Question

Question:The generating unit in Problem 11.1 is initially operating at ${\mathbf{P}}_{\mathbf{mp}\mathbf{.}\mathbf{u}\mathbf{.}}={\mathbf{P}}_{\mathbf{ep}\mathbf{.}\mathbf{u}\mathbf{.}}=\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{per}\mathbf{unit}$, $\mathit{\omega }\mathbf{=}{\mathit{\omega }}_{\mathbf{syn}}$ and $\delta =\mathbf{12}\mathbf{°}$ when a fault reduces the generator electrical power output by . Determine the power angle five cycles after the fault commences. Assume that the accelerating power remains constant during the fault. Also assume that ${\omega }_{\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}\left(\mathbf{t}\right)=\mathbf{1}\mathbf{.}\mathbf{0}$ in the swing equation.

Short answer

Answer

(a)

The value of synchronous radial frequency ${\omega }_{\text{syn}}$ is 377 rad/s.

The value of synchronous angular velocity of rotor ${\omega }_{\text{msyn}}$ is 188.5 rad/s.

(b)

The value of kinetic energy KE is$2.5\times {10}^9\text{joules}$.

(c)

The value of electrical angular acceleration $\alpha \left(t\right)$ is 37.7 rad/s².

The value of mechanical angular acceleration ${\alpha }_m\left(t\right)$ is .18.85 rad/s²

The value of load angle at the end of 5 cycles is $14.62°$.

Step-by-step solution

Write the given data from the question

As reference of problem 11.1

Consider the value of${\text{P}}_{\text{mp.u.}}={\text{P}}_{\text{ep.u.}}=\text{0.7}\text{per}\text{unit}$.

Consider the value of power angle,$\delta =12°$

Consider generator electrical power output by 60% .

Consider the value of frequency is$\text{60}-\text{Hz}$.

Consider the value of base MVA is 500.

Consider the value of voltage is$\text{11.8}\text{kV}$.

Consider -pole steam turbine-generating unit has an H constant of $\text{6 p.u.}$-s.

Determine the formula of ωsyn and ωmsyn.

Write the formula of synchronous electrical radial frequency.

${\omega }_{\mathbf{syn}}=\mathbf{2}\mathbf{\pi f}$ …… (1)

Here, f is frequency.

Write the formula of synchronous angular velocity of rotor.

${\omega }_{\mathbf{msyn}}=\left(\frac{\mathbf{2}}{\mathbf{P}}\right){\mathbf{\omega }}_{\mathbf{syn}}\frac{\mathbf{rad}}{\mathbf{S}}$ …… (2)

Here, P is pole and ${\omega }_{\mathbf{syn}}$ is synchronous electrical radial frequency.

Write the formula of Kinetic energy.

$\mathbf{KE}={\mathbf{HS}}_{\mathbf{rated}}$ …… (3)

Here, H is H-constant and ${\mathbf{S}}_{\mathbf{rated}}$ is rated volt-ampere rating.

Write the formula of electrical angular acceleration$\mathbf{\alpha }\left(\mathbf{t}\right)$.

$\mathbf{\alpha }\left(t\right)\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{ep}\mathbf{.}\mathbf{u}\mathbf{.}}\left(t\right){\mathbf{\omega }}_{\mathbf{syn}}}{\mathbf{2}{\mathbf{H\omega }}_{\mathbf{p}\mathbf{.}\mathbf{u}}\left(t\right)}$ …… (4)

Here, ${\mathbf{P}}_{\mathbf{ep}\mathbf{.}\mathbf{u}\mathbf{.}}$electrical power output of the generator plus electrical losses, per unit, ${\omega }_{\mathbf{syn}}$ is synchronous electrical radial frequency,H is H-constant and${\mathbf{\omega }}_{\mathbf{p}\mathbf{.}\mathbf{u}}\left(\mathbf{t}\right)$ is per-unit rotor angular velocity.

Write the formula of mechanical angular acceleration${\mathbf{\alpha }}_{\mathbf{m}}\left(\mathbf{t}\right)$.

${\mathbf{\alpha }}_{\mathbf{m}}\left(\mathbf{t}\right)=\left(\frac{\mathbf{2}}{\mathbf{P}}\right)\mathbf{\alpha }\left(\mathbf{t}\right)$ …… (5)

Here, P is pole, $\mathbf{\alpha }\left(\mathbf{t}\right)$is electrical angular acceleration.

