Question

As shown in Figure 8.25, a balanced three-phase, positive-sequence source with ${\mathit{V}}_{\mathbf{A}\mathbf{B}}\mathbf{=}\mathbf{480}\mathbf{\angle }\mathbf{0}\mathbf{°}$volts is applied to an unbalanced$\mathit{\Delta }$ load. Note that one leg of the D is open. Determine (a) the load currents ${\mathit{I}}_{\mathbf{A}\mathbf{B}}$and ${\mathit{I}}_{\mathbf{B}\mathbf{C}}$; (b) the line currents ${\mathit{I}}_{\mathbf{A}}\mathbf{,}{\mathit{I}}_{\mathbf{B}}$and ${\mathit{I}}_{\mathbf{C}}$, which feed the$\mathit{\Delta }$ load; and (c) the zero-, positive-, and negative-sequence components of the line currents.

Short answer

$I_{AB}$ is$\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$ and$I_{BC}$ is$\left(-20-j12\right)\text{\hspace{0.33em}A}$.

(b) The line current$I_A,I_B$ and$I_C$ are$\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$,$\left(-40.38-j0.68\right)\text{\hspace{0.33em}A}$ and$\left(20+j12\right)\text{\hspace{0.33em}A}$.

(c) The zero, positive and negative sequence component of the line current are$0\text{\hspace{0.33em}A}$, $13.85-j23.09$and $6.53+j11.77$respectively.

Step-by-step solution

Write the given data from the question:

The voltage,$V_{AB}=480\angle 0°\text{\hspace{0.33em}V}$

The impedance of line AB,$Z_{AB}=18+j10\Omega$

The impedance of line BC,$Z_{BC}=18+j10\Omega$

Determine the formulas to calculate the load current, line current and sequence component of the line current.

The equation to calculate the load current in line AB is given as follows.

…… (1)

${\mathit{I}}_{\mathbf{A}\mathbf{B}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{A}\mathbf{B}}}{{\mathbf{Z}}_{\mathbf{A}\mathbf{B}}}$

The equation to calculate the load current in line BC is given as follows.

${\mathit{I}}_{\mathbf{B}\mathbf{C}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{B}\mathbf{C}}}{{\mathbf{Z}}_{\mathbf{B}\mathbf{C}}}$ …… (2)

The equation to calculate the line current through line A is given as follows.

${\mathit{I}}_{\mathbf{A}}\mathbf{=}{\mathit{I}}_{\mathbf{A}\mathbf{B}}\mathbf{-}{\mathit{I}}_{\mathbf{B}\mathbf{C}}$ …… (3)

The equation to calculate the line current through line B is given as follows.

${\mathit{I}}_{\mathbf{B}}\mathbf{=}{\mathit{I}}_{\mathbf{B}\mathbf{C}}\mathbf{-}{\mathit{I}}_{\mathbf{A}\mathbf{B}}$ …… (4)

The equation to calculate the line current through line C is given as follows.

${\mathit{I}}_{\mathbf{C}}\mathbf{=}{\mathit{I}}_{\mathbf{C}\mathbf{A}}\mathbf{-}{\mathit{I}}_{\mathbf{B}\mathbf{C}}$ …… (5)

The equation to calculate the sequence component of the line current is given as follows.

$\left[\begin{array}{c}{I}_{a0}\\ I_{b0}\\ I_{c0}\end{array}\right]\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_A\\ I_B\\ I_C\end{array}\right]$ …… (6)

Here, ${\mathit{I}}_{\mathbf{A}}\mathbf{,}{\mathit{I}}_{\mathbf{B}}$and${\mathit{I}}_{\mathbf{C}}$are the line current ,${\mathit{I}}_{\mathbf{a}\mathbf{0}}\mathbf{,}{\mathit{I}}_{\mathbf{a}\mathbf{1}}$and${\mathit{I}}_{\mathbf{a}\mathbf{2}}$are the sequence component of the line current.

Calculate the load currents.

