Question

The following unbalanced line-to-ground voltages are applied to the balanced-Y load shown in Figure 3.3: ${\mathit{V}}_{\mathbf{a}\mathbf{g}}\mathbf{=}\mathbf{100}\mathbf{\angle }\mathbf{0}\mathbf{°}$,${\mathit{V}}_{\mathbf{bg}}\mathbf{=}\mathbf{75}\mathbf{\angle }\mathbf{180}\mathbf{°}$and ${\mathit{V}}_{\mathbf{ag}}\mathbf{=}\mathbf{50}\mathbf{\angle }\mathbf{90}\mathbf{°}$volts. The Y load has ${\mathit{Z}}_{\mathbf{∆}}\mathbf{=}\mathbf{3}\mathbf{+}\mathit{j}\mathbf{10}\mathbf{}\mathit{\Omega }$per phase with neutral impedance${\mathit{Z}}_{\mathbf{n}}\mathbf{=}\mathit{j}\mathbf{1}\mathit{\Omega }$. (a) Calculate the line currents role="math" localid="1656398893225" ${\mathit{I}}_{\mathbf{a}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{I}}_{\mathbf{b}}$and $\mathbf{}{\mathit{I}}_{\mathbf{c}}$without using symmetrical components, (b) Calculate the line currents ${\mathit{I}}_{\mathbf{a}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{I}}_{\mathbf{b}}$and ${\mathit{I}}_{\mathbf{c}}$using symmetrical components. Which method is easier?

Short answer

Answer

(a) The line currents $I_a,I_{\mathbf{b}}$and $I_c$without using symmetrical component are role="math" localid="1656399173111" $19.95\angle 132.44°A,15.15\angle 132.44°A$and $8.52\angle 37°A$respectively.

(b) The line current $I_a,I_{\mathbf{b}}$and $I_c$using symmetrical component are $19.97\angle -57.34°,15.15\angle 164.054°A$and $8.535\angle 37.46°A$respectively. The comparison shows that first method is shorter than the second one but complex then the second method.

Step-by-step solution

Write the given data from the question:

The line to ground voltages are,

$$\begin{gathered} V_{ag}=100\angle 0°V \\ {V}_{bg}=75\angle 180°V \\ {V}_{cg}=50\angle 90°V \end{gathered}$$

The impedance per phase is $Z_y=3+j4\Omega$.

The neutral impedance is $Z_n=j1\Omega$.

Determine the formulas to calculate the line currents.

The equation to calculate the sequence component of line to ground voltages 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 line currents without using symmetrical components.

(a) Consider the Y-Y network shown below.

Apply the Kirchhoff’s current law at load neutral point.

$I_n=I_a+I_b+I_c$

Apply the Kirchhoff’s voltage law in loop 1.

$$\begin{gathered} V_{ag}=I_a{Z}_Y+I_n{Z}_n \\ {V}_{ag}=I_a{Z}_Y+\left(I_a+I_b+I_c\right)Z_n \\ {V}_{ag}=\left(Z_Y+Z_n\right)I_a+Z_n\left(I_b+I_c\right) \end{gathered}$$

Substitute $100\angle 0°V$for $V_{ag},3+j4\Omega$for $Z_Y$and $j1\Omega$for $Z_n$into above equation.

$$\begin{gathered} 100\angle 0°=\left(3+j4+j1\right)I_a+j1\left(I_b+I_c\right) \\ 100=\left(3+j5\right)I_a+j1I_b+j1I_c \end{gathered}$$ (6)

Apply the Kirchhoff’s voltage law in loop 2.

$$\begin{gathered} V_{bg}=I_b{Z}_Y+I_n{Z}_n \\ {V}_{bg}=I_b{Z}_Y+\left(I_a+I_b+I_c\right)Z_n \\ {V}_{bg}=\left(Z_Y+Z_n\right)I_b+Z_n\left(I_a+I_c\right) \end{gathered}$$

Substitute $75\angle 180°V$for $V_{bg},3+4\Omega$for $Z_Y$and$j1\Omega$for $Z_n$into above equation.

$$\begin{gathered} 75\angle 180°=\left(3+j4+j1\right)I_b+j1\left(I_a+I_c\right) \\ -75=j1I_a+\left(3+j5\right)I_b+j1I_c \end{gathered}$$ (7)

Apply the Kirchhoff’s voltage law in loop 3.

$$\begin{gathered} V_{ag}=I_c{Z}_Y+I_n{Z}_n \\ {V}_{ag}=I_c{Z}_y+\left(I_a+I_b+I_c\right)Z_n \\ {V}_{cg}=\left(Z_Y+Z_n\right)I_c+Z_n\left(I_a+I_b\right) \end{gathered}$$

Substitute $50\angle 90°V$for $V_{bg},3+j4\Omega$for $Z_Y$and $j1\Omega$for$Z_n$into above equation.

$$\begin{gathered} 50\angle 90°=\left(3+j4+j1\right)I_c+j1\left(I_a+I_c\right) \\ j50=j1I_a+j1I_b+\left(3+j5\right)I_c \end{gathered}$$ (8)

