Question

Solve Example 7.3 parts (a) and (c) without using the superposition principle. First calculate the internal machine voltages $E_g^{"}$and localid="1656758482853" $\mathit{E}{\mathbf{"}}_{\mathbf{m}}$ using the prefault load current. Then determine the subtransient fault, generator, and motor currents directly from Figure 7.4(a). Compare your answers with those of Example 7.3.

Short answer

(a) The value of internal machine voltages $E{"}_g$using the prefault load current is $1.0946+j0.1357perunit$.

The value of internal machine voltages $E{"}_g$using the prefault load current is $0.8998-j0.4568perunit$.

(b) The subtransient fault current $I{"}_f$is$-37.978kA$.

The subtransient current of motor$I{"}_{mf}$ is$-3.784-j7.451kA$.

The subtransient current of generator $I{"}_g$is$3.784-j30.527kA$.

The current obtained from the superposition theorem in the example 7.3 are equal to the current obtained without using the superposition theorem.

Step-by-step solution

Write the given data from the question.

The value of base MVA of generator is$100\times {10}^6$.

The value of base generator voltage is$13.8\times {10}^3$.

The value of line reactance $X_{line}$is $20$.

Determine the formula of internal machine voltages using prefault current and subtransient fault current.

Write the formula of internal machine voltages of generator.

$E{"}_g=V_F+j{I}_{0actual}X{"}_g$ …… (1)

Here, $V_F$is full load voltage, $I_{0actual}$is actual generator current and $X{"}_g$is generator reactance.

Write the formula of internal machine voltages of motor and $X{"}_m$.

$E{"}_m=V_F-j{I}_{0actual}X{"}_m$ …… (2)

Here, $V_F$is full load voltage,$I_{0actual}$is actual generator current and $X{"}_m$is motor reactance.

Write the formula of subtransient generator current.

$I{"}_g=I_{base}I{"}_g\left(inpu\right)$ …… (3)

Here,$I_{base}$is base current and $I{"}_g\left(inpu\right)$is subtransient generator current.

Write the formula of subtransient motor current. $I{"}_m=I_{base}I{"}_m\left(inpu\right)$ …… (4)

Here,$I_{base}$is base current and $I{"}_m\left(inpu\right)$is subtransient motor current.

Write the formula of subtransient fault current.

$I{"}_f=I{"}_g+I{"}_m$ …… (5)

Here, $I{"}_g$is subtransient generator current and $I{"}_m$is subtransient motor current.

(a) Determine the internal machine voltages of generator and motor.

Determine the power angle.

$$\begin{gathered} \Phi =-{\cos }^{-1}\left(powerfactor\right) \\ =-{\cos }^{-1}\left(0.95\right) \\ =-18.{19}^{\circ } \end{gathered}$$

The negative sign shows the power factor is lagging in nature.

Determine the actual generator current.

$I_{0actual}=\frac{S}{\sqrt{3}{V}_{F}V}$ …… (6)

Here, S is bus voltage,$V_F$is full load voltage and V is base generator voltage.

Substitute $100\times {10}^6$for S, 1.05 for $V_F$and $13.8\times {10}^3$for V into equation (6).

$$\begin{gathered} I_{0actual}=\frac{100\times {10}^6}{\sqrt{3}\left(1.05\right)\left(13.8\times {10}^3\right)} \\ =3.984\angle -18.{19}^{\circ }kA \end{gathered}$$

Determine the base current.

$1_b=\frac{BaseMVA}{\sqrt{3}Basegeneratorvoltage}$ …… (7)

Substitute $100\times {10}^6$for Base MVA and $13.8\times {10}^3$for Base generator voltage into equation (7).

$$\begin{gathered} 1_b=\frac{100\times {10}^6}{\sqrt{3}\left(13.8\times {10}^3\right)} \\ =4.1836kA \end{gathered}$$

Determine the actual generator current per unit.

$1_{0pu}=\frac{I_{0actual}}{I_b}$ …… (8)

Here, $1_{0actual}$is actual generator current and $I_b$is base current.

Substitute $3.984\times {10}^3\angle -18.{19}^{\circ }$for $I_{0actual}$and $4.1736\times {10}^3$for $I_b$into equation (8).

$$\begin{gathered} I_{0pu}=\frac{3.984\times {10}^3\angle -18.{19}^{\circ }}{4.1836\times {10}^3} \\ =0.9046-j0.2974perunit \end{gathered}$$
Determine the base reactance of the system.

$X_{base}=\frac{{\left(kV\right)}^2}{MVA}$ …… (9)

Here, kV is generator base voltage and MVA is base MVA of generator.

Substitute $138\times {10}^3$for kV and $100\times {10}^6$for MVA into equation (9).

$$\begin{gathered} X_{base}=\frac{{\left(138\times {10}^3\right)}^2}{100\times {10}^6} \\ =190.44\Omega \end{gathered}$$

Determine the line reactance per unit.

