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Announcements Read Chapters 11 and 12 (sections 12.1 to 12.3)

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Presentation on theme: "Announcements Read Chapters 11 and 12 (sections 12.1 to 12.3)"— Presentation transcript:

0 ECE 476 Power System Analysis
Lecture 24: Transient Stability Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign Special Guest Lecturer: TA Won Jang

1 Announcements Read Chapters 11 and 12 (sections 12.1 to 12.3)
Homework 10 is due today (no quiz) Homework 11 is 11.19, 11.25, 12.3, 12.11, 14.15; it should be done before the final but is not to be turned in Design project due date is Tuesday, December 8 Final exam is Wednesday Dec 16, 7 to 10pm, room 1013; comprehensive, closed book, closed notes with three note sheets and standard calculators allowed

2 Euler’s Method 2

3 Second Order Runge-Kutta Method
Runge-Kutta methods improve on Euler's method by evaluating f(x) at selected points over the time step Simplest method is the second order method in which That is, k1 is what we get from Euler's; k2 improves on this by reevaluating at the estimated end of the time step 3

4 RK2 Versus Euler's RK2 requires twice the function evaluations per iteration, but gives much better results With RK2 the error tends to vary with the cube of the step size, compared with the square of the step size for Euler's The smaller error allows for larger step sizes compared to Eulers 4

5 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 = Use Dt=0.01 second. Ea 5

6 Transient Stability Example, cont'd
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7 Transient Stability Example, cont'd
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8 Transient Stability Example, cont'd
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9 Transient Stability Example, cont'd
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10 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 will be stable after the fault if the Decel Area is greater than the Accel. Area 10

11 Example 11.4: Undamped 11

12 Example 11.4: Damped 12

13 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 13

14 Synchronous Machine Modeling
3 bal. windings (a,b,c) – stator Field winding (fd) on rotor Damper in “d” axis (1d) on rotor 2 dampers in “q” axis (1q, 2q) on rotor

15 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

16 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. 16

17 Generator Torque and Initial Conditions
The generator electrical torque is given by Recall pe = Tep.u (sometimes p.u is assumed=1.0) Solving the differential equations requires determining d; it is determined by noting that in steady-state Then d is the angle of 17

18 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, D = 0, Xd = 2.1, Xq = 2.0, X’d = 0.3, X’q = 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 18

19 Example 11.10, cont. We can then get the initial angle, and initial d and q values 19

20 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 20

21 PowerWorld Solution of 11.10
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22 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 22

23 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. 23

24 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. 24

25 Exciter Block Diagram Example
Consider again the Example 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 = and Vt = 25

26 PowerWorld Example 12.1 Solution
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27 Generator Governors The other key generator control system is the governor, which changes the mechanical power into the generator to maintain a desired speed and hence frequency. Historically centrifugal “flyball” governors have been used to regulate the speed of devices such as steam engines The centrifugal force varies with speed, opening or closing the throttle valve Photo source: en.wikipedia.org/wiki/Centrifugal_governor 27

28 Isochronous Governors
Ideally we would like the governor to maintain the frequency at a constant value of 60 Hz (in North America) This can be accomplished using an isochronous governor. A flyball governor is not an isochronous governor since the control action is proportional to the speed error An isochronous governor requires an integration of the speed error Isochronous governors are used on stand alone generators but cannot be used on interconnected generators because of “hunting” 28

29 Generator “Hunting” Control system “hunting” is oscillation around an equilibrium point Trying to interconnect multiple isochronous generators will cause hunting because the frequency setpoints of the two generators are never exactly equal One will be accumulating a frequency error trying to speed up the system, whereas the other will be trying to slow it down The generators will NOT share the power load proportionally. 29

30 Droop Control The solution is to use what is known as droop control, in which the desired set point frequency is dependent upon the generator’s output R is known as the regulation constant or droop; a typical value is 4 or 5%. 30

31 Governor Block Diagrams
The block diagram for a simple stream unit, the TGOV1 model, is shown below. The T1 block models the governor delays, whereas the second block models the turbine response. 31

32 Example 12.4 System Response
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33 Problem 12.11 33

34 Restoring Frequency to 60 Hz
In an interconnected power system the governors to not automatically restore the frequency to 60 Hz Rather this is done via the ACE (area control area calculation). Previously we defined ACE as the difference between the actual real power exports from an area and the scheduled exports. But it has an additional term ACE = Pactual - Psched – 10b(freqact - freqsched) b is the balancing authority frequency bias in MW/0.1 Hz with a negative sign. It is about 0.8% of peak load/generation 34

35 2600 MW Loss Frequency Recovery
Frequency recovers in about ten minutes

36 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 36

37 Dallman After the Accident
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38 Outside of Dallman 38


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