Question

The maximum power flow for a lossy line is somewhat less than that fora lossless line.

(a) True

(b) False

Short answer

The correct answer is (a) true.

Step-by-step solution

Determine the formula of maximum power.  

Write the formula of maximum power for lossless line.

${\mathbf{P}}_{\mathbf{max}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}$ ……. (1)

Here, ${\mathrm{V}}_{\mathrm{S}}$is sending end voltage, ${\mathrm{V}}_{\mathrm{R}}$is receiving end voltage, and ${\mathrm{X}}^{\text{'}}$is the line reactance.

Write the formula of maximum power for lossy line.

${\mathbf{P}}_{\mathbf{R}\mathbf{}\mathbf{max}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{\mathbf{Z}}\mathbf{-}\frac{{\mathbf{AV}}_{\mathbf{R}}^{\mathbf{2}}}{{\mathbf{Z}}^{\mathbf{\text{'}}}}\mathbf{cos}\mathbf{(}{\mathbf{\theta }}_{\mathbf{Z}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{A}}\mathbf{)}$ ……. (2)

Here, A is transmission line parameter, and ${\mathrm{Z}}^{\text{'}}$is the line impedance.

Determine the correct answer.

On comparing the equations (1) and (2), that is maximum power flow for lossless and lossy line, it is understood that, if impedance ${\mathrm{Z}}^{\text{'}}$is larger than the .

$$\begin{gathered} {\mathrm{X}}^{\text{'}} \\ \mathrm{So},{\mathrm{P}}_{\mathrm{R}\max }<{\mathrm{P}}_{\max } \end{gathered}$$

Hence, the maximum power flow for a lossy line is less than the lossless line.

Therefore, the correct answer is (a) true.