Operational Research in the Energy Industry Part A Topic 4: Power - - PowerPoint PPT Presentation

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Feb 27, 2023 148 likes •360 views

Topic A 4: DC Networks PTDF Networks Operational Research in the Energy Industry Part A Topic 4: Power Transmission Distribution Factors (PTDF) in DC Networks In an electricity network the operator can control the net power removed from the

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SLIDE 1

Networks Topic A 4: DC Networks PTDF

Operational Research in the Energy Industry Part A Topic 4: Power Transmission Distribution Factors (PTDF) in DC Networks

  • In an electricity network the operator can control the net power removed from the net-

work at each bus, i.e. the net loads.

  • At a bus

net load = load - generation, which can be positive or negative.

  • It is useful to be able to express all other variables in terms of the net loads.
  • In the DC approximation to power flow the bus angles and the flows in the lines are

linear functions of the net loads. The coefficients in the linear expressions for the line flows are called the Power Transmission Distribution Factors (PTDF).

  • In AC networks the variables are non-linear functions of the net bus loads, but for small

changes on net loads about a current operating point they can be approximated by affine functions of net loads. This is useful for sensitivity analysis.

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SLIDE 2

Recall: Parallel flows - France to Italy Topic A 4: DC Networks PTDF

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SLIDE 3

Recall: Ex1P-L23: Optimal network for Peak Demand Topic A 4: DC Networks PTDF

  • Design a minimum capacity

network that allows optimal generation at peak demand. – Choose minimum line capacities to take flows

  • Generators’ capacities (in MW)

and prices (in $/MWh) are: Genr Min Max Price G1 200 10 G2 1000 40 G3 600 80

  • This is a tree network
  • It would seem to be more

robust to leave L1 connected. Is that OK? Want to generate 200 MW from G1 1000 MW from G2 300 MW from G3 $66k/h Total Cost

3

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SLIDE 4

Ex1P-L123: Networks with Circuits Topic A 4: DC Networks PTDF

  • The flow of electricity in a line

is linked to the voltages at the buses at its ends.

  • This does not constrain the flow

in a tree network

  • But it is an extra constraint when

there are circuits

  • How do we find the flows in the

lines? Want to generate 200 MW from G1 1000 MW from G2 300 MW from G3 $66k/h Total Cost

4

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SLIDE 5

Solving for power flows in a “DC” model Topic A 4: DC Networks PTDF

  • To simplify notation we will drop the period index t
  • In an electrical network the line flows, bus angles and generation levels must satisfy

both the KCL and KVL laws: Kirchhoff Current Law (KCL):

∑

g∈G|βg=b

pG

g + ∑ l∈L

abl pL

l = PD b

∀ b ∈ B,

Kirchhoff Voltage Law (KVL):

pL

l +V 2

Xl ∑

b∈B

ablδb = 0 ∀ l ∈ L

  • Xl is the reactance of the line

– The value of Xl depend on the units used for power and for voltage. In DC-models we will assume all voltage magnitudes are normalised to V = 1 – In the calculation below power is in GW and V = 1, XL1 = 1

2, XL2 = 1 2 and XL3 = 1

(however the diagrams show the results in MW).

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SLIDE 6

Ex1P-L123: Solving for Flows in Network with Circuit Topic A 4: DC Networks PTDF

KCL B1: 0.2 −pL

L1

−pL

L3

= 0.05 ⇐ ⇒ −pL

L1

−pL

L3

= −0.15 KCL B2: 1.0 +pL

L1

−pL

L2

= 0.45 ⇐ ⇒ pL

L1

−pL

L2

= −0.55 KCL B3: 0.3 +pL

L2

+pL

L3

= 1.00 ⇐ ⇒ pL

L2

+pL

L3

= 0.70

One constraint is redundant (they all sum to zero). Arbitrarily choose B3 as the reference bus and eliminate its constraint, KCL B3. Seting δB3 = 0 in KVL gives combined system:

