Question

For the $\mathbf{500}\mathbf{}\mathbf{KV}$line given in Problem 5.16, (a) calculate the theoretical maximum real power that the line can deliver to the receiving end when rated voltage is applied to both ends; (b) calculate the receiving-end reactive power and power factor at this theoretical loading.

Short answer

(a) The theoretical maximum real power is $2397.5\mathrm{M}.\mathrm{W}$

(b) The receiving end reactive power is $-147.03\mathrm{MW}$and power factor is$0.99$

Step-by-step solution

Write the given data from the question:

The voltage,$\mathrm{V}=500\mathrm{kV}$

Frequency, $\mathrm{f}=60\mathrm{Hz}$

Length,$\mathrm{l}=300\mathrm{km}$

Conductor temperature,$T={50}^{°}C$

The result of the problem 5.16,

(a) The characteristics impedance is .$264.4\angle -1.7°\mathrm{\Omega }$

(b) The quantity $yl$for $300\mathrm{km}$length is .$0.0113+\mathrm{j}0.381\mathrm{pu}.$

(c) The ABCD parameter for the line is,

$$\begin{gathered} \mathrm{A}=0.9285\angle 0.285°\mathrm{pu} \\ \mathrm{B}=98.252\angle 86.69°\mathrm{\Omega } \\ \mathrm{C}=1.408\times {10}^{-3}\angle 90.09°\mathrm{S} \\ \mathrm{D}=0.9285\angle 0.285°\mathrm{pu} \end{gathered}$$

Determine the formulas to calculate the theoretical maximum real power, received end reactive power and power factor.    

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

${\mathrm{P}}_{\max }=\frac{{\mathrm{V}}_{\mathrm{R}}{\mathrm{V}}_{\mathrm{S}}}{{\mathrm{Z}}^{\text{'}}}=\frac{{{\mathrm{AV}}^2}_{\mathrm{R}}}{{\mathrm{Z}}^{\text{'}}}\cos \left({\mathrm{\theta }}_{\mathrm{z}}-{\mathrm{\theta }}_{\mathrm{A}}\right)$ …… (1)

The equation to calculate the receiving end reactive power is given as follows.

${\mathrm{Q}}_{\mathrm{R}}=-\frac{{\mathrm{AV}}_{\mathrm{R}}^2}{\mathrm{Z}}\cos \left({\mathrm{\theta }}_{\mathrm{Z}}-{\mathrm{\theta }}_{\mathrm{A}}\right)$ …… (2)

The equation to calculate the receiving end power factor is given as follows.

$\mathrm{pf}=\mathrm{c}\mathrm{os}\left[{\tan }^{-1}\left(\frac{{\mathrm{Q}}_{\mathrm{R}}}{{\mathrm{P}}_{\mathrm{R}}}\right)\right]$ …… (3)

Calculate the theoretical maximum real power.

(a)

Calculate the theoretical maximum real power.

$$\begin{gathered} \mathrm{Substitute}500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{S}},0.9285\mathrm{for}\mathrm{A},98.25\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}^{\text{'}},86.69° \\ \mathrm{for}{\mathrm{\theta }}_{\mathrm{Z}}\mathrm{and}0.258°\mathrm{for}{\mathrm{\theta }}_{\mathrm{A}}\mathrm{into}\mathrm{equation}\left(1\right). \\ {\mathrm{P}}_{\max }=\frac{500\times {10}^3\times 500\times {10}^3}{98.25}-\frac{0.925\times 500\times {10}^3}{98.25}\cos \left(86.69-0.258\right) \\ {\mathrm{P}}_{\max }=2544.5\times {10}^6-147\times {10}^6 \\ {\mathrm{P}}_{\max }=2397.5\mathrm{MW} \\ \mathrm{Hence}\mathrm{the}\mathrm{theoretical}\mathrm{maximum}\mathrm{real}\mathrm{power}\mathrm{is}2397.5\mathrm{MW}. \end{gathered}$$

Calculate the receiving end reactive and power factor.

(b)

Calculate the receiving end reactive power.

$$\begin{gathered} \mathrm{Substitute}500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},500\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{S}},0.9285\mathrm{for}\mathrm{A},98.25\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}^{\text{'}},86.69° \\ \mathrm{for}{\mathrm{\theta }}_{\mathrm{Z}}\mathrm{and}0.258°\mathrm{for}{\mathrm{\theta }}_{\mathrm{A}}\mathrm{into}\mathrm{equation}\left(1\right). \end{gathered}$$

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{R}}=-\frac{0.9285\times {\left(500\times {10}^3\right)}^2}{98.25}\cos \left(86.69-0.258\right) \\ {\mathrm{Q}}_{\mathrm{R}}=2362.59\cos (86.432) \\ {\mathrm{Q}}_{\mathrm{R}}=147.03\mathrm{MW} \end{gathered}$$

Calculate the receiving power factor.

$$\begin{gathered} \mathrm{Substitute}2397.5\mathrm{MW}\mathrm{for}{\mathrm{P}}_{\mathrm{R}}\mathrm{and}-147.03\mathrm{MW}\mathrm{for}{\mathrm{Q}}_{\mathrm{R}}\mathrm{into}\mathrm{equation}\left(3\right). \\ \mathrm{pf}=\cos \left[{\tan }^{-1}\left(\frac{-147.03}{2397.5}\right)\right] \\ \mathrm{pf}=\cos (3.509) \\ \mathrm{pf}=0.99 \\ \mathrm{Hence}\mathrm{the}\mathrm{receiving}\mathrm{end}\mathrm{reactive}\mathrm{power}\mathrm{is}-147.03\mathrm{MW}\mathrm{and}\mathrm{power}\mathrm{factor}\mathrm{is}0.99. \end{gathered}$$