ee/econ 458, syllabus section v...ga gc gb gd ly lz branch reactance [ p.u.] capacit y [ mw] 1-2 0.2...

© 2005 D. Kirschen 1

EE/Econ 458, Syllabus Section V.C

Power Trading Subject to Transmission Constraints

Leigh Tesfatsion

Last Revised: 3 November 2011


Taking Transmission Network Effects Into Account ( Fig. Source: ABB c/o Jamie Austen, PacifiCorps, SSGwi Conference, 2008 )


Traditional Trade-Off Between Market & Network Models ( Fig. Source: ABB c/o Jamie Austen, PacifiCorps, SSGwi Conference, 2008 )


New agent-based modeling tools are increasingly being

used to study dynamic power market/grid systems … (see www.econ.iastate.edu/tesfatsi/aelect.htm for tools & surveys )


Glossary of Basic Terms for K/S Chapter 6

(Important for EE/Econ 458 Students who did not take EE 303)

• Load: Electric energy user (e.g., company, appliance, household)

• Demand: How much power a load requires (e.g., power to run a

production process, to turn on a light, to heat a home, etc.)

• Bulk Transmission Grid: The portion of the transmission grid operating at relatively high voltage levels (138, 161, 230, 345, 500,

or 765 kV) that is used to transfer electric energy from generators to

major load centers.

• Line: A conductor for transmitting electric energy

• Branch: K/S use “branch” interchangeably with “line”


Glossary of Basic Terms … Continued

• Bus: A physical transmission element that connects two or more branches in a network representation of a transmission grid

• Node: K/S use “node” and “bus” interchangeably and assume all buses/nodes are “pricing nodes” with LMPs. U.S. power markets are

more complicated. (Example: MISO has over 33,000 buses but only

about 1,540 pricing nodes)

• Locational Marginal Price (LMP) at Bus k in

Hour H: The least cost of servicing an additional increment of demand at bus k during hour H while respecting all system constraints.

• Nodal Price: K/S use “nodal price” interchangeably with LMP.


Glossary of Basic Terms … Continued:

• System Operator: Organization charged with the primary responsibility of maintaining bulk transmission grid reliability = [ adequacy (balance of power withdrawals and injections) + security (safe operating range) ]

• Load-Serving Entity: An electric utility, transmitting utility, or Federal power marketing agency that has an obligation under Federal, State, or local law, or under long-term contracts, to provide electric energy to retail customers or to other LSEs with retail customers.

• (Wholesale) Generator: A company that produces and sells electric energy at the wholesale level through the bulk transmission grid.

• Electric Energy Demand: The planned purchase of electric energy by one or more buyers. Can be expressed in various forms, including:

Fixed demand bid = A bid to buy a specified amount of electric energy during a particular time period at a particular bus, regardless of price.

Price-sensitive demand bid = A bid to buy electric energy during a particular time period at a particular bus that expresses the buyer’s maximum willingness to pay for each successive block of energy.


Glossary of Basic Terms … Continued:

• Impedance: Takes the complex form z = r + jx, where r (ohms) = resistance, x (ohms) = reactance, j = √-1, & ohm = 1 V/A.

• Admittance: Inverse of impedance. Takes the complex form y = g + jb, where the conductance g is given by g = r/[r2 + x2] (mhos)

and the susceptance b is given by b = -x/[r2 + x2] (mhos).

• Base Voltage Vo and Base Apparent Power So: Used to convert voltage (kV) and power (MVA) magnitudes into dimensionless

units. Example: Vo = 138 kV, So = 100 MVA.

• Base Impedance Zo: Given a balanced 3-phase network with Vo in

line-to-line kV and So in 3-phase MVA, then Zo = Vo2/So (ohms)

• Per Unit (pu) Normalization: Divide all power system voltage, power, & derived quantities by their base values. Example:

reactance xkm on branch k→m in p.u. = xkm p.u. = xkm/Zo (dimensionless)


Two Fundamental Laws Governing Current and Power

Flows in Electrical Networks (cf. K/S Section 6.2.2.1)

• Kirchhoff’s Current Law (KCL)

The sum of all currents Imk entering a bus k along branches m→k must equal the sum of all currents Ikm exiting bus k along branches k→m, where

m is any other bus and I denotes current in Amperes (A).

KCL implies at any bus k that the inflow/outflow of active (or real) power

(MW) and of reactive power (MVAR) must each separately be in balance.

• Kirchhoff’s Voltage Law (KVL)

The sum of the voltage drops along the branches comprising any closed loop in an electrical network must be zero.

