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 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. 1
Topic A 4: DC Networks PTDF Recall: Parallel flows - France to Italy 2
Topic A 4: DC Networks PTDF Recall: Ex1P-L23: Optimal network for Peak Demand • 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 0 200 10 G2 0 1000 40 G3 0 600 80 • This is a tree network Want to generate • It would seem to be more 200 MW from G1 robust to leave L1 connected. 1000 MW from G2 Is that OK? 300 MW from G3 $66k/h Total Cost 3
Topic A 4: DC Networks PTDF Ex1P-L123: Networks with Circuits • 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 Want to generate lines? 200 MW from G1 1000 MW from G2 300 MW from G3 $66k/h Total Cost 4
Topic A 4: DC Networks PTDF Solving for power flows in a “DC” model • 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 + ∑ ∑ p G a bl p L l = P D ∀ b ∈ B , b l ∈ L g ∈ G | β g = b Kirchhoff Voltage Law (KVL): l + V 2 p L X l ∑ a bl δ b = 0 ∀ l ∈ L b ∈ B • X l is the reactance of the line – The value of X l 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 , X L1 = 1 2 , X L2 = 1 2 and X L3 = 1 (however the diagrams show the results in MW). 5
Topic A 4: DC Networks PTDF Ex1P-L123: Solving for Flows in Network with Circuit − p L − p L − p L − p L = ⇐ ⇒ = − 0 . 15 KCL B1: 0 . 2 0 . 05 L1 L3 L1 L3 + p L − p L p L − p L = ⇐ ⇒ = − 0 . 55 KCL B2: 1 . 0 0 . 45 L1 L2 L1 L2 + p L + p L p L + p L = ⇐ ⇒ = KCL B3: 0 . 3 1 . 00 0 . 70 L2 L3 L2 L3 One constraint is redundant (they all sum to zero). Arbitrarily choose B3 as the reference bus and eliminate its constraint, KCL B3. Seting δ B3 = 0 in KVL gives combined system: p L − 2 δ B1 + 2 δ B2 = KVL L1: 0 L1 p L − 2 δ B2 = KVL L2: 0 L2 p L − δ B1 = KVL L3: 0 L3 − p L − p L = − 0 . 15 KCL B1: L1 L3 p L − p L = − 0 . 55 KCL B2: L1 L2 Eliminate the δ by adding KVL L1 and L2 and subtracting 2 times KVL L3 gives p L + p L − 2 p L = KVL : 0 L1 L2 L3 − p L − p L = − 0 . 15 KCL B1: L1 L3 p L − p L = − 0 . 55 KCL B2: L1 L2 Solving gives p L L1 = − 0 . 0625 , p L L2 = 0 . 4875 , p L L3 = 0 . 2125 KVL L3 and L2 give δ B1 = 0 . 2125 and δ B2 = 0 . 24375 , (and δ B3 = 0 ) 6
Topic A 4: DC Networks PTDF Ex1P-L123: Network with Circuit • 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 p G g , p L l and δ b (Topic A 3 mosel formulation) Line flows are • or eliminate the p L l and δ b − 62 . 5 MW in L1 variables and use only the net 487 . 5 MW in L2 demand variables p B 212 . 5 MW in L3 b (as in current Topic A 4). $66k/h Total Cost 7
Topic A 4: DC Networks PTDF Solving for Flows: Matrix Version When V = 1 the KVL equations become: l + ∑ X l p L a bl δ b = 0 ∀ l ∈ L b ∈ B Let X = diag ( X l ) 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 p L the vector of line flows. Then X p L + ˆ A T ˆ δ = 0 KCL equations are a bl p L l = P D p G g = p B ∑ ∑ b − ∀ b ∈ B b l ∈ L g ∈ G | β g = b where p B b is the net flow measured out of network at bus b . Dropping the reference bus constraint (which is redundant) we get Ap L = ˆ ˆ p B Combining KVL and KCL equations gives �� p L A T ˆ � � � � X 0 = ˆ ˆ p B δ ˆ A 0 8
