lecture 5 power system operation, transmission lines professor tom overbye department of electrical...

Lecture 5 Power System Operation, Transmission Lines

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

2

Reading and Homework

• 1st Exam moved to Oct 11 (in class)• 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 though you should still at least skim those sections.

• HW 1 is 2.9, 22, 28, 32, 48; due Thursday 9/8• For Problem 2.32 you need to use the PowerWorld Software. You can

download the software and cases at the below link; get version 15.http://www.powerworld.com/gloversarma.asp

Direct PowerWorld download page is

http://www.powerworld.com/DemoSoftware/GloverSarmaSimdwnldv15.asp

3

Substation Bus

4

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

5

100 MW Transaction

Bus 2 Bus 1

Bus 3Home Area

Scheduled Transactions

225 MW

113 MVR

150 MW

291 MW 8 MVR

138 MVR

113 MW 56 MVR

1.00 PU

8 MW -2 MVR

-8 MW 2 MVR

-84 MW 27 MVR

85 MW-23 MVR

93 MW-25 MVR

-92 MW

30 MVR

1.00 PU

1.00 PU

0 MW 32 MVR

100 MWAGC ONAVR ON

AGC ONAVR ON

100.0 MW

Scheduled100 MWTransaction from Left to Right

Net tie-lineflow is now100 MW

6

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.

7

Security Constrained Dispatch

Bus 2 Bus 1

Bus 3Home Area

Scheduled Transactions

357 MW

179 MVR

194 MW

448 MW 19 MVR

232 MVR

179 MW 89 MVR

1.00 PU

-22 MW 4 MVR

22 MW -4 MVR

-142 MW 49 MVR

145 MW-37 MVR

124 MW-33 MVR

-122 MW

41 MVR

1.00 PU

1.00 PU

0 MW 37 MVR100%

100%

100 MWOFF AGCAVR ON

AGC ONAVR ON

100.0 MW

Dispatch is no longer optimal due to need to keep line from bus 2 to bus 3 from overloading

8

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.”

9

Seven Bus Case: One-line

Top Area Cost

Left Area Cost Right Area Cost

1

2

3 4

5

6 7

106 MW

168 MW

200 MW 201 MW

110 MW 40 MVR

80 MW 30 MVR

130 MW 40 MVR

40 MW 20 MVR

1.00 PU

1.01 PU

1.04 PU1.04 PU

1.04 PU

0.99 PU1.05 PU

62 MW

-61 MW

44 MW -42 MW -31 MW 31 MW

38 MW

-37 MW

79 MW -77 MW

-32 MW

32 MW-14 MW

-39 MW

40 MW-20 MW 20 MW

40 MW

-40 MW

94 MW

200 MW 0 MVR

200 MW 0 MVR

20 MW -20 MW

AGC ON

AGC ON

AGC ON

AGC ON

AGC ON

8029 $/MWH

4715 $/MWH 4189 $/MWH

Case Hourly Cost 16933 $/MWH

System hasthree areas

Area lefthas onebus

Area right has onebus

Area tophas fivebuses

10

Seven Bus Case: Area View

System has40 MW of“Loop Flow”

Actualflowbetweenareas

Loop flow can result in higher losses

Area Losses

Area Losses Area Losses

Top

Left Right

-40.1 MW

0.0 MW

0.0 MW

0.0 MW

40.1 MW

40.1 MW

7.09 MW

0.33 MW 0.65 MW

Scheduledflow

11

Seven Bus - Loop Flow?

Area Losses

Area Losses Area Losses

Top

Left Right

-4.8 MW

0.0 MW

100.0 MW

0.0 MW

104.8 MW

4.8 MW

9.44 MW

-0.00 MW 4.34 MW

100 MW Transactionbetween Left and Right

Transaction has actually decreasedthe loop flow

Note thatTop’s Losses haveincreasedfrom 7.09MW to9.44 MW

12

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)

13

3 BUS LMPS - OVERLOAD IGNORED

Bus 2 Bus 1

Bus 3

Total Cost

0 MW

0 MW

180 MWMW

10.00 $/MWh

60 MW 60 MW

60 MW

60 MW120 MW

120 MW

10.00 $/MWh

10.00 $/MWh

180 MW120%

120%

0 MWMW

1800 $/hr

Line from Bus 1 to Bus 3 is over-loaded; all buses have same marginal cost

Gen 1’scostis $10per MWh

Gen 2’scostis $12per MWh

14

LINE OVERLOAD ENFORCED

Bus 2 Bus 1

Bus 3

Total Cost

60 MW

0 MW

180 MWMW

12.00 $/MWh

20 MW 20 MW

80 MW

80 MW100 MW

100 MW

10.00 $/MWh

14.01 $/MWh

120 MW 80% 100%

80% 100%

0 MWMW

1921 $/hr

Line from 1 to 3 is no longer overloaded, but nowthe marginal cost of electricity at 3 is $14 / MWh

15

MISO and PJM

MISO and PJM arethe reliabilitycoordinatorscovering theelectric gridin Illinois. ComEd is inPJM, and Ameren is inMISO.