Implies that the voltage drops along parallel branches must be equal.

• Complex power flowing on branch k→m: Skm = Pkm + j Qkm = VkI*km

• Ohm’s Law (AC): Ikm = [Vk – Vm]/zkm (zkm = rkm+ j xkm = impedance)

voltage drop from k to m (cf. K/S p. 143) Current flowing from k to m

Text Box Omit


1 2

zA

zB

Application of Kirchhoff’s Voltage Law (KVL): Power

Transfer Distribution Factors (PTDFs) for Parallel Paths (cf. K/S Section 6.2.2.1)

IB

IA

Impedance

Impedance

Bus 1 Bus 2 I12 I12

Current flowing from Bus 1 into

Bus 2

Current flowing

into Bus 2 from

Bus 1

Current I12 = [IA + IB] (Amperes) flows from Bus 1 into Bus 2 along two

parallel paths with parallel flows IA and IB and impedences zA and zB, (Ohms).

By Ohm’s Law and KVL, IA = [V1 – V2]/zA and IB = [V1 – V2]/zB .

Hence zAIA = zBIB = zB[ I12 – IA] [zA + zB]IA = zB I12 IA = I12 • zB/[zA + zB]

Similarly, IB = I12 • zA/[zA + zB] Text Box Omit


1 2

xA

xB

Power Transfer Distribution Factors (PTDFs) for Parallel

Paths Neglecting Reactive Power Flows: Corollary of KVL (cf. K/S Section 6.2.2.1)

FB

FA

FA

x B

x A x BP F

Bx A

x A x BP

Reactance

Reactance

Path Flows (Neglecting Reactive Power and Losses): K/S pp. 143-144

Bus 1 Bus 2 P P

Injection of active power

at Bus 1

Withdrawal of

active power

at Bus 2

upper path flow lower path flow

PTDFA PTDFB Text Box Omit

© 2011 D. Kirschen and L. Tesfatsion 12

Effects of Transmission Network Constraints

on Electric Power Market Operation

(Kirschen/Strbac Chapter 6, Sections 6.1-6.3.2.9)

Important Acknowledgement:

The following slides are based on materials originally prepared by

Daniel Kirschen (U of Manchester) with edits by Leigh Tesfatsion

(Iowa State U).

Last Revised: 3 November 2011


K/S Chapter 6: Introduction

• Simple Economic Dispatch formulations examined to date assume

a given demand that must be met

all GenCos and LSEs are connected to the same bus

• Not sufficient for real-world restructured wholesale power markets

• Need to consider:

Constraints on transmission lines that restrict power flows

Line congestion and losses

Possibility of strategic demand as well as strategic supply behaviors

• Two forms of trading over a bulk transmission grid will be discussed:

Bilateral (decentralized) trading between a GenCo and an LSE

Pool (centralized) trading among GenCos and LSEs


Bilateral Trading between a GenCo and LSE

• GenCo/LSE agree on price, quantity, & other conditions

• GenCo/LSE must schedule their transactions with the system operator, and pay/receive LMP at their buses

• GenCo/LSE agree to “make whole” any discrepancies between LMPs and their contractual price level

• System operator must maintain reliability of the grid

Buys or sells limited amounts of energy to keep power withdrawals

and injections in balance on the transmission grid (adequacy)

Limits amount of power that GenCos/LSEs can withdraw/inject at

some buses if reliability cannot be ensured by other means


G1

G2

L1

L2

Bus 1 Bus 2

G3

Example of Bilateral Contract Trading:

• Three GenCos {G1,G2,G3} and two LSEs {L1,L2)

• G1 sells 300 MW to L1 (planned generation=demand)

• G2 sells 200 MW to L2 (planned generation=demand)

• Contractual prices are a private matter

• Planned quantities (300 MW & 200 MW) must be reported to system operator to ensure grid reliability.

Line corridor 1→2


G1

G2

L1

L2

Bus 1 Bus 2

G3

Example of Bilateral Contract Trading … Continued

• G1 injects 300 MW at Bus 1 to meet 300 MW demand of L1 at Bus 2

• G2 injects 200 MW at Bus 1 to meet 200 MW demand of L2 at Bus 2

• If line capacity 1→2 ≥ 500 MW No problem

• If line capacity 1→2 < 500 MW Some of these transactions

might have to be curtailed

Line Corridor 1→2


But curtail which one? Generation? Demand?

