Chapter 11, Problem 11.1P

a.

To determine

The value of ${\omega }_{syn}$ and ${\omega }_{msyn}$ .

Answer to Problem 11.1P

The value of synchronous angular radial frequency is ${\omega }_s=377\text{ rad/s}$ .

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

Explanation of Solution

Given:

A three-phase steam turbine-generating unit having

Frequency, $f=60\text{ Hz}$

MVA rating, $S=500\text{ MVA}$

Rated voltage, $V=11.8\text{ kV}$

Poles, $P=4$

Inertia constant, $H=6\text{ pus}$

Formula Used:

The synchronous angular radial frequency is expressed as,

  ${\omega }_s=2\pi f$   ..... (1)

Where $f$ is frequency.

The synchronous angular velocity of rotor is expressed as,

  ${\omega }_{msyn}=\left(\frac{2}{P}\right){\omega }_s$   ..... (2)

Where $P$ is number of poles and ${\omega }_s$ is synchronous angular radial frequency.

Calculation:

To determine the value of synchronous angular radial frequency ${\omega }_s$ , substitute the given data in expression (1)

  $\begin{array}{c}{\omega }_s=2\pi f\\ =2\pi \times 60\text{ rad/s}\\ =377\text{ rad/s}\end{array}$

To determine the value of synchronous angular velocity of rotor ${\omega }_{msyn}$ , substitute the given data in expression (2)

  $\begin{array}{c}{\omega }_{msyn}=\left(\frac{2}{P}\right){\omega }_s\\ =\left(\frac{2}{4}\right)\times 2\pi \times 60\\ =188.5\text{ rad/s}\end{array}$

Conclusion:

The value of synchronous angular radial frequency is ${\omega }_s=377\text{ rad/s}$ .

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

b.

To determine

The kinetic energy in joules stored in the rotating mass a synchronous speed.

Answer to Problem 11.1P

The kinetic energy in joules stored in the rotating mass at synchronous speed is $\text{KE}=\text{3}\times {\text{10}}^9\text{J}$ .

Explanation of Solution

Given:

A three-phase steam turbine-generating unit having

Frequency, $f=60\text{ Hz}$

MVA rating, $S=500\text{ MVA}$

Rated voltage, $V=11.8\text{ kV}$

Poles, $P=4$

Inertia constant, $H=6\text{ pus}$

FormulaUsed:

The kinetic energy in joules stored in the rotating mass at synchronous speed is expressed by

  $\text{Kinetic Eenergy}=HS$

Where $H$ is inertia constant and $S$ is rated KVA.

Calculation:

To calculate the value of kinetic energy in joules stored in the rotating mass at synchronous speed, substitute the values of $H$ and $S$ in above equation.

  $\begin{array}{c}\text{Kinetic Eenergy}=HS\\ =6\times 500\times {10}^6\text{J}\\ \text{=3}\times {\text{10}}^9\text{J}\end{array}$

Conclusion:

The kinetic energy in joules stored in the rotating mass at synchronous speed is $\text{KE}=\text{3}\times {\text{10}}^9\text{J}$ .

c.

To determine

The mechanical angular acceleration ${\alpha }_m$ and electrical angular acceleration $\alpha$ .

Answer to Problem 11.1P

The mechanical angular acceleration is ${\alpha }_m=15.71{\text{ rad/s}}^2$ and the electrical angular acceleration $\alpha =31.42{\text{ rad/s}}^2$ .

Explanation of Solution

Given:

A three-phase steam turbine-generating unit having

Frequency, $f=60\text{ Hz}$

MVA rating, $S=500\text{ MVA}$

Rated voltage, $V=11.8\text{ kV}$

Poles, $P=4$

Inertia constant, $H=6\text{ pus}$

Accelerating power, $P_a=500\text{ MW}$

FormulaUsed:

The power swing equation is expressed as,

  $P_{apu}=\frac{2H{\omega }_{pu}\alpha }{{\omega }_{syn}}$   ..... (3)

Where $H$ is inertia constant, $\alpha$ is electrical angular acceleration, ${\omega }_{syn}$ is synchronous angular radial frequency and ${\omega }_{pu}$ is per unit angular frequency.

The mechanical angular acceleration is expressed as,

  ${\alpha }_m=\frac{2}{P}\alpha$   ...... (4)

Where $P$ is number of poles and $\alpha$ is electrical angular acceleration.

Calculation:

Assume the base power to be 500 MW.

So the per unit accelerating power

  $\begin{array}{c}{P}_{apu}=\frac{\text{Actual power}}{\text{Base power}}\\ =\frac{500}{500}\\ =1\text{ pu}\text{.}\end{array}$

And the per unit angular frequency ${\omega }_{pu}=1$

Now, rearrange the equation (3) for $\alpha$ and solve

  $\begin{array}{c}\alpha =\frac{P_{apu}{\omega }_{syn}}{2H{\omega }_{pu}}\\ =\frac{1\times 377}{2\times 6\times 1}\\ =31.42{\text{ rad/s}}^2\end{array}$

To calculate the value of mechanical angular acceleration, substitute the values in equation (4)

  $\begin{array}{c}{\alpha }_m=\frac{2}{P}\alpha \\ =\frac{2}{4}\times 31.42{\text{rad/s}}^2\\ =15.71{\text{ rad/s}}^2\end{array}$

Conclusion:

The mechanical angular acceleration is ${\alpha }_m=15.71{\text{ rad/s}}^2$ and the electrical angular acceleration $\alpha =31.42{\text{ rad/s}}^2$ .