16

MISO LMPs 8/31/11 at 11:05 AM

www.midwestmarket.org

17

Development of Line Models

Goals of this section are

1) develop a simple model for transmission lines

2) gain an intuitive feel for how the geometry of the transmission line affects the model parameters

18

Primary Methods for Power Transfer

The most common methods for transfer of electric power are

1) Overhead ac

2) Underground ac

3) Overhead dc

4) Underground dc

5) other

19 19

345 kV+ Transmission Growth at a Glance

20 20


21 21


22 22


23 23


24

Magnetics Review

Ampere’s circuital law:

e

F = mmf = magnetomtive force (amp-turns)

= magnetic field intensity (amp-turns/meter)

d = Vector differential path length (meters)

= Line integral about closed path (d is tangent to path)

I =

eF d I

H l

H

l

l

Algebraic sum of current linked by

25

Line Integrals

Line integrals are a generalization of traditional integration

Integration along thex-axis

Integration along ageneral path, whichmay be closed

Ampere’s law is most useful in cases of symmetry, such as with an infinitely long line

26

Magnetic Flux Density

Magnetic fields are usually measured in terms of flux density

0-7

0

= flux density (Tesla [T] or Gauss [G])(1T = 10,000G)

For a linear a linear magnetic material

= where is the called the permeability

=

= permeability of freespace = 4 10

= relative permea

r

r

H m

B

B H

bility 1 for air

27

Magnetic Flux

Total flux passing through a surface A is

=

= vector with direction normal to the surface

If flux density B is uniform and perpendicular to an area A then

=

Ad

d

BA

B a

a

28

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 givenby the “Right-hand” Rule

Easiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose acircle with a radius of x.

29

Single Line Example, cont’d

0

5 00 4

70

5

4

22

25 5

ln 2 10 ln2 4 4

4.46 10 Wb

2 10 2B T Gauss

x

A

IxH I H

xB H

IH dA dx

xI

I

x

For reference, the earth’smagnetic field is about0.6 Gauss (Central US)

30

Flux linkages and Faraday’s law

N

i=1

Flux linkages are defined from Faraday's law

dV = where V = voltage, = flux linkages

The flux linkages tell how much flux is linking an

N turn coil:

=

If all flux links every coil then

i

dt

N

31

Inductance

For a linear magnetic system, that is one where

B = H

we can define the inductance, L, to be

the constant relating the current and the flux

linkage

= L i

where L has units of Henrys (H)

32

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

33

Inductance Example, cont’d

0

0

20

2 (path length is 2 R)

H2

2

H2

e

r

r

r

I d

NI H R

NIB H H

RAB N LI

NINAB NA

R

N AL

R

H l

34

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.

35

Flux Linkages outside of the wire

R0A r

We'll think of the wire as a single loop closed at

infinity. Therefore = since N = 1. The flux linking

the wire out to a distance of R from the wire center is

d length 2Idxx

B a

36

Flux Linkages outside, cont’d

R0A r

R 00r

d length 2

Since length = we'll deal with per unit length values,

assumed to be per meter.

ln2 2

Note, this quantity still goes to infinity as R

Idxx

I Rdx Imeter x r

B a

37

Flux linkages inside of wire

Current inside conductor tends to travel on the outside

of the conductor due to the skin effect. The pentration

of the current into the conductor is approximated using

1the skin depth = where f is

f the frequency in Hz

and is the conductivity in mhos/meter.

0.066 mFor copper skin depth 0.33 inch at 60HZ.

fFor derivation we'll assume a uniform current density.

38

Flux linkages inside, cont’d

Wire cross section

x

r

2

2

2

Current enclosed within distance

x of center I

2 2

e

ex

xI

rI Ix

Hx r

2 30

inside 2 2 40 0

Flux only links part of current

2 82

r r rIx x Ixdx dx I

r r r

39

Line Total Flux & Inductance

0 0

0

0

(per meter) ln2 8

(per meter) ln2 4

L(per meter) ln2 4

Note, this value still goes to infinity as we integrate

R out to infinity

rTotal

rTotal

r

RI I

rR

Ir

Rr

40

Inductance Simplification

0 0 4

0 4

Inductance expression can be simplified using

two exponential identities:

aln(ab)=ln a + ln b ln ln ln ln( )

b

ln ln ln ln2 4 2

ln ln2

r

r

a

r

a b a e

RL R r e

r

L R re

0

4r

ln2 '

Where r' 0.78 for 1r

Rr

r e r

lecture 5 power system operation, transmission lines professor tom overbye department of electrical...

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