Question

Determine the practical line loadability inMWand in per-unit ofSILfor the line in Problem 5.14 if the line length is (a)200 kmor (b) 600 km. Assume${\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{pu}\mathbf{,}\mathbf{}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{pu}\mathbf{,}\mathbf{}{\mathbf{\delta }}_{\mathbf{max}}\mathbf{=}\mathbf{35}\mathbf{°}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{99}$leading power factor at the receiving end.

Short answer

(a) Therefore, the practical loadability is 1499 MW .

(b) Therefore, the practical loadability is 694.2 MW .

Step-by-step solution

Given data.

The line length is(a) $I=200\mathrm{km}$ (b) $I=600\mathrm{km}$.

The sending end per-unit voltage is $V_s=1.0\mathrm{pu}$and the receiving end per-unit voltage is $V_R=0.95\mathrm{pu}$.

The load angle is $\delta =35°$.

The power factor is $pf=0.99$.

The given data from problem 5.14 is,

For $I=200\mathrm{km}$, the transmission line parameters are $A=0.9694\angle 0.154°\mathrm{pu}$and $B=69.54\angle 85.15°\mathrm{\Omega }$.

From B parameters, $Z\text{'}=69.54\Omega ,{\theta }_2=85.15°$and $A=0.9694,{\theta }_A=0.154°$

For $I=600\mathrm{km}$, the transmission line parameters are $A=0.7356\angle 1.685°\mathrm{pu}$and $B=191.8\angle 85.57°\Omega$.

FromB parameters, $Z\text{'}=191.8\Omega ,{\theta }_z=85.57°$and$A=0.7356,{\theta }_A=1.685°$.

The rated voltage is $V_{rated}=500\mathrm{kV}$which is taken as base.

Determine the formulas of real power, load current, sending end voltage and loadability.

Write the formula of real power delivered to the receiving end.

${\mathbf{P}}_{\mathbf{R}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{R}}{\mathbf{V}}_{\mathbf{S}}}{\mathbf{Z}\mathbf{\text{'}}}\mathbf{cos}\left({\theta }_2-\delta \right)\mathbf{-}\frac{{\mathbf{AV}}_{\mathbf{R}}^{\mathbf{2}}}{\mathbf{Z}}\mathbf{cos}\left({\theta }_Z-{\theta }_A\right)$ ……. (1)

Here, $Z\text{'}$is a transmission line impedance, role="math" localid="1655291492007" $V_S$is a sending end voltage, $V_R$is a receiving end voltage, $\delta$is a load angle, ${\theta }_Z$is angle for line impedance, ${\theta }_A$is angle for transmission line parameter $A$.

Write the formula of full-load receiving end current.

${\mathbf{I}}_{\mathbf{RFL}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{R}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{RFL}}\left(\mathrm{pf}\right)}$……. (2)

Here, $pf$is a power factor and $V_{RFL}$is a receiving end voltage.

Write the formula of sending end voltage.

${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{RFL}}\mathbf{+}{\mathbf{BI}}_{\mathbf{RFL}}$ ……. (3)

Here, $A,B$are the transmission line parameters.

Write the formula of practical loadability.

${\mathbf{P}}_{\mathbf{R}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{RFL}}{\mathbf{I}}_{\mathbf{RFL}}\left(\mathrm{pf}\right)$ ……. (4)

Determine the loadability of the system.

(a)

Determine the real power.

Substitute $85.15°\mathrm{for}{\theta }_z,69.54\mathrm{\Omega }\mathrm{for}Z\text{'},1.0\mathrm{pu}\mathrm{for}{V}_S,0.95\mathrm{pu}$for $V_R,35°$for localid="1655292065913" $\delta ,0.9694$for $A,0.154°\mathrm{for}{\theta }_A$in equation (1).

$P_R=\frac{\left(0.95\times 500\right)\left(1.0\times 500\right)}{69.54}\cos \left(85.15°‐35°\right)‐\frac{0.9694{\left(0.95\times 500\right)}^2}{69.54}\cos \left(85.15°‐0.154°\right)$

Determine the full load current at receiving end.

