Question

A three-phase power of$\mathbf{3600}\mathbf{}\mathbf{MW}$is to be transmitted through four identical$\mathbf{60}\mathbf{}\mathbf{Hz}$overhead transmission lines over a distance of$\mathbf{300}\mathbf{}\mathbf{km}$. Based on a preliminary design, the phase constant and surge impedance of the line are$\mathbf{\beta }\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{46}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\frac{\mathbf{rad}}{\mathbf{km}}$and${\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\mathbf{343}\mathbf{}\mathbf{\Omega }$, respectively. Assuming${\mathbf{V}}_{\mathbf{G}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{pu}\mathbf{}\mathbf{,}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{pu}$and a power angle${\mathbf{\delta }}_{\mathbf{max}}\mathbf{=}\mathbf{36}\mathbf{.}\mathbf{87}\mathbf{°}$, determine a suitable nominal voltage level in$\mathbf{kV}$based on the practical line-loadability criteria.

Short answer

Therefore, the nominal voltage is $400\mathrm{kV}$.

Step-by-step solution

Write the given data.

The desired power to transmit is$P=3600\mathrm{MW}$over four overhead transmission lines.

The line length is$I=300\mathrm{km}$.

The phase constant is$\beta =9.45\times {10}^{-4}\frac{\mathrm{rad}}{\mathrm{km}}$.

The sending end per-unit voltage is$V_{Sp.u.}=1.0\mathrm{pu}$and the receiving end per-unit voltage is$V_{Rp.u.}=0.9\mathrm{pu}$.

The load angle is$\delta =36.87°$.

The characteristics impedance is$Z_c=343\Omega$.

Determine the formulas of SIL and power transfer.

Write the formula of Surge Impedance Loading (SIL).

$\mathbf{SIL}\mathbf{=}\frac{{{\mathbf{V}}^{\mathbf{2}}}_{\mathbf{rated}}}{{\mathbf{Z}}_{\mathbf{c}}}$ ……. (1)

Here,$Z_c$is characteristics impedance and$V_{rated}$is line voltage.

Write the formula of steady state stability limit in terms of SIL.

$\mathbf{P}\mathbf{=}{\mathbf{V}}_{\mathbf{Sp}\mathbf{.}\mathbf{u}\mathbf{.}}{\mathbf{V}}_{\mathbf{Rp}\mathbf{.}\mathbf{u}\mathbf{.}}\left(\mathrm{SIL}\right)\frac{\mathbf{sin\delta }}{\mathbf{sin}\left(\beta 1\right)}\mathbf{W}$ ……. (2)

Here, ${\mathrm{V}}_{\mathrm{Sp}.\mathrm{u}.}$is a per-unit sending end voltage,$V_{Rp.u.}$is a receiving end per-unit voltage,$\delta$ is a load angle,$I$ is length of the line and $\lambda$ is wavelength.

Determine the SIL.

The three line power is$P=3600\mathrm{MW}$, so the single line power will be,

$$\begin{gathered} P=\frac{3600\mathrm{MW}}{3} \\ =900\mathrm{MW} \end{gathered}$$

Determine the surge impedance loading.

Substituterole="math" localid="1655888542564" $900\mathrm{MW}\mathrm{for}P,1.0\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{Sp}.\mathrm{u}.,}36.87°\mathrm{for}\mathrm{\delta },9.46\times {10}^{-4}\frac{\mathrm{rad}}{\mathrm{km}}\mathrm{for}\mathrm{\beta }\mathrm{and}300\mathrm{km}\mathrm{for}I$in equation (2).

$900\mathrm{MW}=\left(1.0\mathrm{pu}\right)\left(0.9\mathrm{pu}\right)\left(\mathrm{SIL}\right)\frac{\sin 36.87°}{\sin \left[\left(9.46\times {10}^{-4}{\displaystyle \frac{\mathrm{rad}}{\mathrm{km}}}\right)\left(300\mathrm{km}\right)\right]}$

Determine the rated voltage..

Substitute$466.66\mathrm{MW}\mathrm{for}\mathrm{SIL}\mathrm{and}343\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}$in equation (1).

$$\begin{gathered} 466.66\mathrm{MW}=\frac{{\left({\mathrm{V}}_{\mathrm{rated}}\right)}^2}{343\mathrm{\Omega }} \\ {\mathrm{V}}_{\mathrm{rated}}=\sqrt{\left(466.66\mathrm{MW}\right)\left(343\mathrm{\Omega }\right)} \\ {\mathrm{V}}_{\mathrm{rated}}=400\mathrm{kV} \end{gathered}$$

Therefore, the rated voltage is$400\mathrm{kV}$.