Question

The line in Problem 5.14 has three ACSR $\mathbf{1113}\mathbf{kcmil}$conductors per phase. Calculate the theoretical maximum real power that this line can deliver and compare with the thermal limit of the line. Assume ${\mathbf{V}}_{\mathbf{s}}\mathbf{=}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$per unit and unity power factor at the receiving end.

Short answer

The theoretical practical real power is $1440.44\mathrm{MW}$and the current $1.662\mathrm{kA}$corresponding to theoretical is less than the thermal limit.

Step-by-step solution

Write the given data by the question.

The length of the transmission line, $\mathrm{l}=500\mathrm{km}$

The positive sequence series impedance,$\mathrm{z}=0.03+\mathrm{j}0.35\frac{\mathrm{\Omega }}{\mathrm{km}}$

The positive sequence shunt admittance,$\mathrm{y}=\mathrm{j}4.4\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}$

Per unit sending end voltage,role="math" localid="1655441097049" ${\mathrm{V}}_{\mathrm{s}}=1\mathrm{pu}$

Per unit receiving end voltage, ${\mathrm{V}}_{\mathrm{R}}=1\mathrm{pu}$

The sending ending base voltage,${\mathrm{V}}_{\mathrm{S}.\mathrm{Base}}=500\mathrm{kV}$

Receiving end power factor,$\mathrm{pf}=1$

Determine the formulas to calculate the theoretical maximum real power.

The equation to calculate the propagation constant is given as follows,

$\mathrm{Vl}=\mathrm{l}\sqrt{\mathrm{zy}}$ …… (1)

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

${\mathrm{Z}}_{\mathrm{c}}=\sqrt{\frac{\mathrm{z}}{\mathrm{y}}}$ …… (2)

The equation to calculate the ABCD parameter of the transmission line is given as follows.

localid="1655441993511" $$\begin{gathered} \mathrm{A}=\mathrm{D} \\ \mathrm{A}=\cosh \left(\mathrm{\gamma l}\right)\dots \dots \left(3a\right) \\ \mathrm{B}={\mathrm{Z}}_{\mathrm{C}}\sinh \left(\mathrm{\gamma l}\right)\dots \dots \left(3b\right) \\ \mathrm{C}=\frac{1}{{\mathrm{Z}}_{\mathrm{C}}}\sinh \left(\mathrm{\gamma l}\right)\dots \dots \left(3c\right) \end{gathered}$$

The equation to calculate the per unit reactance is given as follows.

${\mathrm{X}}_{\mathrm{pu}}=\frac{{\mathrm{X}}_{\mathrm{actual}}}{{\mathrm{X}}_{\mathrm{Base}}}$ …… (4)

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

${\mathrm{P}}_{\mathrm{R},\max }=\frac{{\mathrm{V}}_{\mathrm{R}}{\mathrm{V}}_{\mathrm{S}}}{\mathrm{B}}-\frac{{\mathrm{AV}}_{\mathrm{S}}^2}{\mathrm{B}}\cos \left({\mathrm{\theta }}_{\mathrm{B}}-{\mathrm{\theta }}_{\mathrm{A}}\right)$ .....(5)

The equation to calculate the receiving current is given as follows.

${\mathrm{l}}_{\mathrm{R}}=\frac{{\mathrm{P}}_{\mathrm{R}.\max }}{\sqrt{3}{\mathrm{V}}_{\mathrm{R}}\left(\mathrm{p}.\mathrm{f}\right)}\angle -{\cos }^{-1}\left(\mathrm{p}.\mathrm{f}\right)$ …… (6)

Calculate the theoretical real power.

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{\gamma l}. \\ \mathrm{Substitute}500\mathrm{km}\mathrm{for}\mathrm{l},0.03+\mathrm{j}0.35\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{Z}\mathrm{and}\mathrm{j}4.4\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y}\mathrm{into}\mathrm{equation} \\ \left(1\right). \\ \mathrm{\gamma l}=500\sqrt{\left(0.03+\mathrm{j}0.35\right)\left(\mathrm{j}4.4\times {10}^{-6}\right)} \\ \mathrm{\gamma l}=500\sqrt{1.54\times {10}^{-6}\angle 175.1^{°}} \\ \mathrm{\gamma l}=500\times 1.24\times {10}^{-3}\angle 87.{55}^{°} \\ \mathrm{\gamma l}=0.620\angle 87.{55}^{°} \\ \mathrm{Convert}\mathrm{into}\mathrm{cartesian}\mathrm{form}. \\ \mathrm{\gamma l}=0.0265+\mathrm{j}0.619\mathrm{pu} \\ \mathrm{Calculate}\mathrm{the}\mathrm{characteristics}\mathrm{impedance}. \\ \mathrm{Substitute}0.03+\mathrm{j}0.35\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{Z}\mathrm{and}\mathrm{j}4.4\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y}\mathrm{into}\mathrm{equation}\left(2\right). \\ {\mathrm{Z}}_{\mathrm{C}}=\sqrt{\frac{0.03+\mathrm{j}0.35}{\mathrm{j}4.4\times {10}^{-6}}} \\ {\mathrm{Z}}_{\mathrm{C}}=\sqrt{79.84\times {10}^3\angle -4.89°} \\ {\mathrm{Z}}_{\mathrm{C}}=282.6\angle -2.45° \end{gathered}$$

