13 mm. The ultimate strength of the conductor is 8000 kg. Calculate the maximum sag if the factor of safety is 2 and ice weighs 910 \(kg/{m^3}\). (AMIE Winter 2021)
Solution
\(\begin{array}{l}w = 0.85\,kg/m;d = 19.5\,m;t = 13\,mm\\p = 39\,kg/{m^2};{S_f} = 2\\{w_w} = p(d + 2t)\\ = 39(19.5 + 2\,x\,13)\,x\,{10^{ - 3}}\\ = 1.7745\,kg/m\end{array}\)
\(\begin{array}{l}{w_i} = ice\,density\,x\,\pi t(d + t)\\ = 910(\pi )(13\,x\,{10^{ - 3}})(19.5 + 13)\,x\,{10^{ - 3}}\\ = 1.208\,kg/m\end{array}\)
\(\begin{array}{l}{w_t} = \sqrt {{{(w + {w_i})}^2} + {{({w_w})}^2}} \\ = \sqrt {{{(0.85 + 1.208)}^2} + {{(1.7745)}^2}} \\ = 2.7174\end{array}\)
\(\begin{array}{l}{S_f} = \frac{{8000}}{2} = 4000\,kg\\\therefore \,S = \frac{{{w_t}{L^2}}}{{8T}} = \frac{{2.7174x{{275}^2}}}{{8x4000}} = 6.422\,m\end{array}\)
Two generators rated at 10 MVA, 13.2 kV and 15 MVA. 13.2 kV. respectively are connected in parallel to a bus. The bus feeds two motors rated at 8 MVA and 12 MVA respectively. The rated voltage of motors is 12.5 kV. The reactance of each generator is 15% and that of each motor is 20% on its own rating. Assume 50 MVA. 13.8 kV base and draw reactance diagram. (AMIE Winter 2020)
Solution
Using following equation
Per unit impedance referred to new base is equal to
The following figure shows the reactance diagram. The values indicate reactances in per cent. \({E_1},{E_2}\) are e.m.fs of generators, while \({E_3},{E_4}\) are e.m.fs of motors.
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