Lecture 24 Transient Stability Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS.

Lecture 24Transient Stability

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

2

Announcements

Be reading Chapter 11 and Chapter 12 thru 12.3 HW 10 is 11.4, 11.7, 11.10, 11.19, 11.20; due Dec 1 in

class. Project is due Thursday Dec 1 either in class or under my

office door (343 Everitt) Final exam is as given on the UIUC website. That is,

Tuesday Dec 13 from 7 to 10pm here (218 Ceramics). Final is comprehensive, with more emphasis on material since

exam 2. Three notesheets allowed (e.g., ones from previous two exams and

one new notesheet)

3

2007 CWLP Dallman Accident

In 2007 there was an explosion at the CWLP 86 MW Dallman 1 generator. The explosion was eventually determined to be caused by a sticky valve that prevented the cutoff of steam into the turbine when the generator went off line. So the generator turbine continued to accelerate up to over 6000 rpm (3600 normal). High speed caused parts of the generator to shoot out Hydrogen escaped from the cooling system, and eventually escaped

causing the explosion Repairs took about 18 months, costing more than $52 million

4

Dallman After the Accident

5

Outside of Dallman

6

Transient Stability Example

A 60 Hz generator is supplying 550 MW to an infinite bus (with 1.0 per unit voltage) through two parallel transmission lines. Determine initial angle change for a fault midway down one of the lines.H = 20 seconds, D = 0.1. Use t=0.01 second.

Ea

7

Transient Stability Example, cont'd

a

e

We first need to determine the pre-fault values.

Since P = 550 MW (5.5 pu) I = 5.5

E 1.0 0.1 5.5 1.141 28.8

Next to get P ( ) we need to determine the

thevenin equivalent during the fault looking

j

into

the network from the generator

0.05 0.05 0.1 0.08333

0.3333 0th

th

Z j j j j

V

8

Transient Stability Example, cont'd

prefaulte

m

faultede

1 2

1.141 1.0Therefore prefault we have P ( ) sin

0.1and P 5.5 (0) 28.8 (0) 0.50265 radians

1.141 0.3333and during the fault P ( ) sin

0.08333Let x and x . The equations to integ

1 2

2 1 2

1 2

rate are

1 1.141 0.33335.5 sin 0.1

20/ 60 0.08333

(0) 0.50265 (0) 0.0

x x

x x x

x x

9

Transient Stability Example, cont'd

1 2

2 1 29.425 5.5 4.564sin 0.1

0.50265(0)

0

With Euler's Method we get

0.50265 0 0.50265(0.01) 0.01

0 31.11 0.3111

0.50265 0.3111 0.50576(0.02) 0.01

0.3111 30.82 0.

x x

x x x

x

x

x

6193

10

Transient Stability Example, cont'd

0 0.5 1 1.5 2

Simulation time in seconds

0

60

120

180

240G

en

erat

or

ang

le in

deg

rees

clearing at 0.3 seconds

clearing at 0.2 seconds

clearing at 0.1 seconds

11

Equal Area Criteria

• The goal of the equal area criteria is to try to determine whether a system is stable or not without having to completely integrate the system response.

System willbe stable afterthe fault ifthe DecelArea is greaterthan the Accel. Area

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Example 11.4: Undamped

13

Example 11.4: Damped

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Transient Stability Numeric Methods

• Numerical integration is used to solve multi-machine transient stability problems

• Requires a solution of the algebraic network power balance equations and the differential equations

• Two main solution approaches • Partitioned solution, in which the solution of the algebraic

equations alternates with the solution of the differential equations. This approach is used in most commercial packages, but it can suffer from numerical instability

• Simultaneous solution, which uses implicit integration.