(a)

Since the system is balance, the voltage$V_{BC}$ and$V_{CA}$, are

.$\begin{array}{l}{V}_{BC}=480\angle -120°\text{V\hspace{0.33em}}\\ V_{CA}=\text{\hspace{0.33em}480}\angle \text{120}°\text{\hspace{0.33em}V}\end{array}$

Calculate the load current in line AB.

Substitute$480\angle 0°\text{\hspace{0.33em}V}$for$V_{AB}$ and$18+j10\Omega$ for$Z_{AB}$into equation (1).

$\begin{array}{l}{I}_{AB}=\frac{480\angle 0°}{18+j10}\\ I_{AB}=20.38-j11.32\text{\hspace{0.33em}A}\end{array}$

Calculate the load current in line BC.

Substitute$480\angle -120°\text{\hspace{0.33em}V}$for$V_{BC}$ and$18+j10\Omega$ for$Z_{BC}$into equation (2).

$\begin{array}{l}{I}_{BC}=\frac{480\angle 120°}{18+j10}\\ I_{BC}=-20-j12\text{\hspace{0.33em}A}\end{array}$

Hence, the load current$I_{AB}$ is$\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$ and$I_{BC}$ is$\left(-20-j12\right)\text{\hspace{0.33em}A}$.

Calculate the line currents.

(b)

Calculate the line current through line A.

Substitute $\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$for$I_{AB}$and$0$for$I_{CA}$into equation (3).

$\begin{array}{l}{I}_A=\left(20.38-j11.32\right)-0\\ I_A=\left(20.38-j11.32\right)\text{\hspace{0.33em}A}\end{array}$

Calculate the line current through line B.

Substitute$\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$for$I_{AB}$and$\left(-20-j12\right)\text{\hspace{0.33em}A}$for$I_{BC}$into equation (4).

$\begin{array}{l}{I}_B=\left(-20-j12\right)-\left(20.38-j11.32\right)\\ I_B=-20-j12-20.38+j11.32\\ I_B=\left(-40.38-j0.68\right)\text{\hspace{0.33em}A}\end{array}$

Calculate the line current through line C.

Substitute$0$for$I_{CA}$ and$\left(-20-j12\right)\text{\hspace{0.33em}A}$for$I_{BC}$into equation (5).

$\begin{array}{l}{I}_C=0-\left(-20.38-j12\right)\\ I_C=\left(20+j12\right)\text{\hspace{0.33em}A}\end{array}$

Hence the line current$I_A,I_B$ and$I_C$ are$\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$,$\left(-40.38-j0.68\right)\text{\hspace{0.33em}A}$ and$\left(20+j12\right)\text{\hspace{0.33em}A}$.

Calculate the sequence component of the line current.

(c)

Calculate the sequence component of the line current.

Substitute$\left(20.38-j11.32\right)\text{\hspace{0.33em}A}$ for$I_A$, $\left(-40.38-j0.68\right)\text{\hspace{0.33em}A}$for$I_B$ and $\left(20+j12\right)\text{\hspace{0.33em}A}$fori$I_C$nto equation (6).

$\begin{array}{l}\left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle -120°\\ 1& 1\angle -120°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}\left(20.38-j11.32\right)\\ \left(-40.38-j0.68\right)\\ \left(20+j12\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}20.38-j11.32-40.38-j0.68+20+j12\\ 20.38-j11.32+1\angle 120°\left(-40.38-j0.68\right)+1\angle -120°\left(20+j12\right)\\ 20.38-j11.32+1\angle -120°\left(-40.38-j0.68\right)+1\angle 120°\left(20+j12\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}0\\ 41.55-j69.27\\ 19.59+j35.3\end{array}\right]\\ \left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]=\left[\begin{array}{c}0\\ 13.85-j23.09\\ 6.53+j11.77\end{array}\right]\end{array}$

Hence, the zero, positive and negative sequence component of the line current are,$0\text{\hspace{0.33em}A}$ $13.85-j23.09$and$6.53+j11.77$ respectively.