Question

Repeat Problem 8.14 but include balanced three-phase line impedances of $\left(3+j4\right)$ohms per phase between the source and load.

Short answer

The sequence components of line currents $I_0=0.2016\angle 110.786°,I_1=10.896\angle -129.182°\text{\hspace{0.33em}A\hspace{0.33em}}\&I_2=1.113\angle -129.182°\text{\hspace{0.33em}A}$

The line current are$I_a=11.2\angle -53.13°\text{\hspace{0.33em}A},I_b=10\angle -163.13°\text{\hspace{0.33em}A\hspace{0.33em}}\&I_c=11.6\angle 76.87°\text{\hspace{0.33em}A}$

Step-by-step solution

Write the given data from the question.

The line-to-ground voltages are$V_{ag}=280\angle 0°\text{\hspace{0.33em}V},V_{bg}=250\angle -110°\text{\hspace{0.33em}V\hspace{0.33em}}\&V_{cg}=290\angle 130°$

The balanced connected load with impedance per phase, $12+16j\mathrm{\Omega }$.

The balanced three phase line impedance per phase $3+4j\mathrm{\Omega }$.

Determine the formula of sequence components of line currents.

Write the formula of sequence components of line currents ${\mathbf{I}}_0$.

${\mathbf{I}}_0=\frac{{\mathbf{V}}_{\lg 0}}{{\mathbf{Z}}_0\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{4}\mathbf{j}}$ …… (1)

Here, ${\mathbf{V}}_{\lg 0}$is line to ground voltages, ${\mathbf{Z}}_0$is zero sequence impedance.

Write the formula of sequence components of line currents${\mathbf{I}}_1$.

${\mathbf{I}}_1=\frac{{\mathbf{V}}_{\lg 1}}{{\mathbf{Z}}_1\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{4}\mathbf{j}}$ …… (2)

Here, ${\mathbf{V}}_{\lg 1}$is line to ground voltages, ${\mathbf{Z}}_1$is zero sequence impedance.

Write the formula of sequence components of line currents${\mathbf{I}}_2$.

${\mathbf{I}}_2=\frac{{\mathbf{V}}_{\lg 2}}{{\mathbf{Z}}_1\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{4}\mathbf{j}}$ …… (3)

Here, ${\mathbf{V}}_{\lg 2}$is line to ground voltages, ${\mathbf{Z}}_2$is zero sequence impedance.

Write the formula ofline current matrix.

$\left[\begin{array}{c}{\mathbf{I}}_a\\ {\mathbf{I}}_b\\ {\mathbf{I}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathbf{a}}^2& a\\ 1& a& {\mathbf{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_0\\ {\mathbf{I}}_1\\ {\mathbf{I}}_2\end{array}\right]$ …… (4)

Here,${\mathbf{I}}_0,{\mathbf{I}}_1\&{\mathbf{I}}_2$are sequence components of line current, a is identities.

(a) Determine the sequence component of line current.

Determine the sequence component of the line to ground.

$\left[\begin{array}{c}{V}_{lg0}\\ V_{lg1}\\ V_{lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_{ag}\\ V_{bg}\\ V_{cg}\end{array}\right]$ …… (5)

Here, $V_{ag},V_{bg},\&V_{cg}$are line-to-ground voltage and a is identities.

Substitute $280\angle 0°$for $V_{ag}$,$250\angle -110°$ for$V_{bg}$ and$290\angle 130°$ for $V_{cg}$into equation (5).

$\begin{array}{c}\left[\begin{array}{c}{V}_{lg0}\\ V_{lg1}\\ V_{lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle 240°\\ 1& 1\angle 240°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}280\angle 0°\\ 250\angle -110°\\ 290\angle 130°\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{c}280\angle 0°+250\angle -110°+290\angle 130°\\ 280\angle 0°+250\angle \left(120-100\right)°+290\angle \left(240+130\right)°\\ 280\angle 0°+250\angle \left(240-110\right)°+290\angle \left(120+130\right)°\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{c}15.115\angle -57.656°\\ 817.194\angle 6.59°\\ 83.46\angle -76.052°\end{array}\right]\\ =\left[\begin{array}{c}5.04\angle -57.656°\\ 272.4\angle 6.59°\\ 27.82\angle -76.052°\end{array}\right]\text{\hspace{0.33em}V}\end{array}$

Therefore, the sequence components of line to ground voltages are,

$\begin{array}{l}{V}_{lg0}=5.04\angle -57.656°\\ V_{lg1}=272.4\angle 6.59°\\ V_{lg2}=27.82\angle -76.052°\end{array}$

The zero, positive and negative impedances are

$Z_0=Z_1=Z_2=20\angle 53.13°\mathrm{\Omega }$Respectively.

Determine the sequence components of the line current.

Substitute$5.04\angle -57.656°$ for $V_{lg0}$,$20\angle 53.13°$ for$Z_0$ into equation (1).

$\begin{array}{c}{I}_0=\frac{5.04\angle -57.656°}{20\angle 53.13°+3+4j}\\ =0.2016\angle -110.786°\text{\hspace{0.33em}A}\end{array}$

Substitute$272.4\angle 6.59°$ for $V_{lg1}$, $20\angle 53.13°$for$Z_1$ into equation (2).

$\begin{array}{c}{I}_1=\frac{272.4\angle 6.59°}{20\angle 53.13°+3+4j}\\ =10.896\angle -46.54°\text{\hspace{0.33em}A}\end{array}$

Substitute$27.82\angle -76.052°$ for $V_{lg2}$,$20\angle 53.13°$ for$Z_2$ into equation (3).

$\begin{array}{c}{I}_2=\frac{27.82\angle -76.052°}{20\angle 53.13°+3+4j}\\ =1.113\angle -129.182°\text{\hspace{0.33em}A}\end{array}$

Therefore, the values of the sequence components of line currents are,

$\begin{array}{l}{I}_0=0.2016\angle 110.786°\\ I_1=10.896\angle -129.182°\text{\hspace{0.33em}A}\\ I_2=1.113\angle -129.182°\text{\hspace{0.33em}A}\end{array}$

(b) Determine the line current.

Determine the line current.

Substitute$0.2016\angle 110.786°$ for $I_0$, $10.896\angle -129.182°$for$I_1$ and$1.113\angle -129.182°$ for$I_2$ into equation (4).

$\begin{array}{c}\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 240°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle 240°\end{array}\right]\left[\begin{array}{c}0.2016\angle -110.786°\\ 10.896\angle -46.54°\\ 1.1128\angle -129.182°\end{array}\right]\\ =\left[\begin{array}{c}\left(0.2016\angle -110.786°+10.896\angle -46.54°+1.1128\angle -129.182°\right)\\ 0.2016\angle -110.786°+10.896\angle \left(240-46.54\right)°+1.1128\angle \left(120-129.182\right)°\\ 0.2016\angle -110.786°+10.896\angle \left(120-46.54\right)°+1.1128\angle \left(240-129.182\right)°\end{array}\right]\\ =\left[\begin{array}{c}6.72-j8.96\\ -9.57-j2.902\\ 2.634+j11.2968\end{array}\right]\\ =\left[\begin{array}{c}11.2\angle -53.13°\\ 10\angle -163.13°\\ 11.60\angle 76.87°\end{array}\right]\text{\hspace{0.33em}A}\end{array}$

Therefore, the lines current are

$\begin{array}{l}{I}_a=11.2\angle -53.13°\text{\hspace{0.33em}A}\\ I_b=10\angle -163.13°\text{\hspace{0.33em}A}\\ I_c=11.6\angle 76.87°\text{\hspace{0.33em}A}\end{array}$