Substitute 1914 MW for ${\mathrm{P}}_{\mathrm{R}}=0.95\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{R},}0.99\mathrm{for}\mathrm{pf}$ in equation (2).

$$\begin{gathered} {\mathrm{l}}_{\mathrm{RFL}}=\frac{1914\mathrm{MW}}{\sqrt{3}(0.95\times 500)(0.99)} \\ =2.35\mathrm{kA} \end{gathered}$$

Determine the receiving end voltage.

Substitute role="math" localid="1655353311115" $0.9694\angle 0.{154}^0$ for A ,role="math" localid="1655353351026" $69.54\angle 85.{15}^0\mathrm{\Omega }$ for B , $2.35\angle 8.{109}^0\mathrm{kA}$ for${\mathrm{l}}_{\mathrm{RFL}}$ and role="math" localid="1655353525402" $\frac{500}{\sqrt{3}}\angle {\mathrm{\delta }}^0\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{S}}$in equation (3).

role="math" localid="1655353766213" $\frac{500}{\sqrt{3}}\angle {\mathrm{\delta }}^0=(0.9694\angle 0.{154}^0)({\mathrm{V}}_{\mathrm{RFL}}\angle 0^0)+(69.54\angle 85.{15}^0\mathrm{\Omega })(2.35\angle 8.{109}^0\mathrm{kA})$

From solving above equation,

${\mathrm{V}}_{\mathrm{RFL}}=0.8825\mathrm{pu}$

Here,

$$\begin{gathered} \left|\frac{{\mathrm{V}}_{\mathrm{RFL}}}{{\mathrm{V}}_{\mathrm{S}}}\right|=\left|\frac{0.8825\mathrm{pu}}{1\mathrm{pu}}\right| \\ =0.8825\mathrm{pu} \\ =\angle 0.95\mathrm{pu} \end{gathered}$$

Therefore, voltage drop limit not satisfied.

Determine the receiving end current.

Substitute $0.9694\angle 0.{154}^0$ for A , $69.54\angle 85.{15}^0\mathrm{\Omega }$for B , ${\mathrm{l}}_{\mathrm{RFL}}\angle 8.{109}^0\mathrm{kA}$ for ${\mathrm{l}}_{\mathrm{RFL}}$ and $\frac{500}{\sqrt{3}}\angle {\mathrm{\delta }}^0\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{S}}$, $\frac{0.95\times 500}{\sqrt{3}}\angle 0^0\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{RFL}}$ in equation (3).

$\frac{500}{\sqrt{3}}\angle {\mathrm{\delta }}^0=(0.9694\angle 0.{154}^0)\left(\frac{0.95\times 500}{\sqrt{3}}\angle 0^{0}kV\right)+(69.54\angle 85.{15}^0\mathrm{\Omega })(l_{RFL}\angle 8.{109}^0\mathrm{kA})$

From solving above equation,

${\mathrm{l}}_{\mathrm{RFL}}=1.84\mathrm{kA}$

Determine the power.

Substitute 0.99 for pf , 1.84 kA for ${\mathrm{l}}_{\mathrm{RFL}}$, $(0.95\times 500)\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{RFL}}$in equation (4).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{R}}=\sqrt{3}\left(\right(0.95\times 500\left)\mathrm{kV}\right)(1.84\mathrm{kA})(0.99) \\ =1499\mathrm{MW} \end{gathered}$$

Therefore, the practical loadability is 1499 MW .

Determine the loadability of the system.

(b)

Determine the real power.

Substitute $85.{57}^0$for localid="1655355215482" ${\mathrm{\theta }}_{2,}191.8\mathrm{\Omega }$for ${\mathrm{Z}}^{\text{'}}{}_{,}1.0\mathrm{pu}$for ${\mathrm{V}}_{\mathrm{S},}0.95\mathrm{pu}$for ${\mathrm{V}}_{\mathrm{R},}{35}^0$for $\mathrm{\delta }{}_{,}0.7356$for A ,$1.{685}^0$for ${\mathrm{\theta }}_{\mathrm{A}}$in equation (1).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{R}}=\left\{\begin{array}{c}\frac{(0.95\times 500)(1.0\times 500)}{191.8}\cos \left(85.{57}^0-{35}^0\right)\\ -\frac{0.7356{(0.95\times 500)}^2}{191.8}\cos \left(85.{57}^0-1.{685}^0\right)\end{array}\right\} \\ =694.2\mathrm{MW} \end{gathered}$$

Therefore, the practical loadability is$694.2\mathrm{MW}$.