• Could use administrative Transmission Load Relief

(TLR) procedures

• TLR procedures consider:

Firm (i.e., guaranteed) vs. non-firm transactions

Order in which transactions were registered

Historical considerations

• TLR procedures do not consider economic benefits

• Economically inefficient

• How about letting the participants themselves decide?


But curtail which one … Continued

• What about permitting traders to purchase a “right” to use

the network when arranging a bilateral energy trade?

Transmission right = Right to use the transmission system for a particular transaction.

Physical transmission right = Transmission right meant to support the actual transmission of a certain amount of power

over a given transmission line.

Financial transmission right (FTR) for 1 MWh from Bus k to Bus m in Hour H = Entitlement to receive a payment

($/h) equal to [LMPm - LMPk]•1 MWh in hour H if positive, or an

obligation to pay this amount to system operator in hour H if

negative.


G1

G2

L1

L2

Bus 1 Bus 2

G3

Physical Transmission Rights:

• G1 sells 300 MW to L1 at 30.00 $/MWh

• G2 sells 200 MW to L2 at 32.00 $/MWh

• G3 sells electric power at 35.00 $/MWh

• L1 not willing to pay more than 5 $/MWh for transmission rights

• L2 not willing to pay more than 3 $/MWh for transmission rights

• Otherwise, why not buy from G3 instead ?

Line Corridor 1→2

© 2011 D. Kirschen and L. Tesfatsion

20

Problems with Physical Transmission Rights

• Not practical to implement (parallel path problem)

Path that electric power takes through a network from any point

of injection is determined by physical laws and not by the wishes

of market traders

• Market power

The assignment of physical transmission rights can potentially

increase GenCo market power despite parallel path problems

Can permit strategic GenCos to create “load pockets” by limiting

the flow of power into their region


Parallel Path Problem: Three-Bus Example

1 2

3

GC GA

GB

GD

LY

LZ

Branch Reactance

[p.u.]

Capacity

[MW]

1-2 0.2 126

1-3 0.2 250

2-3 0.1 130

Important Technical Remark:

The reactance on a branch

k-m does not depend on the

direction of the power flow on

the branch, i.e., xkm = xmk .


Parallel Path Problem … 3-Bus Example: Continued

1 2

3

GC GA

GB

GD

LY

LZ

I

II

A 400 MW sale between GenCo GB and LSE LY would

require the purchase of transmission rights on all three

branches: 1-3 (direct route) and 1-2 & 2-3 (indirect route)!


400 MW transfer between GB and LY physically possible?

1 2

3

GC GA

GB

GD

LY

LZ

Branch Reactance

[p.u.]

Capacity

[MW]

1-2 0.2 126

1-3 0.2 250

2-3 0.1 130

I

II

sum of reactances for 1-2 and 2-3

Not possible -- exceeds capacities of branches 1-2 and 2-3 !

F I

0 . 2

0.5 400 160 MW

F II

0 . 3

0.5 400 240 MW

sum of all reactances reactance for 1-3

F12

F23 F13


Counter-Flows: 200MW Trade Between GD & LZ

1 2

3

GC GA

GB

GD

LY

LZ

Flow effects resulting from

the 200 MW transaction

between GD and LZ III

IV

F III

0 . 2

200 80 MW

0.5

F IV

0 . 3 200 120 MW

0.5

F12

F23 F13


Resultant Flows:

1 2

3

GC GA

GB

GD

LY

LZ

F12 F23 FIFIII

160 80 80 MW

F13 FII

FIV

240 120 120 MW

The resultant flows are now within the branch capacity limits.

Branch Reactance

[p.u.]

Capacity

[MW]

1-2 0.2 126

1-3 0.2 250

2-3 0.1 130

1→2→3 3→2→1

1→3 3→1


Physical Transmission Rights and Parallel Paths

• Counter-flows create additional physical transmission rights

• Economic efficiency requires that these rights be considered

• Decentralized trading:

System operator only checks overall feasibility

Participants trade physical transmission rights bilaterally

Theory:

• Enough participants market discovers optimum

Practice:

• Complexity and amount of information involved are such that it is

unlikely that this optimum can be found in time to be practical.


