Electric Supply Systems

Demonstrative Video

Problem-1

What is the percentage saving in feeder copper if the line voltage in a 2 -wire d.c. system is raised from 200 volts to 400 volts for the same power transmitted over the same distance and having the same power loss?

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Solution-1

\[\begin{aligned} P & =V_{1} I_{1}& =200 I_{1} \\ P& =V_{2} I_{2} & =400 I_{2} \end{aligned}\] As same power is delivered in both cases, \[\begin{aligned} 200 I_{1}& =400 I_{2} \\ \Rightarrow I_{2} & =(200 / 400) I_{1}&=0 \cdot 5 I_{1} \end{aligned}\] \[\begin{aligned} W_{1}& =2 I_{1}^{2} R_{1} \\ W_{2} & =2 I_{2}^{2} R_{2}& =2\left(0 \cdot 5 \mathrm{I}_{1}\right)^{2} \mathrm{R}_{2} & =0 \cdot 5 \mathrm{I}_{1}^{2} \mathrm{R}_{2} \end{aligned}\]

As power loss in the two cases is the same, \[\begin{aligned} \therefore \quad W_{1} &=W_{2} \\ 2 I_{1}^{2} R_{1} &=0 \cdot 5 T_{1}^{2} R_{2} \end{aligned}\] \[\begin{aligned} R_{2} / R_{1}& =2 / 0 \cdot 5=4 \\ a_{1} / a_{2} & =4 \\ v_{1} / v_{2} & =4 \\ v_{2} / v_{1} & =1 / 4=0 \cdot 25 \end{aligned}\] \(\therefore \quad \%\) age saving in feeder copper \[\begin{aligned} &=\frac{V_{1}-V_{2}}{V_{1}} \times 100\\ &=\left(\frac{V_{1}}{V_{1}}-\frac{V_{2}}{V_{1}}\right) \times 100\\ &=(1-0 \cdot 25) \times 100=75 \% \end{aligned}\]

Problem-2

A d.c. 2 -wire system is to be converted into a.c. 3 -phase, 3 -wire system by the addition of a third conductor of the same cross-section as the two existing conductors. Calculate the percentage additional load which can now be supplied if the voltage between wires and the percentage loss in the line remain unchanged. Assume a balanced load of unity power factor.

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Solution-2

Let \(R\) be the resistance per conductor in each case.

As the % power loss in the two cases is the same,

\[\begin{aligned} \dfrac{2 I_{1}^{2} R}{V I_{1}} \times 100& =\dfrac{3 I_{2}^{2} R}{\sqrt{3} V I_{2}} \times 100 \\ \Rightarrow 2 I_{1} & =\sqrt{3} I_{2} \\ \Rightarrow I_{2} &=\frac{2}{\sqrt{3}} I_{1} \\ \dfrac{P_{2}}{P_{1}} &=\frac{\sqrt{3} V I_{2}}{V I_{1}}=\dfrac{\sqrt{3} V \times \dfrac{2}{\sqrt{3}} I_{1}}{V I_{1}} =2 \\ \Rightarrow P_{2}& =2 P_{1} \end{aligned}\]

i.e. additional power which can be supplied at unity p.f. by 3 -phase, 3 -wire a.c. system =100 %.

Problem-3

A d.c. 3 -wire system is to be converted into a 3 -phase, 4 -wire system by adding a fourth wire equal in \(X\) -section to each outer of the d.c. system. If the percentage power loss and voltage at the consumer’s terminals are to be the same in the two cases, find the extra power at unity power factor that can be supplied by the a.c. system. Assume loads to be balanced.

