Question

A three-phase balanced Y-connected load with series impedances of $(\mathbf{6}\mathbf{+}\mathbf{j}\mathbf{24})\mathbf{\Omega }$ per phase and mutual impedance between any two phases of$\mathbf{j}\mathbf{3}\mathbf{\Omega }$ is supplied by a three-phase unbalanced source with line-to-neutral voltages of ${\mathbf{V}}_{\mathrm{an}}\mathbf{=}\mathbf{200}\angle \mathbf{25}\mathbf{°}$,${\mathbf{V}}_{\mathrm{bn}}=\mathbf{100}\angle -\mathbf{155}\mathbf{°}$ ,${\mathbf{V}}_{\mathrm{cn}}=\mathbf{80}\angle \mathbf{100}\mathbf{°}\mathbf{V}$ . The load and source neutrals are both solidly grounded. Determine:

(a) the load sequence impedance matrix,

(b) the symmetrical components of the line-to-neutral voltages,

(c) the symmetrical componentsof the load currents, and

(d) the load currents

Short answer

(a) The load sequence impedance matrix is $Z_s=\left[\begin{array}{ccc}(6+j30)& 0& 0\\ 0& (6+j21)& 0\\ 0& 0& (6+j21)\end{array}\right]$.

(b) The symmetrical components of the line-to-neutral voltages are $V_0=47.78\angle 57.63°\text{\hspace{0.17em}V}$,$V_1=112.79\angle 359.97°\text{\hspace{0.17em}V}$ and $V_2=61.62\angle 45.88°\text{\hspace{0.17em}V}$.

(c) The symmetrical components of the load currents are $I_0=1.56\angle -21.06°\text{\hspace{0.17em}A}$, $I_1=5.16\angle 285.92°\text{\hspace{0.17em}A}$and $I_2=2.82\angle -28.17°\text{\hspace{0.17em}A}$.

(d) The loads current are $I_a=8.70\angle -51.98°\text{\hspace{0.17em}A}$,$I_b=5.058\angle 136°\text{\hspace{0.17em}A}$ and $I_c=3.126\angle 32.05°\text{\hspace{0.17em}A}$.

Step-by-step solution

Write the given data from the question.

Theseries impedances of three-phase balanced Y-connected load are $(\text{6}+j\text{24})\Omega$.

The mutual impedance between any two phases of$j\text{3\hspace{0.17em}}\Omega$.

The line-to-neutral voltages of three-phase unbalanced source are ${\text{V}}_{\text{an}}=\text{200}\angle \text{25°}$, ${\text{V}}_{\text{bn}}=\text{100}\angle -\text{155°}$, ${\text{V}}_{\text{cn}}=\text{80}\angle \text{100°\hspace{0.17em}V}$.

The load and source neutrals are both solidly grounded.

Determine the formula of load sequence impedance, symmetrical components of the line-to-neutral voltages and load currents.

Write the formula of load sequence impedance matrix.

${\mathbf{Z}}_S\mathbf{=}{\mathbf{A}}^{-1}{\mathbf{Z}}_P\mathbf{A}$ …… (1)

Here, ${\mathbf{Z}}_P$is phase impedance matrix, ${\mathbf{A}}^{-1}$is inverse of matrix and $\mathbf{A}$is adjoin matrix.

Write the formula ofsymmetrical components of the line-to-neutral voltages.

role="math" localid="1656774765183" ${\mathbf{V}}_0=\frac{1}{3}({\mathbf{V}}_{\mathrm{an}}\mathbf{+}{\mathbf{V}}_{\mathrm{bn}}\mathbf{+}{\mathbf{V}}_{\mathrm{cn}})$ …… (2)

Here,${\mathbf{V}}_{\mathrm{an},}{\mathbf{V}}_{\mathrm{bn}}$and ${\mathbf{V}}_{\mathrm{cn}}$are line to ground voltages.

