Underground System of Power Transmission

Demonstrative Video

Two-wire d.c. system.

\[\begin{array}{l} \text { Load current, } I_{1}=P / V_{m} \\ \text { Line losses, } W=2 \mathrm{I}_{1}^{2} R_{1}=2\left(\frac{P}{V_{m}}\right)^{2} \frac{\rho l}{a_{1}} \\ W=\frac{2 P^{2} \rho l}{a_{1} V_{m}^{2}} \end{array}\]

\(\therefore \quad\) Area of \(\mathrm{X}-\) section, \(a_{1}=\frac{2 P^{2} \rho l}{W V_{m}^{2}}\)

Volume of conductor material required \[=2 a_{1} l=2\left(\frac{2 P^{2} \rho l}{W V_{m}^{2}}\right) l=\frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}}=K(s a y)\]

Two-wire d.c. system with mid point earthed

\[\begin{aligned} \text { Load current, } I_{2} &=P / V_{m} \\ \text { Line losses, } W &=2 I_{2}^{2} R_{2}=2\left(\frac{P}{V_{m}}\right)^{2} \rho \frac{l}{a_{2}} \\ W &=\frac{2 P^{2} \rho l}{V_{m}^{2} a_{2}} \end{aligned}\]

Area of \(\mathrm{X}\) -section, \(a_{2}=\frac{2 P^{2} \rho l}{W V_{m}^{2}}\)

Volume of conductor material required \[=2 a_{2} l=2\left(\frac{2 P^{2} \rho l}{W V_{m}^{2}}\right) l=\frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}}=K\]

Three wire d.c. system

\[\begin{array}{l} \text { Load current, } I_{3}=P / V_{m} \\ \text { Line losses, } W=2 I_{3}^{2} R_{3}=2\left(\frac{P}{V_{m}}\right)^{2} \rho \frac{l}{a_{3}} \end{array}\]

\[W=\frac{2 P^{2} \rho l}{V_{m}^{2} a_{3}}\]

Area of \(X\) -section, \(a_{3}=\frac{2 P^{2} \rho l}{W V_{m}^{2}}\)

Assuming the area of \(X\) -section of neutral wire to be half of that of either outers,

Volume of conductor material required \[\begin{array}{l} =2 \cdot 5 a_{3} l=2 \cdot 5\left(\frac{2 P^{2} \rho l}{W V_{m}^{2}}\right) l=\frac{5 P^{2} \rho l^{2}}{W V_{m}^{2}} \\ =1 \cdot 25 K \end{array}\]

\(1\phi\), 2-wire a.c. system

R.M.S value of the voltage = \(V_m / \sqrt{2}\). Assume the p.f. of the load \(=\cos\phi\)

\(\text { Load current, } I_{4} =\frac{P}{V_{m} / \sqrt{2} \cos \phi}=\frac{\sqrt{2} P}{V_{m} \cos \phi}\)
\(\text { Line losses, } W =2 I_{4}^{2} R_{4}\) \[\begin{aligned} &=2\left(\frac{\sqrt{2} P}{V_{m} \cos \phi}\right)^{2} \rho \frac{l}{a_{4}}=\frac{4 P^{2} \rho l}{a_{4} V_{m}^{2} \cos ^{2} \phi} \end{aligned}\]

Area of \(X\) -section, \(a_{4}=\frac{4 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\)
Volume of conductor material required \(=2 a_{4} l\) \[\begin{array}{l} =2\left(\frac{4 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\right) l=\frac{8 P^{2} \rho l^{2}}{W V_{m}^{2} \cos ^{2} \phi} \\ =\frac{2}{\cos ^{2} \phi} \times \frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}} =\frac{2 K}{\cos ^{2} \phi} \end{array}\]

\(1\phi\), 2-wire system with mid-point earthed

R.M.S voltage \(=V_m/\sqrt{2}\)
\(\text { Load current, } I_{5} =\frac{\sqrt{2} P}{V_{m} \cos \phi}\)

\(\begin{aligned} \text { Line losses, } W &=2 I_{5}^{2} R_{5}=2\left(\frac{\sqrt{2} P}{V_{m} \cos \phi}\right)^{2} \rho \frac{l}{a_{5}} =\frac{4 P^{2} \rho l}{a_{5} V_{m}^{2} \cos ^{2} \phi} \end{aligned}\)

Area of \(X\) -section, \(a_{5}=\frac{4 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\)

Volume of conductor material required \[\begin{array}{l} =2 a_{5} l=2\left(\frac{4 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\right) l=\frac{8 P^{2} \rho l^{2}}{W V_{m}^{2} \cos ^{2} \phi} \\ =\frac{2}{\cos ^{2} \phi} \times \frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}} \\ =\frac{2 K}{\cos ^{2} \phi} \end{array}\]

\(1\phi\), 3-wire system

For balanced load system reduces to \(1\phi\) 2-wire except there is additional neutral wire

