Question

If breakers B13 and B22 in Problem 11.13 open later than 3 cycles after the fault commences, determine the critical clearing time.

Short answer

The critical clearing time is$10.6\text{cycles}$

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 is $f=60\text{Hz}$

The infinite bus received poweris $P=1\text{pu}$

The inertial constan $H=3.0$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.

…… (2)

$E\text{'}=V+j{X}_{eq}I$

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

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

The equation to calculate the critical clearing time is given as follows.

$t_{cr}=\sqrt{\frac{4H}{2\pi f\times P}\left({\delta }_{max}-{\delta }_0\right)}$ …… (4)

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

$$\begin{gathered} P=\frac{1.2812\times 1}{0.52}\sin \delta \\ P=2.4638\sin \delta \end{gathered}$$

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

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

Calculate the real power after the fault occurrence.

Substitute 1pu for V , localid="1656226275973" $1.2812\angle 23.94°\text{pu}$for localid="1656226349640" $E\text{'}$and localid="1656226329611" $j0.60\text{pu}$for localid="1656226374669" $X_{eq}$into equation (3).

localid="1656226312081" $$\begin{gathered} P=\frac{1.2812\times 1}{0.60}\sin \delta \\ P=2.1353\sin \delta \end{gathered}$$

Consider the power angle curve before and after the fault.

Calculate the power angle before the fault as.

$$\begin{gathered} \backslash \mathrm{begingathered}1=2.4638\sin {\delta }_0 \\ \sin {\delta }_0=\frac{1}{2.4638} \\ {\delta }_0={\sin }^{-1}\left(\frac{1}{2.4638}\right) \\ {\delta }_0=0.4179\text{rad} \end{gathered}$$

The maximum power angle ${\delta }_{\max }$is obtained at when the mechanical power is equal to the posy fault curve.

$$\begin{gathered} 1-2.1353\sin {\delta }_1=0 \\ {\delta }_1={\sin }^{-1}\frac{1}{2.1353} \\ {\delta }_1=0.4873\text{rad} \\ {\delta }_1=27.92° \end{gathered}$$

At the critical clearing angle, denoted ${\delta }_{cr}$, the fault is extinguished. The power angle then increases to a maximum value

$$\begin{gathered} {\delta }_{\max }=180-{\delta }_1 \\ {\delta }_{\max }=180-29.92 \\ {\delta }_{\max }=152.1° \\ {\delta }_{\max }=2.654\text{rad} \end{gathered}$$

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

$$\begin{gathered} {\int }_{0.4179}^{{\delta }_{cr}}1d\delta ={\int }_{{\delta }_{cr}}^{2.654}\left(2.1353\sin \delta -1\right)d\delta \\ {\left({\delta }_{cr}\right)}_{0.4179}^{{\delta }_{cr}}={\left(2.1353\cos \delta -1\right)}_{{\delta }_{cr}}^{2.654} \\ {\delta }_{cr}-0.4179=-2.1353\left(\cos 2.654-\cos {\delta }_{cr}\right)-\left(2.654-{\delta }_{cr}\right) \\ {\delta }_{cr}-0.4179=2.1353\cos {\delta }_{cr}+1.8863+{\delta }_{cr}-2.654 \end{gathered}$$

Solve further as,

$$\begin{gathered} 2.1353\cos {\delta }_{cr}=-1.8765+2.654-0.4179 \\ 2.1353\cos {\delta }_{cr}=0.3498 \\ \cos {\delta }_{cr}=0.1638 \\ {\delta }_{cr}=1.4062 \end{gathered}$$

Calculate the critical clearing time.

Substitute $3\text{pu - sec}$for H , $1.4062\text{rad}$for${\delta }_{cr}$, $0.4179°$for${\delta }_0$, $60\text{Hz}$for f and 1pu forP into equation (4)

localid="1656225906944" $$\begin{gathered} t_{cr}=\sqrt{\frac{4\times 3}{2\pi \times 60\times 1}\left(1.4062-0.4179\right)} \\ {t}_{cr}=\sqrt{0.0318\times 0.9883} \\ {t}_{cr}=0.1773\text{s} \\ {t}_{cr}=10.6\text{cycles} \end{gathered}$$

Hence, the critical clearing time is$10.6\text{cycles}$.