Write the formula ofsymmetrical components of the line-to-neutral voltages.

role="math" localid="1656774893864" ${\mathbf{V}}_1=\frac{1}{3}({\mathbf{V}}_{\mathrm{an}}\mathbf{+}{\mathbf{aV}}_{\mathrm{bn}}\mathbf{+}{\mathbf{a}}^2{\mathbf{V}}_{\mathrm{cn}})$ …… (3)

Here ${\mathbf{V}}_{\mathrm{an}}\mathbf{,}{\mathbf{V}}_{\mathrm{bn}}$and ${\mathbf{V}}_{\mathrm{cn}}$are line to ground voltages and $\mathbf{a}$, ${\mathbf{a}}^2$are identities .

Write the formula ofsymmetrical components of the line-to-neutral voltages ${\mathbf{V}}_2$.

${\mathbf{V}}_2=\frac{1}{3}({\mathbf{V}}_{\mathrm{an}}\mathbf{+}{\mathbf{a}}^2{\mathbf{V}}_{\mathrm{bn}}\mathbf{+}{\mathbf{aV}}_{\mathrm{cn}})$ …… (4)

Here ${\mathbf{V}}_{\mathrm{an}}\mathbf{,}{\mathbf{V}}_{\mathrm{bn}}$and ${\mathbf{V}}_{\mathrm{cn}}$are line to ground voltages and $\mathbf{a}$, ${\mathbf{a}}^2$are identities .

Write the formula ofsymmetrical components of the load currents ${\mathbf{I}}_0$.

role="math" localid="1656775227566" ${\mathbf{I}}_0=\frac{{\mathbf{V}}_0}{{\mathbf{Z}}_0}$ …… (5)

Here ${\mathbf{V}}_0$is a line-to-neutral voltage and ${\mathbf{Z}}_0$ is load sequence impedance.

Write the formula of symmetrical components of the load currents ${\mathbf{I}}_1$.

role="math" localid="1656775340431" ${\mathbf{I}}_1=\frac{{\mathbf{V}}_1}{{\mathbf{Z}}_1}$ …… (6)

Here,${\mathbf{V}}_1$is a line-to-neutral voltage and ${\mathbf{Z}}_1$is load sequence impedance.

Write the formula ofsymmetrical components of the load currents ${\mathbf{I}}_2$.

role="math" localid="1656775396527" ${\mathbf{I}}_2=\frac{{\mathbf{V}}_2}{{\mathbf{Z}}_2}$ …… (7)

Here, ${\mathbf{V}}_2$is a line-to-neutral voltage and ${\mathbf{Z}}_2$is load sequence impedance.

Write the formula ofloads current ${\mathbf{I}}_a$.

${\mathbf{I}}_a={\mathbf{I}}_0\mathbf{+}{\mathbf{I}}_1\mathbf{+}{\mathbf{I}}_2$ …… (8)

Here,${\mathbf{I}}_0$is zero sequence current,${\mathbf{I}}_1$is positive sequence current and${\mathbf{I}}_2$is negative sequence current.

Write the formula ofloads current ${\mathbf{I}}_a$.

${\mathbf{I}}_a={\mathbf{I}}_0\mathbf{+}{\mathbf{a}}^2{\mathbf{I}}_1\mathbf{+}{\mathbf{aI}}_2$ …… (9)

Here, ${\mathbf{I}}_0$is zero sequence current, ${\mathbf{I}}_1$is positive sequence current, ${\mathbf{I}}_2$is negative sequence current and ${\mathbf{a}}^2\mathbf{,}\mathbf{}\mathbf{a}$are identities.

Write the formula ofloads current ${\mathbf{I}}_c$.

role="math" localid="1656775816193" ${\mathbf{I}}_c={\mathbf{I}}_0\mathbf{+}{\mathbf{aI}}_1\mathbf{+}{\mathbf{a}}^2{\mathbf{I}}_2$ …… (10)

Here,${\mathbf{I}}_0$is zero sequence current, ${\mathbf{I}}_1$is positive sequence current, ${\mathbf{I}}_2$is negative sequence current and $\mathbf{a}\mathbf{,}{\mathbf{a}}^2$are identities.

(a) Determine the load sequence impedance matrix.

