Question

Given the line-to-ground voltage${\mathit{V}}_{\mathbf{a}\mathbf{g}}\mathbf{=}\mathbf{280}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{b}\mathbf{g}}\mathbf{=}\mathbf{250}\mathbf{\angle }\mathbf{-}{\mathbf{110}}^{\mathbf{°}}$, and ${\mathit{V}}_{\mathbf{c}\mathbf{g}}\mathbf{=}\mathbf{290}\mathbf{\angle }{\mathbf{130}}^{\mathbf{°}}$, calculate (a) the sequence components of the line to-ground voltages, denoted ${\mathit{V}}_{\mathbf{L}\mathbf{g}\mathbf{0}}$, ${\mathit{V}}_{\mathbf{L}\mathbf{g}\mathbf{1}}$and ${\mathit{V}}_{\mathbf{L}\mathbf{g}\mathbf{2}}$; (b) line-to-line voltages ${\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{0}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{1}}$, and ${\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{2}}$; and (c) sequence components of the line-to-line voltages ${\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{1}}\mathbf{}\mathbf{,}$and ${\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{2}}$. Also, verifies the following general relation: ${\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{L}\mathbf{L}\mathbf{1}}\mathbf{=}\sqrt{\mathbf{3}}{\mathit{V}}_{\mathbf{L}\mathbf{g}\mathbf{1}}\mathbf{}\mathbf{,}\mathbf{}$and${\mathit{V}}_{\mathbf{L}\mathbf{L}}\mathbf{=}\sqrt{\mathbf{3}}{\mathit{V}}_{\mathbf{L}\mathbf{g}\mathbf{2}}\mathbf{\angle }\mathbf{-}{\mathbf{30}}^{\mathbf{°}}\mathbf{}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{s}$.

Short answer

Answer

(a) The value of sequence component of line to ground voltage is$\left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\left[\begin{array}{c}5.04\angle -57.{656}^{°}\\ 272.4\angle 6.{59}^{°}\\ 27.82\angle -76.{052}^{°}\end{array}\right]V$.

(b) The value of line to line voltage is$\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{cs}\end{array}\right]=\left[\begin{array}{c}434.49\angle 32.{73}^{°}\\ 468.08\angle -77.{55}^{°}\\ 516.61\angle 154.{53}^{°}\end{array}\right]V$.

(c) The value of sequence component of line to line voltage is$\left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\left[\begin{array}{c}0\\ 471.803\angle 36.{59}^{°}\\ 48.425\angle -106.{16}^{°}\end{array}\right]V$.

Step-by-step solution

Write the given data from the question.

The line to ground voltages are $V_{ag}=280\angle 0^{°}V,V_{bg}=250\angle -{110}^{°}V$and$V_{cg}=290\angle {130}^{°}V$.

The line to line voltages are$V_{LLC}=0,V_{II1}=\sqrt{3}{V}_{Ig1}$and$V_{LL2}=\sqrt{3}{V}_{Lg2}\angle -{30}^{°}volts$.

The $a=1\angle {120}^{°}$and$a^2=1\angle {240}^{°}$.

Determine the formula of line to line voltages and line to ground voltages.

Write the formula of sequence component of line to ground voltages.

$\left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_{ag}\\ V_{bg}\\ V_{cg}\end{array}\right]$ …… (1)

Here,${\mathit{V}}_{\mathbf{a}\mathbf{g}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{b}\mathbf{g}}$and ${\mathit{V}}_{\mathbf{cg}}$are line to ground voltages.

Write the formula of line-to-line voltage .

${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}{\mathit{V}}_{\mathbf{a}\mathbf{g}}\mathbf{-}{\mathit{V}}_{\mathbf{b}\mathbf{g}}$ …… (2)

Here, ${\mathit{V}}_{\mathbf{b}\mathbf{g}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{c}\mathbf{g}}$are line to ground voltages.

Write the formula of line-to-line voltage .