Physical Transmission Rights and Market Power

• G3 – the only GenCo at Bus B -- is in a potential “load pocket”

• G3 purchases physical transmission rights from Bus A to Bus B

• G3 does not use or resell these rights

• Effectively reduces transmission capacity from Bus A to Bus B

• Allows G3 to increase price at Bus B charged to L1 and L2

• “Use them or loose them” provision for physical transmission rights is difficult to enforce in a timely manner

G1

G2

L1

L2

Bus A Bus B

G3

Max Capacity = 250 MW


Centralized (Pool) Trading: Day-Ahead Market

(Cf. K/S Section 6.3)

• At the beginning of each day D, LSEs & GenCos submit demand bids & supply offers to an ISO for 24 successive hours of trade in the Day-Ahead Market for day D+1

• ISO’s objective:

Optimally balance generation and load for each hour, conditional on GenCo reported cost and operating capacity attributes

Respect network transmission constraints (e.g., capacity limits)

• No branch congestion/losses in hour H Uniform market price ($/MWh) determined across the grid for hour H

• Branch congestion/losses in hour H LMPs for hour H will typically separate across the grid (different market prices set for LSEs/GenCos located at different buses)


DS= 1500 MW

Syldavia

DB= 500MW

Borduria

Example: Borduria-Syldavia Interconnection (Cf. K/S Section 6.3.1)

• Assume installed generation capacity exceeds demand in each

country by a significant margin.

• Assume no congestion or losses within each country.

• Assume a uniform market price for electric energy is set within

each country at the LMP, i.e., at the least cost of servicing the

next MW of power for that country while respecting all system

constraints.


MW

$/MWh

13

1500

43

DS= 1500 MW

Syldavia

DB= 500MW

Borduria

Borduria/Syldavia Demand and Supply Functions in the

Absence of Any Interconnection Between the Countries

10

15

500 MW

$/MWh

B MCB 10 0.01PB [$ / MWh] S MC S 13 0.02PS [$ / MWh]

S MC S 13 0.02 1500 43 $ /MWh

Borduria Market Price Syldavia Market Price

S

S D

D

B

MC B 10 0 . 01 500 15 $ / MWh


DB= 500MW

Borduria

DS= 1500 MW

Syldavia

Now suppose an interconnection is built …

• Without any interconnection, countries operate where

MCB = 15 $/MWh < MCS = 43 $/MWh.

• Economic effects of introducing an interconnection?

Max Cap = 1600 MW


DB= 500MW

Borduria

DS= 1500 MW

Syldavia

Can Borduria GenCos meet entire demand?

• GenCos in Borduria CAN meet entire demand (2000 MW).

• But then Syldavia MCs = 13 $/MWh < Borduria MCB = 30 $/MWh.

• Having Borduria generate all power is not economically tenable.

• At operating point the load would want to buy from Syldavia instead.

PB 2000MW

PS 0MW

MCB 30$ / MWh

MCS 13$ / MWh

Max Cap = 1600 MW


DB= 500MW

Borduria

DS= 1500 MW

Syldavia

Two-Country Market Clearing Point

B S

PB PS DB DS 500 1500 2000MW

PB 1433MW

PS 567MW

B MCB 10 0.01PB [$ / MWh] S MC S 13 0.02PS [$ / MWh]

Max Cap = 1600 MW

B

S

24 . 30 $ / MWh


More precisely, 5 equations and 5 unknowns:

• 5 Unknowns

Market price π

Bordurian price πB and generation level PB

Syldavian price πS and generation level PS

• 5 Equations in the 5 Unknowns (Can Solve by Substitution)

1. π = πB

2. π = πS

3. PB + PS = 2000 MW

4. πB = 10 + 0.01 PB [ i.e., Bordurian price = MCB(PB) ]

5. πS = 13 + 0.02 PS [ i.e., Syldavian price = MCS(PS) ]


DB= 500MW

Borduria

DS= 1500 MW

Syldavia

Interconnection Flow at the Market Clearing Point

PB 1433MW PS 567MW

Flow of power across the

interconnection B→S

resulting power

flow 933 MW →

F BS P B D B D S P S 933 MW

Max Cap = 1600 MW


Graphical Representation of Market Clearing

= 567 MW

24.30 $/MWh

= 1433 MW

= 2000 MW

= 500 MW = 1500 MW

24.30 $/MWh

= 933 MW

Supply curve for

Syldavia

Supply curve for

Borduria

D B D S

D B D S

Borduria Syldavia

F BS

S

MC S

B

MC B 30.00 $/MWh

13.00 $/MWh

P B

P S


Constrained Transmission

• What if the interconnection can only carry 400 MW?

