Question

For a balanced-$\mathbf{∆}$load with per-phase impedance of localid="1654163226842" ${\mathit{Z}}_{\mathbf{∆}}$, the equivalent Y-load has an open neutral; for the corresponding uncoupled sequence networks, localid="1654163195090" ${\mathit{Z}}_{\mathbf{0}}\mathbf{=}$___________ , ${\mathit{Z}}_{\mathbf{1}}\mathbf{=}$___________ , and${\mathit{Z}}_{\mathbf{2}}\mathbf{=}$ ___________.

Short answer

The zero$\left(Z_0\right)$, positive$\left(Z_1\right)$, and negative$\left(Z_2\right)$sequence impedance are $\mathbf{\infty }\mathbf{,}{\mathit{Z}}_{\mathbf{y}}$, and ${\mathit{Z}}_{\mathbf{y}}$respectively.

Step-by-step solution

Determine the equations to calculate the sequence impedance of delta connected load.

The relation between a sequence voltage and current is ${\mathit{V}}_{\mathbf{s}}\mathbf{=}{\mathit{Z}}_{\mathbf{s}}{\mathit{I}}_{\mathbf{s}}$, where ${\mathit{V}}_{\mathbf{s}}$is the sequence voltage, ${\mathit{Z}}_{\mathbf{s}}$is the sequence impedance, and ${\mathit{I}}_{\mathbf{s}}$is the sequence current.

The matrix that defines the relation between a sequence voltage and current is given below.

$\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}{Z}_y+3Z_n& 0& 0\\ 0& Z_y& 0\\ 0& 0& Z_y\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$

Here, ${\mathit{V}}_{\mathbf{0}\mathbf{,}}{\mathit{V}}_{\mathbf{1}}$, and${\mathit{V}}_{\mathbf{2}}$are the sequence voltage and ${\mathit{I}}_{\mathbf{0}\mathbf{,}}{\mathit{I}}_{\mathbf{1}}$, and ${\mathit{I}}_{\mathbf{2}}$are the sequence currents. ${\mathit{Z}}_{\mathbf{n}}$is the neutral impedance and ${\mathit{Z}}_{\mathbf{y}}$is the star connected load.

Calculate the sequence impedance of delta connected load.

Since the $\mathit{Y}$-load has open neutral, neutral impedance role="math" localid="1654166741459" ${\mathit{Z}}_{\mathbf{n}}\mathbf{=}\mathbf{\infty }$.

From the sequence matrix, the zero-sequence impedance can be written as ${\mathit{Z}}_{\mathbf{0}}\mathbf{=}{\mathit{Z}}_{\mathbf{y}}\mathbf{+}\mathbf{3}{\mathit{Z}}_{\mathbf{n}}$.

Substitute $\mathbf{\infty }$for ${\mathit{Z}}_{\mathbf{n}}$in the above equation.

$$\begin{gathered} {\mathit{Z}}_{\mathbf{0}}\mathbf{=}{\mathit{Z}}_{\mathbf{y}}\mathbf{+}\mathbf{3}\mathbf{(}\mathbf{\infty }\mathbf{)} \\ {\mathit{Z}}_{\mathbf{0}}\mathbf{=}{\mathit{Z}}_{\mathbf{y}}\mathbf{+}\mathbf{(}\mathbf{\infty }\mathbf{)} \\ {\mathit{Z}}_{\mathbf{0}}\mathbf{=}\mathbf{(}\mathbf{\infty }\mathbf{)} \end{gathered}$$

The positive-sequence impedance is ${\mathit{Z}}_{\mathbf{1}}\mathbf{=}{\mathit{Z}}_{\mathbf{y}}$and the negative-sequence impedance is ${\mathit{Z}}_{\mathbf{2}}\mathbf{=}{\mathit{Z}}_{\mathbf{y}}$.

Therefore, the zero $\left(Z_0\right)$, positive$\left(Z_1\right)$and negative$\left(Z_2\right)$sequence impedance are $\mathbf{\infty }$, ${\mathit{Z}}_{\mathbf{y}}$, and ${\mathit{Z}}_{\mathbf{y}}$respectively.