Question

Repeat Problem 8.14 for the load shown in Example 8.4 (Figure 8.6).

Short answer

Answer

The sequence component of the line current $I_0,I_1$and $I_2$are$0.4827\angle -131°A,36.544\angle -19.98°A$and $3.732\angle -102.622°A$respectively. The line current $I_a,I_b$and $I_c$are $37.082\angle -26.41A$,$33.61\angle -137.08°A$and $39.29\angle 103.872°A$respectively.

Step-by-step solution

Write the given data from the question.

The line to ground voltages.

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

The impedance,

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

The neutral impedance,$Z_n=0$

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.

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

Draw the sequence networks.

Calculate the zero-sequence impedance.

$$\begin{gathered} Z_0=3+j4+j6 \\ {Z}_0=3+j10 \\ {Z}_0=10.44\angle 73.3° \end{gathered}$$

Calculate the positive-sequence impedance.

$$\begin{gathered} Z_1=\left(3+j4\right)P\left(-j10\right) \\ {Z}_1=7.454\angle 26.57°\Omega \end{gathered}$$

Calculate the negative-sequence impedance.

$$\begin{gathered} Z_1=\left(3+j4\right)P\left(-j10\right) \\ {Z}_1=7.454\angle 26.57°\Omega \end{gathered}$$

Calculate the sequence component of the line to ground voltage.

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

$$\begin{gathered} \left[\begin{array}{c}{V}_{Lg0}\\ V_{L{g}_1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle 130°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{L{g}_1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle 240°\\ 1& 1\angle 240°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle 130°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{L{g}_1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}280\angle 0°+250\angle -110°+290\angle 130°\\ 280\angle 0°+250\angle \left(120-110\right)°+290\angle \left(240+130\right)°\\ 280\angle 0°+250\angle \left(240-110\right)°+290\angle \left(120+130\right)°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{L{g}_1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}15.115\angle -57.65°\\ 817.194\angle 6.59°\\ 83.82\angle -76.05°\end{array}\right] \end{gathered}$$

Solve further as,

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

Calculate the zero-sequence component of line current

Substitute $5.04\angle -57.65°$for , $V_{Lg0},10.44\angle 73.3°$for $Z_0$and 0 for $Z_n$into equation (2).

$$\begin{gathered} I_0=\frac{5.04\angle -57.65°}{10.44\angle 73.3°+3\left(0\right)} \\ {I}_0=\frac{5.04\angle -57.65°}{10.44\angle 73.3°} \\ {I}_0=0.4827\angle -131°A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $272.4\angle 6.59°$for $V_{Lg1}$and $7.454\angle 26.57°\Omega$for $Z_1$into equation (3).

$$\begin{gathered} I_1=\frac{272.4\angle 6.59°}{7.454\angle 26.57°\Omega } \\ {I}_1=36.544\angle -19.98°A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $27.82\angle -76.05°$for $V_{Lg2}$and $7.454\angle 26.57°\Omega$for $Z_2$into equation (4).

$$\begin{gathered} I_2=\frac{27.82\angle -76.05°}{7.454\angle 26.57°\Omega } \\ {I}_2=3.732\angle -102.622°A \end{gathered}$$

Calculate the line current.

Substitute $0.4827\angle -131°A$for $I_0,36.544\angle -19.98°A$for $I_1$and $3.732\angle -102.622°A$for $I_2$into equation (5).

$$\begin{gathered} \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0.4827\angle -131°\\ 36.544\angle -19.98°\\ 3.732\angle -102.622°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 240°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle 240°\end{array}\right]\left[\begin{array}{c}0.4827\angle -131°\\ 36.544\angle -19.98°\\ 3.732\angle -102.622°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}0.4827\angle -131°+36.544\angle -19.98°+3.732\angle -102.622°\\ 0.4827\angle -131°+36.544\angle \left(240-19.98\right)°+3.732\angle \left(120-102.622\right)°\\ 0.4827\angle -131°+36.544\angle \left(120-19.98\right)°+3.732\angle \left(240-102.622\right)°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}37.082\angle -26.41A\\ 33.61\angle -137.04°A\\ 39.29\angle 103.872°\end{array}\right] \end{gathered}$$

Hence, the sequence component of the line current $I_0,I_1$and $I_2$are $0.4827\angle -131°A,36.544\angle -19.98°A$and $3.732\angle -102.622°A$respectively. The line current $I_a,I_b$and $I_c$are $37.082\angle -26.41A$,$33.61\angle -137.04°A$and $39.29\angle 103.872°A$respectively.