Question

Draw the zero-sequence reactance diagram for the power system shown in Figure 3.38. The zero-sequence reactance of each generator and of the synchronous motor is 0.05 per unit based on equipment ratings. Generator 2 is grounded through a neutral reactor of 0.06 per unit on a 100-MVA, 18-kV base. The zero-sequence reactance of each transmission line is assumed to be three times its positive-sequence reactance. Use the same base as in Problem 3.41.

Short answer

The value of new per unit reactance value of the generator 1 and 2 are 0.00405 per unit and 0.00405 per unit.

The value of new per unit reactance value of the synchronous motor 3 is 0.0033 per unit.

The value of new per unit reactance value of the neutral reactor is 0.0486 per unit.

The neutral reactor value will be three times the value of 0.0486 per unit then the new value of new per unit reactance value of the neutral reactor is 0.1458 per unit.

Step-by-step solution

Write the given data from the question.

The zero-sequence reactance of each generator is 0.05 per unit.

The zero-sequence reactance of synchronous motor 3 is 0.05 per unit.

The generator 2 is grounded through a neutral reactor of 0.06 per unit

The Generator 2 rated at 100-MVA.

The Generator 2 Voltage is 18-kV base.

Determine the formula of per unit reactance of generator, motor and neutral reactor.

Write the formula of new per unit reactance value of the generator 1.

${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{new}}={X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}{\left(\frac{{\mathbf{V}}_{\mathrm{baseold}}}{{\mathbf{V}}_{\mathrm{basenew}}}\right)}^2\left(\frac{{\mathbf{S}}_{\mathrm{basenew}}}{{\mathbf{S}}_{\mathrm{baseold}}}\right)$ …… (1)

Here,${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}$ is per unit old reactance of generator,${\mathbf{V}}_{\mathrm{baseold}}$ is old base voltage,${\mathbf{V}}_{\mathrm{basenew}}$ is new base voltage,${\mathbf{S}}_{\mathrm{basenew}}$ issequence component of new base quantityand${\mathbf{S}}_{\mathrm{baseold}}$ isis sequence component of old base quantity.

Write the formula of new per unit reactance value of the synchronous motor 3.

${X^{\text{'}\text{'}}}_{\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{old}}={X^{\text{'}\text{'}}}_{\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{old}}{\left(\frac{{\mathbf{V}}_{\mathrm{baseold}}}{{\mathbf{V}}_{\mathrm{basenew}}}\right)}^2\left(\frac{{\mathbf{S}}_{\mathrm{basenew}}}{{\mathbf{S}}_{\mathrm{baseold}}}\right)$ …… (2)

Here,${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}$ is per unit old reactance of generator,${\mathbf{V}}_{\mathrm{baseold}}$ is old base voltage,${\mathbf{V}}_{\mathrm{basenew}}$ is new base voltage,${\mathbf{S}}_{\mathrm{basenew}}$ issequence component of new base quantityand${\mathbf{S}}_{\mathrm{baseold}}$ isis sequence component of old base quantity.

Write the formula of new per unit reactance value of the neutral reactor.

${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{new}}={X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}{\left(\frac{{\mathbf{V}}_{\mathrm{baseold}}}{{\mathbf{V}}_{\mathrm{basenew}}}\right)}^2\left(\frac{{\mathbf{S}}_{\mathrm{basenew}}}{{\mathbf{S}}_{\mathrm{baseold}}}\right)$ …… (3)

Here,${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}$ is per unit old reactance of generator,${\mathbf{V}}_{\mathrm{baseold}}$ is old base voltage,${\mathbf{V}}_{\mathrm{basenew}}$ is new base voltage,${\mathbf{S}}_{\mathrm{basenew}}$ issequence component of new base quantityand${\mathbf{S}}_{\mathrm{baseold}}$ isis sequence component of old base quantity.

Write the formula of new per unit reactance value of the transformer.

