Question

For the unbalanced three-phase system described by${\mathit{I}}_{\mathbf{a}}\mathbf{=}\mathbf{10}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}$,${\mathit{I}}_{\mathbf{b}}\mathbf{=}\mathbf{8}\mathbf{\angle }\mathbf{-}{\mathbf{90}}^{\mathbf{°}}\mathbf{}\mathit{A}$and${\mathit{I}}_{\mathbf{c}}\mathbf{=}\mathbf{6}\mathbf{\angle }{\mathbf{150}}^{\mathbf{°}}\mathbf{}\mathit{A}$.

Compute the symmetrical components${\mathit{I}}_{\mathbf{0}}$, ${\mathit{I}}_{\mathbf{1}}$and ${\mathit{I}}_{\mathbf{2}}$.

Short answer

Answer

The positive sequence current is$I_1=7.74\angle 17.{56}^{°}A$.

The negative sequence current is$I_2=1.22\angle -33.{07}^{°}A$.

The zero sequence current is$I_0=2.31\angle -46.{15}^{°}A$.

Step-by-step solution

Write the given data from the question.

Assume that a line current of an un-balanced three-phase network are$I_a=10\angle 0^{°},I_0=8\angle -{90}^{°}A,I_c=6\angle {150}^{°}A$.

Determine the formula of phase sequence current.

Write the formula of positive sequence current.

${\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(I_a+a{I}_b+a^2{I}_c\right)$ …… (1)

Here,${\mathit{I}}_{\mathbf{a}}$, ${\mathit{I}}_{\mathbf{b}}$and ${\mathit{I}}_{\mathbf{c}}$are line current of three phase network.

Write the formula of negative sequence current.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{I}}_{\mathbf{a}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{I}}_{\mathbf{b}}\mathbf{+}{\mathbf{aI}}_{\mathbf{c}}\mathbf{)}$ …… (2)

Here,${\mathit{I}}_{\mathbf{a}}$, ${\mathit{I}}_{\mathbf{b}}$and ${\mathit{I}}_{\mathbf{c}}$are line current of three phase network.

Write the formula of zero sequence current.

${\mathit{I}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{I}}_{\mathbf{a}}\mathbf{+}{\mathbf{I}}_{\mathbf{b}}\mathbf{+}{\mathbf{I}}_{\mathbf{c}}\mathbf{)}$ …… (3)

Here,${\mathit{I}}_{\mathbf{a}}$, ${\mathit{I}}_{\mathbf{b}}$and ${\mathit{I}}_{\mathbf{c}}$are line current of three phase network.

Determine the line current of an un-balanced three-phase network.

First we convert the line voltage in polar form into rectangular form for further solving.

Write the value of line current$I_a$of three phase network.

$$\begin{gathered} I_a=10\angle 0^{°} \\ =10A \end{gathered}$$

Write the value of line current$I_b$of three phase network.

$$\begin{gathered} I_b=8\angle -{90}^{°} \\ =-j8A \end{gathered}$$

Write the value of line current$I_c$ of three phase network.

$$\begin{gathered} I_c=6\angle {150}^{°} \\ =-5.20+j3A \end{gathered}$$

Determine the positive sequence current.

Substitute the value ofrole="math" localid="1656314812696" $I_a,I_b$and $I_c$into equation (1).

$$\begin{gathered} {I}_1=\frac{1}{3}\left(10\angle 0^{°}+8\angle \left(-{90}^{°}+{120}^{°}\right)+6\angle \left({150}^{°}+240\right)\right) \\ =\frac{1}{3}\left(10\angle 0^{°}+8\angle {30}^{°}+6\angle {390}^{°}\right) \\ =\frac{1}{3}\left(10+6.93+j4+5.2+j3\right) \\ =\frac{1}{3}\left(22.13+j7\right) \end{gathered}$$

As further solve as

$I_1=7.38+j2.33or7.74\angle 17.{56}^{°}A$

Determine the negative sequence current.

Substitute the value of$I_a$,$I_b$and $I_c$into equation (2).

$$\begin{gathered} I_2=\frac{1}{3}\left(10\angle 0^{°}+8\angle \left(-{90}^{°}+{240}^{°}\right)+6\angle \left({150}^{°}+{120}^{°}\right)\right) \\ =\frac{1}{3}\left(10\angle 0^{°}+8\angle {150}^{°}+6\angle {270}^{°}\right) \\ =\frac{1}{3}\left(10-6.93+j4-j6\right) \\ =0.023-j0.67or1.22\angle -33.{07}^{°}A \end{gathered}$$

Determine the zero sequence current.

Substitute the value of $I_a$, $I_b$and$I_c$into equation (3).

$$\begin{gathered} I_0=\frac{1}{3}\left(10\angle 0^{°}+8\angle -{90}^{°}+6\angle {150}^{°}\right) \\ =\frac{1}{3}\left(10-j8-5.2+j3\right) \\ =\frac{1}{3}\left(4.80-j5\right) \\ =1.6-j1.67or2.31\angle -46.{15}^{°}A \end{gathered}$$

Therefore, the symmetrical components of line currents are given as$I_0=2.31\angle 46.{15}^{°}A,I_1=7.74\angle 17.{56}^{°}A$and$I_2=1.22\angle -33.{07}^{°}A$.