The phase impedance matrix of a three-phase balanced load with solid neutral grounding is,

$Z_P=\left[\begin{array}{ccc}{Z}_{aa}& Z_m& Z_m\\ Z_m& Z_{bb}& Z_m\\ Z_m& Z_m& Z_{cc}\end{array}\right]$ …… (11)

Here , $Z_{aa}$, $Z_{bb}$and $Z_{cc}$are self-phase impedance and$Z_m$is mutual impedance between any two phases.

Therefore, the phase impedance matrix is,

$Z_P=\left[\begin{array}{ccc}(6+j24)& j3& j3\\ j3& (6+j24)& j3\\ j3& j3& (6+j24)\end{array}\right]\Omega$

Determine the load sequence impedance matrix.

Substitute $\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]$for $A^{-1}$, $\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]$for $A$and $\left[\begin{array}{ccc}(6+j24)& j3& j3\\ j3& (6+j24)& j3\\ j3& j3& (6+j24)\end{array}\right]$for $Z_P$into equation (1).

Also following identities hold true for $a=1\angle 120°$.

$\begin{array}{l}{a}^4=a\\ a^3=1\\ 1+a+a^2=0\end{array}$

Use above identities in next step.

As further solve as

$\begin{array}{c}{Z}_s=\frac{1}{3}\left[\begin{array}{ccc}3(6+j30)& 0& 0\\ 0& 3(6+j21)& 0\\ 0& 0& (6+j21)\end{array}\right]\\ =\left[\begin{array}{ccc}(6+j30)& 0& 0\\ 0& (6+j21)& 0\\ 0& 0& (6+j21)\end{array}\right]\end{array}$

Hence, $Z_0=6+j30$, $Z_1=6+j21$and $Z_2=6+j21$.

Therefore, the load sequence impedance matrix is $\left[\begin{array}{ccc}(6+j30)& 0& 0\\ 0& (6+j21)& 0\\ 0& 0& (6+j21)\end{array}\right]$.

(b) Determine the symmetrical components of the line-to-neutral voltages.

Determine symmetrical components of line-to-neutral voltages.

Substitute$200\angle 25°\text{\hspace{0.17em}V}$for $V_{an}$,$100\angle -155°\text{\hspace{0.17em}V}$for$V_{bn}$and$80\angle 100°\text{\hspace{0.17em}V}$for$V_{cn}$into equation (2).

$\begin{array}{c}{V}_0=\frac{1}{3}(200\angle 25°+100\angle -155°+80\angle 100°)\\ =\frac{1}{3}(181.26+j84.52-90.63-j42.26-13.89+j78.78)\\ =\frac{1}{3}(76.74+j121.04)\\ =25.58+j40.35\text{\hspace{0.17em}or\hspace{0.17em}47.78}\angle \text{57.63}°\end{array}$

Substitute $200\angle 25°\text{\hspace{0.17em}V}$for role="math" localid="1656776896841" $V_{an}$, $100\angle (-155°+120°)$for$a{V}_{bn}$and$80\angle (100°+240°)$for$a^2{V}_{cn}$into equation (3).

$\begin{array}{c}{V}_0=\frac{1}{3}(200\angle 25°+100\angle (-155°+120°)+80\angle (100°+240°))\\ =\frac{1}{3}(200\angle 25°+100\angle -35°+80\angle 340°)\\ =\frac{1}{3}(181.26+j84.52+81.92-j57.36+75.18-j27.36)\\ =\frac{1}{3}(338.36-j0.2)\end{array}$

As further solve as

$V_1=112.79-j0.067\text{\hspace{0.17em}or\hspace{0.17em}112.79}\angle \text{359.97}°$

Substitute $200\angle 25°\text{\hspace{0.17em}V}$ for $V_{an}$, $100\angle (-155°+240°)$for $a^2{V}_{bn}$and $80\angle (100°+120°)$for $a{V}_{cn}$into equation (4).

