Question

Let the terminal voltages at the two ends of the line section shown in Figure 8.24 be given by

$\begin{array}{l}{V}_{an}=\left(182+j70\right)kV{V}_{an\text{'}}=\left(154+j28\right)kV\\ V_{bn}=\left(72.24-j32.62\right)kV{V}_{bn\text{'}}=\left(44.24+j74.62\right)kV\\ V_{cn}=\left(-170.+j88.62\right)kV{V}_{cn\text{'}}=\left(-198.24+j46.24\right)kV\end{array}$

The line impedances are given by:

$Z_{aa}=j60\Omega Z_{ab}=j20\Omega Z_{an}=j80\Omega Z_{an}=0$

(a) Compute the line currents using symmetrical components. (Hint: See Problem 8.27.) (b) Compute the line currents without using symmetrical components.

Short answer

(a) The line current with using the symmetrical currents$I_a,I_b$and $I_c$are , $\left(1274.8-j82.3\right)\text{\hspace{0.33em}A}$and $\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$.

(b) The line current without using the symmetrical currents $I_a,I_b$and $I_c$are$\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$ , $\left(1274.8-j82.3\right)\text{\hspace{0.33em}A}$and $\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$.

Step-by-step solution

Write the given data by the question.

The terminal voltages at both the ends.

_{$\begin{array}{l}{V}_{an}=\left(182+j70\right)\text{\hspace{0.33em}kV}\\ V_{a^{\text{'}}{n}^{\text{'}}}=\left(154+j28\right)\text{\hspace{0.33em}kV}\\ V_{bn}=\left(72.24-j32.62\right)\text{\hspace{0.33em}kV}\\ V_{b^{\text{'}}{n}^{\text{'}}}=\left(44.24+j74.62\right)\text{\hspace{0.33em}kV}\end{array}$}

Also,

$\begin{array}{l}{V}_{cn}=\left(-170.+j88.62\right)\text{\hspace{0.33em}kV}\\ V_{c^{\text{'}}{n}^{\text{'}}}=\left(-198.24+j46.24\right)\text{\hspace{0.33em}kV}\end{array}$

The line impedances are,

$\begin{array}{l}{Z}_{aa}=j60\Omega \\ Z_{ab}=j20\Omega \\ Z_{an}=j80\Omega \\ Z_{an}=0\end{array}$

Determine the formulas to calculate the line current using symmetrical component and without using symmetrical component.

The equation between the phase current, phase voltage and phase impedance is given as follows.

$V_p=Z_p{I}_p$ …… (1)

Here$V_p$is the phase voltage,$Z_p$is the phase impedance and$I_p$is the phase current.

The equation to calculate the sequence impedance is given as follows.

$Z_s=A^{-1}{Z}_{P}A$ …… (2)

The equation to calculate the sequence voltage is given as follows.

$V_s=A^{-1}{V}_p$ …… (3)

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

$I_s=Z_s^{-1}{V}_s$ …… (4)

The equation to calculate the phase current is given as follows.

$I_p=A{I}_s$ …… (5)

Calculate the line current using the symmetrical component.

(a)

For the three-phase system with neutral, the sum of all the current is equal to zero.

$\begin{array}{c}{I}_a+I_b+I_c+I_n=0\\ I_n=-\left(I_a+I_b+I_c\right)\end{array}$ …… (6)

Apply the Kirchhoff’s voltage law for the phase which includes self and mutual impedances.

$\begin{array}{c}{V}_{an}=Z_{a0}{I}_a+Z_{ab}{I}_b+Z_{ab}{I}_c+Z_{an}{I}_n+V_{a^{\text{'}}{n}^{\text{'}}}-\left(Z_{nn}{I}_n+Z_{an}{I}_a+Z_{an}{I}_b+Z_{an}{I}_c\right)\\ V_{an}-V_{a^{\text{'}}{n}^{\text{'}}}=\left(Z_{a0}-Z_{an}\right)I_a+\left(Z_{ab}-Z_{an}\right)\left(I_b+I_c\right)+\left(Z_{an}-Z_{nn}\right)I_n\end{array}$

Substitute$-\left(I_a+I_b+I_c\right)$for$I_n$into above equation.