${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{new}}={X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}{\left(\frac{{\mathbf{V}}_{\mathrm{baseold}}}{{\mathbf{V}}_{\mathrm{basenew}}}\right)}^2\left(\frac{{\mathbf{S}}_{\mathrm{basenew}}}{{\mathbf{S}}_{\mathrm{baseold}}}\right)$ …… (4)

Here,${X^{\text{'}\text{'}}}_{\mathbf{pu}\mathbf{.}\mathbf{old}}$ is per unit old reactance of generator,${\mathbf{V}}_{\mathrm{baseold}}$ is old base voltage, ${\mathbf{V}}_{\mathrm{basenew}}$is new base voltage,${\mathbf{S}}_{\mathrm{basenew}}$ is sequence component of new base quantity and${\mathbf{S}}_{\mathrm{baseold}}$ is is sequence component of old base quantity.

 Determine the zero-sequence reactance diagram for the power system shown in Figure 3.38.

Reference of figure 3.38 in the text book.

Consider the base value of the power system is 100-MVA and 20 kV.

Determine the new per-unit reactance for generator 1.

Substitute 0.05 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 18 for $V_{\text{baseold}}$, 20 for $V_{\text{basenew}}$, 100 for $S_{\text{basenew}}$and 1000 for $S_{\text{baseold}}$into equation (1).

localid="1656771321961" $\begin{array}{c}\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}=\left(0.05\right){\left(\frac{18\text{\hspace{0.33em}kV}}{20\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{1000\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.05\right){\left(0.9\right)}^2\left(0.1\right)\\ =\left(0.005\right)\left(0.81\right)\\ =0.00405\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the new per-unit reactance for generator 1 is 0.00405 per unit.

The generator 1 and 2 are similar so new per-unit reactance for generator 2 is 0.00405 per unit.

Determine the new per-unit reactance of transformer 1.

Substitute 0.1 for localid="1656774757785" $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 20 for $V_{\text{baseold}}$and 20 for $V_{\text{basenew}}$, 100 for $S_{\text{basenew}}$and 1000 for $S_{\text{baseold}}$into equation (4).

localid="1656771481873" $\begin{array}{c}\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}=\left(0.1\right){\left(\frac{20\text{\hspace{0.33em}kV}}{20\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{1000\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.1\right){\left(1\right)}^2\left(0.1\right)\\ =\text{0.01\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the new per-unit reactance of transformer 1 is 0.01 per unit.

The transformers 1, 2, 3, 4 are similar so new per-unit reactance for transformer 2, 3, 4 are 0.01 per unit

Determine the transmission lines of base voltages.

${\text{V}}_{\text{base,H}}=\left({\text{V}}_{\text{base,L}}\right)\left(\frac{{\text{V}}_{\text{H}}}{{\text{V}}_{\text{L}}}\right)$

Here, $\left({\text{V}}_{\text{base,L}}\right)$is low base voltage, $V_{\text{H}}$is high voltage and $V_L$is low voltage.

Substitute 20 kV for $\left({\text{V}}_{\text{base,L}}\right)$, 500 kV for $V_{\text{H}}$and 20 kV for $V_L$in above equation.

$\begin{array}{c}{V}_{\text{base,H}}=\left(20\text{\hspace{0.33em}kV}\right)\left(\frac{500\text{\hspace{0.33em}kV}}{20\text{\hspace{0.33em}kV}}\right)\\ =500\text{\hspace{0.33em}kV}\end{array}$

Hence, the transmission line of base voltages is 500 kV.

Determine the transmission line of base reactance.

$X_{\text{base}}=\frac{{\left(V_{\text{base}}\right)}^2}{S_{\text{base}}}$

Here, $V_{\text{base}}$is base voltage and $S_{\text{base}}$is base sequence component.

Substitute 500 kV for $V_{\text{base}}$and 100 MVA for $S_{\text{base}}$into above equation.

$\begin{array}{c}{X}_{\text{base}}=\frac{{\left(500\text{\hspace{0.33em}kV}\right)}^2}{100\text{\hspace{0.33em}MVA}}\\ =\frac{\left(500\text{\hspace{0.33em}kV}\right)\left(500\text{\hspace{0.33em}kV}\right)}{100\text{\hspace{0.33em}MVA}}\\ =2500\mathrm{\Omega }\end{array}$

Hence, the transmission line of base reactance is $2500\mathrm{\Omega }$.

