Question

A three-phase power of is transmitted to a substation located $\mathbf{500}\mathbf{}\mathbf{km}$from the source of power. With ${\mathbf{V}}_{\mathbf{S}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{,}\mathbf{}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{.}$$\mathbf{\lambda }\mathbf{=}\mathbf{5000}\mathbf{}\mathbf{km}\mathbf{,}\mathbf{}{\mathbf{Z}}_{\mathbf{C}}\mathbf{=}\mathbf{500}\mathbf{}\mathbf{\Omega }$and $\mathbf{\delta }\mathbf{=}\mathbf{36}\mathbf{.}{\mathbf{87}}^{\mathbf{°}}$, determine a nominal voltage level for the lossless transmission line based on Eq. (5.4.29) of the text. Using this result, find the theoretical three-phase maximum power that can be transferred by the lossless transmission line.

Short answer

The nominal voltage is $500.32\mathrm{kV}$ and maximum theoretical power is$459.92\mathrm{MW}.$

Step-by-step solution

Write the given data from the question.

$$\begin{gathered} \mathrm{The}\mathrm{delivered}\mathrm{power},\mathrm{P}=460\mathrm{MW} \\ \mathrm{The}\mathrm{sending}\mathrm{end}\mathrm{per}\mathrm{unit}\mathrm{voltage},{\mathrm{V}}_{\mathrm{S}.\mathrm{pu}}=1\mathrm{pu} \\ \mathrm{The}\mathrm{receiving}\mathrm{end}\mathrm{per}\mathrm{unit}\mathrm{voltage},{\mathrm{v}}_{\mathrm{R}.\mathrm{pu}}=0.9\mathrm{pu} \\ \mathrm{The}\mathrm{wavelength},\mathrm{\lambda }=5000\mathrm{km} \\ \mathrm{Characteristics}\mathrm{impedance},{\mathrm{Z}}_{\mathrm{C}}=500\mathrm{\Omega } \\ \mathrm{The}\mathrm{angle},\mathrm{\delta }=36.{87}^{°} \\ \mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{transmission}\mathrm{line},\mathrm{l}=500\mathrm{km} \end{gathered}$$

Determine the formulas to calculate the nominal voltage and theoretical three phase maximum power.

The equation to calculate the value of electric length is given as follows.

$\mathbf{\beta l}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{\lambda }}\mathbf{l}$ …… (1)

The equation to calculate the real power delivered is given as follows.

$\mathbf{P}\mathbf{=}{\mathbf{V}}_{\mathbf{s}\mathbf{.}\mathbf{pu}}{\mathbf{V}}_{\mathbf{R}\mathbf{,}\mathbf{pu}}\left(\mathrm{SIL}\right)\left(\frac{\mathrm{sin\delta }}{\sin \left(\mathrm{\beta l}\right)}\right)$ …… (2)

The equation to calculate the surge impedance is given as follows.

$\mathit{S}\mathit{I}\mathit{L}\mathbf{=}\frac{{\mathbf{\left(}{\mathbf{V}}_{\mathbf{L}\mathbf{,}\mathbf{B}\mathbf{a}\mathbf{s}\mathbf{e}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{C}}}$ ....(3)

The equation to calculate the series reactance is given as follows.

${\mathbf{X}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{Z}}_{\mathbf{C}}\mathbf{sin}\mathbf{\left(}\mathbf{\beta l}\mathbf{\right)}$ ....(4)

The equation to calculate the theoretical power is given as follows.

${\mathbf{P}}_{\mathbf{max}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}\mathbf{sin}\mathbf{\left(}\mathbf{\delta }\mathbf{\right)}$ …(5)

Calculate the nominal voltage and theoretical three phase maximum power.

Calculate the electric length of transmission line.

$$\begin{gathered} \mathrm{Substitute}500\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{and}500\mathrm{km}\mathrm{\lambda }\mathrm{into}\mathrm{equation}\left(1\right). \\ \mathrm{\beta l}=\frac{360}{500\times {10}^3}\times 500\times {10}^3 \\ \mathrm{\beta l}={36}^{°} \\ \mathrm{Calculate}\mathrm{the}\mathrm{Surge}\mathrm{impedance}\mathrm{loading}. \\ \mathrm{Substitute}1\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{s}.\mathrm{pu}},460\mathrm{MW}\mathrm{for},\mathrm{P},36.{87}^{°}\mathrm{for}\mathrm{\delta }\mathrm{into}\mathrm{equation}\left(2\right). \\ 460\times {10}^6=1\times 0.9\times \mathrm{SIL}\times \left(\frac{\sin 36.87}{\sin 36}\right) \\ 460\times {10}^6=0.9\times \mathrm{SIL}\times 1.020 \\ \mathrm{SIL}=\frac{460\times {10}^6}{0.9187} \\ \mathrm{SIL}=500.70\mathrm{MW} \\ \mathrm{Calculate}\mathrm{the}\mathrm{nominal}\mathrm{voltage}\mathrm{for}\mathrm{transmission}\mathrm{line}. \\ \mathrm{Substitute}500.70\mathrm{MW}\mathrm{for}\mathrm{SIL}\mathrm{and}500\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{into}\mathrm{equation}\left(3\right). \\ 500.70\times {10}^6=\frac{{\left({\mathrm{V}}_{\mathrm{L}.\mathrm{Base}}\right)}^2}{500} \\ {\left({\mathrm{V}}_{\mathrm{L}.\mathrm{Base}}\right)}^2=500.65\times {10}^6\times 500 \\ {\left({\mathrm{V}}_{\mathrm{L}.\mathrm{Base}}\right)}^2=\sqrt{250325\times {10}^6} \\ {\left({\mathrm{V}}_{\mathrm{L}.\mathrm{Base}}\right)}^2=500.32\mathrm{kV} \end{gathered}$$

Calculate the series reactance.

$$\begin{gathered} \mathrm{Substitute}500\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{and}{36}^{°}\mathrm{for}\mathrm{\beta l}\mathrm{into}\mathrm{equation}\left(4\right). \\ {\mathrm{X}}^{\text{'}}=500\times \sin \left(36\right) \\ {\mathrm{X}}^{\text{'}}=293.9\mathrm{\Omega } \\ \mathrm{Calculate}\mathrm{the}\mathrm{theoretical}\mathrm{maximum}\mathrm{power}. \\ \mathrm{Substitute}500.32\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{s}}\mathrm{and}0.9\times 500.32\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},293.9\mathrm{\Omega }\mathrm{and}{36}^{°} \\ \mathrm{for}\mathrm{\beta l}\mathrm{into}\mathrm{equation}\left(5\right). \\ {\mathrm{P}}_{\max }=\frac{500.32\times 0.9\times 500.32}{293.9}\sin 36 \\ {\mathrm{P}}_{\max }=\frac{135173.177}{293.9} \\ {\mathrm{P}}_{\max }=459.92\mathrm{MW} \\ \mathrm{Hence}\mathrm{the}\mathrm{nominal}\mathrm{voltage}\mathrm{is}500.32\mathrm{kV}\mathrm{and}\mathrm{maximum}\mathrm{theoretical}\mathrm{power}\mathrm{is} \\ 459.92\mathrm{MW}. \end{gathered}$$