KVL L1: pL

L1

−2δB1 +2δB2 = KVL L2: pL

L2

−2δB2 = KVL L3: pL

L3

− δB1 = KCL B1: −pL

L1

−pL

L3

= −0.15 KCL B2: pL

L1

−pL

L2

= −0.55

Eliminate the δ by adding KVL L1 and L2 and subtracting 2 times KVL L3 gives

KVL : pL

L1

+pL

L2

−2pL

L3

= KCL B1: −pL

L1

−pL

L3

= −0.15 KCL B2: pL

L1

−pL

L2

= −0.55

Solving gives pL

L1 = −0.0625, pL L2 = 0.4875, pL L3 = 0.2125

KVL L3 and L2 give δB1 = 0.2125 and δB2 = 0.24375, (and δB3 = 0)

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SLIDE 7

Ex1P-L123: Network with Circuit Topic A 4: DC Networks PTDF

  • Flow in L3 violates the line limit
  • The optimal generation is in-

feasible now that edge L1 has been added. How can we find the optimal flow? Solve the OPF problem

  • Use all KCL and KVL, line and

generator constraints

  • use either all variables

pG

g, pL l and δb

(Topic A 3 mosel formulation)

  • or eliminate the pL

l and δb

variables and use only the net demand variables pB

b

(as in current Topic A 4). Line flows are

−62.5 MW in L1 487.5 MW in L2 212.5 MW in L3

$66k/h Total Cost

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SLIDE 8

Solving for Flows: Matrix Version Topic A 4: DC Networks PTDF When V = 1 the KVL equations become:

XlpL

l + ∑ b∈B

ablδb = 0 ∀l ∈ L

Let X = diag(Xl) and ˆ

A be the bus-line adjacency matrix with reference bus row deleted.

Let ˆ

δ be the vector of bus angles with the reference bus omitted.

Let pL the vector of line flows. Then

X pL + ˆ AT ˆ δ = 0

KCL equations are

∑

l∈L

abl pL

l = PD b −

∑

g∈G|βg=b

pG

g = pB b

∀ b ∈ B

where pB

b is the net flow measured out of network at bus b.

Dropping the reference bus constraint (which is redundant) we get

ˆ ApL = ˆ pB

Combining KVL and KCL equations gives

  • X

ˆ AT ˆ A pL ˆ δ

  • =
  • ˆ

pB

  • 8
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SLIDE 9

Power Transmission Distribution Factors (PTDF) Topic A 4: DC Networks PTDF Provided all line reactances are non zero X−1 = diag( 1

Xl) is non singular, it follows that:

  • X

ˆ AT ˆ A pL ˆ δ

  • =
  • ˆ

pB

  • ⇐

⇒

  • I

X−1 ˆ AT − ˆ AX−1 ˆ AT pL ˆ δ

  • =
  • ˆ

pB

  • Also since ˆ

A has full rank ˆ AX−1 ˆ AT is non singular so the equations have the unique solution: ˆ δ = −(AX−1AT)−1 ˆ pB = FA ˆ pB

(1)

pL = X−1 ˆ AT(AX−1AT)−1 ˆ pB = FP ˆ pB,

(2) where FA = −(AX−1AT)−1 is the (|B|−1)×(|B|−1) matrix of angle load sensitivities, and FP = X−1 ˆ

AT(AX−1AT)−1 is the |L| × (|B| − 1) matrix of PTDFs, both relative to the

reference bus bref. If the demand in bus b1 increases by 1 unit and this is supplied by increasing supply at bref by 1 unit, then the change in flow in line l is FPTD

lb1

and the change of angle at bus b is FA

bb1.

If the net load at bus b1 increases by 1 unit and the net load at bus b2 decreases by 1 unit, then there will be no change of net demand at bus bref. The change of flow in line l is FPTD

l,b1 −FPTD l,b2 and the change of angle at bus b is FA bb1 −FA bb2.

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SLIDE 10

Ex1P-L123: KVL PTDF example Topic A 4: DC Networks PTDF Find the PTDF for the previous 3 bus loop.