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Solution-3

\[\begin{aligned} \frac{2 I_{1}^{2} R}{2 V I_{1}} \times 100 & =\frac{3 I_{2}^{2} R}{3 V I_{2}} \times 100 \\ \Rightarrow I_{1} &=I_{2} \\ \frac{P_{2}}{P_{1}} &=\frac{3 V I_{2}}{2 V I_{1}}=\frac{3 V I_{1}}{2 V I_{1}}=1 \cdot 5\\ \Rightarrow P_{2}& =1 \cdot 5 P_{1} \end{aligned}\]

i.e., extra power that can be supplied at unity power factor by 3 -phase, 4 -wire a.c. system \(=50 \%\)

Problem-4

A single phase a.c. system supplies a load of \(200 \mathrm{kW}\) and if this system is converted to 3 -phase, 3 -wire a.c. system by running a third similar conductor; calculate the 3 -phase load that can now be supplied if the voltage between the conductors is the same. Assume the power factor and transmission efficiency to be the same in the two cases.

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Solution-4

Let \(R\) be the resistance per conductor and \(\cos \phi\) the power factor in each case. single phase 2 -wire system.

0.48 1-phase, 2-wire a.c. system \[\begin{aligned} P_{1} &=V I_{1} \cos \phi \\ W_{1} &=2 I_{1}^{2} R \\ \% \text { P.L} &=\frac{2 I_{1}^{2} R}{V I_{1} \cos \phi} \times 100 \end{aligned}\]

0.48 3-phase, 3-wire a.c. system \[\begin{aligned} P_{2}&=\sqrt{3} V I_{2} \cos \phi\\ W_{2} &=3 I_{2}^{2} R \\ \% \text {P.L } &=\frac{3 I_{2}^{2} R}{\sqrt{3} V I_{2} \cos \phi} \times 100 \end{aligned}\]

As the transmission efficiency in the two cases is the same, therefore, percentage power loss will also be the same \(i . e .\) \[\begin{aligned} \frac{2 I_{1}^{2} R}{V I_{1} \cos \phi} \times 100& =\frac{3 I_{2}^{2} R}{\sqrt{3} V I_{2} \cos \phi} \times 100 \\ \Rightarrow 2 I_{1}& =\sqrt{3} I_{2}\\ \Rightarrow I_{2} &=\frac{2}{\sqrt{3}} I_{1} \\ \frac{P_{2}}{P_{1}} &=\frac{\sqrt{3} V I_{2} \cos \phi}{V I_{1} \cos \phi}=\frac{\sqrt{3} V \frac{2}{\sqrt{3}} I_{1} \cos \phi}{V I_{1} \cos \phi}=2\\ P_{2}& =2 P_{1}=2 \times 200=400 \mathrm{kW} \end{aligned}\]

It may be seen that 3 -phase, 3 -wire system can supply \(100 \%\) additional load.

Problem-5

A \(50 \mathrm{km}\) long transmission line supplies a load of \(5 \mathrm{MVA}\) at 0.8 p.f. lagging at 33 \(k V .\) The efficiency of transmission is \(90 \%\). Calculate the volume of aluminium conductor required for the line when

  1. 1-phase, 2-wire system is used

  2. 3-phase, 3-wire system is used

The specific resistance of aluminium is \(2 \cdot 85 \times 10^{-8}~ \Omega \mathrm{m}\)

Solution-5

\[\begin{aligned} P &=\text { MVA } \times \cos \phi=5 \times 0 \cdot 8=4~ \mathrm{MW}=4 \times 10^{6}~ \mathrm{W} \\ W &=10 \% \text { of power transmitted }\\ &=(10 / 100) \times 4 \times 10^{6}=4 \times 10^{5} ~\mathrm{W} \\ l&=50 ~\mathrm{km} =50 \times 10^{3} ~\mathrm{m} \end{aligned}\]

1-phase, 2-wire system \[\begin{aligned} \text { Apparent power } &=V I_{1} \\ \therefore \quad & I_{1}=\frac{\text { Apparent power }}{V}=\frac{5 \times 10^{6}}{33 \times 10^{3}}=151 \cdot 5 \mathrm{A} \end{aligned}\]