Write the formula of load angle at the end of cycles.

$\delta \left(\mathbf{t}\right)=\mathbf{6}\mathbf{.}\mathbf{5975}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{2094}$ …… (6)

Here, t is time taken for 5 cycles.

(a) Determine the value of ωsyn and ωmsyn.

Determine the value of synchronous electrical radial frequency.

Substitute for into equation (1)

$$\begin{gathered} {\omega }_{\text{syn}}=2\pi \left(60\right) \\ =377rad/s \end{gathered}$$

Therefore,the value of synchronous radial frequency is.

Determine the value of angular velocity of rotor.

Substitute 4 for P, 377 for W_syn into equation (2).

$$\begin{gathered} {\omega }_{\text{msyn}}=\left(\frac{2}{4}\right)\left(377\right) \\ =\left(0.5\right)\left(377\right) \\ =188.5rad/s \end{gathered}$$

Therefore, the value of synchronous angular velocity of rotor ${\omega }_{\text{msyn}}$ is 188.5 rad/s.

(b) Determine the value of kinetic energy KE.

Determine the value of the kinetic energy.

Substitute $\text{5}\text{p.u,}-\text{s}$ for 500 MVA and $S_{\text{rated}}$ for into equation (3).

$$\begin{gathered} \text{KE}=\left(5\right)\left(500\text{MVA}\right) \\ =2.5\times {10}^9\text{joules} \end{gathered}$$

Therefore, the value of kinetic energy KE is .$2.5\times {10}^9\text{joules}$

(c) Determine the value of electrical angular acceleration , mechanical angular acceleration  and moment of inertia .

Consider,

$P_e\left(t\right)=0\text{MW}$

Determine the value of the$P_{ep.u.}\left(t\right)$.

$P_{ep.u.}\left(t\right)=\frac{P_m\left(t\right)-P_e\left(t\right)}{S_{\text{rated}}}$

Here, $P_m\left(t\right)$is mechanical power, $P_e\left(t\right)$is electrical power and is three-phase volt ampere rating of the generator.

Substitute 500MW for$P_m\left(t\right)$, for and 0 MW for $P_e\left(t\right)$ into above equation.

$$\begin{gathered} P_{ep.u.}\left(t\right)=\frac{500\text{M}-0}{500\text{M}} \\ =\frac{500\text{M}}{500\text{M}} \\ =1 \end{gathered}$$

Determine the expression for swing equation.

$\frac{2H}{{\omega }_{syn}}{\omega }_{p.u.}\left(t\right)\frac{d^2\delta \left(t\right)}{d^{2}t}=P_{ep.u.}\left(t\right)$

Here, H is H-constant, is synchronous electrical radial frequency, is per-unit rotor angular velocity, is power angle and electrical power output of the generator plus electrical losses, per unit.

Here, $\frac{d^2\delta \left(t\right)}{d^{2}t}$ represent $\alpha \left(t\right)$.

$\alpha \left(t\right)=\frac{P_{ep.u.}\left(t\right){\omega }_{\text{syn}}}{2\text{H}{\omega }_{\text{p.u.}}\left(t\right)}$

Substitute 1 for $P_{ep.u.}\left(t\right)$, 376.99rad/s for ,${\omega }_{\text{syn}}$ or $5\text{p.u},-\text{s}$and 1 for ${\omega }_{\text{p.u.}}$into equation (4).

$$\begin{gathered} \alpha \left(t\right)=\frac{\left(1\right)\left(377\right)}{2\left(5\right)\left(1\right)} \\ =\frac{377}{10} \\ =37.7 rad/s^2 \end{gathered}$$

Therefore, the value of electrical angular acceleration $\alpha \left(t\right)$ is 37.7 rad/s².

Determine the value of${\alpha }_m\left(t\right)$.

Substitute 37.7 for $\alpha \left(t\right)$and 4 for P into equation (5).

$$\begin{gathered} {\alpha }_m\left(t\right)=\left(\frac{2}{4}\right)\left(37.7\right) \\ =\left(0.5\right)\left(37.7\right) \\ =18.85rad/s^2 \end{gathered}$$

Therefore, the value of mechanical angular acceleration ${\alpha }_m\left(t\right)$is 18.85 rad/s.

