1 / 44

ECE 476 POWER SYSTEM ANALYSIS

ECE 476 POWER SYSTEM ANALYSIS. Lecture 4 Power System Operation, Transmission Line Modeling Professor Tom Overbye Department of Electrical and Computer Engineering. Reading and Homework. For lectures 4 through 6 please be reading Chapter 4

Download Presentation

ECE 476 POWER SYSTEM ANALYSIS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ECE 476POWER SYSTEM ANALYSIS Lecture 4 Power System Operation, Transmission Line Modeling Professor Tom Overbye Department of Electrical andComputer Engineering

  2. Reading and Homework • For lectures 4 through 6 please be reading Chapter 4 • we will not be covering sections 4.7, 4.11, and 4.12 in detail • HW 1 is 2.7, 12, 21, 26 is due now • HW 2 is 2.32, 43, 47 • (You can download the latest educational version of PowerWorld (version 13) at http://www.powerworld.com/gloversarma.asp • The Problem 2.32 case will also be on the website

  3. Interesting Graph: Electricity Prices Real isinflationadjustedto 2000prices;nominalis unad-justed Source: EIA 2007 Annual Energy Review, Fig 8.10

  4. Operating (Balancing Authority) Areas • Transmission lines that join two areas are known as tie-lines. • The net power out of an area is the sum of the flow on its tie-lines. • The flow out of an area is equal to total gen - total load - total losses = tie-flow

  5. Area Control Error (ACE) • The area control error is the difference between the actual flow out of an area, and the scheduled flow. • Ideally the ACE should always be zero. • Because the load is constantly changing, each utility must constantly change its generation to “chase” the ACE.

  6. Automatic Generation Control • Most utilities use automatic generation control (AGC) to automatically change their generation to keep their ACE close to zero. • Usually the utility control center calculates ACE based upon tie-line flows; then the AGC module sends control signals out to the generators every couple seconds.

  7. Three Bus Case on AGC Generation is automatically changed to match change in load Net tie flow is close to zero

  8. Generator Costs • There are many fixed and variable costs associated with power system operation. • The major variable cost is associated with generation. • Cost to generate a MWh can vary widely. • For some types of units (such as hydro and nuclear) it is difficult to quantify. • For thermal units it is much easier. These costs will be discussed later in the course.

  9. Economic Dispatch • Economic dispatch (ED) determines the least cost dispatch of generation for an area. • For a lossless system, the ED occurs when all the generators have equal marginal costs. IC1(PG,1) = IC2(PG,2) = … = ICm(PG,m)

  10. Power Transactions • Power transactions are contracts between areas to do power transactions. • Contracts can be for any amount of time at any price for any amount of power. • Scheduled power transactions are implemented by modifying the area ACE:ACE = Pactual,tie-flow - Psched

  11. 100 MW Transaction Scheduled 100 MW Transaction from Left to Right Net tie-line flow is now 100 MW

  12. Security Constrained ED • Transmission constraints often limit system economics. • Such limits required a constrained dispatch in order to maintain system security. • In three bus case the generation at bus 3 must be constrained to avoid overloading the line from bus 2 to bus 3.

  13. Security Constrained Dispatch Dispatch is no longer optimal due to need to keep line from bus 2 to bus 3 from overloading

  14. Multi-Area Operation • If Areas have direct interconnections, then they may directly transact up to the capacity of their tie-lines. • Actual power flows through the entire network according to the impedance of the transmission lines. • Flow through other areas is known as “parallel path” or “loop flows.”

  15. Seven Bus Case: One-line System has three areas Area top has five buses Area left has one bus Area right has one bus

  16. Seven Bus Case: Area View Actual flow between areas System has 40 MW of “Loop Flow” Scheduled flow Loop flow can result in higher losses

  17. Seven Bus - Loop Flow? Note that Top’s Losses have increased from 7.09MW to 9.44 MW Transaction has actually decreased the loop flow 100 MW Transaction between Left and Right

  18. Pricing Electricity • Cost to supply electricity to bus is called the locational marginal price (LMP) • Presently some electric makets post LMPs on the web • In an ideal electricity market with no transmission limitations the LMPs are equal • Transmission constraints can segment a market, resulting in differing LMP • Determination of LMPs requires the solution on an Optimal Power Flow (OPF)

  19. 3 BUS LMPS - OVERLOAD IGNORED Gen 2’s cost is $12 per MWh Gen 1’s cost is $10 per MWh Line from Bus 1 to Bus 3 is over-loaded; all buses have same marginal cost

  20. LINE OVERLOAD ENFORCED Line from 1 to 3 is no longer overloaded, but now the marginal cost of electricity at 3 is $14 / MWh

  21. LMPs at 7:40 AM Today Source: www.midwestmarket.org

  22. LMPs at 7:55 AM Today

  23. Development of Line Models • Goals of this section are • develop a simple model for transmission lines • gain an intuitive feel for how the geometry of the transmission line affects the model parameters

  24. Primary Methods for Power Transfer • The most common methods for transfer of electric power are • Overhead ac • Underground ac • Overhead dc • Underground dc • other

  25. Magnetics Review • Ampere’s circuital law:

  26. Line Integrals • Line integrals are a generalization of traditional integration Integration along the x-axis Integration along a general path, which may be closed Ampere’s law is most useful in cases of symmetry, such as with an infinitely long line

  27. Magnetic Flux Density • Magnetic fields are usually measured in terms of flux density

  28. Magnetic Flux

  29. Magnetic Fields from Single Wire • Assume we have an infinitely long wire with current of 1000A. How much magnetic flux passes through a 1 meter square, located between 4 and 5 meters from the wire? Direction of H is given by the “Right-hand” Rule Easiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose a circle with a radius of x.

  30. Single Line Example, cont’d For reference, the earth’s magnetic field is about 0.6 Gauss (Central US)

  31. Flux linkages and Faraday’s law

  32. Inductance • For a linear magnetic system, that is one where • B = m H • we can define the inductance, L, to be • the constant relating the current and the flux • linkage • l = L i • where L has units of Henrys (H)

  33. Inductance Example • Calculate the inductance of an N turn coil wound tightly on a torodial iron core that has a radius of R and a cross-sectional area of A. Assume • 1) all flux is within the coil • 2) all flux links each turn

  34. Inductance Example, cont’d

  35. Inductance of a Single Wire • To development models of transmission lines, we first need to determine the inductance of a single, infinitely long wire. To do this we need to determine the wire’s total flux linkage, including • 1. flux linkages outside of the wire • 2. flux linkages within the wire • We’ll assume that the current density within the wire is uniform and that the wire has a radius of r.

  36. Flux Linkages outside of the wire

  37. Flux Linkages outside, cont’d

  38. Flux linkages inside of wire

  39. x r Flux linkages inside, cont’d Wire cross section

  40. Line Total Flux & Inductance

  41. Inductance Simplification

  42. R Two Conductor Line Inductance • Key problem with the previous derivation is we assumed no return path for the current. Now consider the case of two wires, each carrying the same current I, but in opposite directions; assume the wires are separated by distance R. To determine the inductance of each conductor we integrate as before. However now we get some field cancellation Creates a clockwise field Creates counter- clockwise field

  43. Two Conductor Case, cont’d R R Rp Direction of integration Key Point: As we integrate for the left line, at distance 2R from the left line the net flux linked due to the Right line is zero! Use superposition to get total flux linkage. Left Current Right Current

  44. Two Conductor Inductance

More Related