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Two-Axis Synchronous Machine Model

• Classical model is appropriate only for the most basic studies; no longer widely used in practice

• More realistic models are required to couple in other devices such as exciters and governors

• A more realistic synchronous machine model requires that the machine be expressed in a reference frame that rotates at rotor speed

• Standard approach is d-q reference frame, in which the major (direct or d-axis) is aligned with the rotor poles and the quadrature (q-axis) leads the direct axis by 90

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D-q Reference Frame

• Machine voltage and current are “transformed” into the d-q reference frame using the rotor angle, • Terminal voltage in network (power flow) reference frame are

VT = Vr - Vi

sin cos

cos sindr

qi

VV

VV

sin cos

cos sind real

q imag

V V

V V

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Two-Axis Model Equations

• Numerous models exist for synchronous machines. The following is a relatively simple model that represents the field winding and one damper winding; it also includes the generator swing eq.

' 'q q a q d dE V R I X I ' '

d d a d q qE V R I X I

'

' ''

1( )q

q d d d fddo

dEE X X I E

dt T

'

' ''

1( )d

d q q qqo

dEE X X I

dt T

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Generator Torque and Initial Conditions

• The generator electrical torque is given by

• Recall pe = Tep.u (sometimes p.u is assumed=1.0)

• Solving the differential equations requires determining ; it is determined by noting that in steady-state

Then is the angle of 2 2( )e d d q q a d qT V I V I R I I

T qE V jX I

E

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Example 11.10

• Determine the initial conditions for the Example 11.3 case with the classical generator replaced by a two-axis model with H = 3.0 per unit-seconds, = 0, = 2.1, = 2.0, = 0.3, = 0.5, all per unit using the 100 MVA system base

• First determine the current out of the generator from the initial conditions, then the terminal voltage

1.0526 18.20 1 0.3288I j

1.0 0 0.22 1.0526 18.20

1.0946 11.59 1.0723 0.220TV j

j

20

Example 11.10, cont.

We can then get the initial angle and initial dq values

1.0946 11.59 2.0 1.052 18.2 2.814 52.1

52.1

E j

0.7889 0.6146 1.0723 0.7107

0.6146 0.7889 0.220 0.8326d

q

V

V

0.7889 0.6146 1.000 0.9909

0.6146 0.7889 0.3287 0.3553d

q

I

I

21

Example 11.10, cont.

• The initial state variable are determined by solving with the differential equations equal to zero.

• The transient stability solution is then solved by numerically integrating the differential equations, coupled with solving the algebraic equations '

'

0.8326 0.3 0.9909 1.1299

0.7107 (0.5)(0.3553) 0.5330

1.1299 (2.1 0.3)(0.9909) 2.9135

q

d

fd

E

E

E

22

PowerWorld Solution of 11.10

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Generator Exciters and Governors

• The two-axis synchronous model takes as an input the field voltage and the mechanical power. The next section discusses how these values are controlled

24

Generator Exciters

• The purpose of the exciter is to maintain the generator terminal voltage (or other close by voltage) at a specified value.• Input is the sensed voltage• Output is the field voltage to the machine, Efd

• Physically several technologies are used. • Older generators used dc machines with brushes transferring the power• With the newer brushless (or static) exciters power is obtained from an

“inverted” synchronous generator whose field voltage is on the stator and armature windings are on rotor; output is rectified to create dc.

•

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Exciter Block Diagrams

• Block diagrams are used to setup the transient stability models. The common IEEE Type 1 exciter is shown below (neglecting saturation); this is a dc type exciter. Initial state values are determined by knowing Efd and the terminal voltage Vt.

fV

1

1 rsT 1a

a

K

sT

maxVR

minVR

Vref

1

e eK sT

E fdRV

1f

f

sK

sT

VtV

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Exciter Block Diagram Example

• Consider again the Example 11.10 case, with an IEEE T1 exciter with Tr = 0, Ka = 100, Ta = 0.05, Vrmax = 5, Vrmin = -5, Ke = 1, Te = 0.26, Kf = 0.01 and Tf = 1.0. Determine the initial states. Initial value of Efd = 2.9135 and Vt = 1.0946

1.0 2.9135 2.9135r e fdV K E

2.9135

1.0946 1.1237100

ref t f a r

rref t f

a

V V V K V

VV V V

K

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PowerWorld Example 12.1 Solution