ECE 476 Power System Analysis

Lecture 15: Power Flow Sensitivities, Economic Dispatch

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Read Chapter 6, Chapter 12.4 and 12.5 • HW 7 is 6.50, 6.52, 6.59, 12.20, 12.26; due October

22 in class (no quiz)

2

Balancing Authority Areas

• An balancing authority area (use to be called operating areas) has traditionally represented the portion of the interconnected electric grid operated by a single utility

• 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 3

Area Control Error (ACE)

• The area control error (ace) is the difference between the actual flow out of an area and the scheduled flow, plus a frequency component

• Ideally the ACE should always be zero.• Because the load is constantly changing,

each utility must constantly change its generation to “chase” the ACE.

int schedace 10P P f

4

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.

5

Power Transactions

• Power transactions are contracts between generators and loads 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 value of Psched used in the ACE calculation

6

PTDFs

• Power transfer distribution factors (PTDFs) show the linear impact of a transfer of power.

• PTDFs calculated using the fast decoupled power flow B matrix

1 ( )

Once we know we can derive the change in

the transmission line flows

Except now we modify several elements in ( ),

in portion to how the specified generators would

participate in the pow

θ B P x

θ

P x

er transfer7

Nine Bus PTDF Example

10%

60%

55%

64%

57%

11%

74%

24%

32%

A

G

B

C

D

E

I

F

H

300.0 MW 400.0 MW 300.0 MW

250.0 MW

250.0 MW

200.0 MW

250.0 MW

150.0 MW

150.0 MW

44%

71%

0.00 deg

71.1 MW

92%

Figure shows initial flows for a nine bus power system

8

Nine Bus PTDF Example, cont'd

43%

57% 13%

35%

20%

10%

2%

34%

34%

32%

A

G

B

C

D

E

I

F

H

300.0 MW 400.0 MW 300.0 MW

250.0 MW

250.0 MW

200.0 MW

250.0 MW

150.0 MW

150.0 MW

34%

30%

0.00 deg

71.1 MW

Figure now shows percentage PTDF flows from A to I

9

Nine Bus PTDF Example, cont'd

6%

6% 12%

61%

12%

6%

19%

21%

21%

A

G

B

C

D

E

I

F

H

300.0 MW 400.0 MW 300.0 MW

250.0 MW

250.0 MW

200.0 MW

250.0 MW

150.0 MW

150.0 MW

20%

18%

0.00 deg

71.1 MW

Figure now shows percentage PTDF flows from G to F

10

WE to TVA PTDFs

11

Line Outage Distribution Factors (LODFS)

• LODFs are used to approximate the change in the flow on one line caused by the outage of a second line– typically they are only used to determine the change in

the MW flow– LODFs are used extensively in real-time operations– LODFs are state-independent but do dependent on the

assumed network topology

,l l k kP LODF P

12

Flowgates

• The real-time loading of the power grid is accessed via “flowgates”

• A flowgate “flow” is the real power flow on one or more transmission element for either base case conditions or a single contingency– contingent flows are determined using LODFs

• Flowgates are used as proxies for other types of limits, such as voltage or stability limits

13

NERC Regional Reliability Councils

NERCis theNorthAmericanElectricReliabilityCouncil

Image: http://www.nerc.com/AboutNERC/keyplayers/Documents/NERC_Regions_Color.jpg 14

NERC Reliability Coordinators

Image: www.nerc.com/pa/rrm/TLR/Pages/Reliability-Coordinators.aspx 15

Generation Dispatch

• Since the load is variable and there must be enough generation to meet the load, almost always there is more generation capacity available than load

• Optimally determining which generators to use can be a complicated task due to many different constraints– For generators with low or no cost fuel (e.g., wind and solar

PV) it is “use it or lose it”– For others like hydro there may be limited energy for the year – Some fossil has shut down and start times of many hours

• Economic dispatch looks at the best way to instantaneously dispatch the generation

16

Generator types

• Traditionally utilities have had three broad groups of generators– baseload units: large coal/nuclear; always on at max.– midload units: smaller coal that cycle on/off daily– peaker units: combustion turbines used only for several

hours during periods of high demandWind and solarPV can be quite variable;usually they areoperated at max.available power

17

Example California Wind Output

Source: www.megawattsf.com/gridstorage/gridstorage.htm

18

Image shows wind output for a month by hour and day

Thermal versus Hydro Generation

• The two main types of generating units are thermal and hydro, with wind rapidly growing

• For hydro the fuel (water) is free but there may be many constraints on operation– fixed amounts of water available– reservoir levels must be managed and coordinated– downstream flow rates for fish and navigation

• Hydro optimization is typically longer term (many months or years)

• In 476 we will concentrate on thermal units and some wind, looking at short-term optimization

19

Block Diagram of Thermal Unit

To optimize generation costs we need to develop

cost relationships between net power out and operating

costs. Between 2-6% of power is used within the

generating plant; this is known as the auxiliary power20

Modern Coal Plant

Source: Masters, Renewable and Efficient Electric Power Systems, 200421

Turbine for Nuclear Power Plant

Source: http://images.pennnet.com/articles/pe/cap/cap_gephoto.jpg22

Basic Gas Turbine

Compressor

Fuel100%

Fresh air

Combustion chamber

Turbine

Exhaustgases 67%

Generator

ACPower 33%

1150 oC

550 oC

Brayton Cycle: Working fluid is always a gas

Most common fuel is natural gas

Maximum Efficiency

550 2731 42%

1150 273

Typical efficiency is around 30 to 35%

23

Combined Cycle Power Plant

Efficiencies of up to 60% can be achieved, with even highervalues when the steam is used for heating. Fuel is usually natural gas

24

Generator Cost Curves

• Generator costs are typically represented by up to four different curves– input/output (I/O) curve– fuel-cost curve– heat-rate curve– incremental cost curve

• For reference– 1 Btu (British thermal unit) = 1054 J– 1 MBtu = 1x106 Btu– 1 MBtu = 0.293 MWh– 3.41 Mbtu = 1 MWh

25

I/O Curve

• The IO curve plots fuel input (in MBtu/hr) versus net MW output.