$\begin{array}{l}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}=\left(Z_{a0}-Z_{an}\right)I_a+\left(Z_{ab}-Z_{an}\right)\left(I_b+I_c\right)-\left(Z_{an}-Z_{nn}\right)\left(I_a+I_b+I_c\right)\\ V_{an}-V_{a^{\text{'}}{n}^{\text{'}}}=\left\{\begin{array}{l}\left(Z_{a0}-Z_{an}-Z_{an}+Z_{nn}\right)I_a+\left(Z_{ab}-Z_{an}-Z_{an}+Z_{nn}\right)I_b\\ +\left(Z_{ab}-Z_{an}-Z_{an}+Z_{nn}\right)I_c\end{array}\right\}\\ V_{an}-V_{a^{\text{'}}{n}^{\text{'}}}=\left(Z_{a0}+Z_{nn}-2Z_{an}\right)I_a+\left(Z_{ab}+Z_{an}-2Z_{an}\right)I_b+\left(Z_{ab}+Z_{nn}-2Z_{an}\right)I_c\end{array}$….. (7)

Solve for phase bas,

$V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}=\left(Z_{ab}+Z_{nn}-2Z_{an}\right)I_a+\left(Z_{aa}+Z_{an}-2Z_{an}\right)I_b+\left(Z_{ab}+Z_{nn}-2Z_{an}\right)I_c$ ….. (8)

Solve for phase c as,

$V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}=\left(Z_{ab}+Z_{nn}-2Z_{an}\right)I_a+\left(Z_{ab}+Z_{an}-2Z_{an}\right)I_b+\left(Z_{aa}+Z_{nn}-2Z_{an}\right)I_c$ ….. (9)

Write the equation (7), (8) and (9) into the matrix form,

$\left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]=\left[\begin{array}{ccc}{Z}_{a0}+Z_{nn}-2Z_{an}& Z_{ab}+Z_{an}-2Z_{an}& Z_{ab}+Z_{nn}-2Z_{an}\\ Z_{ab}+Z_{nn}-2Z_{an}& Z_{aa}+Z_{an}-2Z_{an}& Z_{ab}+Z_{nn}-2Z_{an}\\ Z_{ab}+Z_{nn}-2Z_{an}& Z_{ab}+Z_{an}-2Z_{an}& Z_{aa}+Z_{nn}-2Z_{an}\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

Substitute $j60\Omega$for$Z_{aa}$, $j20\Omega$for$Z_{ab}$,$j80\Omega$for $Z_{nn}$and$0$for $Z_{an}$into above equation.

$\begin{array}{c}\left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]=\left[\begin{array}{ccc}j60+j80-2\left(0\right)& j20+j80-2\left(0\right)& j20+j80-2\left(0\right)\\ j20+j80-2\left(0\right)& j60+j80-2\left(0\right)& j20+j80-2\left(0\right)\\ j20+j80-2\left(0\right)& j20+j80-2\left(0\right)& j60+j80-2\left(0\right)\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]\\ \left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]=\left[\begin{array}{ccc}j140& j100& j100\\ j100& j140& j100\\ j100& j100& j140\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]\end{array}$ …… (10)

Compare the above matrix with the equation (1).

$\begin{array}{l}{V}_p=\left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]\\ Z_p=\left[\begin{array}{ccc}j140& j100& j100\\ j100& j140& j100\\ j100& j100& j140\end{array}\right]\\ I_p=\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]\end{array}$

Calculate the sequence impedance.

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}j140& j100& j100\\ j100& j140& j100\\ j100& j100& j140\end{array}\right]$for$Z_s$and $\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]$for $A$into equation (2).

$\begin{array}{l}{Z}_s=\frac{j}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{ccc}140& 100& 100\\ 100& 140& 100\\ 100& 100& 140\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\\ Z_s=\frac{j}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{ccc}340& 40& 40\\ 340& 40a^2& 40a\\ 340& 40a& 40a^2\end{array}\right]\\ Z_s=j\left[\begin{array}{ccc}340& 0& 0\\ 0& 40& 0\\ 0& 0& 40\end{array}\right]\end{array}$

Calculate the sequence voltages.

