ECE 740 Optimal Power Flow 1
ED vs OPF Economic Dispatch (ED) ignores the effect the dispatch has on the loading on transmission lines and on bus voltages. OPF couples the ED calculation with power flow calculations they are solved simultaneously. The power loss in the system is simply part of the power flow calculation hence not need to use approximate loss formulas. The result of OPF is generation dispatch at minimum total cost while not violating line flows and voltage constraints. 2
Illustration Two Bus System (B2OPF) 201 MW 99 MW 399 MW A Limit: 120 MW B 300 MW 300 MW 16.00 15.00 14.00 Gen A Gen B Generator A F(P) = 239.4+7P+0.002P 2 Mbtu/hr 100 P 600 MW Fuel Cost: 1.67 $/Mbtu 13.00 12.00 0 175 350 525 700 Generator Power (MW) Current generator operating points Generator B F(P) = 414+7.9P+0.001P 2 Mbtu/hr 100 P 700 MW Fuel Cost: 1.67 $/Mbtu 3
Illustration Two Bus System (B2OPF) With no overloads the OPF matches the economic dispatch Transmission line is not overloaded Marginal cost of supplying power to each bus (locational marginal costs) is the same. 4
Illustration- load increase at Bus A With the line loaded to its limit (120 MW), the additional load at Bus A must be supplied locally, causing the localized marginal costs to diverge. Solution when ignoring line flow limit: PA= 223 MW, PB = 457 MW, λ = 3.19 $/MWh, line flow: 157 MW, hourly cost: 9,507 $/hr 5
Optimal Power Flow Formulation Minimize cost function (operating cost) while taking into account the following equality and inequality constraints. Equality constraints: bus real and reactive power balance (power flow equations) Generator voltage set points Area MW interchange Inequality constraints: transmission line/transformer/interface flow limits generator MW and MVAR limits bus voltage magnitudes. Control variables: Active power output of the generating units Voltage at the generating units Position of the transformer taps Position of the phase shifter taps Status of the switched capacitors and reactors 6
Mathematical Formulation of the OPF 7
OPF Challenges Size of the problem in a large power system Problem is non-linear Problem is non-convex Some of the variables are discrete Position of transformer and phase shifter taps Status of switched capacitors or reactors Solution methodologies: Gradient (Newton s) Method: slow convergence and difficulty in handling inequality constraints Linear Programming: use piecewise linear cost function, and DC power flow to handle constraints. Fast convergence, but the solution may be somewhat sub-optimal. 8
Sequential LP OPF Sequential Linear Programming iterates the solution of the linearized problem as follows: Algorithm: 1. Linearize the problem around an operating point 2. Find the solution to this linearized optimization 3. Perform a full ac power flow at that solution to find the new operating point 4. Repeat as necessary. 9
Example - 3 Bus Example (B3LP) Consider a three bus case (bus 1 is system slack), with all buses connected through 0.1 pu reactance lines, each with a 100 MVA limit. Let the generator marginal costs be: Bus 1: 10 $ / MWhr; Range = 0 to 400 MW, Bus 2: 12 $ / MWhr; Range = 0 to 400 MW, Bus 3: 20 $ / MWhr; Range = 0 to 400 MW, Assume a single 180 MW load at bus 3. 10
Solution with Line Limits NOT Enforced 60 MW 60 MW Bus 2 Bus 1 10.00 $/MWh 0.0 MW 10.00 $/MWh 0 MW 60 MW Total Cost 60 MW 1800 $/hr Bus 3 0 MW 120 MW 120% 120% 120 MW 10.00 $/MWh 180 MW 180.0 MW Line from Bus 1 to Bus 3 is overloaded; all buses have same marginal cost (but not allowed to dispatch to overload line!) 11
Solution with Line Limits Enforced 20 MW 20 MW Bus 2 Bus 1 10.00 $/MWh 60.0 MW 12.00 $/MWh 0 MW 80 MW Total Cost 80 MW 1920 $/hr Bus 3 0 MW 100 MW 100% 100% 100 MW 14.00 $/MWh 180 MW 120.0 MW LP OPF re-dispatches to remove violation. Bus marginal costs are now different. Prices will be different at each bus. 12
Verify Bus 3 Marginal Cost 19 MW 19 MW Bus 2 Bus 1 10.00 $/MWh 62.0 MW 12.00 $/MWh 0 MW 81 MW 81% 81% Total Cost 81 MW 1934 $/hr Bus 3 0 MW 100 MW 100% 100% 100 MW 14.00 $/MWh 181 MW 119.0 MW One additional MW of load at bus 3 raised total cost by 14 $/hr, as G2 went up by 2 MW and G1 went down by 1MW. 13
Why is Bus 3 LMP = $14 /MWh? All lines have equal impedance. Power flow in a simple network distributes inversely to impedance of path. For bus 1 to supply 1 MW to bus 3, 2/3 MW would take direct path from 1 to 3, while 1/3 MW would loop around from 1 to 2 to 3. Likewise, for bus 2 to supply 1 MW to bus 3, 2/3MW would go from 2 to 3, while 1/3 MW would go from 2 to 1to 3. 14
Why is Bus 3 LMP $ 14 / MWh? With the line from 1 to 3 limited, no additional power flows are allowed on it. To supply 1 more MW to bus 3 we need: Extra production of 1 MW: P g1 + P g2 = 1 MW No more flow on line 1 to 3: 2/3 P g1 + 1/3 P g2 = 0; Solving requires we increase P g2 by 2 MW and decrease P g1 by 1 MW for a net increase of $14/h for the 1 MW increase. That is, the marginal cost of delivering power to bus 3 is $14/MWh. 15
Both Lines into Bus 3 Congested 0 MW 0 MW Bus 2 Bus 1 10.00 $/MWh 100.0 MW12.00 $/MWh 0 MW 100 MW 100% 100% 100 MW 100% 100% 100 MW Total Cost 100 MW 2280 $/hr Bus 3 4 MW 20.00 $/MWh 204 MW 100.0 MW For bus 3 loads above 200 MW, the load must be supplied locally. 16
PowerWorld Videos https://www.powerworld.com/training/online-training/marketmodeling/linear-programming https://www.powerworld.com/training/online-training/marketmodeling/optimal-power-flow 17
Assignment Verify the results of Example 8C (pages 372-376) 18