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{value}\mathrm{of}\sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right), \\ \sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=\frac{{\mathrm{e}}^{0.0265+\mathrm{j}0.619}-{\mathrm{e}}^{0.0265+\mathrm{j}0.619}}{2} \\ \sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=\frac{1.026\angle 0.619-0.9738\angle -6.19}{2} \\ \sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=\frac{1.162\angle 1.534}{2}\mathrm{radian} \\ \sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=0.581\angle 87.89° \end{gathered}$$

$$\begin{gathered} \mathrm{Calculate}\cosh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right). \\ \cosh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=\frac{{\mathrm{e}}^{0.0265+\mathrm{j}0.619}+{\mathrm{e}}^{-0.0265-\mathrm{j}0.619}}{2} \\ \cosh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=\frac{1.026\angle 0.619+0.9738\angle -0.619}{2} \\ \cosh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)\frac{1.63\angle 0.019}{2}\mathrm{radian} \\ \cosh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)=0.815\angle 1.089° \end{gathered}$$

Calculate the ABCD parameter of the transmission line.

Calculate the A and D parameter,

$$\begin{gathered} \mathrm{Substitute}0.815\angle 1.089°\mathrm{for}\cosh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)\mathrm{into}\mathrm{equation}\left(3\mathrm{a}\right). \\ \mathrm{A}=\mathrm{D} \\ \mathrm{A}=0.815\angle 1.089°\mathrm{pu} \end{gathered}$$

Calculate B parameter,

$$\begin{gathered} \mathrm{Substitute}282.6\angle -2.45°\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{and}0.581\angle 87.89°\mathrm{for}\sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)\mathrm{into}\mathrm{equation} \\ \left(3\mathrm{b}\right). \\ \mathrm{B}={\mathrm{Z}}^{\text{'}} \\ \mathrm{B}=282.6\angle -2.45°\left(0.581\angle 87.89°\right) \\ \mathrm{B}=164.19\angle 85.44°\mathrm{\Omega } \end{gathered}$$

Calculate the C parameter,

$$\begin{gathered} \mathrm{Substitute}282.6\angle -2.45°\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{and}0.581\angle 87.89°\mathrm{for}\sinh \left(\overrightarrow{\mathrm{\gamma }}\mathrm{l}\right)\mathrm{into}\mathrm{equation}\left(3\mathrm{c}\right) \\ \mathrm{C}=\frac{0.581\angle 87.89°}{282.6\angle -2.45°} \\ \mathrm{C}=2.056\angle 90.34°\mathrm{mS} \end{gathered}$$

Calculate the theoretical line load ability.

$$\begin{gathered} \mathrm{Substitute}500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{s}},500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{s}},85.44°\mathrm{for}{\mathrm{\theta }}_{\mathrm{B}}\mathrm{and}1.089°\mathrm{for}{\mathrm{\theta }}_{\mathrm{A}}\mathrm{into} \\ \mathrm{equation}\left(5\right). \\ {\mathrm{P}}_{\mathrm{R}.\max }=\frac{500\times 500}{164.19}-\frac{0.815\times {500}^2}{164.19}\cos {\left(85.44-1.089\right)}^{°} \\ {\mathrm{P}}_{\mathrm{R}.\max }=1522.62-122.15 \\ {\mathrm{P}}_{\mathrm{R}.\max }=1440.44\mathrm{MW} \end{gathered}$$

Calculate the receiving end current.

$$\begin{gathered} \mathrm{Substitute}1440.44\mathrm{MW}\mathrm{for}{\mathrm{P}}_{\mathrm{R}.\max },500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}\mathrm{AND}1\mathrm{for}\mathrm{PF}\mathrm{into}\mathrm{equation}\left(6\right). \\ {\mathrm{l}}_{\mathrm{R}}=\frac{14440.44\times {10}^6}{\sqrt{3}\times 500\times {10}^3\left(1\right)}\angle {\cos }^{-1}\left(1\right) \\ {\mathrm{l}}_{\mathrm{R}}=\frac{1440.44\times {10}^3}{866.02}\angle 0° \\ {\mathrm{l}}_{\mathrm{R}}=1.663\mathrm{kA} \end{gathered}$$

From Table A.4, the thermal limit for 3 ACSR $1113\mathrm{kcmil}$conductors is$3.33kA$

Therefore, the theoretical real power is 1440.44mw and the current 1.662kA corresponding to theoretical is less than the thermal limit.