Question

Let an unbalanced, three-phase, Y-connected load (with phase impedances of ${\mathit{Z}}_{\mathbf{a}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{Z}}_{\mathbf{b}}$and ${\mathit{Z}}_{\mathbf{c}}$) be connected to a balanced three-phase supply, resulting in phase voltages of ${\mathit{V}}_{\mathbf{a}}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{b}}$, and ${\mathit{V}}_{\mathbf{c}}$across the corresponding phase impedances.

Choosing ${\mathit{V}}_{\mathbf{a}\mathbf{b}}$ as the reference, show that

${\mathit{V}}_{\mathbf{a}\mathbf{b}\mathbf{,}\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{;}\mathbf{}\mathbf{}{\mathit{V}}_{\mathbf{a}\mathbf{b}\mathbf{,}\mathbf{1}}\mathbf{=}\sqrt{\mathbf{3}}{\mathit{V}}_{\mathbf{a}\mathbf{,}\mathbf{1}}{\mathit{e}}^{\mathbf{j}{\mathbf{30}}^{\mathbf{°}}}\mathbf{}\mathbf{;}\mathbf{}{\mathit{V}}_{\mathbf{a}\mathbf{b}\mathbf{,}\mathbf{2}}\mathbf{=}\sqrt{\mathbf{3}}{\mathit{V}}_{\mathbf{a}\mathbf{,}\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{j}{\mathbf{30}}^{\mathbf{°}}}$

Short answer

Answer

The value of sequence components of the line voltages with $V_{ab}$as the reference are:

$$\begin{gathered} V_{ab,0}=0 \\ {V}_{ab,1}=\sqrt{3}{V}_{a,1}{e}^{j{30}^{°}} \\ {V}_{ab,2}=\sqrt{3}{V}_{a,2}{e}^{j{30}^{°}} \end{gathered}$$

Step-by-step solution

Write the given data from the question.

The voltage $V_a,V_b$and $V_c$are phase voltages.

The voltage $V_{ab},V_{bc}$and $V_{ca}$are line voltages.

The given line voltages are $V_{ab,0}=0;V_{ab,1}=\sqrt{3}{V}_{a,1}{e}^{j{30}^{°}}$and

$V_{ab,2}=\sqrt{3}{V}_{a,2}{e}^{-j{30}^{°}}$ and $a^2=1\angle {240}^{°}$.

Determine the formula of symmetrical components of the line voltages.

Write the formula of sequence component of line voltage of phase.

${\mathit{V}}_{\mathbf{a}\mathbf{,}\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(V_{ab}+V_{bc}+V_{ca}\right)$ …… (1)

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

Write the formula of positive sequence component of line voltage of phase.

${\mathit{V}}_{\mathbf{a}\mathbf{,}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{V}}_{\mathbf{a}}\mathbf{+}\mathit{a}{\mathbf{V}}_{\mathbf{b}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{c}}\mathbf{)}$ …… (2)

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

${\mathit{V}}_{\mathbf{ab}\mathbf{,}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{V}}_{\mathbf{a}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{b}}\mathbf{+}\mathit{a}{\mathbf{V}}_{\mathbf{c}}\mathbf{)}$ …… (3)

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

Determine the sequence components of the line voltages.

The relation between line voltages and phase voltages are:

$$\begin{gathered} V_{ab}=V_a-V_b \\ {V}_{bc}=V_b-V_c \\ {V}_{ca}=V_c-V_a \end{gathered}$$

Assume $V_{ab}$as reference from given sequence line voltages.

Determine the sequence components of line voltages of phase a.

Substitute$V_a-V_b$for$V_{ab},V_b-V_c$for $V_{bc}$and $V_c-Va$for $V_{ca}$into equation (1).

Determine line-to-line voltageis given by.

Substituteforandforinto equation (3).

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

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

Substitute role="math" localid="1656136810016" $290\angle {130}^{°}$for $V_{cg}$and $280\angle 0^{°}$for $V_{ag}$into equation (4).

$$\begin{gathered} V_{ca}=290\angle {130}^{°}-280\angle 0^{°} \\ =516.61\angle 154.{53}^{°}V \end{gathered}$$

Therefore, the line-to-line voltages obtained are$\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]=\left[\begin{array}{c}434.49\angle 32.{73}^{°}\\ 468.08\angle -77.{55}^{°}\\ 516.61\angle 154.{53}^{°}\end{array}\right]V°$.

Use this relation,

$$\begin{gathered} a^3=1\angle {\left(120\times 3\right)}^{°} \\ =1\angle {360}^{°} \\ =1\angle 0^{°} \end{gathered}$$

Determine the positive sequence components of line voltages of phase.

Substitutefor,forandforinto equation (2).

$$\begin{gathered} V_{ab,1}=\frac{1}{3}\left(\left(V_a-V_b\right)+a\left(V_b-V_c\right)+a^2\left(V_c-V_a\right)\right) \\ =\frac{1}{3}\left(\left(V_a+a{V}_b+a^2{V}_c\right)-\left(a^2{V}_a+V_b+a{V}_c\right)\right) \\ =\frac{1}{3}\left(\left(V_a+a{V}_b+a^2{V}_c\right)-\left(a^2{V}_a+a^3{V}_b+a^3\times a{V}_c\right)\right) \\ =\frac{1}{3}\left(\left(V_a+a{V}_c+a^2{V}_c\right)-a^2\left(V_a+a{V}_b+a^2{V}_c\right)\right) \end{gathered}$$

As further solve as

$$\begin{gathered} V_{ab,1}=\frac{1}{3}\left(\left(1-a^2\right)V_{a,1}\right) \\ =\left(\sqrt{3}\angle {30}^{°}\right)V_{a,1} \\ =\sqrt{3}{e}^{j{30}^{°}{V}_{a,1}} \end{gathered}$$

Now, determine the negative sequence components of line voltages of phase.

Substitutefor,forandforinto equation (3).

$$\begin{gathered} V_{ab,1}=\frac{1}{3}\left(\left(V_a-V_b\right)+a^2\left(V_b-V_c\right)+a\left(V_c-V_a\right)\right) \\ =\frac{1}{3}\left(\left(V_a+a^2{V}_b+a{V}_c\right)-\left(a{V}_a+V_b+a^2{V}_c\right)\right) \\ =\frac{1}{3}\left(\left(V_a+a^2{V}_b+a{V}_c\right)-\left(V_a+a^2{V}_b+a{V}_c\right)\right) \\ =\left(1-a\right)V_{a,2} \end{gathered}$$

Solve further as,

$V_{ab,2}=\sqrt{3}\angle -{30}^{°}{V}_{a,2}$

Therefore, the value of sequence components of the line voltages with $V_{ab}$as the reference are:

$$\begin{gathered} V_{ab,0}=0 \\ {V}_{ab,1}=\sqrt{3}{V}_{a,1}{e}^{j{30}^{°}} \\ {V}_{ab,2}=\sqrt{3}{V}_{a,2}{e}^{j{30}^{°}} \end{gathered}$$