Question

The generator in Figure 11.4 is initially operating in the steady-state condition given in Example 11.3 when circuit breaker B12 inadvertently opens. Use the equal-area criterion to calculate the maximum value of the generator power angle$\mathit{\delta }$. Assume${\mathit{\omega }}_{\mathbf{p}\mathbf{u}}\left(t\right)\mathbf{=}\mathbf{1}$ in the swing equation.

Short answer

The maximum value of the generator power angle is $42.62°$.

Step-by-step solution

Write the given data from the question:

The voltage at bus$V=1\text{\hspace{0.33em}pu}$.

The Power factor is $pf=095$.

The frequency of the system $f=60\text{\hspace{0.33em}Hz}$.

The Infinite bus received power $P=1\text{\hspace{0.33em}pu}$.

Determine the equation to calculate the power angle.

The equation to calculate the current is given as follows.

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{V}\mathbf{\times }\mathbf{p}\mathbf{f}}\mathbf{\angle }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\left(pf\right)$ …… (1)

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

${\mathit{E}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{V}\mathbf{+}\mathit{j}{\mathit{X}}_{\mathbf{e}\mathbf{q}}\mathit{I}$ …… (2)

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

$\mathit{P}\mathbf{=}\frac{{\mathbf{E}}^{\mathbf{\text{'}}}\mathbf{V}}{{\mathbf{X}}_{\mathbf{e}\mathbf{q}}}\mathbf{sin}\mathit{\delta }$ …… (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.

$\begin{array}{c}{X}_{eq}=j0.30+j0.10+j0.20\parallel \left(j0.10+j0.20\right)\\ X_{eq}=j0.40+j0.20\parallel j0.30\\ X_{eq}=j0.40+j0.12\\ X_{eq}=j0.52\text{\hspace{0.33em}pu}\end{array}$

Calculate the value of the current,

Substitute 1.0 p.u. for P, 1.0 p.u. for V, and 0.95 for pf into equation (1).

$\begin{array}{c}I=\frac{1}{1\times 0.95}\angle {\cos }^{-1}\left(0.95\right)\\ I=1.05\angle -18.19°\text{\hspace{0.33em}pu}\end{array}$

Calculate the interval voltage of the generator,

Substitute 1 p.u. for V, $1.05\angle -18.19°\text{\hspace{0.33em}pu}$ for I and $j0.520\text{\hspace{0.33em}pu}$ for X_eq into equation (2).

$\begin{array}{c}{E}^{\text{'}}=1+j520\left(1.05\angle 18.19°\right)\\ E^{\text{'}}=1.2812\angle 23.94°\text{\hspace{0.33em}pu}\end{array}$

Calculate the load angle.

Substitute 1 p.u. for V , $1.2812\angle 23.94°\text{\hspace{0.33em}pu}$ for E' and $j0.520\text{\hspace{0.33em}pu}$for X_eq into equation (3).

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

Consider the equivalent circuit after the fault occurrence.

Calculate the equivalent reactance after the fault.

$\begin{array}{l}{X}_{eq}=j0.4+j0.3\\ X_{eq}=j0.7\end{array}$

Calculate the real power after the fault occurrence.

Substitute 1 p.u. for V , $1.2812\angle 23.94°\text{\hspace{0.33em}pu}$ for E' and $j0.70\text{\hspace{0.33em}pu}$for X_eq into equation (3).

$\begin{array}{l}P=\frac{1.2812\times 1}{0.70}\sin \delta \\ P=1.8303\sin \delta \end{array}$

Consider the power angle curve before and after the fault.

Calculate the power angle before the fault as.

$\begin{array}{c}1=2.4638\sin {\delta }_0\\ \sin {\delta }_0=\frac{1}{2.4638}\\ {\delta }_0={\sin }^{-1}\left(\frac{1}{2.4638}\right)\\ {\delta }_0=0.4179\text{\hspace{0.33em}rad}\end{array}$

Calculate the power angle after the fault as.

$\begin{array}{c}1=1.8304\sin {\delta }_1\\ \sin {\delta }_1=\frac{1}{1.8303}\\ {\delta }_1={\sin }^{-1}\left(\frac{1}{1.8303}\right)\\ {\delta }_1=0.578\text{\hspace{0.33em}rad}\end{array}$

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

$\begin{array}{c}{\int }_{0.4179}^{0.5780}\left(1-1.8303\sin \delta \right)d\delta ={\int }_{0.5780}^{{\delta }_2}\left(1.8303\sin \delta -1\right)d\delta \\ {\left[\delta -1.8303\cos \delta \right]}_{0.4179}^{0.5780}={\left[1.8303\cos \delta -\delta \right]}_{0.4179}^{0.5780}\\ \left[\begin{array}{l}\left(0.5780-0.4179\right)+\\ 1.8303\left(\cos 0.5780-\cos 0.4179\right)\end{array}\right]=\left[\begin{array}{l}1.8303\left(\cos 0.5780-\cos {\delta }_2\right)\\ -\left({\delta }_2-0.5780\right)\end{array}\right]\\ 0.1601-0.1398=1.8303\left(0.8375-\cos {\delta }_2\right)-\left({\delta }_2-0.5780\right)\end{array}$

Solve further as,

$\begin{array}{c}0.0203=1.5329-1.8303{\delta }_2-{\delta }_2+0.5780\\ 1.8303\cos {\delta }_2+{\delta }_2=2.0907\end{array}$

Solve the above equation for .

$\begin{array}{l}{\delta }_2=0.7439\text{\hspace{0.33em}rad}\\ {\delta }_2=42.62°\end{array}$

Hence, the maximum value of the generator power angle is $42.62°$.