ˆ A =

  • −1

−1 1 −1

  • , X =

 

1 2 1 2

1  , X−1 =   2 2 1   X−1 ˆ AT =   2 2 1     −1 1 −1 −1   =   −2 2 −2 −1  , ˆ AX−1 ˆ AT =

  • 3

−2 −2 4

  • −FA

= ( ˆ AX−1 ˆ AT)−1 = 1 8

  • 4

2 2 3

  • (3)

FP = X−1 ˆ AT( ˆ AX−1 ˆ AT)−1 = 1 8   −2 2 −2 −1  

  • 4

2 2 3

  • = −1

4   2 −1 2 3 2 1  

(4)

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SLIDE 11

Ex1P-L123: Check Loop Flow Example Topic A 4: DC Networks PTDF Using (1), (2), (3) and (4) gives

pL = −1

4

  2 −1 2 3 2 1  

  • −0.15

−0.55

  • =

  −0.0625 0.4875 0.2125   δ = −1

8

  • 4

2 2 3

  • −0.15

−0.55

  • =
  • 0.2125

0.24375

  • which agree with earlier results.

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SLIDE 12

Ex1P-L123: Eliminating line flow variables using PTDFs Topic A 4: DC Networks PTDF Equation (2) expresses the flow in a line in terms of the net flow out of each bus in the network excluding the reference bus. We can use this to eliminate the line flow variables from the problem and express all constraints in terms of the generator power variables. Using(2)

  pL

L1

pL

L2

pL

L3

  = 1 4   −2 1 −2 −3 −2 −1  

  • pB

B1

pB

B2

  • = −1

4   2pB

B1 − pB B2

2pB

B1 +3pB B2

2pB

B1 + pB B2

 

(5) We can now express net demands in terms of generation and line limits in terms of net demands:

pB

B1

= 0.05− pG

G1

pB

B2

= 0.45− pG

G2

−0.10 ≤ −1

4(2pB

B1 − pB BB)

≤ 0.10

(6)

−0.55 ≤ −1

4(2pB

B1 +3pB B2)

≤ 0.55

(7)

−0.15 ≤ −1

4(2pB

B1 + pB B2)

≤ 0.15

(8)

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SLIDE 13

Ex1P-L123: OPF in Net Demand Variables Topic A 4: DC Networks PTDF Because there are no line losses, total power balances, so total generation equals total load:

pB

B1 + pB B2 + pB B3 = 0

(9) Using this to eliminate pG

G3 from the objective

and G3 generator limit constraints gives the OPF problem in terms of pB

B1 and pB B2:

min 70pB

B1 +40pB B2 +98.5

L1: −0.1 ≤ −1

4(2pB

B1 − pB B2) ≤ 0.1

L2: −0.55 ≤ −1

4(2pB

B1 +3pB B2) ≤ 0.55

L3: −0.15 ≤ −1

4(2pB

B1 + pB B2) ≤ 0.15

B1: −0.15 ≤ pB

B1 ≤ 0.05

B2: −0.55 ≤ pB

B2 ≤ 0.45

B3: −1.0 ≤ pB

B1 + pB B2 ≤ −0.4

Optimal: pB

B1 = −0.05, pB B2 = −0.5, Objective = 75

(Units are GW and $k/GWh.)

B+

P

B1 B+

P

B2

p

B2 B

−0.25 −0.30 −0.35 −0.40 −0.45 −0.50 −0.55

p

B1 B

−0.15 −0.10 −0.05 0.00 0.05

L+

P

L2 L+

P

L3 B+

P

B3 L−

P

L1 B−

P

B1

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SLIDE 14

Ex1P-L123: Optimal Flow and Generation in Network with Circuit Topic A 4: DC Networks PTDF

  • OPF solution and (9)

pB

B1 = −0.05, pB B2 = −0.5

pB

B3 = −pB B1 − pB B2 = 0.55

  • Net Demand

= Demand - Generation

pG

G1 = 0.05− pB B1 = 0.1

pG

G2 = 0.45− pB B2 = 0.95

pG

G3 = 1.00− pB B3 = 0.45

  • From the PTDF formulae (5)

pL

L1 = −1

4(2pB

B1 − pB B2)

= −0.1 pL

L2 = −1

4(2pB

B1 +3pB B2)

= 0.4 pL

L3 = −1

4(2pB

B1 + pB B2)

= 0.15

Optimal Feasible generation 100 MW from G1 950 MW from G2 450 MW from G3 $75k/h Total Cost

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SLIDE 15

Reconfiguring Networks: “Line Switching” Topic A 4: DC Networks PTDF We have seen that in some circumstances adding an extra line to a network can increase the operational cost. Conversely removing or switching out a line can improve the cost. e.g. in Ex1P-L123 removing line L1 returns to the optimal Ex1P-L23 network. There is an analogous situation in traffic flow, called Braess’s Paradox, where the closure

  • f a road sometime improves average journey times.

