Question

The generator in Figure 11.4 is initially operating in the steady-state condition given in Example 11.3 when a temporary three-phase-to-ground short circuit occurs at point F. Three cycles later, circuit breakers B13 and B22 permanently open to clear the fault. Use the equal-area criterion to determine the maximum value of the power angle$\delta$.

Short answer

The maximum value of the power angle is $44.98°$.

Step-by-step solution

Write the given data from the question.

The Voltage at bus $V=1\text{pu}$.

The Power factor$pf=0.95$ .

The frequency of the system$f=60\text{Hz}$ .

The Infinite bus received power is $P=1\text{pu}$.

The Inertial constant$H=3.0$ per unit second.

Determine the equation to calculate the power angle.

The equation to calculate the current is given as follows.

$I=\frac{P}{V\times pf}\angle co{s}^{-1}\left(pf\right)$ …… (1)

The equation to calculate the internal voltage of the generator is given as follows.

$E\text{'}=V+j{X}_{eq}I$ …… (2)

The equation to calculate the real power is given as follows.

$P=\frac{E\text{'}V}{X_{eq}}sin\delta$ …… (3)

Calculate the power angle.

Consider the single line diagram of the system.

The equivalent circuit of the system before the fault.

Calculate the equivalent reactance of the system.

$$\begin{gathered} X_{eq}=j0.30+j0.10+j0.20\parallel \left(j0.10+j0.20\right) \\ {X}_{eq}=j0.40+j0.20\parallel j0.30 \\ {X}_{eq}=j0.40+j0.12 \\ {X}_{eq}=j0.52\text{pu} \end{gathered}$$
Calculate the value of the current,

Substitute $1.0\text{pu}$for P , $1.0\text{pu}$for V , and $0.95$for $pf$into equation (1).

$$\begin{gathered} I=\frac{1}{1\times 0.95}\angle {\cos }^{-1}\left(0.95\right) \\ I=1.05\angle -18.19°\text{pu} \end{gathered}$$

Calculate the interval voltage of the generator,

Substitute 1pu for V ,$1.05\angle -18.19°\text{pu}$for l and $j0.520\text{pu}$for $X_{eq}$into equation (2).

$$\begin{gathered} E\text{'}=1+j520\left(1.05\angle 18.19°\right) \\ E\text{'}=1.2812\angle 23.94°\text{pu} \end{gathered}$$

Calculate the load angle.

Substitute 1pu for V , $1.2812\angle 23.94°\text{pu}$for $E\text{'}$and$j0.520\text{pu}$ for $X_{eq}$ into equation (3).

width="142">P=1.2812×10.52sinδP=2.4638sinδ

Calculate the equivalent reactance when breakers B13 and B22 are parentally open.

localid="1656249347032" $$\begin{gathered} X_{eq}=j0.3+j0.1+j0.2 \\ {X}_{eq}=j0.6\text{pu} \end{gathered}$$

Calculate the real power after the fault occurrence.

Substitute 1pu for V ,$1.2812\angle 23.94°\text{pu}$for $E\text{'}$and$j0.60\text{pu}$for $X_{eq}$into equation (3).

localid="1656249427088" $$\begin{gathered} P=\frac{1.2812\times 1}{0.60}\sin \delta \\ P=2.1353\sin \delta \end{gathered}$$

Consider the power angle curve before and after the fault.

Calculate the power angle before the fault as.

localid="1656249434699" $$\begin{gathered} 1=2.4638\sin {\delta }_0 \\ \sin {\delta }_0=\frac{1}{2.4638} \\ {\delta }_0={\sin }^{-1}\left(\frac{1}{2.4638}\right) \\ {\delta }_0=0.4179\text{rad} \end{gathered}$$

Write the swing equation,

localid="1656249441632" $\frac{2H}{{\omega }_{syn}}{\omega }_{pu}\left(t\right)\frac{d^2\delta \left(t\right)}{d{t}^2}=p_{mpu}\left(t\right)-p_{epu}\left(t\right)$....(4)

localid="1656249449099" $\frac{2H}{{\omega }_{syn}}{\omega }_{pu}\left(t\right)\frac{d^2\delta \left(t\right)}{d{t}^2}=p_{apu}\left(t\right)$

Here,${\omega }_{syn}$is the synchronous speed, $p_{mpu}$is mechanical per unit power, $p_{epu}$is electrical per unit power and $p_{apu}$is the accelerating per unit power.

With the given conditions,

localid="1656249457407" $\frac{2H}{{\omega }_{syn}}\frac{d^2\delta \left(t\right)}{d{t}^2}=p_{apu}\left(t\right)0⩽t⩽0.05s$

Apply the integration,

localid="1656249464657" $\frac{d\delta \left(t\right)}{dt}=\frac{{\omega }_{syn}{p}_{mpu}}{2H}t+0$

Again, apply the integration,

localid="1656249475252" $\delta \left(t\right)=\frac{{\omega }_{syn}{p}_{mpu}}{4H}{t}^2+{\delta }_0$

Substitute $0.05s$for t , for t ,3 3 for H, for into above equation.

localid="1656246912239" $$\begin{gathered} {\delta }_1=\frac{2\pi \left(60\right)}{4\left(3\right)}{\left(0.05\right)}^2+0.4169 \\ {\delta }_1=10\pi \left(0.0025\right)+0.4179 \\ {\delta }_1=0.4964\text{rad} \end{gathered}$$

Equate the area of the curve before and after the fault,

localid="1656246920472" $$\begin{gathered} {\int }_{{\delta }_0}^{{\delta }_1}1d\delta ={\int }_{{\delta }_1}^{{\delta }_2}\left(2.1353\sin \delta -1\right)d\delta \\ {\int }_{0.4179}^{0.4964}1d\delta ={\int }_{0.4964}^{{\delta }_2}\left(2.1353\sin \delta -1\right)d\delta \\ 0.4964-0.4179=-2.1353\left(\cos {\delta }_2-\cos 0.4964\right)-\left({\delta }_2-0.4964\right) \\ 0.0785=2.1353\left(0.8793-\cos {\delta }_2\right)-\left({\delta }_2-0.4964\right) \end{gathered}$$

Solve further as,

localid="1656246927095" $$\begin{gathered} 0.0785=1.8775-2.1353\cos {\delta }_2-{\delta }_2=0.4964 \\ 2.1353\cos {\delta }_2+{\delta }_2=2.2955 \end{gathered}$$

Solve the equation for localid="1656246941361" ${\delta }_2$,

localid="1656246948912" $$\begin{gathered} {\delta }_2=0.785\text{rad} \\ {\delta }_2=44.98° \end{gathered}$$

Hence,the maximum value of the power angle is localid="1656246959963" $44.98°$