$X_{linep.u.}=\frac{X_{line}}{X_{base}}$ …… (10)

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

Substitute 20 for $X_{line}$and 190.44 for $X_{base}$into equation (10).

$$\begin{gathered} X_{linep.u.}=\frac{20}{190.44} \\ =0.1050p.u. \end{gathered}$$

Draw a diagram of synchronous generator where fault occurs at bus 1 of feeding the motor.

Figure 1

Determine the internal machine per unit voltage of behind generator.

Substitute 1.05 for $V_F$, 0.9046-0.2974 for $I_{0actual}$and 0.15 for $X{"}_g$into equation (1).

$$\begin{gathered} E{"}_g=1.05+j\left(0.9046-j0.2974\right)\left(0.15\right) \\ =1.05+0.04461+j0.1357 \\ =1.0946+j0.1357perunit \end{gathered}$$

Determine the internal machine per unit voltage of behind motor.

Substitute 1.05 for $V_F$, 0.9046-j0.2974 for $1_{0actual}$and 0.505 for $X{"}_{mf}$into equation (2).

$$\begin{gathered} E{"}_{mf}=1.05-j\left(0.9046-j0.2974\right)\left(0.505\right) \\ =1.05-0.15-j0.4568 \\ =0.8998-j0.4568perunit \end{gathered}$$

(b) Determine the subtransient fault, generator and motor current.

Determine the subtransient fault current of motor reactance.

$X{"}_{mf}=X_{line}\left(inpu\right)+X_{T1}+X_{T2}+X{"}_m$ …… (11)

Here,$X_{line}\left(inpu\right)$is per unit line reactance,$X_{T1}$is first transformer reactance,$X_{T2}$is second transformer reactance and $X{"}_m$is motor reactance.

Substitute 0.1050 for$X_{line}\left(inpu\right)$, 0.1 for$X_{T1}$, 0.1 for $X_{T2}$and 0.20 for $X{"}_m$into equation (11).

$$\begin{gathered} X{"}_m=0.1050+0.1+0.1+0.20 \\ =0.505p.u. \end{gathered}$$

Determine the subtransient current of generator in per unit.

$I{"}_g=\frac{E{"}_g}{jX{"}_g}$ …… (12)

Here,$E{"}_g$is generator voltage and $X{"}_g$is generator reactance.

Substitute 1.0946+j0.1357 for $E{"}_g$and 0.15 for $X{"}_g$
into equation (12).

$$\begin{gathered} I{"}_g=\frac{1.0946+j0.1357}{j0.15} \\ =0.9046-j7.297 \\ =7.353\angle -82.{93}^{\circ } \end{gathered}$$

Determine the subtransient motor current in per unit.

$I{"}_m=\frac{E{"}_m}{jX{"}_{mf}}$ …… (13)

Here,$E{"}_m$is motor voltage and $X{"}_{mf}$is motor reactance.

Substitute 0.8998-j0.4568 for $E{"}_m$and 0.505 for $X{"}_{mf}$into equation (13).

$$\begin{gathered} I{"}_m=\frac{0.8998-j0.4568}{j0.505} \\ =-0.9046-j1.781 \\ =1.998\angle 243.1^{\circ }perunit \end{gathered}$$

Determine the subtransient fault current in per unit.

$I{"}_f=I{"}_g+I{"}_m$ …… (14)

Here, $I{"}_g$is subtransient current of generator and $I{"}_m$is subtransient current of motor.

Substitute 0.9046-j7.297 for $I{"}_g$and 0.9046-j1.781 for $I{"}_m$into equation (14).

$$\begin{gathered} I{"}_f=0.9046-j7.297-0.9046-j1.781 \\ =-j9.078perunit \end{gathered}$$

Now, determine the subtransient current of generator.

Substitute 4.1836 for $I_{base}$and 0.9046-j7.297 for $I{"}_g$into equation (3).

Determine the subtransient current of motor.

Substitute 4.1836 for $I_{base}$and -0.9046-j1.781 for $I{"}_m\left(inpu\right)$into equation (4).

$$\begin{gathered} I{"}_m=\left(4.1836\right)\left(-0.9046-j1.781\right) \\ =-3.784-j7.451kA \end{gathered}$$

Determine the subtransient current of fault current.

Substitute 3.784-J30.527 for $I{"}_g$and -3.784-j7.451 for $I{"}_m$into equation (5).

$$\begin{gathered} I{"}_f=3.784-j30.527-3.784-j7.451 \\ =-37.978kA \end{gathered}$$

Therefore, the subtransient of generator, motor and fault current are

$$\begin{gathered} I{"}_f=3.784-j30.527kA \\ I{"}_m=-3.784-j7.451kA \\ I{"}_g=-37.978kA \end{gathered}$$

The current obtained from the superposition theorem in the example 7.3 are equal to the current obtained without using the superposition theorem.