Question

A 60 Hz generator is supplying (and ) to an infinite bus (with per unit voltage) through two parallel transmission lines. Each transmission line has a per unit impedance ( base) of . The per unit transient reactance for the generator is , the per unit inertia constant for the generator (H) is seconds, and damping is per unit (all with a base). At time 5 0, one of the transmission lines experiences a balanced three-phase short to ground one third () of the way down the line from the generator to the infinite bus. (a) Using the classical generator model, determine the prefault internal voltage magnitude and angle of the generator. (b) Express the system dynamics during the fault as a set of first order differential equations. (c) Using Euler’s method, determine the generator internal angle at the end of the second timestep. Use an integration step size of one cycle.

Short answer

(a) The magnitude of the internal voltage and angle are $1.053\text{pu}$and$18.26°$ respectively.

(b) The expression for system dynamics during the fault is $\frac{{\omega }_{syn}-D\frac{d{\delta }_t}{dt}}{2H{\omega }_{pu}}$.

(c) The generator power angle at end of the second time step${\delta }_{0.0333}$ is$0.3213\text{rad}$ .

Step-by-step solution

Write the given data from the question.

Frequency of the system,$f=60\text{Hz}$

Voltage at bus,$V_{bus}=1.0\text{pu}$

Power supplied by the generator is $400\text{MW}$.

Transmission line impedance,$Z_L=j0.09$

Transient reactance,$X=j0.0375$

Damping factor,$D=0.1\text{pu}$

Determine the equation to calculate the magnitude of pre-fault voltage and angle, system first order differential equation.

The equation to calculate the current in the system is given as follows.

$I=\frac{P}{V_{bus}\left(pf\right)}\angle co{s}^{-1}\left(p.f\right)$ …… (1)

Here is the power factor.

The equation to calculate the pre-fault internal voltage is given as follows.

$E\text{'}=V_{bus}+j{X}_{eq}I$ …… (2)

Calculate the magnitude of pre-fault voltage and angle.

(a)

Draw the equivalent circuit of the given parameter.

Calculate the equivalent reactance of the system,

$$\begin{gathered} X_{eq}=j0.0375+\left(j0.09\parallel j0.09\right) \\ {X}_{eq}=j0.0375+j0.045 \\ {X}_{eq}=j0.0825\text{pu} \end{gathered}$$

Since the reactive power is zero, the power factor is also zero.

The power delivered on$100\text{MVA}$base,

$$\begin{gathered} P=\frac{400}{100} \\ P=4\text{pu} \end{gathered}$$

Calculate the current in the system.

Substitute $4\text{pu}$for $P$,and $1\text{pu}$for $V_{bus}$,into equation (1).

$$\begin{gathered} I=\frac{4}{1\times 1}\angle {\cos }^{-1}\left(0\right) \\ I=4\angle 0°\text{pu} \end{gathered}$$

Calculate the pre-fault internal voltage and angle.

Substitute $1\text{pu}$for $V_{bus}$, $4\angle 0°\text{pu}$for l and $j0.0825$for$X_{eq}$ into equation (2).

$$\begin{gathered} E\text{'}=1+j0.0825\times 4\angle 0° \\ E\text{'}=1+j0.33 \\ E\text{'}=1.053\angle 18.26° \end{gathered}$$

Hence the magnitude of the internal voltage and angle are $1.053\text{pu}$and $18.26°$respectively.

Determine the first order differential equation for the system.

(b)

The equation for the speed of the motor is given as follows.

$\frac{d{\delta }_t}{dt}=\omega -{\omega }_{syn}$

Differentiate the above equation with respect to t .

$\frac{d^2{\delta }_t}{d{t}^2}=\frac{d\omega }{dt}$

Substitute $\frac{P_{apu}{\omega }_{syn}}{2H{\omega }_{pu}}$for $\frac{d^2{\delta }_t}{d{t}^2}$into above equation.

$$\begin{gathered} \frac{d{\omega }_t}{dt}=\frac{P_{apu}{\omega }_{syn}}{2H{\omega }_{pu}} \\ \frac{d{\omega }_t}{dt}=\frac{{\omega }_{syn}-D\frac{d{\delta }_t}{dt}}{2H{\omega }_{pu}} \end{gathered}$$

Hence the expression for system dynamics during the fault is $\frac{{\omega }_{syn}-D\frac{d{\delta }_t}{dt}}{2H{\omega }_{pu}}$

determine the generator internal angle at the end of the second timestep.