• Borduria must cut back its production so that at most 400 MW is

transmitted to Syldavia over the interconnection:

PB = 500 MW + 400 MW = 900 MW

• Syldavia must then generate any additional power needed to

meet it’s local load of 1500 MW:

PS = 1500 MW - 400 MW = 1100 MW

• Results in persistent market price differential of [35 – 19] $/MWh

= 16 $/MWh between Borduria and Syldavia

• The two bus prices 19 $/MWh and 35 $/MWh are called

Locational marginal Prices (LMP) or nodal prices

B MCB 10 0.01 900 19 $ / MWh

S MC S 13 0.02 1100 35 $ /MWh

Market prices in

Borduria & Syldavia

set by price = MC


= 1100 MW

35 $/MWh

= 900 MW

= 2000 MW

= 500 MW = 1500 MW

= 400 MW

16 $/MWh

Graphical Representation: Constrained Transmission

S MC S

B MC B

FBS

D B D S

D B D S

Borduria Syldavia

13 $/MWh

19 $/MWh

P B

P S


Separate markets Single market Single market

with conges tion

PB [MW] 500 1,433 900

B [$/MWh] 15 24.33 19

RB [$/h] 7,500 34,865 17,100

E B [$/h] 7,500 12,165 9,500

PS [MW] 1500 567 1100

S [$/MWh] 43 24.33 35

RS [$/h] 64,500 13,795 38,500

E S [$/h] 64,500 36,495 52,500

FBS [MW] 0 933 400

RTOTAL R B R S 72,000 48,660 55,600

E TOTAL E B E S 72,000 48,660 62,000

Summary of Results for Borduria and Syldavia

500 1433 900

15.00 24.33 19

7,500 34,865 17,100 (buyer payments)

(seller revenues)

(seller revenues)

(buyer payments)

7,500 12,165 9,500

1,500 567 1,100

43 24.33 35

64,500 13,795 38,500

64,500 36,495 52,500

0 933 400

72,000 48,660 55,600

72,000 48,600 62,000

Why? Why?


Winners and Losers from Interconnection

• Winners? (Knowing prices, payments, revenues not enough! )

Economies of both countries? (Is total net surplus higher?)

Bordurian power sellers ? (Is their total net seller surplus higher?)

Syldavian power buyers ? (Is their total net buyer surplus higher?)

• Losers ?

Bordurian power buyers ? (Is their total net buyer surplus lower?)

Syldavian power sellers ? (Is their total net seller surplus lower?)

• Congestion in the interconnection reduces its benefits


ISO Net Surplus (or “Congestion Rent”) Defined: (K/S also refer to it as “merchandizing surplus”, Section 6.3.1.3)

E TOTAL B DB S D S

RTOTAL B PB S PS B (DB FBS ) S (DS FBS )

E TOTAL RTOTAL S D S B DB S PS B PB

S (D S PS ) B (DB PB )

S FBS B ( FBS )

( S B ) FBS

Buyers’ Payments (Collected by ISO):

Sellers’ Revenues (Paid by ISO):

ISO Net Surplus (“Congestion Rent”):

Tends to increase as

price separation increases


ISO Net Surplus as a Function of Power Flow FBS

0

10000

20000

30000

40000

50000

60000

70000

80000

0 100 200 300 400 500 600 700 800 900 1000

Flow on the Interconnection [MW]

Pa

ym

ents

and

Re

ve

nu

es [

$/h

]

Buyers’ payments

Sellers’ revenues


ISO Net Surplus Summary:

• ISO Net Surplus = [buyer payments – seller revenues]

collected in pool-based trading by an independent

system operator (ISO), who also acts as market operator

• ISO Net Surplus is typically positive in loss-less systems

whenever one or more lines are congested.

• Should not be kept by the ISO because it gives a

perverse incentive to the ISO to maintain or encourage

grid congestion.

• Should not be returned directly to buyers and sellers

users because that would blunt the economic incentive

provided by nodal pricing.


Pool Trading in Three-Bus Examples: K/S 6.3.2

1 2

3

C A

B

D

50 MW 60 MW

300 MW

Assumptions:

Constant branch capacity limits restricting branch power flows. Resistance of lines is negligible and can be ignored.

GenCos report their true capacities and marginal costs to the ISO.

The load’s 60MW demand is fixed (price insensitive).

= GenCo

= Load


Pool Trading in a Three-Bus Systems: K/S 6.3.2

1 2

3

C A

B

D

Branch Reactance

[p.u.]

Capacity

[MW]

1-2 0.2 126

1-3 0.2 250

2-3 0.1 13050 MW 60 MW

300 MW

Gene r a t o r C a pac it y

[ MW ]

M a rg i na l Co s t

[ $ / MWh ]

A 140 7 .50

B 285 6.00

C 90 14.00

D 85 10.00

Note: Each GenCo assumed

to have a constant MC!


