Question

Question:A three-phase,60-Hz , 500-MVA ,11.8 KV , 4 -pole steam turbine-generating unit has an H constant of 5p.u-s. Determine: (a) ${\omega }_{\mathbf{syn}}$ and ${\omega }_{msyn}$; (b) the kinetic energy in joules stored in the rotating masses at synchronous speed; (c) the mechanical angular acceleration ${\mathbf{\alpha }}_{\mathbf{m}}$ and electrical angular acceleration if the unit is operating at synchronous speed with an accelerating power of 500 MW.

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 .

Step-by-step solution

Write the given data from the question.

Consider the value of frequency is 60-Hz .

Consider the value of base MVA is 500.

Consider the value of voltage is 11.8 KV .

Consider -pole steam turbine-generating unit has an H constant of 5 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 ${\omega }_{\mathbf{syn}}$ issynchronous 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 speed.

Write the formula of electrical angular acceleration.

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

Here, P_e.p.u electrical power output of the generator plus electrical losses, per unit,${\omega }_{\mathbf{syn}}$ is synchronous electrical radial frequency, is H-constant and 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.

(a) Determine the value of ωsyn and ωmsyn.

Determine the value of synchronous electrical radial frequency.

Substitute 60-Hz for f into equation (1)

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

Therefore,the value of synchronous radial frequency ${\omega }_{\text{syn}}$ is 377.

Determine the value of angular velocity of rotor.

Substitute 4 forP , for ${\omega }_{\text{syn}}$ into equation (2).

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

Therefore, the value of synchronous angular velocity of rotor is188.5 .

(b) Determine the value of kinetic energy .

Determine the value of the kinetic energy.

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

role="math" localid="1655891169519" $2.5\times {10}^9\text{joules}$

Therefore, the value of kinetic energy KE is 2.5.

(c) Determine the value of electrical angular acceleration  and mechanical angular acceleration αmt

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 500 MW for$P_m\left(t\right)$, 0MW for $P_e\left(t\right)$ and $S_{\text{rated}}$ for 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, is H-constant, issynchronous electrical radial frequency, isper-unit rotor angular velocity, is power angle andelectrical power output of the generator plus electrical losses, per unit.

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

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

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

Substitute 1 for P_e.p.u, 376.99 rad/s for ${\omega }_{\text{syn}}$, 5 p.u.s or 1 and ${\omega }_{\text{p.u.}}$ for into equation (4).

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

Therefore, the value of electrical angular acceleration is.

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

Substitute for and for into equation (5).

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

Therefore, the value of mechanical angular acceleration is 18.85 rad/s² .