Question

Repeat Problem 8.14 for a balanced $\mathbf{∆}$-load consisting of $\left(12+j16\right)$ohms per phase.

Short answer

Answer

The sequence component of the line current $I_0,I_1$and $I_2$are 0,$40.86\angle -46.54°A$and $4.173\angle -129.182°A$respectively. The line current $I_a,I_b$and $I_c$are $41.601\angle -52.25°A,37.05\angle 164.054°A$and $44.28\angle 76.74°A$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 of $∆$-connected load,

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

The neutral is open circuited, $Z_n=\infty$.

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

The equation to convert the $∆$- connected load to Y- connected load is given as follows.

${\mathit{Z}}_{\mathbf{Y}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{∆}}}{\mathbf{3}}$ …… (1)

Here ${\mathit{Z}}_{\mathbf{Y}}$is the impedance of star connected load and ${\mathit{Z}}_{\mathbf{∆}}$is the impedance of delta connected load.

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

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

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

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

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

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

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

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Lg}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{2}}}$ …… (5)

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

$\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{a}}\\ {\mathbf{I}}_{\mathbf{b}}\\ {\mathbf{I}}_{\mathbf{c}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{2}}\end{array}\mathbf{\right]}$ …… (6)

Calculate the sequence component of line voltages and line current.

Calculate the Y- connected load impedance for zero, positive and negative sequence.

Substitute $20\angle 53.13\Omega$for $Z_{∆}$into equation (1).

$$\begin{gathered} Z_Y=\frac{20\angle 53.13°}{3} \\ {Z}_Y=6.66\angle 53.13°\Omega \end{gathered}$$

Consider the diagram of Y-connected load.

Draw the sequence networks.

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 $V_{ag}$into equation (2).

$$\begin{gathered} \left[\begin{array}{c}{I}_{Lg0}\\ I_{Lg1}\\ I_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle -130°\end{array}\right] \\ \left[\begin{array}{c}{I}_{Lg0}\\ I_{Lg1}\\ I_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle 240°\\ 1& 1\angle 240°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle 130°\end{array}\right] \\ \left[\begin{array}{c}{I}_{Lg0}\\ I_{Lg1}\\ I_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}280\angle 0°+250\angle -110°+290\angle 130°\\ 280\angle 0°+250\angle \left(120-110\right)°+290\angle \left(240+130\right)°\\ 280\angle 0°+250\angle \left(2400-110\right)°+290\angle \left(120+130\right)°\end{array}\right] \\ \left[\begin{array}{c}{I}_{Lg0}\\ I_{Lg1}\\ I_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}15.115\angle -57.65°\\ 817.194\angle 6.59°\\ 83.82\angle -76.05°\end{array}\right] \end{gathered}$$

Solve further as,

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

Calculate the zero-sequence component of line current

Substitute $5.04\angle -57.65°$for $V_{Lg0},6.66\angle 53.13°\Omega$for $Z_0$and $\infty$for $Z_n$into equation (3).

$$\begin{gathered} I_0=\frac{5.04\angle -57.65°}{6.66\angle 53.13+3\left(\infty \right)} \\ {I}_0=\frac{5.04\angle -57.65°}{\infty } \\ {I}_0=0A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $272.4\angle 6.59°$for $V_{Lg1}$and $6.66\angle 53.13°\Omega$for $Z_1$into equation (4).

$$\begin{gathered} I_1=\frac{272.4\angle -6.59°}{6.66\angle 53.13°} \\ {I}_1=40.86\angle -46.54°A \end{gathered}$$

Calculate the positive-sequence component of line current

Substitute $27.82\angle -76.05°$for $V_{Lg2}$and $6.66\angle 53.13°\Omega$for $Z_2$into equation (5).

$$\begin{gathered} I_2=\frac{27.82\angle -76.05°}{6.66\angle 53.13°} \\ {I}_2=4.173\angle -129.182°A \end{gathered}$$

Calculate the line current.

Substitute 0 for $I_0,40.86\angle -46.54°A$for $I_1$and $4.173\angle -129.182°A$for $I_2$into equation (6).

$$\begin{gathered} \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 40.86\angle -46.54°A\\ 4.173\angle -129.182°A\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 240°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle 240°\end{array}\right]\left[\begin{array}{c}0\\ 40.86\angle -46.54°A\\ 4.173\angle -129.182°A\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}0+40.86\angle -46.54°+4.173\angle -129.182°\\ 0+40.86\angle \left(240-46.54°\right)+4.173\angle \left(120-129.182\right)°\\ 0+40.86\angle \left(120-46.54°\right)+4.173\angle \left(240-129.182\right)°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}41.601\angle -52.25°A\\ 37.05\angle 164.054°A\\ 44.28\angle 76.74°A\end{array}\right] \end{gathered}$$

Hence, the sequence component of the line current $I_0,I_1$and $I_2$are 0,$40.86\angle -46.54°A$and $4.173\angle -129.182°A$respectively. The line current $I_a,I_b$and $I_c$are $41.601\angle -52.25°A,37.05\angle 164.054°A$and $44.28\angle 76.74°A$respectively.