Question

(a) Given the symmetrical components to be${\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathbf{10}\mathbf{\angle }{\mathbf{0}}^{\mathbf{0}}$,${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathbf{80}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}\mathit{V}$role="math" localid="1656305320298" ${\mathit{V}}_2\mathbf{=}\mathbf{40}\mathbf{\angle }\mathbf{-}{\mathbf{30}}^{\mathbf{°}}\mathit{V}$.Determine the unbalanced phase voltages ${\mathit{V}}_{\mathbf{a}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{b}}$, and ${\mathit{V}}_{\mathbf{c}}$.(b) Using the results of part (a), calculate the line-to-line voltages ${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{b}\mathbf{c}}$ and ${\mathit{V}}_{\mathbf{ca}}$. Then determine the symmetrical components of these line-to-line voltages, the symmetrical components of the corresponding phase voltages, and the phase voltages. Compare them with the result of part (a). Comment on why they are different, even though either set results in the same line-to-line voltages

Short answer

Answer

The unbalanced phase voltages of $V_{an},V_{bn}$and $V_{cn}$are$115.66\angle 9.{96}^{°}V$,and$96.026\angle 167.{98}^{°}V$.

The line-to-line voltage $V_{ab},V_{bc}$and $V_{cb}$are $120\angle {30}^{°},120\angle -{30}^{°}V$and$207.84\angle -{180}^{°}V$.

The symmetrical components of line-to-line voltages ${\left(V^1\right)}_0,{\left(V^1\right)}_1$,and ${\left(V^1\right)}_2$are 0 V, $\left(\sqrt{3}\angle {30}^{°}\right)V_{1}V$and $\left(\sqrt{3}\angle -30\right)V_{2}V$.

The symmetrical components of phase voltages ${\left(V^1\right)}_0,{\left(V^1\right)}_1,{\left(V^1\right)}_1$,and ${\left(V^1\right)}_2$are $0V,138.56\angle {60}^{°}V$and $69.28\angle -{60}^{°}V$.

The unbalanced phase voltages of $V_a,V_b$and$V_c$are$105.83\angle 10.{89}^{°}$,$40\angle {90}^{°}$and$105.83\angle 196\angle {106}^{°}$.

Step-by-step solution

Write the given data from the question.

Assume that a symmetrical components of voltage are $V_0=10\angle 0^{°}$,$V_1=80\angle {30}^{°}V$and$V_2=40\angle -{30}^{°}V$.

Determine the formula of unbalance phase voltage.

Write the formula of unbalance phase voltage.

$\left[\begin{array}{c}{V}_{an}\\ V_{bn}\\ V_{cn}\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]$ …… (1)

Here,${\mathit{V}}_{\mathbf{an}}$,${\mathit{V}}_{\mathbf{bn}}$and${\mathit{V}}_{\mathbf{cn}}$are phase voltages and ${\mathit{V}}_{\mathbf{0}}$,${\mathit{V}}_{\mathbf{1}}$and ${\mathit{V}}_{\mathbf{2}}$are phase sequence component.

Write the formula of line-to-line voltage.

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

Here,${\mathit{V}}_{\mathbf{a}\mathbf{n}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{b}\mathbf{n}}$are phase voltages.

${\mathit{V}}_{\mathbf{b}\mathbf{c}}\mathbf{=}{\mathit{V}}_{\mathbf{b}\mathbf{n}}\mathbf{-}{\mathit{V}}_{\mathbf{c}\mathbf{n}}$ …… (3)

Here, ${\mathit{V}}_{\mathbf{b}\mathbf{n}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{c}\mathbf{n}}$are phase voltages.

${\mathit{V}}_{\mathbf{c}\mathbf{n}}\mathbf{=}{\mathit{V}}_{\mathbf{c}\mathbf{n}}\mathbf{-}{\mathit{V}}_{\mathbf{a}\mathbf{n}}$ …… (4)

Here, ${\mathit{C}}_{\mathbf{c}\mathbf{n}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{C}}_{\mathbf{a}\mathbf{n}}$are phase voltages.

