Question

Repeat Problem 8.14 with the load neutral open.

Short answer

Answer

The sequence component of the line current $I_0,I_1$and $I_2$are 0,$13.62\angle -46.54°A$and $1.39\angle -129.182°A$respectively. The line current $I_a,I_b$and $I_c$are $13.86\angle -52.25°A,12.35\angle 164.054°A$and $14.76\angle 76.74°A$respectively.

Step-by-step solution

Write the given data from the question.

The line to ground voltages are,

$$\begin{gathered} V_{ag}=280\angle 0°V \\ {V}_{bg}=250\angle -110°V \\ {V}_{cg}=290\angle 130°V \end{gathered}$$

The impedance is,

$$\begin{gathered} Z=12+j16 \\ Z=20\angle 53.13° \end{gathered}$$

Determine the formulas to calculate the sequence component of line current and line current.

The equation to calculate the sequence component of the line to ground voltage is given as follows.

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

The equation to calculate the zero-sequence component of line current is given as follows.

${\mathit{I}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Lg}\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}\mathbf{3}{\mathbf{Z}}_{\mathbf{n}}}$ …… (2)

The equation to calculate the positive-sequence component of line current is given as follows.

${\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Lg}\mathbf{1}}}{{\mathbf{Z}}_{\mathbf{1}}}$ …… (3)

The equation to calculate the negative sequence component of line current is given as follows.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Lg}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{2}}}$ …… (4)

The equation to calculate the line current is given as follows.

$\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{a}}\\ {\mathbf{I}}_{\mathbf{b}}\\ {\mathbf{I}}_{\mathbf{c}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{2}}\end{array}\mathbf{\right]}$ …… (5)

Calculate the sequence component of line current and line current.

Calculate the sequence component of the line to ground voltage.

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

$$\begin{gathered} \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle 130°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle 240°\\ 1& 1\angle 240°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle 130°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}280\angle 0°+250\angle -110°+290\angle 130°\\ 280\angle 0°+250\left(120-110\right)°+290\left(240+130\right)°\\ 280\angle 0°+250\left(240-110\right)°+290\left(120+130\right)°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}15.115\angle -57.65°\\ 817.194\angle 6.59°\\ 83.82\angle -76.05°\end{array}\right] \end{gathered}$$

Solve further as,

$\left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\left[\begin{array}{c}5.04\angle -57.65°\\ 272.4\angle 6.59°\\ 27.82\angle -76.05\end{array}\right]$

The zero, positive and negative sequence impedance.

$$\begin{gathered} Z_0=Z_1 \\ {Z}_0=Z_2 \\ {Z}_0=Z \\ Z=20\angle 53.13° \end{gathered}$$

Since the neutral is open circuited, therefore the neutral impedance,$Z_n=\infty$.

Calculate the zero-sequence component of line current

Substitute $5.04\angle -57.65°$for $V_{Lg0},20\angle 53.13°$for $Z_0$and $\infty$for $Z_n$into equation (2).

$$\begin{gathered} I_0=\frac{5.04\angle -57.65°}{20\angle 53.13+3\left(\infty \right)} \\ {I}_0=\frac{5.04\angle -57.65°}{\infty } \\ {I}_0=0A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $27.82\angle -76.05°$for $V_{Lg2}$and role="math" localid="1656419534769" $20\angle 53.13°$for $Z_1$into equation (3).

$$\begin{gathered} I_1=\frac{272.4\angle 6.59°}{20\angle 53.13} \\ {I}_1=13.62\angle -46.54°A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $27.82\angle -76.05°$for $V_{Lg2}$and $20\angle 53.13°$for $Z_2$into equation (4).

$$\begin{gathered} I_2=\frac{27.82\angle -76.05°}{20\angle 53.13} \\ {I}_2=1.391\angle -129.182°A \end{gathered}$$

Calculate the line current.

Substitute 0 A for $I_0,13.62\angle -46.54°A$for $I_1$and $14.76\angle 76.74°A$for $I_2$into equation (5).

$$\begin{gathered} \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 13.62\angle -54°\\ 1.391\angle -129.182°\end{array}\right] \\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 240°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle 240°\end{array}\right]\left[\begin{array}{c}0\\ 13.62\angle -46.54°\\ 1.391\angle -129.182°\end{array}\right] \\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}0+13.62\angle -46.54°+1.391\angle -129.182°\\ 0+13.62\left(240-46.54\right)°+1.391\angle \left(120-129.182\right)°\\ 0+13.62\left(120-46.54\right)°+1.391\angle \left(240-129.182\right)°\end{array}\right] \\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}13.86\angle -52.25°A\\ 12.35\angle 164.054°A\\ 14.76\angle 76.74°A\end{array}\right] \end{gathered}$$

Hence, the sequence component of the line current role="math" localid="1656419159235" $I_0,I_1$and $I_2$are 0,$13.62\angle -46.54°A$and $1.391\angle -129.182°A$respectively. The line current $I_a,I_b$and $I_c$are $13.86\angle -52°A,12.35\angle 164.054°A$and $14.76\angle 76.74°A$respectively.