Determine the expression for the per unit swing equation.

$\text{2H}\frac{{\omega }_{p.u}\left(t\right)}{{\omega }_{\text{syn}}}\frac{{\text{d}}^2\delta \left(t\right)}{\text{d}{t}^2}=P_{\text{mp.u.}}\left(t\right)-P_{\text{ep.u.}}\left(t\right)$

Here, is H-constant, issynchronous electrical radial frequency, isper-unit rotor angular velocity, is power angle,electrical power output of the generator plus electrical losses, per unit and is mechanical power supplied by prime mover minus mechanical losses in.

Substitute 1 for ${\omega }_{\text{p.u.}}\left(t\right)$ into above equation.

$\frac{2\text{H}}{{\omega }_{\text{syn}}}\frac{d^2\delta \left(t\right)}{d{t}^2}=P_{\text{mp.u.}}\left(t\right)-P_{\text{ep.u.}}\left(t\right)$ …… (7)

Determine the synchronous frequency.

${\omega }_{\text{syn}}=2\pi frad/s$

Here, is supply frequency.

Substitute for into above equation.’

$$\begin{gathered} {\omega }_{\text{syn}}=2\pi \left(60\right)rad/s \\ =377rad/s \end{gathered}$$

Initially operating at $P_{\text{mp.u.}}=P_{\text{ep.u.}}=0.7$

When an electrical problem decreases electrical output by 60%, it is called an electrical fault.

$$\begin{gathered} P_{\text{ep.u.}}=\left(0.7\right)\left(1-0.6\right) \\ =\left(0.7\right)\left(0.4\right) \\ =0.28 \end{gathered}$$

Recall equation (7).

Substitute 2 for h, 377 for ${\omega }_{\text{syn}}$, 0.7 for $P_{\text{mp.u.}}$and 0.28 for into equation (7).

$$\begin{gathered} \frac{\left(2\right)\left(6\right)}{377}\frac{d^2\delta }{d{t}^2}=0.7-0.28 \\ \left(\frac{12}{377}\right)\frac{d^2\delta }{d{t}^2}=0.42 \\ \frac{d^2\delta }{d{t}^2}=\left(0.42\right)\left(\frac{377}{12}\right) \\ \frac{d^2\delta }{d{t}^2}=13.195 \end{gathered}$$ …… (8)

Determine the initial condition.

$$\begin{gathered} \delta \left(0\right)=12° \\ =\left(12\right)\left(\frac{\pi }{180}\right)\text{rad} \\ =0.2094\text{rad} \end{gathered}$$

Then, the differential is:

$\frac{d\delta }{dt}=0$

Apply integration equation (8).

$$\begin{gathered} \frac{d\delta \left(t\right)}{dt}=13.195t+\frac{d\delta \left(0\right)}{dt} \\ \frac{d\delta \left(t\right)}{dt}=13.195t+0 \\ \frac{d\delta \left(t\right)}{dt}=13.195t \end{gathered}$$ …… (9)

Integrate equation (9) using the initial conditions.

$$\begin{gathered} \delta \left(t\right)=13.195\frac{t^2}{2}+\delta \left(0\right) \\ \delta \left(t\right)=6.5975t^2+\delta \left(0\right) \end{gathered}$$

Substitute 2094 for into above equation.

$\delta \left(t\right)=6.5975t^2+0.2094$

Determine the time taken for 5 cycles.

$t=\frac{n}{f}\sec$

Here, is number of cycles and () is frequency.

Substitute for and for.

$$\begin{gathered} f=\frac{5}{60} \\ =0.0833\sec \end{gathered}$$

Determine the load angle at the end of cycles.

Substitute 0.0833 for into equation (6).

$$\begin{gathered} \delta \left(0.0833\right)=6.5975t^2+0.2094 \\ =6.5975{\left(0.0833\right)}^2+0.2094 \\ =0.0458+0.2094 \\ =0.2552\text{rad} \end{gathered}$$

Convert load angle into degrees from radians.

$$\begin{gathered} \delta \left(0.0833\right)=\left(0.2552\text{rad}\right)\left(\frac{180°}{\pi }\right) \\ =14.62° \end{gathered}$$

Therefore, the value of load angle at the end of 15 cycles is 14.62⁰.