Write the formula of symmetrical component of line voltages.

${\left(V^1\right)}_{\mathbf{0}}\mathbf{=}{\mathit{V}}_{\mathbf{a}\mathbf{0}}\mathbf{-}{\mathit{V}}_{\mathbf{b}\mathbf{0}}$ …… (5)

Here,${\mathit{V}}_{\mathbf{a}\mathbf{0}}$and${\mathit{V}}_{\mathbf{a}\mathbf{1}}$is sequence component of phase a and b respectively.

${\left(V^1\right)}_{\mathbf{1}}\mathbf{=}{\mathit{V}}_{\mathbf{a}\mathbf{1}}\mathbf{-}{\mathit{V}}_{\mathbf{b}\mathbf{1}}$ …… (6)

Here, ${\mathit{V}}_{\mathbf{a}\mathbf{1}}$and ${\mathit{V}}_{\mathbf{b}\mathbf{1}}$is sequence component of phase a and b respectively.

${\left(V^1\right)}_{\mathbf{2}}\mathbf{=}{\mathit{V}}_{\mathbf{a}\mathbf{2}}\mathbf{-}{\mathit{V}}_{\mathbf{b}\mathbf{2}}$ …… (7)

Here, ${\mathit{V}}_{\mathbf{a}\mathbf{2}}$and${\mathit{V}}_{\mathbf{b}\mathbf{2}}$is sequence component of phase a and b respectively.

Write the formula of symmetrical component of phase voltages${\mathbf{\left(}{\mathbf{V}}^{\mathbf{1}}\mathbf{\right)}}_{\mathbf{0}}$.

${\left(V^1\right)}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(V_{ab}+V_{bc}+V_{ca}\right)$ …… (8)

Here, ${\mathit{V}}_{\mathbf{a}\mathbf{b}}$, ${\mathit{V}}_{\mathbf{bc}}$and${\mathit{V}}_{\mathbf{ca}}$are line-to-line voltages.

Write the formula of symmetrical component of phase voltages${\mathbf{\left(}{\mathbf{V}}^{\mathbf{1}}\mathbf{\right)}}_{\mathbf{1}}$.

…… (9)

Here,,andare line-to-line voltages.

Write the formula of symmetrical component of phase voltages.

${\mathbf{\left(}{\mathbf{V}}^{\mathbf{1}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(V_{ab}+a{V}_{bc}+a^2{V}_{ca}\right)$ …… (10)

Here,${\mathit{V}}_{\mathbf{a}\mathbf{b}}$,${\mathit{V}}_{\mathbf{bc}}$and ${\mathit{V}}_{\mathbf{ca}}$are line-to-line voltages.

(a) Determine the unbalanced phase voltages.

Determine an unbalanced phase voltage.

Substitute $1\angle {120}^{°}$for a,$1\angle {240}^{°}$for $a^2$and $10\angle 0^{°}$for$V_0$,$80\angle {30}^{°}$for $V_1$and $40\angle -{30}^{°}$for$V_2$into equation (1).

$$\begin{gathered} \left[\begin{array}{c}{V}_{an}\\ V_{bn}\\ V_{cn}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle {240}^{°}& 1\angle {120}^{°}\\ 1& 1\angle {120}^{°}& 1\angle {240}^{°}\end{array}\right]\left[\begin{array}{c}10\angle 0^{°}\\ 80\angle {30}^{°}\\ 40\angle -{30}^{°}\end{array}\right] \\ =\left[\begin{array}{c}10\angle 0^{°}+80\angle {30}^{°}+40\angle -{30}^{°}\\ 10\angle 0^{°}+80\angle {270}^{°}+40\angle {90}^{°}\\ 10\angle 0^{°}+80\angle {150}^{°}+40\angle {210}^{°}\end{array}\right] \\ =\left[\begin{array}{c}113.92+20j\\ 10-40j\\ -93.92+20j\end{array}\right] \\ =\left[\begin{array}{c}115.56\angle 9.{96}^{°}\\ 41.23\angle -75.{96}^{°}\\ 96.026\angle 167.{98}^{°}\end{array}\right]V \end{gathered}$$

Therefore, the unbalanced phase voltages for the given system.

$$\begin{gathered} V_{an}=115.66\angle 9.{96}^{°}V \\ {V}_{bn}=41.23\angle -75.{96}^{°}V \\ {V}_{cn}=96.026\angle 167.{98}^{°}V \end{gathered}$$

(b) Determine the line-to-line voltages and symmetrical components of phase and line voltages.

