EBK POWER SYSTEM ANALYSIS AND DESIGN
EBK POWER SYSTEM ANALYSIS AND DESIGN
6th Edition
ISBN: 9781305886957
Author: Glover
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 11, Problem 11.1P

a.

To determine

The value of ωsyn and ωmsyn .

a.

Expert Solution
Check Mark

Answer to Problem 11.1P

The value of synchronous angular radial frequency is ωs=377 rad/s .

The value of synchronous angular velocity of rotor is ωmsyn=188.5 rad/s .

Explanation of Solution

Given:

A three-phase steam turbine-generating unit having

Frequency, f=60 Hz

MVA rating, S=500 MVA

Rated voltage, V=11.8 kV

Poles, P=4

Inertia constant, H=6 pus

Formula Used:

The synchronous angular radial frequency is expressed as,

  ωs=2πf   ..... (1)

Where f is frequency.

The synchronous angular velocity of rotor is expressed as,

  ωmsyn=(2P)ωs   ..... (2)

Where P is number of poles and ωs is synchronous angular radial frequency.

Calculation:

To determine the value of synchronous angular radial frequency ωs , substitute the given data in expression (1)

  ωs=2πf=2π×60 rad/s=377 rad/s

To determine the value of synchronous angular velocity of rotor ωmsyn , substitute the given data in expression (2)

  ωmsyn=(2P)ωs=(24)×2π×60=188.5 rad/s

Conclusion:

The value of synchronous angular radial frequency is ωs=377 rad/s .

The value of synchronous angular velocity of rotor is ωmsyn=188.5 rad/s .

b.

To determine

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

b.

Expert Solution
Check Mark

Answer to Problem 11.1P

The kinetic energy in joules stored in the rotating mass at synchronous speed is KE=3×109J .

Explanation of Solution

Given:

A three-phase steam turbine-generating unit having

Frequency, f=60 Hz

MVA rating, S=500 MVA

Rated voltage, V=11.8 kV

Poles, P=4

Inertia constant, H=6 pus

FormulaUsed:

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

  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.

  Kinetic Eenergy=HS=6×500×106J=3×109J

Conclusion:

The kinetic energy in joules stored in the rotating mass at synchronous speed is KE=3×109J .

c.

To determine

The mechanical angular acceleration αm and electrical angular acceleration α .

c.

Expert Solution
Check Mark

Answer to Problem 11.1P

The mechanical angular acceleration is αm=15.71 rad/s2 and the electrical angular acceleration α=31.42 rad/s2 .

Explanation of Solution

Given:

A three-phase steam turbine-generating unit having

Frequency, f=60 Hz

MVA rating, S=500 MVA

Rated voltage, V=11.8 kV

Poles, P=4

Inertia constant, H=6 pus

Accelerating power, Pa=500 MW

FormulaUsed:

The power swing equation is expressed as,

  Papu=2Hωpuαωsyn   ..... (3)

Where H is inertia constant, α is electrical angular acceleration, ωsyn is synchronous angular radial frequency and ωpu is per unit angular frequency.

The mechanical angular acceleration is expressed as,

  αm=2Pα   ...... (4)

Where P is number of poles and α is electrical angular acceleration.

Calculation:

Assume the base power to be 500 MW.

So the per unit accelerating power

  Papu=Actual powerBase power=500500=1 pu.

And the per unit angular frequency ωpu=1

Now, rearrange the equation (3) for α and solve

  α=Papuωsyn2Hωpu=1×3772×6×1=31.42 rad/s2

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

  αm=2Pα=24×31.42rad/s2=15.71 rad/s2

Conclusion:

The mechanical angular acceleration is αm=15.71 rad/s2 and the electrical angular acceleration α=31.42 rad/s2 .

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EBK POWER SYSTEM ANALYSIS AND DESIGN

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