Question

The generator in Figure 11.4 is initially operating in the steady-state condition given in Problem 11.8 when a three-phase-to-ground bolted short circuit occurs at bus 3. Determine an equation for the electrical power delivered by the generator versus power angle $\delta$ during the fault.

Short answer

(a)

The value of equivalent reactance between the generator voltage and the infinite bus voltage is 0.52 per unit.

The value of current is $0.8421\angle -18.195°\text{\hspace{0.33em}A}$.

(b)

The value of generator internal voltage is $1.21\angle 20.11°\text{\hspace{0.33em}p.u.}$

(c)

The value of real power delivered by the synchronous generator to the infinite bus is $2.231\sin \delta$.

The value of the electrical power delivered by the generator versus power angle $\delta$during the fault is $0.6763\sin \delta \text{\hspace{0.33em}per\hspace{0.33em}unit}$.

Step-by-step solution

Write the given data from the question.

As reference to question 11-8P in the textbook.

As reference of figure 11.4.

As reference to example 11.3 in the textbook.

Consider the value of infinite bus receives 0.8 per unit of real power.

Consider the value of infinite bus receives 1.05 per unit terminal voltage.

All the reactances are given in per-unit on a common system base.

A three-phase-to-ground bolted short circuit occurs at bus 3.

Step 2: Determine the formula of equivalent impedance Xeq, current I, generator internal voltage E', real power delivered and electrical power delivered by the generator versus power angle δ.

Write the formula ofequivalent reactance between the generator voltage and the infinite bus voltage.

${\mathbf{X}}_{\mathrm{eq}}={X^{\text{'}}}_d\mathbf{+}{\mathbf{X}}_{\mathrm{TR}}\mathbf{+}\left[{\mathbf{X}}_{12}\mathbf{P}\left({\mathbf{X}}_{13}\mathbf{+}{\mathbf{X}}_{23}\right)\right]$ …… (1)

Here, X^'_d represent the generator transient reactance, X_TR represent the transformer reactance; X₁₂ represent the reactance between the bus 1 and bus 2, X₁₃ represent the reactance between the bus 1 and bus 3 and X₂₃ represent the reactance between the bus 2 and bus 3.

Write the formula of current.

$\mathbf{I}=\frac{P}{{\mathbf{V}}_{\mathrm{bus}}\left(\mathbf{p}\mathbf{.}\mathbf{f}\right)}\angle -{\mathbf{cos}}^{-\mathbf{1}}\left(\mathbf{p}\mathbf{.}\mathbf{f}\right)$ …… (2)

Here, P is power, V_bus is bus voltage and p.f is power factor.

Write the formula ofgenerator internal voltage.

${\mathbf{E}}^{\mathbf{\text{'}}}={\mathbf{V}}_{\mathrm{bus}}\mathbf{+}{\mathbf{jX}}_{\mathrm{eq}}\mathbf{I}$ …… (3)

Here, V_bus is bus voltage, X_eq equivalent reactance between the generator voltage and the infinite bus voltage and I is current flowing in the infinite bus.

Write the formula ofreal power delivered by the synchronous generator to the infinite bus.

${\mathbf{P}}_e=\frac{{\mathbf{E}}^{\mathbf{\text{'}}}{\mathbf{V}}_{\mathrm{bus}}}{{\mathbf{X}}_{\mathrm{eq}}}\mathbf{sin}\delta$ …… (4)

Here, E' is generator internal voltage, V_bus is bus voltage, X_eq equivalent reactance between the generator voltage and the infinite bus voltage and $\mathit{\delta }$ is generator power angle.

Write the formula of electrical power delivered by the generator versus power angle $\delta$during the fault.

${\mathbf{P}}_e=\frac{{\mathbf{E}}^{\mathbf{\text{'}}}{\mathbf{V}}_{\mathrm{Th}}}{{\mathbf{X}}_{\mathrm{Th}}}\mathbf{sin}\delta$ …… (5)

Here, E' is generator internal voltage, V_th is thevenin voltage, X_th is thevenin voltage and $\delta$is generator power angle.

(a) Determine the value of value of equivalent reactance between the generator voltage and the infinite bus voltage and current.

As illustrated in figure 1, create a single-line schematic of a straightforward generating and distribution unit with a problem.

Draw an impedance diagram that corresponds to the figure 1 diagram.

The equivalent reactance between the generator voltage and the infinite bus voltage is shown in figure 2.

Substitute 0.3 for X^'_d, 0.10 for X_TR, 0.20 for X₁₂, 0.10 for X₁₃and 0.20 for X₂₃ into equation (1).

$\begin{array}{c}X=0.3+0.10+0.20\parallel \left(0.10+0.20\right)\\ =0.3+0.10+0.12\\ =0.52\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the equivalent reactance between the generator voltage and the infinite bus voltage is 0.52 per unit.