26

Fuel-cost Curve

• The fuel-cost curve is the I/O curve scaled by fuel cost. A typical cost for coal is $ 1.70/Mbtu.

27

Heat-rate Curve

• Plots the average number of MBtu/hr of fuel input needed per MW of output.

• Heat-rate curve is the I/O curve scaled by MW

Best for most efficient units are

around 9.0

28

Incremental (Marginal) cost Curve

• Plots the incremental $/MWh as a function of MW.• Found by differentiating the cost curve

29

Mathematical Formulation of Costs

• Generator cost curves are usually not smooth. However the curves can usually be adequately approximated using piece-wise smooth, functions.

• Two representations predominate– quadratic or cubic functions– piecewise linear functions

• In 476 we'll assume a quadratic presentation

2( ) $/hr (fuel-cost)

( )( ) 2 $/MWh

i Gi i Gi Gi

i Gii Gi Gi

Gi

C P P P

dC PIC P P

dP

30

Coal Usage Example 1

• A 500 MW (net) generator is 35% efficient. It is being supplied with Western grade coal, which costs $1.70 per MBtu and has 9000 Btu per pound. What is the coal usage in lbs/hr? What is the cost?

At 35% efficiency required fuel input per hour is

500 MWh 1428 MWh 1 MBtu 4924 MBtuhr 0.35 hr 0.29 MWh hr

4924 MBtu 1 lb 547,111 lbshr 0.009MBtu hr

4924 MBtu $1.70Cost = 8370.8 $/hr or $16.74/MWh

hr MBtu

31

Coal Usage Example 2

• Assume a 100W lamp is left on by mistake for 8 hours, and that the electricity is supplied by the previous coal plant and that transmission/distribution losses are 20%. How coal has been used?

With 20% losses, a 100W load on for 8 hrs requires

1 kWh of energy. With 35% gen. efficiency this requires

1 kWh 1 MWh 1 MBtu 1 lb1.09 lb

0.35 1000 kWh 0.29 MWh 0.009MBtu

32

Incremental Cost Example

21 1 1 1

22 2 2 2

1 11 1 1

1

2 22 2 2

2

For a two generator system assume

( ) 1000 20 0.01 $ /

( ) 400 15 0.03 $ /

Then

( )( ) 20 0.02 $/MWh

( )( ) 15 0.06 $/MWh

G G G

G G G

GG G

G

GG G

G

C P P P hr

C P P P hr

dC PIC P P

dP

dC PIC P P

dP

33

Incremental Cost Example, cont'd

G1 G2

21

22

1

2

If P 250 MW and P 150 MW Then

(250) 1000 20 250 0.01 250 $ 6625/hr

(150) 400 15 150 0.03 150 $6025/hr

Then

(250) 20 0.02 250 $ 25/MWh

(150) 15 0.06 150 $ 24/MWh

C

C

IC

IC

34

Economic Dispatch: Formulation

• The goal of economic dispatch is to determine the generation dispatch that minimizes the instantaneous operating cost, subject to the constraint that total generation = total load + losses

T1

m

i=1

Minimize C ( )

Such that

m

i Gii

Gi D Losses

C P

P P P

Initially we'll ignore generatorlimits and thelosses

35

Unconstrained Minimization

• This is a minimization problem with a single inequality constraint

• For an unconstrained minimization a necessary (but not sufficient) condition for a minimum is the gradient of the function must be zero,

• The gradient generalizes the first derivative for multi-variable problems:

1 2

( ) ( ) ( )( ) , , ,

nx x x

f x f x f xf x

( ) f x 0

36

Minimization with Equality Constraint

• When the minimization is constrained with an equality constraint we can solve the problem using the method of Lagrange Multipliers

• Key idea is to modify a constrained minimization problem to be an unconstrained problem

That is, for the general problem

minimize ( ) s.t. ( )

We define the Lagrangian L( , ) ( ) ( )

Then a necessary condition for a minimum is the

L ( , ) 0 and L ( , ) 0

T

x λ

f x g x 0

x λ f x λ g x

x λ x λ37

Economic Dispatch Lagrangian

G1 1

G

For the economic dispatch we have a minimization

constrained with a single equality constraint

L( , ) ( ) ( ) (no losses)

The necessary conditions for a minimum are

L( , )

m m

i Gi D Gii i

Gi

C P P P

dCP

P

P

1

( )0 (for i 1 to m)

0

i Gi

Gi

m

D Gii

PdP

P P

38

Economic Dispatch Example

D 1 2

21 1 1 1

22 2 2 2

1 1

1

What is economic dispatch for a two generator

system P 500 MW and

( ) 1000 20 0.01 $/

( ) 400 15 0.03 $/

Using the Largrange multiplier method we know

( )20 0

G G

G G G

G G G

G

G

P P

C P P P hr

C P P P hr

dC PdP

1

2 22

2

1 2

.02 0

( )15 0.06 0

500 0

G

GG

G

G G

P

dC PP

dP

P P

39

Economic Dispatch Example, cont’d

1

2

1 2

1

2

1

2

We therefore need to solve three linear equations

20 0.02 0

15 0.06 0

500 0

0.02 0 1 20

0 0.06 1 15

1 1 500

312.5 MW

187.5 MW

26.2 $/MWh

G

G

G G

G

G

G

G

P

P

P P

P

P

P

P

40