EE532 Power System Dynamics and Transients

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EE532 Power System Dynamics and Transients
EUMP Distance Education Services

• Satish J Ranade
• Classical Analysis
• Equal Area Criterion
• Lecture 4

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First swing stability-Equal Area Criterion

Pm Pe

• A generator connected to an infinite bus through
  a line. Initially PmPe

Stability is governed by the Swing Equation
Swing Equation Power Angle Equation
d2d/dt2 (pf/H) (Pm-Pe)
dd /dt ?-?syn
Pe E V sin (d) /(XXL)
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First swing stability-Equal Area
Criterion Stable Equilibriumsmall increase in
mechanical power
At D ? is decreasing but gt 0 d increases further
say to point E By now suppose ? is back to zero
and decreasing Thus ? becomes lt 0 as the
generator continues to slow Since ?lt0 d
decreases towards B First swing stable!
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First swing stability-Equal Area
Criterion Stable Equilibriumsmall increase in
mechanical power
P
Pe
E
C
D
Pm1
B
A
Pm
d
do
d1
First swing Stable
?-?syn
0
d
d1
d0
Time
0
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First swing stability-Equal Area
Criterion Stable Equilibriumsmall increase in
mechanical power
P
Pe
E
C
D
Pm1
B
A
Pm
d
do
d1
?-?syn
0
d
d1
Equal Speeds at Points A and E
d0
Time
0
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First swing stability-Equal Area CriterionBasic
Principle

• Two points on the trajectory
• A speed?1 angle d1 time t1
• B speed?1 angle d2 time t2gtt1
• d/dt(dd/dt)2 (dd/dt)(d2d/dt2)

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First swing stability-Equal Area CriterionBasic
Principle
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First swing stability-Equal Area CriterionBasic
Principle
If ?(t1 ) ?(t2)
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First swing stability-Equal Area CriterionBasic
Principle
Relates to changes in KE If change in KE can be
recovered 1 machine-infinite bus system will be
stable
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First swing stability-Equal Area
CriterionBasic Principle
P
Pe
E
C
Pm1
A
Pm
d
do
dx
d2
dx
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First swing stability-Equal Area
CriterionApplication

• Establish initial conditions
• Define sequence of events and network for each
  event
• Develop Power angle curves
• Apply EAC

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First swing stability-Equal Area Criterion
Example 1
Stability under small change in mechanical power
Xd0.2 XL0.2 pu.
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First swing stability-Equal Area Criterion
Example 1
Stability under small change in mechanical power
A 10 MVA, 0.8 pf lagging, 4160 V, 60Hz,
three-phase generator supplies 50 rated power
at .8 pf lagging to a 4160 V infinite bus.
Determine if the generator is first-swing stable
if the prime mover power is increased by 10
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Equal Area Criterion
Example 1 Stability under small change in
mechanical power
Xd0.2 XL0.2 pu.
1. Initial
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First swing stability-Equal Area Criterion
Example 1 Stability under small change in
mechanical power
Xd0.2 XL0.2 pu.
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First swing stability-Equal Area Criterion
Example 1 Stability under small change in
mechanical power
Xd0.2 XL0.2 pu.
2. Power angle equation before and after change
in Pm Pe Pmax Sin(d) Pmax EV8 /(XdXL)
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First swing stability-Equal Area Criterion Small
change in mechanical power
P
Pe
E
C
Pm1
A
Pm
d
do
dx
d2
dx
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Equal Area Criterion-Small change in mechanical
power
EAC
Remember
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Equal Area CriterionExample 2-Fault
2
1
8
3
The infinite bus receives 1 pu real power at 0.95
power factor lagging
A fault at bus 3 is cleared by opening lines from
1-3 and 2-3 when the generator power angle
dReaches 40 deg. Is the system first swing
stable?
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Equal Area CriterionExample 2-Fault
Example 2
2
1
8
3
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Equal Area CriterionExample 2-Fault
Example 2
Data (Resistances are zero)
Xd0.3 Ztj0.1 Z1Z3j0.2 Z2j0.1 V81 pu
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Equal Area CriterionExample 2-Fault
Example 2
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
1. Initial
S(1/0.95)/acos(0.95) I(S/V8)
1.05/-18.2o Xeq (XdXt)X1(X2X3)0.52 pu
E/d V8 jXeq I 1.28/23.95o
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Equal Area CriterionExample 2-Fault

• 2. Events
• Prefault Steady State
• Fault three phase fault at bus 3
• Post fault Line 2-3 1-3 open

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Equal Area CriterionExample 2-Fault
3a. Pre-fault Power Angle curve
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
Xeq (XdXt)X1(X2X3)0.52 pu E/d V8
jXeq I 1.28/23.95o V81/0 Pe E V8 sin d /Xeq
2.46 sin d Note d 23.95o Pe1
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Equal Area CriterionExample 2-Fault
3b. During-fault Power Angle curve
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
1
Zt
Zth
jXd
Vth0.33 V 8 Zth j 0.066
Vth -
E -
Vt -
I
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Equal Area CriterionExample 2-Fault
3c. During-fault Power Angle curve
Xeq0.466
Pe E V8 sin d /Xeq 0.915 sin d
1
Zt
Zth
jXd
Vth0.33 V 8 Zth j 0.066 Xd 0.2 Zt0.2
Vth -
E -
Vt -
I
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Equal Area CriterionExample 2-Fault
3c. Post Fault Line 1-3 2-3 out
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
3
Xeq .2.2.2
Pe E V8 sin d /Xeq 2.14 sin d
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Equal Area CriterionExample 2-Fault
Deceleration
4. Trajectories and ares
P
Pe
Acceleration
Pm
d
dm
do
dcl
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Equal Area CriterionExample 2-Fault
Apply EAC
P
Pe
Pm
d
dm
do
dcl
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Equal Area CriterionExample 2-Fault
Solve by trial and error Want Error 0
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Equal Area CriterionExample 2-Fault
Rotor swings to 55 degrees then swings back-
STABLE
P
Pe
Pm
d
do
dcl
dm
55 40 24
23.95 40 55
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Equal Area CriterionExample 2-Fault
Basic Question How Quickly can the fault be
cleared Relay Time Breaker Operation Arc
Extinction If breaker fails then Backup
operates gt10 Leads to notion of critical
clearing time more the better
3-5
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Equal Area CriterionExample 2-Fault with delayed
clearing
Fault cleared when rotor angle is 120 deg no
solution
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Equal Area CriterionExample 2-Fault
Rotor swings past 156 degrees UN STABLE
P
Pe
Pm
d
do
dcl
dm
156 120 24
23.95 40 55
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Equal Area CriterionExample 2-Fault-Critical
clearing angle
For stability dm lt p- d1 solve for dcl
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Equal Area CriterionExample 2-Fault- Critical
Clearing
Rotor swings past 156 degrees UN STABLE
P
Pe
Pm
d
d0 d1
dmp- d1
dcl112.9
23.95
156 112 24
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Summary Equal Area Criterion

• For simple systems
• 1 machine infinite bus
• 2 machines
• Provides convenient way to determine first swing
  stability
• For 2 points on the P-d curve that have equal
  speed
• Accelerating are Decelerating area

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Summary Equal Area Criterion

• 1 machine to infinite bus
• Machine must return to synchronous speed
• EAC indicates stable or unstable
• Introduced notion of critical clearing

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Next

• Critical Clearing Angle v. Critical Clearing
  Time See example 13.8
• Numerical Solution
• Multimachine System