Determine the value of current.

Substitute 0.8 for P, 1.0 for V_bus and 0.95 for p.f. into equation (2).

Here, p.f. is taken from example 11.3 in the textbook.

$\begin{array}{c}I=\frac{0.8}{\left(1.0\right)\left(0.95\right)}\angle -{\cos }^{-1}\left(0.95\right)\\ =0.8421\angle -18.195°\text{\hspace{0.33em}A}\end{array}$

(b) Determine the value of generator internal voltage.

Determine the expression for generator internal voltage.

Substitute 1.00 for V_bus, 0.52 for X_eq and $0.8421\angle -18.195°$ for I into equation (3).

$\begin{array}{c}{E}^{\text{'}}=\left(1.0\right)+j\left(0.52\right)\left(0.8421\angle -18.195°\right)\\ =1.1367+j0.416\\ =1.21\angle 20.11°\text{\hspace{0.33em}p.u.}\end{array}$

Therefore, the value of generator internal voltage is$1.21\angle 20.11°\text{\hspace{0.33em}p.u.}$

(c) Determine the value of electrical power delivered by the generator versus power angle δ during the fault.

Determine the expression for power delivered by synchronous generator to the infinite bus.

Substitute 0.8 per unit for P_e, 1.21 per unit for E', 1.0 per unit for V_bus and 0.52 per uimt for X into equation (4).

$\begin{array}{c}0.8=\frac{\left(1.21\right)\left(1.0\right)}{0.52}\sin \delta \\ \sin \delta =0.3438\\ \delta ={\sin }^{-1}\left(0.3438\right)\\ \delta =20.11°\end{array}$

Therefore, the value of power angle with respect to the infinite bus is $20.11°$.

According to figure 2, bus 3 experiences a three-phase-to-ground bolted short circuit problem.

Re-draw the circuit diagram.

Determine the thevenin equivalent reactance of the fault network as seen from the fault location is shown in figure 3 (Bus 3).

$X_{\text{eq}}=\left[\left({X^{\text{'}}}_{\text{d}}+X_{\text{TR}}\right)\parallel X_{12}\right]+X_{13}+X_{23}$

Here, X^'_d represent the generator transient reactance, X_TR represent the transformer reactance; X₁₂ represent the reactance between the bus 1 and bus 2, X₁₃ represent the reactance between the bus 1 and bus 3 and X₂₃ represent the reactance between the bus 2 and bus 3.

Substitute 0.3 for X^'_d, 0.10 for X_TR, 0.20 for X₁₂, 0.10 for X₁₃and 0.20 for X₂₃ into above equation.

$\begin{array}{c}{X}_{\text{eq}}=\left[\left(0.30+0.10\right)\parallel \left(0.2\right)\right]+\left(0.1+0.2\right)\\ =\left[\left(0.4\right)\parallel \left(0.2\right)\right]+\left(0.3\right)\\ =\frac{\left(0.4\right)\left(0.2\right)}{0.4+0.2}+0.3\\ =0.433\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the value of equivalent reactance is 0.433 per unit.

As seen from the fault location (Bus 3) to the left of the circuit, record the Thevenin voltage.

$V_{Th}=E^{\text{'}}\left(\frac{X_{13}}{X_{13}+{X^{\text{'}}}_d+X_{TR}}\right)$

Here, E' is generator internal voltage, X₁₃ represent the reactance between the bus 1 and bus 3 , X^'_d represent the generator transient reactance and X_TR represent the transformer reactance.

Substitute $1.21\angle 20.11°$ for E' , 0.10 for X₁₃ , 0.3 for X^'_d and 0.1 for X_TR into above equation.

$\begin{array}{c}{V}_{\text{Th}}=\left(1.21\angle 20.11°\right)\left(\frac{0.10}{0.1+0.3+0.1}\right)\\ =\left(1.21\angle 20.11°\right)\left(\frac{0.1}{0.5}\right)\\ =0.242\angle 20.11°\text{\hspace{0.17em}per\hspace{0.33em}unit}\end{array}$

Therefore, the thevenin’s voltage is $0.242\angle 20.11°\text{\hspace{0.33em}per\hspace{0.33em}unit}$.

Determine the expression for the electrical power delivered by the generator versus power angle $\delta$ during the fault.

Substitute 1.21 for E' , 0.242 for V_Th and 0.433 for X_Th into equation (5).

$\begin{array}{c}{P}_e=\frac{\left(1.21\right)\left(0.242\right)}{0.433}\sin \delta \\ =0.6763\sin \delta \text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the electrical power delivered by the generator versus power angle $\delta$during the fault is $0.6763\sin \delta \text{\hspace{0.33em}per\hspace{0.33em}unit}$.