(c)

Write the equation for power angle by using the Euler’s iterative method.

${\delta }_{t+\Delta t}={\delta }_t+\frac{d{\delta }_t}{dt}\Delta t$

${\delta }_{t+\Delta t}={\delta }_t+\Delta t\left({\omega }_t-{\omega }_{syn}\right)$ …… (3)

Write the equation for angular velocity by using the Euler’s iterative method.

${\omega }_{t+\Delta t}={\omega }_t+\frac{d{\omega }_t}{dt}\Delta t$

$w_{t+∆t}=w_t+∆t\left(\frac{w_{xn}-D{\displaystyle \frac{d{\delta }_t}{}}}{2H{W}_{tw}}\right)$......(4)

Calculate the iterative interval,

$$\begin{gathered} \Delta t=\frac{1}{60} \\ \Delta t=0.01667\text{sec} \end{gathered}$$

Initial values of iteration.

$$\begin{gathered} {\omega }_0={\omega }_{syn} \\ {\omega }_0=2\pi \times 60 \\ {\omega }_0=377\frac{\text{rad}}{\text{s}} \end{gathered}$$

The power angle,

$$\begin{gathered} {\delta }_0=18.26° \\ {\delta }_0=0.3187\text{rad} \end{gathered}$$

Substitute for into equation (3).

${\delta }_{0+\Delta t}={\delta }_0+\Delta t\left({\omega }_0-{\omega }_{syn}\right)$

Substitute$0.3187$for ${\delta }_0$, $377\frac{\text{rad}}{\text{s}}$for ${\omega }_0$ and $0.01667\text{sec}$for into above equation.

$$\begin{gathered} {\delta }_{0.01667}=0.3187+0.01667\left(377-377\right) \\ {\delta }_{0.01667}=0.3187+0.01667\left(0\right) \\ {\delta }_{0.01667}=0.3187 \end{gathered}$$

Substitute for into equation (4).

${\omega }_{0+\Delta t}={\omega }_t+\Delta t\left(\frac{{\omega }_{syn}-0}{2H{\omega }_{pu}}\right)$

Substitute 1 for ${\omega }_{pu}$,20 for H ,$377\frac{\text{rad}}{\text{s}}$for${\omega }_0$and$0.01667\text{sec}$for$\Delta t$into equation.

$$\begin{gathered} {\omega }_{0.01667}=377+0.01667\left(\frac{377}{2\times 20\left(1\right)}\right) \\ {\omega }_{0.01667}=377+0.157 \\ {\omega }_{0.01667}=377.157 \end{gathered}$$

Substitute$0.01667$for t into equation (3).

${\delta }_{0.01667+\Delta t}={\delta }_{0.01667}+\Delta t\left({\omega }_{0.01667}-{\omega }_{syn}\right)$

Substitute $0.3187$for ${\delta }_0$, $377.157\frac{\text{rad}}{\text{s}}$for ${\omega }_{0.01667}$ and $0.01667\text{sec}$for $\Delta t$into above equation.

$$\begin{gathered} {\delta }_{0.01667}=0.3187+0.01667\left(377.157-377\right) \\ {\delta }_{0.01667}=0.3187+0.0026 \\ {\delta }_{0.01667}=0.3213 \end{gathered}$$

Substitute $0.01667$for t into equation (4).

${\omega }_{0,01667+\Delta t}={\omega }_{0.01667}+\Delta t\left(\frac{{\omega }_{syn}+0.1\left(0.157\right)}{2H{\omega }_{pu}}\right)$

Substitute 1 for ${\omega }_{pu}$, $20$for H ,$377.157\frac{\text{rad}}{\text{s}}$for ${\omega }_{0.01667}$and $0.01667\text{sec}$for $\Delta t$into equation.

$\begin{array}{rcl}{\omega }_{0.01667+0.01667}& =& 377.157+0.01667\left(\frac{377}{2\times 20\left(1\right)}\right)\\ {\omega }_{0.0333}& =& 377.157+0.157\\ {\omega }_{0.0333}& =& 377.314\\ & & \end{array}$

Hence, the generator power angle at end of the second time step ${\delta }_{0.0333}$is$0.3213\text{rad}$.