Problem Solution: First consider simple economic dispatch with no binding transmission constraints

1 2

3

C A

B

D

50 MW 60 MW

300 MW

125 MW

285 MW

0 MW

0 MW

F13 F23

F12

Total Demand (MW): 410 = [50+300+60]

Cheapest GenCos B&A:

Capacity B = 285 MW Capacity A = 140 MW

Tot Cap A+B = 425 MW

= enough to meet demand

PB=285 (max), PA=125 Are branch limits exceeded?


Solving for Power Flows: What Do We Know?

• Overall power balance equation (assuming no branch losses):

(a) P1 + P2 + P3 = Demand1 + Demand2 + Demand3

• Power balance equation for each bus, from KCL (assuming no branch losses) – compare K/S (6.30)-(6.32)

(b) P1 – Demand1 = F12 + F13

(c) P2 – Demand2 = F21 + F23 (Note: F21 = - F12)

(d) P3 – Demand3 = F31 + F32 (Note: F31 = - F13 , and F32 = - F23)

NOTE: Summing equations (b), (c), and (d) gives equation (a), so any one of these 4 equations can be omitted with any loss of information.


Solving for Power Flows for Given Power Injections & Demands:

• Overall power balance equation (assuming no branch losses):


410 + 0 + 0 = 50 + 60 + 300 (MW)

• Power balance equation for each bus, from KCL (assuming no branch losses) – compare K/S (6.30)-(6.32)

(b) P1 – Demand1 = 360 = F12 + F13

(c) P2 – Demand2 = - 60 = F21 + F23 (Note: F21 = - F12)

(d) P3 – Demand3 = -300 = F31 + F32 (Note: F31 = - F13 , and F32 = - F23)

NOTE: Since equations (a), (b), and (c) sum to zero, we have only two

independent equations in 3 unknowns {F12,F13,F23} ! We need an

additional equation.


1

60 MW

2

3

300 MW

360 MW

1

60 MW

2

3

60 MW

1 2

3

300 MW

300 MW

Calculating Branch Flows Via Superposition:

F12 = F1A + F2

A = 120+36 = 156 MW

F1

A = PTDF1A • 300

= ( x13/ [ ∑ xiJ ] ) • 300

= (.2/.5) • 300 = 120MW F2

A = PTDF2A• 60 = ([x13+x32]/ [ ∑ xiJ ]) • 60MW

= ([.2+.1])/.5) • 60 = 36MW

Additional equation from linear

superposition (Fkm = PTDFkm• Pk):

F12

F1A

F2A


Solving for Power Flows for Given Bus Power Injections & Demands:

• Overall Power Balance Equation (Assuming No Losses):


410 + 0 + 0 = 50 + 60 + 300 (MW)

• Power Balance Equation for Each Bus from KCL (Assuming No Losses) – compare K/S (6.30)-(6.32)

(b) P1 – Demand1 = 360 = F12 + F13

(c) P2 – Demand2 = - 60 = F21 + F23 (Note: F21 = - F12)

(d) P3 – Demand3 = -300 = F31 + F32 (Note: F31 = - F13 , and F32 = - F23)

• Implied Solution: Compare K/S, (6.45)-(6.47), p. 160

F12 = 156 MW (superposition) Violates branch 1-2 Max Cap = 126!

F13 = [ 360 – 156 ] = 204 MW from equation (b)

F23 = [ F31 + 300 ] = [ - 204 + 300 ] = 96 MW from equation (d)


Power Flows Implied by Simple Economic Dispatch

1 2

3

C A

B

D

50 MW 60 MW

300 MW

125 MW

285 MW

0 MW

0 MW

156 MW

96 MW 204 MW


Branch 1-2 is overloaded by 30 MW !

1 2

3

C A

B

D

50 MW 60 MW

300 MW

125 MW

285 MW

0 MW

0 MW

156 MW

96 MW 204 MW

Max Cap = 126 MW


1

1 MW

2

3

1 MW

Correcting the Economic Dispatch: Find the Least-Cost Dispatch

Modification that will Remove the Branch Overload

Suppose 1 MW of power is injected at Bus 2

and withdrawn at Bus 1:

0.6 MW

0.4 MW

PTDF21 = [ X13 + x23]/[x12 + x13 + x23] = 0.3/0.5 = 0.6

PTDF231 = [ X12]/[x12 + x13 + x23] = 0.2/0.5 = 0.4

X12 = 0.2

X13 = 0.2 X23 = 0.1


1

60

MW

2

3

300 MW

360 MW 156 MW

204 MW 96 MW

1

50 MW

2

3

50 MW 30 MW

20MW

1

10

MW

2

3

300 MW

310 MW 126 MW

184 MW 116 MW

Suppose 50 MW of power is injected at Bus 2 and

withdrawn at Bus 1: Solve by Superposition

NOTE: 50 x 0.6 = 30

50 x 0.4 = 20

This gives a feasible solution !


