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A Power System Example
Starrett Mini-Lecture #3
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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
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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
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Accelerating Power, Pacc
Pacc = Pmech - Pelec
Steady State => No acceleration Pacc = 0 => Pmech = Pelec
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Classical Generator Model
Generator connected to Infinite bus through 2 lossless transmission lines
E’ and xd’ are constants is governed by the swing equation
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Simplifying the system . . .
Combine xd’ & XL1 & XL2
jXT = jxd’ + jXL1 || jXL2
The simplified system . . .
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Recall the power-angle curve
Pelec = E’ |VR| sin( ) XT
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Use power-angle curve
Determine steady state (SEP)
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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
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Power angle curves
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Graphical illustration of the fault study
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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
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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
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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
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For the system to be stable for a given clearing angle , there must be sufficient area under the curve for A2 to “cover” A1.
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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?
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Clearing at 80 degrees
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Clearing at 120 degrees
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What would plots of vs. t look like for these 2 cases?