a power system example

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A Power System Example Starrett Mini-Lecture #3

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A Power System Example. Starrett Mini-Lecture #3. Power System Equations. Start with Newton again .... T = I a We want to describe the motion of the rotating masses of the generators in the system. The swing equation. 2H d 2 d = P acc w o dt 2 P = T w - PowerPoint PPT Presentation

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Page 1: A Power System Example

A Power System Example

Starrett Mini-Lecture #3

Page 2: A Power System Example

Power System Equations

Start with Newton again ....T = I

We want to describe the motion of the rotating masses of the generators in the system

Page 3: A Power System Example

The swing equation

2H d2 = Pacc

o dt2

P = T = d2/dt2, acceleration is the second

derivative of angular displacement w.r.t. time

= d/dt, speed is the first derivative

Page 4: A Power System Example

Accelerating Power, Pacc

Pacc = Pmech - Pelec

Steady State => No acceleration Pacc = 0 => Pmech = Pelec

Page 5: A Power System Example

Classical Generator Model

Generator connected to Infinite bus through 2 lossless transmission lines

E’ and xd’ are constants is governed by the swing equation

Page 6: A Power System Example

Simplifying the system . . .

Combine xd’ & XL1 & XL2

jXT = jxd’ + jXL1 || jXL2

The simplified system . . .

Page 7: A Power System Example

Recall the power-angle curve

Pelec = E’ |VR| sin( ) XT

Page 8: A Power System Example

Use power-angle curve

Determine steady state (SEP)

Page 9: A Power System Example

Fault study

Pre-fault => system as given Fault => Short circuit at infinite bus

Pelec = [E’(0)/ jXT]sin() = 0 Post-Fault => Open one transmission line

XT2 = xd’ + XL2 > XT

Page 10: A Power System Example

Power angle curves

Page 11: A Power System Example

Graphical illustration of the fault study

Page 12: A Power System Example

Equal Area Criterion

2H d2 = Pacc

o dt2

rearrange & multiply both sides by 2d/dt

2 d d2 = o Pacc d dt dt2 H dt

=>d {d}2 = o Pacc ddt {dt } H dt

Page 13: A Power System Example

Integrating,

{d}2 = o Pacc d{dt} H dt

For the system to be stable, must go through a maximum => d/dt must go through zero. Thus . . . m

o Pacc d = 0 = { d2

H { dt } o

Page 14: A Power System Example

The equal area criterion . . .

For the total area to be zero, the positive part must equal the negative part. (A1 = A2)

Pacc d = A1 <= “Positive” Area

Pacc d = A2 <= “Negative” Area

cl

o

m

cl

Page 15: A Power System Example

For the system to be stable for a given clearing angle , there must be sufficient area under the curve for A2 to “cover” A1.

Page 16: A Power System Example

In-class Exercise . . .

Draw a P- curve

For a clearing angle of 80 degrees is the system stable? what is the maximum angle?

For a clearing angle of 120 degrees is the system stable? what is the maximum angle?

Page 17: A Power System Example

Clearing at 80 degrees

Page 18: A Power System Example

Clearing at 120 degrees

Page 19: A Power System Example

What would plots of vs. t look like for these 2 cases?