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

Substitute$115.66\angle 9.{96}^{°}$for $V_n$and $41.23\angle -75.{96}^{°}$for $V_{bn}$into equation (2).

$$\begin{gathered} V_{ab}=\left(115.66\angle 9.{96}^{°}\right)-\left(41.23\angle -75.{96}^{°}\right) \\ =103.923-j60 \\ =120\angle -{30}^{°}V \end{gathered}$$

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

Substitute $41.23\angle -{7596}^{°}$for$V_{bn}$and $96.026\angle 167.{98}^{°}$for $V_{cn}$into equation (3).

$$\begin{gathered} V_{bc}=41.23\angle -75.{96}^{°}-96.026\angle 167.{98}^{°} \\ =103.923-j60 \\ =120\angle -{30}^{°}V \end{gathered}$$

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

Substitute $96.026\angle 167.{98}^{°}$for$V_{cn}$and $115.66\angle 9.{96}^{°}$for into equation (4).

$$\begin{gathered} V_{ca}=96.026\angle 167.{98}^{°}-115.66\angle 9.{96}^{°} \\ =-207.837-j0.0069 \\ =207.84\angle -{180}^{°}V \end{gathered}$$

Determine the symmetrical components of phase voltages${\left(V^1\right)}_0$.

Substitute $120\angle {30}^{°}$for$V_{ab}$,$120\angle -{30}^{°}$for $V_{bc}$and$207.84\angle -{180}^{°}$for $V_{ca}$into equation (8).

$$\begin{gathered} {\left(V^1\right)}_0=\frac{1}{3}\left(120\angle {30}^{°}+120\angle -{30}^{°}+207.84\angle -{180}^{°}\right) \\ =\frac{1}{3}\left(0\right) \\ =0V \end{gathered}$$

Determine the symmetrical components of phase voltages ${\left(V^1\right)}_1$.

Substitute$120\angle {30}^{°}$for$V_{ab}$,$1\angle {120}^{°}$for,$a,120\angle -{30}^{°}$for$V_{bc}$,$1\angle {240}^{°}$for $a^2$and $207.84\angle -{180}^{°}$for $V_{ca}$into equation (9).

$$\begin{gathered} {\left(V^1\right)}_1=\frac{1}{3}\left(120\angle {30}^{°}+120\angle {\left(120-30\right)}^{°}+207.84\angle {\left(-180+240\right)}^{°}\right) \\ =\frac{1}{3}\left(207.843+j360\right) \\ =69.28+j120 \\ =138.56\angle {60}^{°}V \end{gathered}$$

Determine the symmetrical components of phase voltages ${\left(V^1\right)}_2$.

Substitute $120\angle {30}^{°}$for$V_{ab},1\angle {240}^{°}$for $a^2,120\angle {30}^{°}$for $V_{bc}$,$1\angle {120}^{°}$for a and $207.84\angle -{180}^{°}$for$V_{ca}$into equation (10).

$$\begin{gathered} {\left(V^1\right)}_2=\frac{1}{3}\left(120\angle {30}^{°}+120\angle {\left(240-30\right)}^{°}+207.84\angle {\left(-180+120\right)}^{°}\right) \\ =\frac{1}{3}\left(103.92-180j\right) \\ =34.64-j60 \\ =69.28\angle -{60}^{°}V \end{gathered}$$

Determine the symmetrical components of line voltages${\left(V^1\right)}_0$.

Substitute 0 for $V_{a0}$and 0 for $V_{b0}$into equation (5).

$$\begin{gathered} {\left(V^1\right)}_0=V_0-V_0 \\ =0V \end{gathered}$$

Determine the symmetrical components of line voltages ${\left(V^1\right)}_1$.