1

1 MW

2

3

1 MW

Alternatively, suppose 1 MW of power is injected at

Bus 3 and withdrawn at Bus 1:

0.6 MW

0.4 MW

X12 = 0.2

X13 = 0.2 X23 = 0.1

PTDF321 = 0.2/0.5 = 0.4 PTDF31 = 0.3/0.5 = 0.6


1

60 MW

2

3

300 MW

360 MW 156 MW

204 MW 96 MW

1

60 MW

2

3

225 MW

285 MW 126 MW

159 MW 66 MW

1

75 MW

2

3

75 MW

30 MW

45 MW

Suppose an additional 75 MW of power is injected at

Bus 3 and withdrawn at Bus 1: Solve by Superposition

NOTE: 75 x 0.4 = 30

75 x 0.6 = 45

This also is a feasible solution !


Compare the Total Avoidable Costs Resulting from

Each of the Two Feasible Solutions: (See K/S Section 6.3.2.2) Non-Feasible Simple Economic Dispatch (NOTE: Here MC is constant):

MCA•125 + MCB•285 + MCC•0 + MCD•0 = 937.50 + 710 = 2,647.50 $/h

Cost with [-50] MW Redispatch of GenCo A at Bus 1 and [+50] MW Redispatch of GenCo C at Bus 2:

MCA•75 + MCB•285 + MCC•50 = 562.50 + 1710 + 700 = 2,972.50 $/h

Cost with [-75] MW Redispatch of GenCo A at Bus 1 and a [+75] MW Redispatch of GenCo D at Bus 3:

MCA•50 + MCB•285 + MCD•75 = 375 + 1710 + 750 = 2,835.00 $/h

Least-Cost Solution: Redispatch GenCos A & D

Cost of Having a Max Cap of 126 MW on Line 1-2:

[ 2,835.00 – 2,647.50 ] = 187.50 $/h


Transmission-Constrained Economic Dispatch Solution:

1 2

3

C A

B

D

50 MW 60 MW

50 MW

285 MW

0 MW

75 MW

126 MW

66 MW 159 MW

300 MW


Locational Marginal Prices (Nodal Prices):

• Least cost of servicing an additional MW of demand at a

particular bus at a particular time subject to system

feasibility constraints

• These system feasibility constraints typically include

power balance constraints, GenCo operating capacity

constraints, and branch capacity constraints.

• K/S refer to branch capacity constraints as security

constraints.


LMP: Transmission-Constrained Economic Dispatch

3-Bus Example

• Bus 1 LMP?

• GenCo A at Bus 1 is not at its max or min operating capacity limits (i.e., Genco A is marginal )

• GenCo A cheaper than all other GenCos not capacity constrained

• Therefore the LMP at Bus 1 is

1 2

3

C A

B

D

50 MW 60 MW

50 MW

285 MW

0 MW

75 MW

126

MW

66 MW

159 MW

300 MW

1

MC A 7 . 50 $ / MWh


LMP Example … Continued

• Bus 3 LMP ?

• Genco A cheaper than GenCo D

• But increasing A’s generation would overload line 1-2

• D cheaper than C and is at Bus 3. Therefore LMP at Bus 3 is

1 2

3

C A

B

D

50 MW

60 MW

50 MW

285 MW

0 MW

75 MW

126

MW

66 MW

159 MW

300 MW

3

MC D 10 $ / MWh


LMP Example … Continued

• Bus 2 LMP ?

• GenCo C is very expensive

• GenCo B is already at its max operating capacity

• Need to consider redispatch of marginal GenCos A and/or D

1 2

3

C A

B

D

50 MW 60 MW

50 MW

285 MW

0 MW

75 MW

126

MW

66 MW

159 MW

300 MW


1

1 MW

2

3

1 MW

0.6 MW

0.4 MW

1

1 MW

2

3

0.2 MW

0.8 MW

1 MW

Bus 2 LMP ... Continued

Option 1: 1 MW for Bus 2

generated at Bus 1

by GenCo A

Option 2: 1 MW for Bus 2

generated at Bus 3

by GenCo D

Either option increases flow on branch 1-2, violating max cap!