Write the equation (6), (7) and (8) into matrix from,

$$\begin{gathered} \left[\begin{array}{c}100\\ -75\\ j50\end{array}\right]=\left[\begin{array}{ccc}3+j5& j1& j1\\ j1& 3+j5& j1\\ j1& j1& 3+j5\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}3+j5& j1& j1\\ j1& 3+j5& j1\\ j1& j1& 3+j5\end{array}\right]\left[\begin{array}{c}100\\ 75\angle 180°\\ j50\angle 90°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}0.176\angle -56.5°& 0.0263\angle 150.12°& 0.0263\angle 150.12°\\ 0.0263\angle 150.12°& 0.176\angle -56.5°& 0.0263\angle 150.12°\\ 0.0263\angle 150.12°& 0.0263\angle 150.12°& 0.176\angle -56.5°\end{array}\right]\left[\begin{array}{c}100\\ 75\angle 180°\\ j50\angle 90°\end{array}\right] \end{gathered}$$

Solve further as,

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}19.95\angle -57.34°\\ 15.15\angle 132.44°\\ 8.52\angle 37.47°\end{array}\right]$

Hence, the line currents $I_a,I_b$and $I_c$without using symmetrical component are $19.95\angle -57.34°A,15.15\angle 132.44°$and $8.52\angle 37.47°A$respectively.

Calculate the line current using symmetrical current.

(b)

Calculate the sequence component of the line to ground voltage.

Substitute $100\angle 0°V$for $V_{ag},75\angle 180°V$for $V_{bg}$and $50\angle 90°V$for $V_{cg}$into equation (1).

$$\begin{gathered} \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}100\angle 0°V\\ 75\angle 180°\\ 50\angle 90°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle 240°\\ 1& 1\angle 240°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}100\angle 0°V\\ 75\angle 180°\\ 50\angle 90°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}100\angle 0°+75\angle 180°+50\angle 90°\\ 100\angle 0°+75\angle \left(120-180\right)°+50\angle \left(240+90\right)°\\ 100\angle 0°+75\angle \left(240-180\right)°+50\angle \left(120+90\right)°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}100\angle 0°+75\angle 180°+50\angle 90°\\ 100\angle 0°+75\angle -60°+50\angle 330°\\ 100\angle 0°+75\angle 60°+50\angle 210°\end{array}\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}55.9\angle 63.43°\\ 201.94\angle -26.45°\\ 102.32\angle 22.98°\end{array}\right] \\ \left[\begin{array}{c}{V}_{Lg0}\\ V_{Lg1}\\ V_{Lg2}\end{array}\right]=\left[\begin{array}{c}18.63\angle 63.43°\\ 67.31\angle -26.45°\\ 34.1\angle 22.98°\end{array}\right] \end{gathered}$$

Consider the zero-sequence network.

Calculate the zero-sequence component of line current

Substitute $18.63\angle 63.43°$for , $V_{Lg0},3+j4\Omega$for $Z_0$and $j1\Omega$for $Z_n$into equation (2).

$$\begin{gathered} I_0=\frac{18.63\angle 63.43°}{3+j4+3\left(j1\right)} \\ {I}_0=\frac{18.63\angle 63.43°}{3+j7} \\ {I}_0=\frac{18.63\angle 63.43°}{7.61\angle 66.80} \\ {I}_0=2.446\angle -3.37°A \end{gathered}$$

Consider positive sequence network.

Calculate the positive-sequence component of line current

Substitute $67.31\angle -36.45°$for $V_{Lg1}$and role="math" localid="1656403124616" $3+j4\Omega$for $Z_1$into equation (3).

$$\begin{gathered} I_1=\frac{67.31\angle -26.45°}{3+j4} \\ {I}_1=\frac{67.31\angle -26.45°}{5\angle 53.13} \\ {I}_1=13.46\angle -79.58°A \end{gathered}$$

Consider negative sequence network.

Calculate the positive-sequence component of line current

Substitute $34.1\angle 22.98°$for $V_{Lg2}$and $3+j4\Omega Z_2$for $Z_2$into equation (4).

$$\begin{gathered} I_2=\frac{34.1\angle 22.98°}{3-j4} \\ {I}_2=\frac{34.1\angle 22.98°}{5\angle 53.13°} \\ {I}_2=6.82\angle -30.15°A \end{gathered}$$

Calculate the line current.

Substitute $2.446\angle -3.37°A$for $I_0,13.46\angle -79.58°A$for $I_1$and $6.82\angle -30.15°A$for $I_2$into equation (5).

$$\begin{gathered} \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}2.446\angle -3.37°\\ 13.46\angle -79.58°\\ 6.82\angle -30.15°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 240°& 1\angle 240°\\ 1& 1\angle 120°& 1\angle 240°\end{array}\right]\left[\begin{array}{c}2.446\angle -3.37°\\ 13.46\angle -79.58°\\ 6.82\angle -30.15°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}2.446\angle -3.37°+13.46\angle -79.58°+6.82\angle -30.15°\\ 2.446\angle -3.37°+13.46\angle \left(240-79.58\right)°+6.82\angle \left(120-30.15\right)°\\ 2.446\angle -3.37°+13.46\angle \left(140-79.58\right)°+6.82\angle \left(240-30.15\right)°\end{array}\right] \\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}19.97\angle -57.34°A\\ 15.15\angle 164.054°A\\ 8.535\angle 37.46°A\end{array}\right] \end{gathered}$$

Hence, the line current $I_a,I_b$and $I_c$using symmetrical component are $19.97\angle -57.34°A,15.15\angle 164.054°$and $A8.535\angle 37.46°A$respectively.

The comparison shows that first method is shorter than the second one but complex then the second method.