${\mathit{V}}_{\mathbf{c}\mathbf{s}}\mathbf{=}{\mathit{V}}_{\mathbf{c}\mathbf{g}}\mathbf{-}{\mathit{V}}_{\mathbf{a}\mathbf{g}}$ …… (3)

Here,,are line to ground voltages.

Write the formula of line-to-line voltage.

…… (4)

Here, $V_{cg},V_{ag}$are line to ground voltages.

Write the formula of sequence component of line to line voltages.

$\left[\begin{array}{c}{V}_{LL0}\\ V_{LL1}\\ V_{LL2}\end{array}\right]\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{cb}\end{array}\right]$ …… (5)

Here, ${\mathit{V}}_{\mathbf{ab}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{bc}}$and ${\mathit{V}}_{\mathbf{c}\mathbf{a}}$are line to line voltages.

(a) Determine the sequence components of line to ground voltages.

Assume $V_{ag}$as reference from given line-to-ground voltages.

Determine the sequence components of line to ground voltages.

Substitute $280\angle 0^{°}$for$V_{ag},250\angle -{110}^{°}$for $V_{bg}$and $290\angle {130}^{°}$for$V_{cg}$into equation (1).

$$\begin{gathered} \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle {120}^{°}& 1\angle {240}^{°}\\ 1& 1\angle {240}^{°}& 1\angle {120}^{°}\end{array}\right]\left[\begin{array}{c}280\angle 0^{°}\\ 250\angle -{110}^{°}\\ 290\angle {130}^{°}\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{c}280\angle 0^{°}+250\angle -{110}^{°}+290\angle {130}^{°}\\ 280\angle 0^{°}+250\angle {\left(120-110\right)}^{°}+290\angle {\left(240+130\right)}^{°}\\ 280\angle 0^{°}+250\angle {\left(240-110\right)}^{°}+290\angle {\left(120+130\right)}^{°}\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{c}15.115\angle -57.{656}^{°}\\ 817.194\angle 6.{59}^{°}\\ 83.46\angle -76.{052}^{°}\end{array}\right] \\ =\left[\begin{array}{c}5.04\angle -57.6.{656}^{°}\\ 272.4\angle 6.{59}^{°}\\ 27.82\angle -76.{052}^{°}\end{array}\right]V \end{gathered}$$

Therefore, sequence components of line to ground voltages are $\left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\left[\begin{array}{c}5.04\angle -57.{656}^{°}\\ 272.4\angle 6.{59}^{°}\\ 27.82\angle -76.{052}^{°}\end{array}\right]V$.

(b) Determine the line-to-line voltages.

Determine line-to-line voltage $V_{ab}$is given by.

Substitute$280\angle 0^{°}$for $V_{ag}$and $250\angle -{110}^{°}$for $V_{bg}$into equation (2).

$$\begin{gathered} V_{ab}=\left(280\angle 0^{°}\right)-\left(250\angle -{110}^{°}\right) \\ =434.49\angle 32.{73}^{°}V \end{gathered}$$

Determine line-to-line voltage $V_{bc}$is given by.

Substitute $250\angle -{110}^{°}$for $V_{bg}$and $290\angle {130}^{°}$for $V_{ag}$into equation (3).

$$\begin{gathered} V_{bc}=250\angle {110}^{°}-290\angle {130}^0 \\ =468.08\angle -77.{55}^{°}V \end{gathered}$$

Determine line-to-line voltageis given by.

Substituteforandforinto equation (4).

Therefore, the line-to-line voltages obtained are.

(c) Determine the sequence component of line to line voltage is.

Determine the value of sequence component of line to line voltage.

Substitutefor,forandforinto equation (5).

Therefore, the sequence component of line to line voltage is.

Assume the result of part (a),

,and

Assume the result of part (c),

,and

By using the result of part (a) and part (b), obtained the following relation.

Consider the value of.

Consider the value of.

Consider the value of.

Hence, it is verified.