**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

\(\begin{array}{l}{\mathop{\rm Re}\nolimits} ac\tan ce\,of\,{G_1} = 15{\left( {\frac{{13.2}}{{13.8}}} \right)^2}\left( {\frac{{50}}{{10}}} \right) = 68.62\% \\{\mathop{\rm Re}\nolimits} ac\tan ce\,of\,{G_2} = 15{\left( {\frac{{13.2}}{{13.8}}} \right)^2}\left( {\frac{{50}}{{15}}} \right) = 45.75\% \\{\mathop{\rm Re}\nolimits} ac\tan ce\,of\,{M_1} = 20\left( {\frac{{12.5}}{{13.8}}} \right)\left( {\frac{{50}}{8}} \right) = 102.56\% \\{\mathop{\rm Re}\nolimits} ac\tan ce\,of\,{M_2} = 20{\left( {\frac{{12.5}}{{13.8}}} \right)^2}\left( {\frac{{50}}{{12}}} \right) = 68.37\% \end{array}\)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.

**AMIE/BTech/Junior Engineer exams**is available at

**https://amiestudycircle.com**

## Comments