Substitute$\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]$ for $A^{-1}$and $\left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]$ for $V_p$into equation (3).

$V_s=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]$

Substitute $\left(182+j70\right)kV$for$V_{an}$, $\left(72.24-j32.62\right)kV$for $V_{bn}$, $\left(-170.+j88.62\right)kV$for$V_{cn}$, $\left(154+j28\right)kV$for$V_{a^{\text{'}}{n}^{\text{'}}}$,$\left(44.24+j74.62\right)kV$for $V_{b^{\text{'}}{n}^{\text{'}}}$and $\left(-198.24+j46.24\right)kV$for$V_{c^{\text{'}}{n}^{\text{'}}}$into equation (3).

$\begin{array}{l}{V}_s=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}\left(182+j70\right)-\left(154+j28\right)\\ \left(72.24-j32.62\right)-\left(44.24+j74.62\right)\\ \left(-170.+j88.62\right)-\left(-198.24+j46.24\right)\end{array}\right]\times {10}^3\\ V_s=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}28+j42\\ 28+j107.24\\ 28+j42\end{array}\right]\times {10}^3\\ V_s=\frac{1}{3}\left[\begin{array}{c}28+j42+28+j107.24+28+j42\\ 28+j42+a\left(28+j107.24\right)+a^2\left(28+j42\right)\\ 28+j42+a^2\left(28+j107.24\right)+a\left(28+j42\right)\end{array}\right]\times {10}^3\\ V_s=\frac{1}{3}\left[\begin{array}{c}84+j191.24\\ 28\left(1+a+a^2\right)+j42\left(1+a^2\right)+107.24a\\ 28\left(1+a+a^2\right)+j42\left(1+a\right)+107.24a^2\end{array}\right]\times {10}^3\end{array}$

Substitute$0$for$1+a+a^2$and $-a$for$1+a^2$into above equation.

$\begin{array}{l}{V}_s=\frac{1}{3}\left[\begin{array}{c}84+j191.24\\ 28\left(0\right)+j42\left(-a\right)+j107.24a\\ 28\left(0\right)+j42\left(a^2\right)+j107.24a^2\end{array}\right]\times {10}^3\\ V_s=\frac{1}{3}\left[\begin{array}{c}84+j191.24\\ j42\left(-a\right)+j107.24a\\ j42\left(a^2\right)+j107.24a^2\end{array}\right]\times {10}^3\\ V_s=\frac{1}{3}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\times {10}^3\end{array}$

Calculate the sequence current.

Substitute $\frac{1}{3}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\times {10}^3$for$V_s$and $j\left[\begin{array}{ccc}340& 0& 0\\ 0& 40& 0\\ 0& 0& 40\end{array}\right]$for $Z_s$into equation (4).

$\begin{array}{l}{I}_s=\frac{j}{3}{\left[\begin{array}{ccc}340& 0& 0\\ 0& 40& 0\\ 0& 0& 40\end{array}\right]}^{-1}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\\ I_s=\frac{j\times {10}^3}{3}\frac{adj\left[\begin{array}{ccc}340& 0& 0\\ 0& 40& 0\\ 0& 0& 40\end{array}\right]}{\left|\left[\begin{array}{ccc}340& 0& 0\\ 0& 40& 0\\ 0& 0& 40\end{array}\right]\right|}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\\ I_s=\frac{j\times {10}^3}{3}\frac{\left[\begin{array}{ccc}1600& 0& 0\\ 0& 13600& 0\\ 0& 0& 13600\end{array}\right]}{544000}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\\ I_s=\frac{j}{3}\frac{\left[\begin{array}{ccc}1600& 0& 0\\ 0& 13600& 0\\ 0& 0& 13600\end{array}\right]}{544}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\end{array}$

Solve further as,

$\begin{array}{l}{I}_s=j\frac{\left[\begin{array}{ccc}1600& 0& 0\\ 0& 13600& 0\\ 0& 0& 13600\end{array}\right]}{1632}\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\\ I_s=j\left[\begin{array}{ccc}0.980& 0& 0\\ 0& 8.333& 0\\ 0& 0& 8.333\end{array}\right]\left[\begin{array}{c}84+j191.24\\ j65.24a\\ j65.24a^2\end{array}\right]\\ I_s=\left[\begin{array}{c}187.5+j82.3\\ 543.7a\\ 543.7a^2\end{array}\right]\end{array}$

Calculate the phase current.