$\begin{array}{c}{V}_0=\frac{1}{3}(200\angle 25°+100\angle (-155°+240°)+80\angle (100°+120°))\\ =\frac{1}{3}(200\angle 25°+100\angle 85°+80\angle 220°)\\ =\frac{1}{3}(181.26+j84.52+8.72-j99.62+61.28-j51.42)\\ =\frac{1}{3}(181.26+j84.52+8.72+j99.62-61.28-j51.42)\end{array}$

As further solve as

$V_2=42.9+j44.24\text{\hspace{0.17em}or\hspace{0.17em}61.62}\angle \text{45.88}°$

Therefore, the symmetrical components of line-to-neutral voltages are:

$\begin{array}{l}{V}_0=47.78\angle 57.63°\text{\hspace{0.17em}V}\\ V_1=112.79\angle 359.97°\text{\hspace{0.17em}V}\\ V_2=61.62\angle 45.88°\text{\hspace{0.17em}V}\end{array}$

(c) Determine the symmetrical components of the load currents.

Substitute$47.78\angle 57.63°$for$V_0$and$6+j30$ for$Z_0$into equation (5).

$\begin{array}{c}{I}_0=\frac{47.78\angle 57.63°}{6+j30}\\ =\frac{47.78\angle 57.63°}{30.59\angle 78.69°}\\ =1.56\angle (57.63°-78.69°)\\ =1.56\angle -21.06°\text{\hspace{0.17em}or\hspace{0.17em}1.46-j0.56\hspace{0.17em}A}\end{array}$

Substitute $112.79\angle 359.97°$for$V_1$and$6+j21$for$Z_1$into equation (6).

$\begin{array}{c}{I}_1=\frac{112.79\angle 359.97°}{6+j21}\\ =\frac{112.79\angle 359.97°}{21.84\angle 74.05°}\\ =5.16\angle (359.97°-74.05°)\end{array}$

Substitute$61.62\angle 45.88°$for$V_2$and$6+j21$for$Z_2$into equation (7).

$\begin{array}{c}{I}_2=\frac{61.62\angle 45.88°}{6+j21}\\ =\frac{61.62\angle 45.88°}{21.84\angle 74.05°}\\ =2.82\angle (45.88°-74.05°)\\ =2.82\angle -28.17°\text{\hspace{0.17em}or\hspace{0.17em}2.49-j1.33\hspace{0.17em}A}\end{array}$

Therefore, thesymmetrical components of the load currents are:

$\begin{array}{l}{I}_0=1.56\angle -21.06°\text{\hspace{0.17em}A}\\ I_1=5.16\angle 285.92°\text{\hspace{0.17em}A}\\ I_2=2.82\angle -28.17°\text{\hspace{0.17em}A}\end{array}$

(d) Determine the load currents.

Substitute$1.56\angle -21.06°$for $I_0$,$5.16\angle 285.92°$for$I_1$and$2.82\angle -28.17°$for$I_2$into equation (8).

$\begin{array}{l}{I}_a=1.56\angle -21.06°+5.16\angle 285.92°+2.82\angle -28.17°\\ =8.70\angle -51.98°\text{\hspace{0.17em}A}\end{array}$

Substitute $1.56\angle -21.06°$for $I_0$, $5.16\angle (285.92°+240°)$for $a^2{I}_1$and $2.82\angle (-28.17°+120°)$for $a{I}_2$into equation (9).

$\begin{array}{c}{I}_b=1.56\angle -21.06°+5.16\angle (285.92°+240°)+2.82\angle (-28.17°+120°)\\ =5.058\angle 136°\text{\hspace{0.17em}A}\end{array}$

Substitute$1.56\angle -21.06°$ for $I_0$,$5.16\angle (285.92°+120°)$ for$a{I}_1$ and$2.82\angle (-28.17°+240°)$ for$a^2{I}_2$ into equation (10).

$\begin{array}{c}{I}_c=1.56\angle -21.06°+5.16\angle (285.92°+120°)+2.82\angle (-28.17°+240°)\\ =3.126\angle 32.05°\text{\hspace{0.17em}A}\end{array}$

Therefore, the load currents are:

$\begin{array}{l}{I}_a=8.70\angle -51.98°\text{\hspace{0.17em}A}\\ I_b=5.058\angle 136°\text{\hspace{0.17em}A}\\ I_c=3.126\angle 32.05°\text{\hspace{0.17em}A}\end{array}$