Suppose \(a_{1}\) is the area of cross-section of aluminium conductor. \[\text { Line loss, } W=2 I_{1}^{2} R_{1}=2 I_{1}^{2}\left(\rho \frac{l}{a_{1}}\right)\] \[\begin{aligned} \therefore \quad a_{1} &=\frac{2 I_{1}^{2} \rho l}{W}=\frac{2 \times(151 \cdot 5)^{2} \times\left(2 \cdot 85 \times 10^{-8}\right) 50 \times 10^{3}}{4 \times 10^{5}} \\ &=1 \cdot 635 \times 10^{-4} \mathrm{m}^{2} \end{aligned}\] Volume of conductor required \(=2 a_{1} l=2 \times\left(1 \cdot 635 \times 10^{-4}\right) \times 50 \times 10^{3}=16 \cdot 35 \mathrm{m}^{3}\)

3 -phase, 3-wire system \[\text { Line current, } I_{2}=\frac{\text { Apparent power }}{\sqrt{3} V}=\frac{5 \times 10^{6}}{\sqrt{3} \times 33 \times 10^{3}}=87 \cdot 5 \mathrm{A}\] Suppose \(a_{2}\) is the area of cross-section of the conductor in this case. \[\begin{aligned} \text { Line } \operatorname{loss}, W &=3 I_{2}^{2} R_{2}=3 I_{2}^{2}\left(\rho \frac{l}{a_{2}}\right) \\ \therefore \quad a_{2} &=\frac{3 I_{2}^{2} \rho l}{W}=\frac{3 \times(87 \cdot 5)^{2} \times\left(2 \cdot 85 \times 10^{-8}\right) \times 50 \times 10^{3}}{4 \times 10^{5}} \\ &=0 \cdot 818 \times 10^{-4} \mathrm{m}^{2} \end{aligned}\] Volume of conductor required \(=3 a_{2} l=3 \times\left(0 \cdot 818 \times 10^{-4}\right) \times 50 \times 10^{3}=12 \cdot 27 \mathrm{m}^{3}\)

Note that volume of conductor (and hence weight) required is less in case of 3 -phase, 3 -wire system.

Problem-6

A sub-station supplies power at \(11 \mathrm{kV}\), 0 8 p.f. lagging to a consumer through a single phase transmission line having total resistance (both go and return) of 0.15 \(\Omega .\) The voltage drop in the line is \(15 \%\). If the same power is to be supplied to the same consumer by two wire d.c. system by a new line having a total resistance of 0.05 \(\Omega\) and if the allowable voltage drop is \(25 \%\), calculate the d.c. supply voltage.

Solution-6

single phase system

Let \(I_{1}\) be the load current. Then, \[\begin{aligned} \text { Voltage drop } &=I_{1} R_{1}=I_{1} \times 0 \cdot 15 \text { volts } \\ \text { Also voltage drop } &=\frac{15}{100} \times 11000=1650 \text { volts } \end{aligned}\] From eqs. \((i)\) and \((i i), I_{1} \times 0 \cdot 15=1650\) \[\therefore \quad I_{1}=\frac{1650}{0 \cdot 15}=11000 \mathrm{A}\] Power received by consumer \(=\) Apparent power \(\times \cos \phi\) \[=\frac{(11,000 \times 11,000) \times 0 \cdot 8}{1000} \mathrm{kW}=9 \cdot 68 \times 10^{4} \mathrm{kW}\]

D.C. two-wire system

The power to be supplied by d.c. 2 -wire system is \[9 \cdot 68 \times 10^{4} \mathrm{kW}=9 \cdot 68 \times 10^{7} \mathrm{W}\]