Substitute 1 for $V_{a1},1\angle {240}^{°}$for $a^2$and$V_1$for $V_{b1}$into equation (6).

$$\begin{gathered} {\left(V^1\right)}_1=V_1-a^2{V}_1 \\ =\left(1-1\angle {240}^{°}\right)V_1 \\ =\left(\sqrt{3}{30}^{°}\right)V_{1}V \end{gathered}$$

Determine the symmetrical components of line voltages ${\left(V^1\right)}_2$.

Substitute 1 for $V_{a2},1\angle {240}^{°}$for$a^2$and $V_2$for $V_{b2}$into equation (7).

$$\begin{gathered} {\left(V^1\right)}_2=V_2-a{V}_2 \\ =\left(1-1\angle {120}^{°}\right)V_2 \\ =\left(\sqrt{3}\angle -30\right)V_{2}V \end{gathered}$$

Determine the value.

$V_1=\frac{{\left(V^1\right)}_1}{\sqrt{3}}\angle -{30}^{°}$ …… (11)

Here, ${\left(V^1\right)}_1$is symmetrical component of phase voltage.

Substitute $138.56\angle {60}^{°}$for ${\left(V^1\right)}_1$into equation (11).

$$\begin{gathered} V_1=\frac{138.56\angle {\left(60-30\right)}^{°}}{\sqrt{3}} \\ =80\angle {30}^{°}V \end{gathered}$$

Determine the value$V_2$.

$V_2=\frac{{\left(V^1\right)}_2}{\sqrt{3}}\angle {30}^{°}$ …… (12)

Here, ${\left(V^1\right)}_2$is symmetrical component of phase voltage.

Substitute $69.28\angle -{60}^{°}$for ${\left(V^1\right)}_2$into equation (12).

$$\begin{gathered} V_2=\frac{69.28\angle {\left(-60+30\right)}^{°}}{\sqrt{3}} \\ =40\angle -{30}^{°}V \end{gathered}$$

Again determine an unbalanced phase voltage.

Substitute $1\angle {120}^{°}$for $a,1\angle {240}^{°}$for $a^2$and 0 for$V_0,80\angle {30}^{°}$, for $V_1$and $40\angle -{30}^{°}$for$V_2$into equation (1).

$$\begin{gathered} \left[\begin{array}{c}{V}_{an}\\ V_{bn}\\ V_{cn}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle {240}^{°}& 1\angle {120}^{°}\\ 1& 1\angle {120}^{°}& 1\angle {240}^{°}\end{array}\right]\left[\begin{array}{c}0\\ 80\angle {30}^{°}\\ 40\angle -{30}^{°}\end{array}\right] \\ =\left[\begin{array}{c}0+80\angle {30}^{°}+40\angle -{30}^{°}\\ 0+80\angle {270}^{°}+40\angle {90}^{°}\\ 0+80\angle {150}^{°}+40\angle {210}^{°}\end{array}\right] \\ =\left[\begin{array}{c}105.83\angle 10.{893}^{°}\\ 40\angle -{90}^{°}\\ 105.83\angle 169.{106}^{°}\end{array}\right] \end{gathered}$$

Therefore, the unbalanced phase voltages for the given system.

$$\begin{gathered} V_{an}=105.83\angle 10.{89}^{°}V \\ {V}_{bn}=40\angle -{90}^{°}V \\ {V}_{cn}=105.83\angle 169.{106}^{°}V \end{gathered}$$

The above values which are different from part (a) of unbalanced phase voltages.

$$\begin{gathered} V_{an}=115.66\angle 9.{96}^{°}V \\ {V}_{bn}=41.23\angle -75.{96}^{°}V \\ {V}_{cn}=96.026\angle 167.{98}^{°}V \end{gathered}$$

Here, we note that Zero sequence component always zero for symmetrical line to line voltages but Zero sequence component of symmetrical phase voltages are not exist. Both positive and negative sequence components are remaining same in both line to line and phase voltages. The symmetrical components of phase voltage to represent the unbalanced system without any zero sequence component.