Question

The voltages given in Problem 8.10 are applied to a balanced-Y load consisting of $(12+j16)$ohms per phase. The load neutral is solidly grounded. Draw the sequence networks and calculate ${\mathbf{l}}_{\mathbf{0}}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{l}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{2}}$, the sequence components of the line currents. Then calculate the line currentslocalid="1655457516799" role="math" ${\mathbf{l}}_{\mathbf{a}}\mathbf{,}{\mathbf{l}}_{\mathbf{b}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{c}}$.

Short answer

$$\begin{gathered} \mathrm{The}\mathrm{sequence}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{current}{l}_0,l_1\mathrm{and}0.152\angle -110.78°\mathrm{A}, \\ 13.62\angle -46.54°\mathrm{A}\mathrm{and}1.391\angle -129.182°\mathrm{A}\mathrm{respectively}.\mathrm{The}\mathrm{line}\mathrm{current}{l}_a,l_{b}and \\ {l}_c\mathrm{are}14\angle -53.13°\mathrm{A},12.5\angle 163.12°\mathrm{A}\mathrm{and}14.5\angle 76.86°\mathrm{A}\mathrm{respectively}. \end{gathered}$$

Step-by-step solution

Write the given data from the question.

The line to ground voltages.

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

The impedance,

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

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.

$\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{Lg}\mathbf{0}}\\ {\mathbf{V}}_{\mathbf{Lg}\mathbf{1}}\\ {\mathbf{V}}_{\mathbf{Lg}\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{ag}}\\ {\mathbf{V}}_{\mathbf{bg}}\\ {\mathbf{V}}_{\mathbf{cg}}\end{array}\mathbf{\right]}\mathbf{}$ …… (1)

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

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

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

${\mathbf{l}}_1\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Lg}\mathbf{1}}}{{\mathbf{Z}}_{\mathbf{1}}}$ …… (3)

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

${\mathbf{l}}_2\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Lg}\mathbf{1}}}{{\mathbf{Z}}_2}$ …… (4)

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

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

Calculate the sequence component of line current and line current.

Draw the sequence networks

Calculate the sequence component of the line to ground voltage.

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

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

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 sequence, positive and negative sequence.

$\begin{array}{l}{Z}_0=Z_1\\ Z_0=Z_2\\ Z_0=Z\\ Z=20\angle 53.{13}^0\end{array}$

Calculate the zero-sequence component of line current

Substitute $5.04\angle -57.65°\mathrm{for}{\mathrm{V}}_{\mathrm{Lgo}},20\angle 53.13°\mathrm{for}{\mathrm{Z}}_0\mathrm{and}0\mathrm{for}{\mathrm{Z}}_{\mathrm{n}}$into equation (2).

$$\begin{gathered} l_0=\frac{5.04\angle -57.65°}{20\angle 53.13+3\left(0\right)} \\ {l}_0=\frac{5.04\angle -57.65°}{20\angle 53.13} \\ {l}_0=0.252\angle -110.78°A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $272.4\angle 6.59°\mathrm{for}{\mathrm{V}}_{\mathrm{Lg}1}\mathrm{and}20\angle 53.13°\mathrm{for}{\mathrm{Z}}_0$into equation (3).

$$\begin{gathered} l_1=\frac{272.4\angle 6.59°}{20\angle 53.13} \\ {l}_1=13.62\angle -46.54°A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $27.82\angle -76.05°\mathrm{for}{\mathrm{V}}_{\mathrm{Lg}2}\mathrm{and}20\angle 53.13°\mathrm{for}{\mathrm{Z}}_0$into equation (4).

$$\begin{gathered} l_2=\frac{27.82\angle -76.05°}{20\angle 53.13} \\ {L}_2=1.391\angle -129.182°\mathrm{A} \end{gathered}$$

Calculate the line current. $\mathrm{Substitute}0.252\angle -110.78°\mathrm{A}\mathrm{for}{l}_0,1362\angle -46.54°\mathrm{A}\mathrm{for}{l}_1\mathrm{and}1.391\angle -129.182°\mathrm{A}\mathrm{for}{l}_2$into equation (5).

$$\begin{gathered} \left[\begin{array}{c}{l}_a\\ l_b\\ l_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0.252\angle -110.78°\\ 13.62\angle -46.54°\\ 1.391\angle -129.182°\end{array}\right] \\ \left[\begin{array}{c}{l}_a\\ l_b\\ l_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 240°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle 240°\end{array}\right]\left[\begin{array}{c}0.252\angle -110.78°\\ 13.62\angle -46.54°\\ 1.391\angle -129.182°\end{array}\right] \\ \left[\begin{array}{c}{l}_a\\ l_b\\ l_c\end{array}\right]=\left[\begin{array}{c}0.25\angle -110.78°+13.62\angle -46.54°+1.391\angle -129.182°\\ 0.25\angle -110.78°+13.62\angle \left(240-46.54\right)°+1.391\angle \left(120-129.182\right)°\\ 0.25\angle -110.78°+13.62\angle \left(240-46.54\right)°+1.391\angle \left(120-129.182\right)°\end{array}\right] \\ \left[\begin{array}{c}{l}_a\\ l_b\\ l_c\end{array}\right]=\left[\begin{array}{c}14\angle -53.13°A\\ 12.5\angle 163.12°A\\ 14.5\angle 76.86°A\end{array}\right] \end{gathered}$$

Hence the sequence component of the line current $$\begin{gathered} \mathrm{Hence}\mathrm{the}\mathrm{sequence}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{current}{l}_0,l_1\mathrm{and}{l}_2\mathrm{are} \\ 0.252\angle -110.78°\mathrm{A},13.62\angle -46.54°\mathrm{A}\mathrm{and}1.39\angle -129.182°\mathrm{A}\mathrm{respectively}.\mathrm{The} \\ \mathrm{line}\mathrm{current}{\mathrm{l}}_{\mathrm{a}},{\mathrm{l}}_{\mathrm{b}}\mathrm{and}{\mathrm{l}}_{\mathrm{c}}\mathrm{are}14\angle -53.13°\mathrm{A},12.5\angle 163.12°\mathrm{A}\mathrm{and}14.5\angle 76.86°\mathrm{A} \\ \mathrm{respectively}. \end{gathered}$$