Question

One line of a three-phase generator is open-circuited, while the other

two are short-circuited to ground. The line currents are ${\mathbf{I}}_{\mathbf{a}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}{\mathbf{l}}_{\mathbf{b}}\mathbf{=}\mathbf{1200}\mathbf{\angle }{\mathbf{150}}^{\mathbf{\circ }}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{I}}_{\mathbf{C}}\mathbf{=}\mathbf{1200}\mathbf{\angle }\mathbf{+}{\mathbf{30}}^{\mathbf{\circ }}$. Find the symmetrical components of these currents. Also find the current into the ground.

Short answer

The Symmetrical components of the line currents are obtained as,

$$\begin{gathered} I_0=400\angle {90}^{\circ }A \\ I1=800\angle {270}^{\circ }A \\ I2=400\angle {90}^{\circ }A \end{gathered}$$

The fault current flowing through the ground is,

$I_f=1200\angle {90}^{\circ }A$

Step-by-step solution

Given data

The line currents are,

$$\begin{gathered} I_a=0 \\ {I}_b=1200\angle {150}^{\circ } \\ {I}_c=1200\angle +{30}^{\circ } \end{gathered}$$

Determine the formulas for positive sequence currentI1, negative sequence currentI2, and zero sequence currentI3with respect to the line currents flowing in the  generatorsIa ,Ib and Ic .

Formula for zero sequence current$I_0$.

${\mathbf{I}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(I_a+l_b+I_c\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\left(1\right)$

Formula for positive sequence current$I_1$.

${\mathbf{I}}_1\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{I}}_{\mathbf{a}}\mathbf{+}\mathit{a}{\mathbf{l}}_{\mathbf{b}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathbf{I}}_{\mathbf{c}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Formula for negative sequence current$I_2$.

${\mathbf{I}}_1\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{I}}_{\mathbf{a}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{b}}\mathbf{+}\mathit{a}{\mathbf{I}}_{\mathbf{c}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

 Step 3: Convert the line currentsIa,lbandIcpolar to rectangular form 

The expression for the line current $I_a$.

$$\begin{gathered} I_a=0\angle 0^{\circ } \\ =0A......\left(4\right) \end{gathered}$$

The expression for the line current $I_b$.

$$\begin{gathered} I_b=1200\angle {150}^{\circ } \\ =1039.23+j600A......\left(5\right) \end{gathered}$$

The expression for the line current $I_c$.

$$\begin{gathered} I_c=1200\angle +{30}^{\circ } \\ =1039.23+j600A......\left(6\right) \end{gathered}$$

Calculate the zero sequence current.

Put the current values of equations (4),(5) and(6) into the equation (1)

$$\begin{gathered} I_0=\frac{1}{3}\left(I_a+I_b+I_c\right) \\ =\frac{1}{3}\left(0\angle 0^{\circ }+1200\angle {150}^{\circ }+1200\angle {30}^{\circ }\right) \\ =\frac{1}{3}\left(0-1039.23+j600+1039.23+j600\right) \\ =\frac{1}{3}\left(j1200\right) \end{gathered}$$

Further, calculate the zero sequence current $I_0$.

$$\begin{gathered} I_0=j400 \\ =400\angle {90}^{\circ }\mathrm{A} \end{gathered}$$

Calculate the positive sequence current.

Put the current values of equations (4), (5) and (6) into the equation (2)

$$\begin{gathered} I_1=\frac{1}{3}\left(I_a+a{I}_b+a^2{I}_c\right) \\ =\frac{1}{3}\left(0\angle 0^{\circ }+1200\angle ({150}^{\circ }+{120}^{\circ })+1200\angle ({30}^{\circ }+{240}^{\circ })\right) \\ =\frac{1}{3}\left(0\angle 0^{\circ }+1200\angle {270}^{\circ }+1200\angle {270}^{\circ }\right) \\ =\frac{1}{3}\left(0-j1200-j1200\right) \end{gathered}$$

Further, calculate the positive sequence current$I_0$.

$$\begin{gathered} I_1=\frac{1}{3}\left(-j2400\right) \\ =-j800 \\ =800\angle {270}^{\circ }\mathrm{A}. \end{gathered}$$

Calculate the negative sequence current

Put the current values of equations (4), (5) and(6) into the equation (3)

$$\begin{gathered} I_2=\frac{1}{3}\left(I_a+a^2{I}_b+a{I}_c\right) \\ =\frac{1}{3}\left(0\angle 0^{\circ }+1200\angle ({150}^{\circ }+{120}^{\circ })+1200\angle ({30}^{\circ }+{240}^{\circ })\right) \\ =\frac{1}{3}\left(0\angle 0^{\circ }+1200\angle {390}^{\circ }+1200\angle {150}^{\circ }\right) \\ =\frac{1}{3}\left(0+1039.23+j600-1039.23+j600\right). \end{gathered}$$

Further, calculate the negative sequence current $I_2$.

$$\begin{gathered} I_2=\frac{1}{3}\left(j200\right) \\ =j400 \\ =400\angle {90}^{\circ }\mathrm{A}. \end{gathered}$$

Therefore, symmetrical components of the sequence currents give as

$$\begin{gathered} I_0=400\angle {90}^{\circ }\mathrm{A}. \\ {I}_1=800\angle {270}^{\circ }\mathrm{A}. \\ {I}_2=400\angle {90}^{\circ }\mathrm{A}. \end{gathered}$$

Determine fault current If

The two outgoing lines from generator are short-circuited to ground, the current into the ground is fault current $I_f$.

The summation of short-circuited line currents gives the fault current. So,

$$\begin{gathered} I_f=I_b+I_c \\ =1200\angle {150}^{\circ }+1200\angle {30}^{\circ } \\ =-1039.23+j600+1039.23+j600 \\ =j1200\mathrm{A}. \end{gathered}$$

Therefore, fault current flowing in the ground is$I_f=j1200\mathrm{A}.$