Let \(\mathrm{V}\) volts be the supply voltage. \[\begin{aligned} \therefore \quad \text { Load current, } I_{2} &=\frac{9 \cdot 68 \times 10^{7}}{V} \\ \text { Voltage drop } &=I_{2} R_{2}=\left(\frac{9 \cdot 68 \times 10^{7}}{V}\right) \times 0.05 \\ \text { Allowable voltage drop } &=\frac{25}{100} \times \mathrm{V}=0.25 \mathrm{V} \\ \therefore \quad V^{2} &=\frac{9 \cdot 68 \times 10^{7} \times 0 \cdot 05}{0 \cdot 25}=1936 \times 10^{4} \\ \therefore \quad V &=\sqrt{1936 \times 10^{4}}=4400 \mathrm{V} \end{aligned}\]

Problem-7

A 3 -phase, 4 -wire system is used for lighting. Compare the amount of copper required with that needed for a 2 -wire \(D . C .\) system with same line voltage. Assume the same losses and balanced load. The neutral is one half the cross-section of one of the respective outers.

Solution-7

Two-wire D.C

Let \(V=\) voltage between conductors \(P=\) power delivered, \(R_{1}=\) resistance/conductor Current \[\begin{aligned} I_{1} &=P V \\ \text { power loss } &=2 I_{1}^{2} R_{1}=2 P^{2} R_{1} / V^{2} \end{aligned}\]

Three-phase, 4-wire

Let \(V\) be the line-to-neutral voltage and \(R_{2}\) resistance of each phase wire. \[\begin{aligned} P &=3 V I_{2} \cos \phi=3 V I_{2} \quad-\mathrm{if} \cos \phi=1 \\ \text { Power loss } &=3 I_{2}^{2} R_{2}=3(P / 3 V)^{2} R_{2}=P^{2} R_{2} / 3 V^{2} \end{aligned}\] Since power losses in both systems are equal \(\therefore\) \[2 P^{2} R_{1} / V^{2}=P^{2} R_{2} / 3 V^{2} \quad \text { or } \quad R_{1} / R_{2}=1 / 6\] If \(A_{1}\) and \(A_{2}\) are the cross-sectional areas of conductors in the two systems, then \(A_{1} / A_{2}=6 .\) Because \(R \propto l / A\) \(\therefore \quad\) Cu reqd. for 2 -wire system \(=2 A_{1} l\) Cu reqd. for \(3-\phi, 4\) -wire system \(=\left(3 A_{2} l+A_{2} l / 2\right)\) \[\therefore \quad \frac{\text { Cu for } 3-\phi \text { system }}{\text { Cu for d.c. system }}=\frac{3.5 A_{2} l}{2 A_{1} l}=\frac{3.5}{2} \times \frac{1}{6}=0.292\] Cu for \(3-\phi\) system =0.292 \(\times\) Cu for d.c. system.

Problem-8

Estimate the weight of copper required to supply a load of \(100 \mathrm{MW}\) at UPF by a 3-phase, 380 kV system over a distance of 100 km. The neutral point is earthed. The resistance of mission can be assumed to be 90 percent.

Solution-8

Power loss in the line \(=(1-0.9) \times 100=10~ \mathrm{MW}\) \[\text { Line current }=100 \times 10^{6} / \sqrt{3} \times 380 \times 10^{3} \times 1=152~ \mathrm{A}\] since \(I^{2} R\) loss in 3 -conductors is \(10 \times 10^6 ~\mathrm{W},\) loss per conductor is \(=10 \times 10^{6} / 3=3.333 \times 10^{6}~ \mathrm{W}\) Resistance per conductor \(=10 \times 10^{6} / 3 \times 152^{2}=144.3 \Omega\) Resistance per conductor per \(\mathrm{km}=144.3 / 100=1.443 ~\Omega\) \[\therefore \quad \text { Conductor cross-section }=0.045 / 1.443=0.03~ \mathrm{m}^{3}\] Volume of copper per meter \(=0.03 \times 100=3 \mathrm{cm}^{3}\) Weight of copper for 3 -conductor for \(100 \mathrm{km}\) length \(=3 \times(3 \times 0.01) \times 100 \times 1000=9000 \mathrm{kg}\)