Determine the transmission line per unit reactance.

$X_{\text{pu}}=\frac{X}{X_{\text{base}}}$

Here, $X$is reactance and $X_{\text{base}}$is base reactance.

Substitute 50 for $X$and 2500 for $X_{\text{base}}$into above equation.

$\begin{array}{c}{X}_{\text{pu}}=\frac{50}{2500}\\ =0.02\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the transmission line per unit reactance is 0.02 per unit.

Determine the $50\mathrm{\Omega }$of transmission line of new per unit reactance.

localid="1656772864524" $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{new}}=\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}{\left(\frac{{\text{V}}_{\text{baseold}}}{{\text{V}}_{\text{basenew}}}\right)}^{\text{2}}\left(\frac{{\text{S}}_{\text{basenew}}}{{\text{S}}_{\text{baseold}}}\right)$ …… (5)

Here, localid="1656772940926" $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$is per unit old reactance of generator, ${\text{V}}_{\text{baseold}}$is old base voltage, ${\text{V}}_{\text{basenew}}$is new base voltage, ${\text{S}}_{\text{basenew}}$is sequence component of new base quantity and ${\text{S}}_{\text{baseold}}$is sequence component of old base quantity.

Substitute 0.02 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 500 kV for ${\text{V}}_{\text{baseold}}$, 500 kV for ${\text{V}}_{\text{basenew}}$, 100 MVA for ${\text{S}}_{\text{basenew}}$and 100 MVA for${\text{S}}_{\text{baseold}}$into equation (5).

$\begin{array}{c}{X^{\text{'}\text{'}}}_{\text{pu.new}}=\left(0.02\right){\left(\frac{500\text{\hspace{0.33em}kV}}{500\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{100\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.02\right){\left(1\right)}^2\left(1\right)\\ =0.02\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the new per unit reactance for $50\mathrm{\Omega }$transmission line is 0.02 per unit.

The zero sequence value of$50\mathrm{\Omega }$transmission line is three times the value of positive sequence value. Then the value of zero sequence new per unit reactance of$50\mathrm{\Omega }$transmission line is 0.06 per unit.

Determine the$25\mathrm{\Omega }$of transmission line of per unit reactance.

$X_{\text{pu}}=\frac{X}{X_{\text{base}}}$

Here,$X$is reactance and$X_{base}$is base reactance.

Substitute 25 for$X$and 2500 for$X_{base}$into above equation.

$\begin{array}{c}{X}_{\text{pu}}=\frac{25}{2500}\\ =0.01\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, $25\mathrm{\Omega }$the of transmission line of per unit reactance 0.01 pu.

Determine,$25\mathrm{\Omega }$the of transmission line of new per unit reactance.

Substitute 0.01 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 500 kV for ${\text{V}}_{\text{baseold}}$, 500 kV for $V_{\text{basenew}}$, 100 MVA for${\text{S}}_{\text{basenew}}$ and 100 MVA for${\text{S}}_{\text{baseold}}$ into equation (5).

$\begin{array}{c}{X^{\text{'}\text{'}}}_{\text{pu.new}}=\left(0.01\right){\left(\frac{500\text{\hspace{0.33em}kV}}{500\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{100\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.01\right){\left(1\right)}^2\left(1\right)\\ =0.01\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the$25\mathrm{\Omega }$ of transmission line have new per unit reactance is 0.01 per unit.

The zero sequence of$25\mathrm{\Omega }$ transmission line is three times the value of positive sequence value. The value of zero sequence new per unit reactance of$25\mathrm{\Omega }$ transmission line is 0.03 per unit.

Determine the$2\mathrm{\Omega }$ of transmission line of per unit reactance.

$X_{\text{pu}}=\frac{X}{X_{\text{base}}}$

Here, X is reactance and X_base is base reactance.

