Question

It is desired to transmit $\mathbf{2000}\mathbf{MW}$from a power plant to a load centre located $\mathbf{300}\mathbf{}\mathbf{km}$from the plant. Determine the number of $\mathbf{60}\mathbf{Hz}$, three-phase,uncompensated transmission lines required to transmit this powerwith one line out of service for the following cases: (a)role="math" localid="1655878639508" $$\begin{gathered} \mathbf{345}\mathbf{kV}\mathbf{}\mathbf{lines}\mathbf{,}\mathbf{}{\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\mathbf{300}\mathbf{\Omega }\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{500}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{lines}\mathbf{,}\mathbf{}{\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\mathbf{275}\mathbf{\Omega }\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{765}\mathbf{kV}\mathbf{}\mathbf{lines}\mathbf{,}\mathbf{}{\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\mathbf{260}\mathbf{\Omega }\mathbf{.} \\ \mathbf{}\mathbf{assumethat}\mathbf{}{\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{,}\mathbf{}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{,}\mathbf{}{\mathbf{and\sigma }}_{\mathbf{max}}\mathbf{=}\mathbf{35}\mathbf{°}\mathbf{.} \end{gathered}$$

Short answer

(a) Therefore, the number of lines are $5$.

(b) Thenumber of lines are $3$.

(c) The number of lines are $2$.

Step-by-step solution

Write the given data.

The desired power to transmit is $P_{rated}=2000\mathrm{MW}$.

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

The sending end per-unit voltage is $V_{spu.}=1.0\mathrm{pu}$ and the receiving end per-unit voltage is $V_{\mathrm{Rp}.\mathrm{u}.}=0.95\mathrm{pu}.$.

The load angle is $\delta =35°$.

Assume that the wavelength is$\lambda =5000\mathrm{km}$.

(a) The rated voltage is $V_{rated}=345kV$and characteristics impedance is $Z_c=300\Omega$.

(b) The rated voltage is $V_{rated}=500kV$and characteristics impedance is $Z_c=275\Omega$.

(c) The rated voltage is $V_{rated}=765kV$and characteristics impedance is $Z_c=260\Omega$.

Determine the formulas of SIL, power transfer and number of lines.

Write the formula of Surge Impedance Loading (SIL).

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

$\mathbf{P}\mathbf{=}{\mathbf{V}}_{\mathbf{sp}\mathbf{.}\mathbf{u}\mathbf{.}}{\mathbf{V}}_{\mathbf{Rp}\mathbf{.}\mathbf{u}\mathbf{.}}\left(\mathrm{SIL}\right)\frac{\mathbf{sin\delta }}{\mathbf{sin}\left({\displaystyle \frac{2\mathrm{\pi l}}{\lambda }}\right)}\mathbf{w}$ ……. (2)

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

Write the formula to determine number of lines.

$\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{lines}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{rated}}}{\mathbf{P}}\mathbf{+}\mathbf{1}$ ……. (3)

Here, $P_{rated}$ is the desired power.

Determine the number of lines in (a).

(a)

Determine the surge impedance loading.

Substitute

$245\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{rated}}\mathrm{and}300\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}$in equation (1).

$$\begin{gathered} SIL=\frac{{\left(345\mathrm{kV}\right)}^2}{300\mathrm{\Omega }} \\ =396.8\mathrm{MW} \end{gathered}$$

Determine the steady state stability limit.

$\mathrm{Substitute}396.8\mathrm{MW}\mathrm{for}\mathrm{SIL},1.0\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{sp}.\mathrm{u}.}0.95\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{Rp}.\mathrm{u}},35°\mathrm{for}\mathrm{\delta },5000\mathrm{km}\mathrm{for}\mathrm{\lambda },\mathrm{and}300\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in}\mathrm{equation}\left(2\right).$

$$\begin{gathered} P=\left(1.0\mathrm{pu}\right)\left(396.8\mathrm{MW}\right)\left(0.95\mathrm{pu}\right)\frac{\sin 35°}{\sin \left[{\displaystyle \frac{2\mathrm{\pi }\left(300\mathrm{km}\right)}{5000\mathrm{km}}}\right]} \\ =\left(1.48\right)\left(396.8\mathrm{MW}\right) \\ =587.3\frac{\mathrm{MW}}{\mathrm{line}} \end{gathered}$$

Determine the number of lines in a three-phase system.