Volume of conductor material required \[\begin{array}{l} =2 \cdot 5 * a_{4} l \\ =2 \cdot 5\left(\frac{4 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\right) \\ =\frac{10 P^{2} \rho l^{2}}{W V_{m}^{2} \cos ^{2} \phi} \\ =\frac{2 \cdot 5}{\cos ^{2} \phi} \times \frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}} \\ =\frac{2 \cdot 5 K}{\cos ^{2} \phi} \end{array}\]

\(2\phi\), 4-wire system

system can be considered as two independent \(1\phi\) systems, each transmitting \(1/2^ *\text{total power}\)
Outer voltage (AB or CD) is twice that of \(1\phi\) 2 wire
Hence, current \(I_7\) in each conductor will be half that of \(1\phi\) 2 wire
Consequently, area of \(X-\)section of each conductor is also half but as there are 4-wires, so volume of conductor material is same as in that of \(1\phi\) 2 wire

Volume of conductor material required. \[=\frac{2 K}{\cos ^{2} \phi}\]

\(2\phi\), 3-wire system

Maxm. voltage between either outer and neutral wire \(=V_m/\sqrt{2}\)

R.M.S. voltage between outer and neutral wire \[=\frac{V_{m} / \sqrt{2}}{\sqrt{2}}=\frac{V_{m}}{2}\]

Current in each outer, \(I_{8}=\frac{P / 2}{V_{m} / 2 \cos \phi}=\frac{P}{V_{m} \cos \phi}\)

Current in neutral wire \[=\sqrt{I_{8}^{2}+I_{8}^{2}}=\sqrt{2} I_{8}\]

Assuming the current density to be constant, the area of \(\mathrm{X}\) -section of neutral wire will be \(\sqrt{2}\) times that of either of the outers.

\(\therefore \quad \text { Resistance of neutral wire } =\frac{R_{8}}{\sqrt{2}}=\frac{\rho}{\sqrt{2}} a_{8}\)

\[\begin{array}{l} \begin{aligned} \text { Line losses, } W &=2 I_{8}^{2} R_{8}+\left(\sqrt{2} I_{8}\right)^{2} \frac{R_{8}}{\sqrt{2}}=I_{8}^{2} R_{8}(2+\sqrt{2}) \\ &=\left(\frac{P}{V_{m} \cos \phi}\right)^{2} \times \rho \frac{l}{a_{8}}(2+\sqrt{2}) \\ \therefore \quad W &=\frac{P^{2} \rho l}{a_{8} V_{m}^{2} \cos ^{2} \phi}(2+\sqrt{2}) \end{aligned} \end{array}\]

\(\therefore\) Area of \(\mathrm{X}\) -section, \(a_{8}=\frac{P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}(2+\sqrt{2})\)

Volume of conductor material required \[\begin{array}{l} =2 a_{8} l+\sqrt{2} a_{8} l=a_{8} l(2+\sqrt{2}) \\ =\frac{P^{2} \rho l^{2}}{W V_{m}^{2} \cos ^{2} \phi}(2+\sqrt{2})^{2} \\ =\frac{2 \cdot 194 K}{\cos ^{2} \phi} \quad\left[\because K=\frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}}\right] \end{array}\]

\(3\phi\), 3-wire system.

Maximum voltage between each phase and neutral is \(V_m/\sqrt{3}\)
RMS voltage per phase \(=\frac{V_{m}}{\sqrt{3}} \times \frac{1}{\sqrt{2}}=\frac{V_{m}}{\sqrt{6}}\)
Power transmitted per phase \(=P/3\)

Load current / phase, \(I_{9}=\frac{P / 3}{V_{m} / \sqrt{6} \cos \phi}=\frac{\sqrt{6} P}{3 V_{m} \cos \phi}\)

Line losses \(W=3 I_{9}^{2} R_{9} = 3\left(\frac{\sqrt{6} P}{3 V_{m} \cos \phi}\right)^{2} \times \rho \frac{l}{a_{9}}\) \[\begin{aligned} W &=\frac{2 P^{2} \rho l}{a_{9} V_{m}^{2} \cos ^{2} \phi} \end{aligned}\]

Area of \(X\) -section, \(a_{9}=\frac{2 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\)
Volume of conductor material required \[\begin{array}{l} =3 a_{9} l=3\left(\frac{2 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\right) l=\frac{6 P^{2} \rho l^{2}}{W V_{m}^{2} \cos ^{2} \phi} \\ =\frac{1 \cdot 5}{\cos ^{2} \phi} \times \frac{4 P^{2} \rho l^{2}}{W V_{m}^{2}} \\ =\frac{1 \cdot 5 K}{\cos ^{2} \phi} \end{array}\]

\(3\phi\), 4-wire system

If loads are balanced, then neutral wire carries no current
system reduces to \(3\phi\), 3-wire except there is additional neutral wire

Volume of the conductor material required \[\begin{array}{l} =3 \cdot 5 a_{9} l \\ =3 \cdot 5\left(\frac{2 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\right) l \\ =\frac{7 P^{2} \rho l^{2}}{W V_{m}^{2} \cos ^{2} \phi}=\frac{7}{\cos ^{2} \phi} \times \frac{P^{2} \rho l^{2}}{W V_{m}^{2}} \\ =\frac{1 \cdot 75 K}{\cos ^{2} \phi} \end{array}\]