Substitute 2 for X and 2500 for X_base into above equation.

$\begin{array}{c}{X}_{\text{pu}}=\frac{2}{2500}\\ =0.0008\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the of $2\mathrm{\Omega }$transmission line of per unit reactance.

Determine the$2\mathrm{\Omega }$of transmission line of new per unit reactance.

Substitute 0.0008 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 500 kV for ${\text{V}}_{\text{baseold}}$, 500 kV for $V_{\text{basenew}}$, 100 MVA for${\text{S}}_{\text{basenew}}$and 100 MVA for${\text{S}}_{\text{baseold}}$into equation (5).

$\begin{array}{c}{X^{\text{'}\text{'}}}_{\text{pu.new}}=\left(0.0008\right){\left(\frac{500\text{\hspace{0.33em}kV}}{500\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{100\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.0008\right){\left(1\right)}^2\left(1\right)\\ =0.0008\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the$2\mathrm{\Omega }$transmission line have new per unit reactance is 0.0008 per unit.

The zero sequence of$2\mathrm{\Omega }$transmission line is three times the value of positive sequence value. Then the value of zero sequence new per unit reactance of$2\mathrm{\Omega }$transmission line is 0.0024 per unit.

Determine the new per unit reactance value of the synchronous motor 3.

Substitute 0.05 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 20 kV for ${\text{V}}_{\text{baseold}}$, 20 kV for $V_{\text{basenew}}$, 100 MVA for${\text{S}}_{\text{basenew}}$and 1500 MVA forlocalid="1656785727067" ${\text{S}}_{\text{baseold}}$into equation (2).

$\begin{array}{c}{X^{\text{'}\text{'}}}_{\text{pu.new}}=\left(0.05\right){\left(\frac{20\text{\hspace{0.17em}kV}}{20\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{1500\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.05\right){\left(1\right)}^2\left(0.0667\right)\\ =0.0033\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence,the new per unit reactance value of the synchronous motor 3 is 0.0033 per unit.

Determine the new per unit reactance of transformer 5.

Substitute 0.1 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 20 kV for ${\text{V}}_{\text{baseold}}$, 20 kV for $V_{\text{basenew}}$, 100 MVA for${\text{S}}_{\text{basenew}}$and 1500 MVA for${\text{S}}_{\text{baseold}}$into equation (4).

$\begin{array}{c}{X^{\text{'}\text{'}}}_{\text{pu.new}}=\left(0.1\right){\left(\frac{20\text{\hspace{0.17em}kV}}{20\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{1500\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.1\right){\left(1\right)}^2\left(0.0667\right)\\ =0.00667\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the new per unit reactance of transformer 5 is 0.00667 per unit.

Determine the new per unit reactance of neutral reactor.

Substitute 0.06 for $\mathrm{X}{\text{'}\text{'}}_{\mathrm{pu}.\mathrm{old}}$, 18 kV for ${\text{V}}_{\text{baseold}}$, 20 kV for $V_{\text{basenew}}$, 100 MVA for${\text{S}}_{\text{basenew}}$and 100 MVA for${\text{S}}_{\text{baseold}}$into equation (3).

$\begin{array}{c}{X^{\text{'}\text{'}}}_{\text{pu.new}}=\left(0.06\right){\left(\frac{18\text{\hspace{0.17em}kV}}{20\text{\hspace{0.33em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.33em}MVA}}{100\text{\hspace{0.33em}MVA}}\right)\\ =\left(0.06\right){\left(0.9\right)}^2\left(1\right)\\ =\left(0.06\right)\left(0.81\right)\\ =0.0486\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Hence, the new per unit reactance of neutral reactor is 0.0486 per unit.

The neutral reactor value will be three times the of new per unit neutral reactance reactor value as:

$3\left(\text{0.0486\hspace{0.33em}per\hspace{0.33em}unit}\right)=\text{0.1458\hspace{0.17em}per\hspace{0.33em}unit}$

Now draw the sequence reactance diagram for the power system.