Question

A 40-km, 220-kV, 60-Hz, three-phase overhead transmission line has a per-phase resistance of 0.15 V/km, a per-phase inductance of 1.3263 mH/km, and negligible shunt capacitance. Using the short line model, find the sending-end voltage, voltage regulation, sending-end power, and transmission line efficiency when the line is supplying a three-phase load of (a) 381 MVA at 0.8 power factor lagging and at 220 kV and (b) 381 MVA at 0.8 power factor leading and at 220 kV.

Short answer

(a) For 381 MVA at 0.8 lagging power factor

Sending-end voltage and Voltage regulation are $144.33\angle 4.{93}^{\circ }\mathrm{kV}$and $0.136$.

Sending-end power and Transmission line efficiency are $433\angle 41.8^{\circ }\mathrm{MVA}$and 0.944.

(b) For 381 MVA at 0.8 leading power factor

Sending-end voltage and Voltage regulation are$121.39\angle 9.{29}^{\circ }\mathrm{kV}$and role="math" localid="1655446153847" $-0.0443$.

Sending-end power and Transmission line efficiency are$3618\angle -27.{58}^{\circ }\mathrm{MVA}$ and 0.944

Step-by-step solution

Given data.

Consider a$220\mathrm{kV}$ three-phase overhead transmission line of 60-Hz frequency.

Consider a per-phase resistance of three-phase overhead transmission line $0.15\mathrm{V}/\mathrm{km}$.

Consider a per-phase inductance of three-phase overhead transmission line role="math" localid="1655447250113" $1.3263\mathrm{mH}/\mathrm{km}$.

Determine the formula for calculate the lagging and leading power factor.

Write the formula of sending-end voltage.

$\overline{{\mathbf{V}}_{\mathbf{S}}}\mathbf{=}\overline{{\mathbf{V}}_{\mathbf{R}}}\mathbf{+}\overline{\sout{\mathbf{Z}}{\mathbf{I}}_{\mathbf{R}}}$ …… (1)

Here, $\overline{{\mathbf{V}}_{\mathbf{S}}}$is the sending end voltage,$\overline{{\mathbf{V}}_{\mathbf{R}}}$ is the receiving end voltage,$\overline{{\mathbf{I}}_{\mathbf{R}}}$ is the receiving end current and$\overline{\sout{\mathbf{Z}}}$ is the impedance.

Write the formula of voltage regulation.

$\mathbf{VR}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{RNL}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}$ …… (2)

Here, $\mathbf{VR}$is the voltage regulation,${\mathbf{V}}_{\mathbf{RNL}}$ is no-load receiving-end voltage and${\mathbf{V}}_{\mathbf{RFL}}$ is full-load receiving-end voltage.

Write the formula of sending end power.

role="math" localid="1655453406130" $\overline{{\mathbf{S}}_{\mathbf{S}\mathbf{\left(}\mathbf{3}\mathbf{\varphi }\mathbf{\right)}}}\mathbf{=}\mathbf{3}\overline{{\mathbf{V}}_{\mathbf{S}}}\mathbf{\times }{\mathbf{I}}_{\mathbf{S}}^{\overline{\mathbf{★}}}$ …… (3)

Here, role="math" localid="1655457144348" $\overline{\overline{{\mathbf{S}}_{\mathbf{S}\mathbf{\left(}\mathbf{3}\mathbf{\varphi }\mathbf{\right)}}}}$is the sending end voltage, $\overline{{\mathbf{V}}_{\mathbf{S}}}$is sending end voltage and $\overline{{{\mathbf{I}}_{\mathbf{S}}}^{\overline{\mathbf{★}}}}$is sending end current.

Write the formula of transmission line efficiency.

$\mathbf{\eta }\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{R}}\mathbf{}\mathbf{\left(}\mathbf{3}\mathbf{\varphi }\mathbf{\right)}}{{\mathbf{P}}_{\mathbf{S}}\mathbf{}\mathbf{\left(}\mathbf{3}\mathbf{\varphi }\mathbf{\right)}}$

Here, $\mathbf{\eta }$ is transmission efficiency,${\mathbf{P}}_{\mathbf{s}}\mathbf{\left(}\mathbf{3}\mathbf{\varphi }\mathbf{\right)}$ is sending end power and${\mathbf{P}}_{\mathbf{R}}\mathbf{\left(}\mathbf{3}\mathbf{\varphi }\mathbf{\right)}$ is receiving end power.

:( a) for lagging power factor.

Calculate the series impedance per phase.