Topic A 4: DC Networks PTDF Power Transmission Distribution Factors (PTDF) Provided all line reactances are non zero X − 1 = diag ( 1 X l ) is non singular, it follows that: �� p L �� p L X − 1 ˆ A T A T ˆ � � � � � � � � 0 I 0 X = ⇐ ⇒ = AX − 1 ˆ ˆ ˆ ˆ p B − ˆ A T p B δ ˆ δ ˆ A 0 0 AX − 1 ˆ A T is non singular so the equations have the unique solution: Also since ˆ A has full rank ˆ = F A ˆ − ( AX − 1 A T ) − 1 ˆ ˆ p B p B δ = (1) p L = = F P ˆ X − 1 ˆ A T ( AX − 1 A T ) − 1 ˆ p B , p B (2) where F A = − ( AX − 1 A T ) − 1 is the ( | B |− 1 ) × ( | B |− 1 ) matrix of angle load sensitivities, and F P = X − 1 ˆ A T ( AX − 1 A T ) − 1 is the | L | × ( | B | − 1 ) matrix of PTDFs, both relative to the reference bus b ref . If the demand in bus b 1 increases by 1 unit and this is supplied by increasing supply at b ref and the change of angle at bus b is F A by 1 unit, then the change in flow in line l is F PTD bb 1 . lb 1 If the net load at bus b 1 increases by 1 unit and the net load at bus b 2 decreases by 1 unit, then there will be no change of net demand at bus b ref . l , b 2 and the change of angle at bus b is F A bb 1 − F A The change of flow in line l is F PTD l , b 1 − F PTD bb 2 . 9
Topic A 4: DC Networks PTDF Ex1P-L123: KVL PTDF example Find the PTDF for the previous 3 bus loop. 1 0 0 2 0 0 � � 2 − 1 − 1 0 , X − 1 = ˆ 1 A = , X = 0 2 0 0 0 − 1 2 1 0 0 0 1 0 0 1 − 1 − 2 2 0 0 1 2 � � − 2 3 A T = A T = X − 1 ˆ AX − 1 ˆ = ˆ − 1 − 2 0 2 0 0 0 , − 2 4 − 1 − 1 0 0 1 0 0 � � A T ) − 1 = 1 4 2 AX − 1 ˆ − F A ( ˆ = (3) 2 3 8 − 2 − 1 2 2 � � A T ) − 1 = 1 = − 1 4 2 X − 1 ˆ AX − 1 ˆ A T ( ˆ F P = − 2 0 2 3 (4) 2 3 8 4 − 1 0 2 1 10
Topic A 4: DC Networks PTDF Ex1P-L123: Check Loop Flow Example Using (1), (2), (3) and (4) gives − 1 − 0 . 0625 2 � � − 0 . 15 p L = − 1 = 2 3 0 . 4875 4 − 0 . 55 2 1 0 . 2125 � �� � � � − 0 . 15 4 2 0 . 2125 − 1 δ = = 8 − 0 . 55 2 3 0 . 24375 which agree with earlier results. 11
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) p L 2 p B B1 − p B − 2 1 � p B � L1 B2 = 1 = − 1 p L 2 p B B1 + 3 p B B1 − 2 − 3 (5) p B L2 B2 4 4 p L 2 p B B1 + p B B2 − 2 − 1 L3 B2 We can now express net demands in terms of generation and line limits in terms of net demands: − 1 4 ( 2 p B B1 − p B − 0 . 10 ≤ BB ) ≤ 0 . 10 (6) p B 0 . 05 − p G = B1 G1 − 1 4 ( 2 p B B1 + 3 p B − 0 . 55 ≤ B2 ) ≤ 0 . 55 (7) p B 0 . 45 − p G = B2 G2 − 1 4 ( 2 p B B1 + p B − 0 . 15 ≤ B2 ) ≤ 0 . 15 (8) 12
Topic A 4: DC Networks PTDF Ex1P-L123: OPF in Net Demand Variables Because there are no line losses, total power balances, so total generation equals total load: p B B1 + p B B2 + p B B3 = 0 (9) p B B2 Using this to eliminate p G G3 from the objective p B −0.15 −0.10 −0.05 0.00 0.05 B1 and G3 generator limit constraints gives the −0.25 OPF problem in terms of p B B1 and p B B2 : −0.30 70 p B B1 + 40 p B B2 + 98 . 5 min P B+ − 0 . 1 ≤ − 1 4 ( 2 p B B1 − p B B2 ) ≤ 0 . 1 B3 L1: −0.35 − 0 . 55 ≤ − 1 4 ( 2 p B B1 + 3 p B P B+ B2 ) ≤ 0 . 55 L2: P B− B1 B1 −0.40 − 0 . 15 ≤ − 1 4 ( 2 p B B1 + p B B2 ) ≤ 0 . 15 L3: P L+ L3 − 0 . 15 ≤ p B B1 ≤ 0 . 05 B1: −0.45 − 0 . 55 ≤ p B B2 ≤ 0 . 45 B2: P L− L1 − 1 . 0 ≤ p B B1 + p B B2 ≤ − 0 . 4 B3: −0.50 Optimal: p B B1 = − 0 . 05 , p B B2 = − 0 . 5 , Objective = 75 −0.55 (Units are GW and $k/GWh.) P B+ L+ P B2 L2 13
Recommend
More recommend
