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
• Numerical Solution Multi-machine Systems
• Lecture 5

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Equal Area CriterionExample 3(13.11)-Fault
2
1
8
3
The infinite bus receives 1 pu real power at 0.95
power factor lagging
A fault at bus 1 is cleared by opening lines from
1-3 and 2-3 Find critical clearing angle and
time
3
Equal Area CriterionExample 3(13.11)-Fault
2
1
8
3
Find critical clearing angle and
time Approach Use EAC to find Critical clearing
angle Solve swing equation during fault --
can solve since Pe0 -- find time where
ddcrit In general case we will need to simulate
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Equal Area CriterionExample 3-Fault
1. Initial conditions
2
1
8
3
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Equal Area CriterionExample 3-Fault
1. Initial conditions
Data (Resistances are zero)
Xd0.3 Ztj0.1 Z1Z3j0.2 Z2j0.1 V81 pu
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Equal Area CriterionExample 3-Fault
Example 2
1. Initial conditions
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 3-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 3-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 3-Fault
3b. During-fault Power Angle curve
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
Pe0
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Equal Area CriterionExample 3-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 3-Fault
Deceleration
4. Trajectories and areas
P
Pe
Acceleration
Pm
d
do
dcl
dclrdcrit
dmax
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EAC
P
Pe2.14sin(d)
Pm1
d
do
dclrdcrit
dmax
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CCT
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Time domain Simulation -- Generally cannot get
Analytical Solution to Swing equation
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First swing stability-Numerical Solution

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-Numerical Solution
Nonlinear ODE in state variable form
d ? /dt (pf/H) (Pm-Pmax sin d)
dd /dt ?-?syn
PmaxEV/(XdXL)
Usually cannot get a closed form solution
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First swing stability-Numerical Solution
Numerical solution find d(t) and ?(t)
d ? /dt (pf/H) (Pm-Pmax sin d)
dd /dt ?-?syn
Divide time into intervals tn-2,tn-1,tn, Predic
t dn from dn-1, d n-2,
Step size
dn
dn-1
dn-2
tn-2 tn-1 tn t
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First swing stability-Numerical Solution
Numerical solution find d(t) and ?(t)
d ? /dt (pf/H) (Pm-Pmax sin d)
dd /dt ?-?syn
Euler Method Uniform time step h dn dn-1 h
dd/dtttn-1
dd/dtttn-1
dn-1
h
tn-1 tn-2 tn1 t
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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
• SImulate

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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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First swing stability-Numerical Solution-Small
change in Pm
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First swing stability-Numerical Solution-Small
change in Pm
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Equal Area Criterion-Small change in mechanical
power
EAC
Remember
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First swing stability-Numerical Solution-Small
change in Pm
The EAC in the previous slide says angle swings
to 9.77 deg and then swings back Oscillates
around the new equilibrium of 8.949 deg
( Step size of 0.001 is a little big
oscillation is growing Due to numerical
instability)
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First swing stability-Numerical Solution-Small
change in PmEffect of Damping Damper windings
provide relative speed damping. Other effects
provide absolute damping.This will make swing
settle
Swing equation with relative speed damping
d ? /dt (pf/H) (Pm-Pmax sin d-D (?- ?syn)
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Example 2-Fault
First swing stability-Numerical Solution-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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Example 2-Fault
Apply EAC
P
Pe
Pm
d
dm
do
dcl
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First swing stability-Numerical Solution-Fault
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First swing stability-Numerical Solution-Fault
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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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First swing stability-Numerical Solution-Fault
tgttrict.35
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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
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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Stability of Numerical Solutions

• Can become unstable due to
• Roundoff
• Approximation

dd/dtttn-1
Approximation Error
dn-1
h
tn-1 tn-2 tn1 t
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Stability of Numerical Solutions

• Can control
• Roundoff ( Reduce airthmetic operations)
• Approximation(Use more advanced method, but will
  increase arithmetic operations)
• Need to chose a reasonable step size ( or
  adaptively vary step size)
• Numerical stability Means the the global error
  remains bounded
• (See Crows Text for EE531/ Also see Kundur)

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Stability of Numerical Solutions
For the small change in Pm(Example 1) Euler is
unstable even with h.001
Note x- axis units n h time in seconds
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Additional Methods

• Taylor Series-based Second order method

First Estimate
Slope 1
Slope 2
Final Estimate
Estimated
Actual
h
Average
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Additional Methods
Modified Euler Example for Fault Case
(Glover/Sarma p.626)
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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Additional Methods
Modified Euler Example for Fault Case
Notation Revisited dx/dt f(x,t)
d2d/dt2 (pf/H) (Pm-Pe)
Swing Equation Power Angle Equation
dd /dt ?-?syn
Pe E V sin (d) /(XdXL)
Time t n h nstep number h step size

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Additional Methods
Modified Euler Example for Fault Case
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Additional Methods
Modified Euler Example for Fault Case
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Additional Methods
Modified Euler Example for Fault Case
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Additional Methods
Compare Euler Example for Fault Case h0.01