Question

Determine the theoretical maximum real power that the series compensated line in Problem 5.57 can deliver when ${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$. Compare your result with that of Problem 5.38.

Short answer

The maximum power in problem 5.38 is 1708 MW and is 46.7% less than 2498 MW .

Step-by-step solution

Given data.

From problems 5.57,

The A and B parameters are,

$$\begin{gathered} \mathrm{A}=0.9248\angle 0.{64}^0\mathrm{pu} \\ =\left|\mathrm{A}\right|\angle {\mathrm{\theta }}_{\mathrm{A}}\mathrm{pu} \end{gathered}$$

Also,

$$\begin{gathered} \mathrm{B}=86.66\angle 82.{31}^0\mathrm{\Omega } \\ =\mathrm{Z}\angle {\mathrm{\theta }}_{\mathrm{B}}\mathrm{\Omega } \end{gathered}$$

Determine the formula of maximum power.

Write the formula of maximum power.

${\mathrm{P}}_{\max }=\frac{{\mathrm{V}}_{\mathrm{S}}{\mathrm{V}}_{\mathrm{R}}}{\mathrm{Z}}-\frac{{\mathrm{AV}}_{\mathrm{R}}^2}{\mathrm{Z}}\cos ({\mathrm{\theta }}_{\mathrm{B}}-{\mathrm{\theta }}_{\mathrm{A}})$ .................(1)

Here, ${\mathrm{V}}_{\mathrm{R}}$and ${\mathrm{V}}_{\mathrm{S}}$are receiving and sending end voltages.

Determine the maximum power.

Substitute 500 kV for ${\mathrm{V}}_{\mathrm{S}}$ , ${\mathrm{V}}_{\mathrm{R}}$, $0.9248\angle 0.{64}^0$for $\mathrm{A}\angle {\mathrm{\theta }}_{\mathrm{A}}$and $86.66\angle 82.{31}^0\mathrm{\Omega }$for $\mathrm{Z}\angle {\mathrm{\theta }}_{\mathrm{B}}$in equation (1).

$$\begin{gathered} {\mathrm{P}}_{\max }=\frac{500\times 500}{86.66}-\frac{0.9248{\left(500\right)}^2}{86.66}\cos \left(82.{31}^0-0.{64}^0\right) \\ =2498\mathrm{MW} \end{gathered}$$

Therefore, the maximum power is $2498\mathrm{MW}$.

The maximum power in problem 5.38 is $1708\mathrm{MW}$and is 46.7% less than $2498\mathrm{MW}$.