The optimal network configuration depends on the distribution of loads. Load patterns are different at different times, so a network configuration that is optimal for

  • ne load pattern many be sub-optimal for another. This leads to the

Line Switching Problem: Find the lines to disconnect in the network to minimize the generating cost.

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SLIDE 16

Line Switching Problem Topic A 4: DC Networks PTDF The following issues need to be considered when line switching:

  • Frequent switching will reduce the life of the switches and the network may be at risk

during switching, so keeping the number of switchings low is a goal.

  • The network configuration should be good for the range of loads expected until the next
  • reconfiguration. e.g. if the network is separately configured for peak and base loads

then each configuration should work for the range of loads expected within the peak or within the base.

  • Switching out a line may make the network more vulnerable to component failures,

(by e.g. lighting striking a line). e.g. if the network is configured to be a tree, then loss of any remaining line will discon- nect part of the network. So if there is no generator in that part, all the load in it will be lost.

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SLIDE 17

Line Switching Solution Approaches Topic A 4: DC Networks PTDF

  • Brute force: Try all combination of one or two lines switched out,

resolve the OPF problem and choose the best.

  • Use a marginal analysis to find how a small change in line impedance would effect cost.

Extrapolate from that to get an estimate of the effect of making the impedance infinite (which is equivalent to removing the line). Then check just the lines which are estimated to produce a good improvement.

  • Formulate and solve the problem of selecting the lines to switch as a MILP

.

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SLIDE 18

Phase Angle Regulators Topic A 4: DC Networks PTDF Line switching is an all or nothing approach to overcoming constraints introduced by KVL. A better solution is to use a Phase Angle Regulator (PAR) (alternatively called a Phase- Shifting Transformer or (in the UK) a Quad Booster). A PAR in a line allows

  • an arbitrary phase difference to be introduced between the end of a line and its attached

bus

  • as a result the KVL is no longer a constraint for that line, and
  • flow in the line can be set independently of the phase of the buses at the ends of the

line Unfortunately the capital cost of a PAR is high — a 1 GMvar (≈ GW) PAR costs about $10m. Question: How many PARs would be needed in a network to eliminate KVL constraints? Answer: None are needed in a Tree network. So the number needed is the number of lines that have to be removed from the network to leave it as a tree, i.e. |L|−(|B|−1).

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SLIDE 19

Phase Angle Regulators Topic A 4: DC Networks PTDF Line switching is an all or nothing approach to overcoming constraints introduced by KVL. A better solution is to use a Phase Angle Regulator (PAR) (alternatively called a Phase- Shifting Transformer or (in the UK) a Quad Booster). A PAR in a line allows

  • an arbitrary phase difference to be introduced between the end of a line and its attached

bus

  • as a result the KVL is no longer a constraint for that line, and
  • flow in the line to be set independently of the phase of the buses at the ends of the line

Unfortunately the capital cost of a PAR is high — a 1 GMvar (≈ GW) PAR costs about $10m. Question: How many PARs would be needed in a network to eliminate KVL constraints? Answer: None are needed in a Tree network. So the number needed is the number of lines that have to be removed from the network to leave it as a tree, i.e. |L|−(|B|−1).

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SLIDE 20

Tasks: Topic A 4: DC Networks PTDF Suggested Tasks:

  • 1. Run your Xpress model (for the full set of variables) on the example in this lecture and

verify you get the same results.

  • 2. Formulate the problem of finding the best line to switch out as a MILP using the full set
  • f variables.
  • 3. Repeat the net demand calculations done on slides 10 to 14 for a different example –

e.g. the values given in Assignment 2.

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