$\overline{\sout{Z}}=(R+j\omega L)I$

Here, $\overline{\sout{Z}}$is the total impedance, R is resistance per phase, $\omega$is frequency, L is inductance per phase and $I$is length of line.

Substitute 0.15 for R ,2II (60) for, $\omega$, $1.3263\times {10}^{-3}$for L and 40 for $I$into equation (1).

$$\begin{gathered} \overline{\sout{\mathrm{Z}}}=(0.15+\mathrm{j}2\mathrm{II}(60)1.3263\times {10}^{-3} \\ =6+\mathrm{j}20\mathrm{\Omega } \end{gathered}$$

Calculate the receiving end voltage per phase

$$\begin{gathered} \overline{{\mathrm{V}}_{\mathrm{R}}}=\frac{220}{\sqrt{3}}\angle 0^{\circ } \\ =127\angle 0^{\circ }\mathrm{kV} \end{gathered}$$

Calculate the complex power at the receiving end.

$\overline{S_{R\left(3\varphi \right)}}=381\times {10}^6\angle {\cos }^{-1}0.83)$

Rewrite as,

$$\begin{gathered} \overline{\overline{{\mathrm{S}}_{\mathrm{R}\left(3\mathrm{\varphi }\right)}}}=381\times {10}^6\angle 36.{87}^{\circ } \\ =(304.8+\mathrm{j}228.6)\mathrm{MVA} \end{gathered}$$

Determine the current per phase at receiving end.

$I_R=\frac{S_{R3\varphi }^{*}}{3V_R}$

Here,$\overline{\overline{{\mathrm{S}}_{\mathrm{S}\left(3\mathrm{\varphi }\right)}}}$ is complex power at receiving-end side and${\mathrm{V}}_{\mathrm{R}}$ is receiving-end voltage.

Substitute $381\angle 36.{87}^{\circ }$for${\mathrm{S}}_{\mathrm{R}3\mathrm{\varphi }}^{★}$ and$127\angle 0^{\circ }$ for${\mathrm{V}}_{\mathrm{R}}$ into equation.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{R}}=\frac{(381\angle -36.{87}^{\circ })}{3\times 127\angle 0^{\circ }} \\ =1000\angle -36.{87}^{\circ }\mathrm{A} \end{gathered}$$

Substitute $127\angle 0^{\circ }$for $\overline{{\mathrm{V}}_{\mathrm{R}}}$, $6+\mathrm{j}20$for $\overline{\sout{\mathrm{Z}}}$and $1000\angle -36.{87}^{\circ }$for $\overline{{\mathrm{I}}_{\mathrm{R}}}$into equation (1).

role="math" localid="1655462479662" $$\begin{gathered} {\mathrm{V}}_{\mathrm{S}}=127\angle 0^{\circ }+(6+\mathrm{j}20)(1000\angle -36.87){10}^{-3} \\ =144.33\angle 4.{93}^{\circ }\mathrm{kV} \end{gathered}$$

Determine the line to line sending end voltage.

role="math" localid="1655462418852" $V_{S(L-L)}=\sqrt{3}\left(V_{SLN}\right)$

Here,$V_{SLN}$ is no-load sending-end voltage.

Substitute (144.33) for ${\mathrm{V}}_{\mathrm{SLN}}$ into equation.

role="math" localid="1655462143643" $$\begin{gathered} {\mathrm{V}}_{\mathrm{S}(\mathrm{L}-\mathrm{L})}=\sqrt{3}(144.33) \\ =250{\mathrm{kV}}_{\mathrm{LL}} \end{gathered}$$

Calculate the power at the sending end side

$\overline{{\mathrm{S}}_{\mathrm{S}\left(3\mathrm{\varphi }\right)}}=3\overline{{\mathrm{V}}_{\mathrm{S}}}\times {\mathrm{I}}_{\mathrm{S}}^{\overline{★}}$

Here,$\overline{{\mathrm{V}}_{\mathrm{S}}}$ is sending-end voltage and$\overline{{\mathrm{I}}_{\mathrm{S}}^{★}}$ is sending-end current.

