Underground Systems

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

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

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

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

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

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,

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

• \(\text { Load current, } I_{4} =\frac{P}{V_{m} / \sqrt{2} \cos \phi}=\frac{\sqrt{2} P}{V_{m} \cos \phi}\)

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

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

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

• 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, \(a_{5}=\frac{4 P^{2} \rho l}{W V_{m}^{2} \cos ^{2} \phi}\)

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

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

• 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

\[=\frac{2 K}{\cos ^{2} \phi}\]

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

\[=\frac{V_{m} / \sqrt{2}}{\sqrt{2}}=\frac{V_{m}}{2}\]

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

\[=\sqrt{I_{8}^{2}+I_{8}^{2}}=\sqrt{2} I_{8}\]

Current in neutral wire

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.

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

• \[\begin{aligned} W &=\frac{2 P^{2} \rho l}{a_{9} V_{m}^{2} \cos ^{2} \phi} \end{aligned}\]
  Line losses

• If loads are balanced, then neutral wire carries no current

• system reduces to \(3\phi\), 3-wire except there is additional neutral wire

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