Substitute $587.3\frac{\mathrm{MW}}{\mathrm{line}}\mathrm{for}\mathrm{P}2000\mathrm{MW}\mathrm{for}{\mathrm{P}}_{\mathrm{rated}}$in equation (3).

$$\begin{gathered} \mathrm{number}\mathrm{of}\mathrm{lines}=\frac{2000\mathrm{MW}}{587.3{\displaystyle \frac{\mathrm{MW}}{\mathrm{line}}}} \\ =4.4\mathrm{lines} \\ =5\mathrm{lines} \end{gathered}$$

Therefore, the number of lines are $5$.

Determine the number of lines in (b).

(b)

Determine the surge impedance loading.

Substitute $500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{rated}}\mathrm{and}275\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}$in equation (1).

$$\begin{gathered} \mathrm{SIL}=\frac{{\left(500\mathrm{kV}\right)}^2}{275\mathrm{\Omega }} \\ =909.1\mathrm{MW} \end{gathered}$$

Determine the steady state stability limit.

localid="1655881460504" $$\begin{gathered} \mathrm{Substitute}909.1\mathrm{MW}\mathrm{for}\mathrm{SIL},1.0\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{sp}.\mathrm{u}},0.95\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{Rp}.\mathrm{u}},35°\mathrm{for}\mathrm{\delta }, \\ 5000\mathrm{km}\mathrm{for}\mathrm{\lambda }\mathrm{and}300\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in}\mathrm{equation}\left(2\right). \end{gathered}$$

$$\begin{gathered} \mathrm{P}=\left(1.0\mathrm{pu}\right)\left(909.1\mathrm{MW}\right)\left(0.95\mathrm{pu}\right)\frac{\sin 35°}{\sin \left[{\displaystyle \frac{2\mathrm{\pi }\left(300\mathrm{km}\right)}{5000\mathrm{km}}}\right]} \\ =\left(1.48\right)\left(909.1\mathrm{MW}\right) \\ =1345.6\frac{\mathrm{MW}}{\mathrm{line}} \end{gathered}$$

Determine the number of lines in a three-phase system.

Substitute $1345.6\frac{\mathrm{MW}}{\mathrm{line}}\mathrm{for}\mathrm{P}\mathrm{and}2000\mathrm{MW}\mathrm{for}{\mathrm{P}}_{\mathrm{rated}}$in equation (3).

$$\begin{gathered} \mathrm{number}\mathrm{of}\mathrm{lines}=\frac{2000\mathrm{MW}}{1345.6{\displaystyle \frac{\mathrm{MW}}{\mathrm{line}}}}+1 \\ =2.49\mathrm{lines} \\ =3\mathrm{lines} \end{gathered}$$

Therefore, the number of lines are$3$.

Determine the number of lines in (c).

(c)

Determine the surge impedance loading.

Substitute $765\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{rated}}\mathrm{and}260\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}$in equation (1).

$$\begin{gathered} \mathrm{SIL}=\frac{{\left(765\mathrm{kV}\right)}^2}{260\mathrm{\Omega }} \\ =2250.9\mathrm{MW} \end{gathered}$$

Determine the steady state stability limit.

$$\begin{gathered} \mathrm{Substitute}2250.9\mathrm{MW}\mathrm{for}\mathrm{SIL},1.0\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{sp}.\mathrm{u}},0.95\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{Rp}.\mathrm{u}},35°\mathrm{for}\mathrm{\delta }, \\ 5000\mathrm{km}\mathrm{for}\mathrm{\lambda }\mathrm{and}300\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in}\mathrm{equation}\left(2\right). \end{gathered}$$

$$\begin{gathered} \mathrm{P}=\left(1.0\mathrm{pu}\right)\left(2250.9\mathrm{MW}\right)\left(0.95\mathrm{pu}\right)\frac{\sin 35°}{\sin \left[{\displaystyle \frac{2\mathrm{\pi }\left(300\mathrm{km}\right)}{5000\mathrm{km}}}\right]} \\ =\left(1.48\right)\left(2250.9\mathrm{MW}\right) \\ =3331.3\frac{\mathrm{MW}}{\mathrm{line}} \end{gathered}$$

Determine the number of lines in a three-phase system.

$\mathrm{Substitute}3331.3\frac{\mathrm{MW}}{\mathrm{line}}\mathrm{for}\mathrm{P}\mathrm{and}2000\mathrm{MW}\mathrm{for}{\mathrm{P}}_{\mathrm{rated}}\mathrm{in}\mathrm{equation}\left(3\right).$

$$\begin{gathered} \mathrm{number}\mathrm{of}\mathrm{lines}=\frac{2000\mathrm{MW}}{333.1{\displaystyle \frac{\mathrm{MW}}{\mathrm{line}}}}+1 \\ =1.6\mathrm{lines} \\ =2\mathrm{lines} \end{gathered}$$

Therefore, the number of lines are $2$.