Branch Reactance

[p.u.]

Capacity

[MW]

1-2 0.2 126

1-3 0.2 250

2-3 0.1 130


LMP at Bus 2 … Continued

Consider simultaneous redispatch ΔP1 for GenCo A at

Bus 1 and ΔP3 for GenCo D at Bus 3 satisfying:

(i) ΔP1 + ΔP3 = ΔP2 ; (ii) ΔP2 = 1 MW ; (iii) ΔF12 = 0 ;

(iv) ΔF12 = [ 0.6 • ΔP1 + 0.2 • ΔP3 ]

1

1 MW

2

3

F 12 = 0

P 3

P 1

P 2


1

1 MW

2

3

0.2 MW

0.8 MW

1 MW

LMP Solution Using Superposition

P1 P3 P2 1 MW

0.6 P1 0.2 P3 F12 0 MW

P1 0.5 MW

P3 1.5 MW

2 1.5 MCD 0.5 MCA 11.25 $ / MWh

1

1 MW

2

3

1 MW

0.6 MW

0.4 MW

(10 $/MWh =

LMP at Bus 1)

(7.50 $/MWh =

LMP at Bus 3)

(LMP at

Bus 2)


General LMP Observations

A GenCo is marginal if it is operating at a point p that lies

strictly between its min and max operating capacity limits:

CapL < p < CapU.

The LMP at a bus where there is a marginal GenCo is determined by this GenCo’s marginal cost.

The LMP at a bus with no marginal GenCo is a linear

combination of the marginal costs of the marginal GenCos.

For the 3-bus example at hand:

GenCos A and D are marginal.

GenCos B & C are not marginal, but for different reasons.

• GenCo B is an inframarginal unit at its max operating capacity

• GenCo C is an extramarginal unit too expensive to dispatch, hence it

is at its min operating capacity limit of 0


Outcome Summary for the Three-Bus Example:

Bu s 1 Bu s 2 Bu s 3 S ys t e m

Purchase [ MW ] 50 60 300 410

P rodu ct ion [ MW ] 335 0 75 410

LMP [ $ / M W h] 7 . 50 11 . 25 10 . 00 -

[ $ / h] 375 . 00 675 . 00 3 , 000.00 4 , 050.00

P rodu c er r evenue s [ $ / h] 2 , 512.50 0 . 00 750 . 00 3 , 262.50

[ $ / h ] 787 . 50

Buyer Payments

ISO Net Surplus*

* In ISO-managed wholesale power markets, ISO Net Surplus is also known as “congestion rent” or “merchandising surplus.” The latter usage is

somewhat confusing for those familiar with standard economic terminology

since it represents an amount of money collected by the market operator and

not by “merchants” (private traders).


Power flows from

high price to low

price!

Counter-Intuitive Flows: (cf. K/S 6.3.2.5)

1 2

3

C A

B

D

50 MW 60 MW

50 MW

285 MW

0 MW

75 MW

126 MW

66 MW 159 MW

300 MW

2=11.25 $/MWh

3=10.00 $/MWh

3=7.50 $/MWh


Counter-Intuitive Prices and Strategic Behavior:

• LMPs at buses without a marginal GenCo can be higher or

lower than LMPs at other buses.

• LMPs at such buses can even be negative!

• Determining LMPs in realistically sized and structured

grids is complicated, requiring specialized software.

• Strategically placed GenCos can exert control over LMPs

through appropriate choice of reported supply offers.

• Transmission branch congestion can facilitate the exercise

of GenCo market power (i.e., it can help GenCos exert

advantageous control over LMPs).

© 2011 D. Kirschen and L. Tesfatsion

70

General Method for Computing LMPs:

Bid/Offer-Based Optimal Power Flow (OPF)

• Optimisation Problem:

Objective: Maximisation of Total Net Surplus

Constraints: Power flow equations (bus balance constraints, generation operating capacity limits, branch capacity limits, …)

The Lagrange multiplier associated with the bus power balance constraint for any bus k gives the LMP solution value for this bus.

OPF solution often approximated by a “dc optimal power flow” formulation (discussed in K/S Section 6.3.4.4)

• In North American Restructured Wholesale Power Markets:

Hourly LMPs are typically calculated each operating day D for all 24 hours of a “day-ahead market” on day D+1

LMPs are also calculated/updated every five minutes for a “real-time market” (managed spot market) that remains open throughout day D.

ee/econ 458, syllabus section v...ga gc gb gd ly lz branch reactance [ p.u.] capacit y [ mw] 1-2 0.2...

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