Substitute $\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]$ for $A$and $\left[\begin{array}{c}187.5+j82.3\\ 543.7a\\ 543.7a^2\end{array}\right]$for $I_s$into equation (5).

$\begin{array}{l}{I}_p=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}187.5+j82.3\\ 543.7a\\ 543.7a^2\end{array}\right]\\ I_p=\left[\begin{array}{c}187.5+j82.3+543.7a+543.7a^2\\ 187.5+j82.3+543.7a^3+543.7a^3\\ 187.5+j82.3+543.7a^2+543.7a^4\end{array}\right]\\ I_p=\left[\begin{array}{c}-356.3-j82.3\\ 1274.8-j82.3\\ -356.3-j82.3\end{array}\right]\end{array}$

Hence the line current with using the symmetrical currents$I_a,I_b$ and $I_c$are$\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$ , $\left(1274.8-j82.3\right)\text{\hspace{0.33em}A}$and $\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$.

Calculate the line current without using symmetrical component.

$\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$(b)

By using the equation (10), calculate the line current.

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]={\left[\begin{array}{ccc}j140& j100& j100\\ j100& j140& j100\\ j100& j100& j140\end{array}\right]}^{-1}\left[\begin{array}{c}{V}_{an}-V_{a^{\text{'}}{n}^{\text{'}}}\\ V_{bn}-V_{b^{\text{'}}{n}^{\text{'}}}\\ V_{cn}-V_{c^{\text{'}}{n}^{\text{'}}}\end{array}\right]$

Substitute $\left(182+j70\right)kV$ for$V_{an}$, $\left(72.24-j32.62\right)kV$for$V_{bn}$,$\left(-170.+j88.62\right)kV$for$V_{cn}$, for $V_{a^{\text{'}}{n}^{\text{'}}}$, $\left(44.24+j74.62\right)kV$for$V_{b^{\text{'}}{n}^{\text{'}}}$ and $\left(-198.24+j46.24\right)kV$for $V_{c^{\text{'}}{n}^{\text{'}}}$into above equation.

$\begin{array}{l}\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]={\left[\begin{array}{ccc}j140& j100& j100\\ j100& j140& j100\\ j100& j100& j140\end{array}\right]}^{-1}\left[\begin{array}{c}\left(182+j70\right)-\left(154+j28\right)\\ \left(72.24-j32.62\right)-\left(44.24+j74.62\right)\\ \left(-170.+j88.62\right)-\left(-198.24+j46.24\right)\end{array}\right]\times {10}^3\\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}\frac{3}{170}& -\frac{1}{136}& -\frac{1}{136}\\ -\frac{1}{136}& \frac{3}{170}& -\frac{1}{136}\\ -\frac{1}{136}& -\frac{1}{136}& \frac{3}{170}\end{array}\right]\left[\begin{array}{c}28+j42\\ 28+j107.24\\ 28+j42\end{array}\right]\times {10}^3\\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}\frac{3}{170}\left(28+j42\right)-\frac{1}{136}\left(28+j107.24\right)-\frac{1}{136}\left(28+j42\right)\\ -\frac{1}{136}\left(28+j42\right)+\frac{3}{170}\left(28+j107.24\right)-\frac{1}{136}\left(28+j42\right)\\ -\frac{1}{136}\left(28+j42\right)-\frac{1}{136}\left(28+j107.24\right)+\frac{3}{170}\left(28+j42\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}-356.3-j82.3\\ 1274.8-j82.3\\ -356.3-j82.3\end{array}\right]\end{array}$

Hence, the line current without using the symmetrical currents$I_a,I_b$ and $I_c$are $\left(-356.3-j82.3\right)\text{\hspace{0.33em}A}$, $\left(1274.8-j82.3\right)\text{\hspace{0.33em}A}$and .