Substitute$144.33\angle 4.{93}^{\circ }$ for$\overline{{\mathrm{V}}_{\mathrm{S}}}$ and$1000\angle 36.{87}^{\circ }$ forrole="math" localid="1655453038712" $\overline{{\mathrm{I}}_{\mathrm{S}}^{★}}$ into equation.

role="math" localid="1655461744773" $$\begin{gathered} \overline{{\mathrm{S}}_{\mathrm{S}\left(3\mathrm{\varphi }\right)}}=3(144.33\angle 4.{93}^{\circ })(1000\angle 36.{87}^{\circ }){10}^{-3} \\ =322.8\mathrm{MW}+\mathrm{j}288.6\mathrm{MVAR} \\ =433\angle 41.8^{\circ }\mathrm{MVA} \end{gathered}$$

Substitute 250 for $V_{RNL}$and 220 for $V_{RFL}$into equation (2).

role="math" localid="1655461486198" $$\begin{gathered} \mathrm{VR}=\frac{250-220}{220} \\ =0.136 \end{gathered}$$

Calculate the efficiency of transmission line.

$\eta =\frac{P_R\left(3\varphi \right)}{P_S\left(3\varphi \right)}$

Here, $P_{S\left(3\varphi \right)}$is sending end power and$P_{R\left(3\varphi \right)}$ is receiving end power.

Substitute 304.8 MW forrole="math" localid="1655461266160" $P_{R\left(3\varphi \right)}$ and 328.8 MW for $P_{S\left(3\varphi \right)}$.

role="math" localid="1655461212789" $$\begin{gathered} \mathrm{\eta }\%=\frac{304.8}{322.8}\times 100 \\ =94.4\% \end{gathered}$$

:( b) for leading power factor.

Therefore, the receiving end current from part (a) is,

$\overline{I_R}=1000\angle -36.{87}^{\circ }A$

Calculate the sending end voltage.

role="math" localid="1655463157879" $\overline{{\mathrm{V}}_{\mathrm{S}}}=\overline{{\mathrm{V}}_{\mathrm{R}}}+\overline{\sout{\mathrm{Z}}}\overline{{\mathrm{I}}_{\mathrm{R}}}$

Here,$\overline{{\mathrm{V}}_{\mathrm{R}}}$ is receiving end voltage,$\overline{{\mathrm{I}}_{\mathrm{R}}}$ is receiving-end current and $\overline{\sout{\mathrm{Z}}}$is total impedance.

Substitute $127\times {10}^3\angle 0^{\circ }$for $\overline{V_R}$, $6+\mathrm{j}20$for $\overline{\sout{Z}}$and $1000\angle -36.87$for $\overline{I_R}$into equation (11).

role="math" localid="1655462658011" $$\begin{gathered} \overline{{\mathrm{V}}_{\mathrm{S}}}=(127\times {10}^3\angle 0^{\circ })+(6+\mathrm{j}20)(1000\angle -36.87) \\ =121.39\angle 9.{29}^{\circ }\mathrm{kV} \end{gathered}$$

Determine the line to line sending end voltage.

$V_{S(L-L)}=\sqrt{3}\left(V_{SLN}\right)$

Here, $V_{S(L-L)}$ is the sending end line to line voltage.

Substitute $121.39\times {10}^3$ for$V_{SLN}$ into equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{S}(\mathrm{L}-\mathrm{L})}=\sqrt{3}\times 121.39\times {10}^3 \\ =210.26\mathrm{kV} \end{gathered}$$

Substitute $121.39\angle 9.{29}^{\circ }$for $\overline{V_S}$ and $1\angle -36.{87}^{\circ }$for $I_S^{*}$into equation (3).

role="math" localid="1655461971569" $$\begin{gathered} {\mathrm{S}}_{\mathrm{S}\left(3\mathrm{\varphi }\right)}=3(121.39\angle 9.{29}^{\circ })(1\angle 36.{87}^{\circ }) \\ =322.8\mathrm{MW}+\mathrm{j}168.6\mathrm{Mvar} \\ =3618\angle -27.{58}^{\circ }\mathrm{MVA} \end{gathered}$$

Substitute 210.26 for$V_{RNL}$ and 220 for$V_{RFL}$ into equation (2).

$$\begin{gathered} \mathrm{VR}=\frac{210.26-220}{220} \\ =0.0443 \end{gathered}$$

Determine the transmission line efficiency.

$\eta =\frac{P_R\left(3\varphi \right)}{P_S\left(3\varphi \right)}$

Here, $P_S\left(3\varphi \right)$is sending end power and $P_S\left(3\varphi \right)$is receiving end power.

Substitute 304.8 MW for $P_{R\left(3\varphi \right)}$and 328.8 MW for $P_{S\left(3\varphi \right)}$.

$$\begin{gathered} \mathrm{\eta }\%=\frac{304.8}{322.8}\times 100 \\ =94.4\% \end{gathered}$$