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1 System planning 2013 Lecture L8: Short term planning of hydro systems Chapter 5.1-5.3 Contents: –General about short term planning –General about hydropower –Hydropower generation, discharge and efficiency –Hydrological coupling –Example: hydro planning problem

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2 Course goals To pass the course, the students should show that they are able to formulate short-term planning problems of hydro-thermal power systems. To receive a higher grade the students should also show that they are able to create specialized models for short-term planning problems.

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3 General about short term planning What is short term planning? Time frame: 24 hours – 1 week In this course: Hourly planning. Minimize cost, maximize profit Results: –Operation plans for the power plants –Trading on the market Exists limitations when planning operation: –Technical –Economical –Legal In this course: Deterministic modeling

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4 General about short term planning Optimization problem! maximizeincome during period + future income – costs during period – future costs regardingphysical limitations economical restrictions legal restrictions General short term planning problem:

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5 General about short term planning In this course: Hydro-thermal systems. Quite different characteristics and modeling.

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6 General about hydropower Generates power by using the difference in potential energy between an upper and a lower water level. Often reservoirs, but also run-of-the-river hydro plants

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7 Hydropower station with low head General about hydropower

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8 Hydropower station with high head General about hydropower

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9 Variables in hydropower models: –Discharge, Q –Spillage, S –Reservoir contents, M All measured in hour equivalents, HE –Discharge, spillage:1 HE = 1 m 3 /s during 1 h. –Reservoir contents:1 HE = Volume represented by the flow 1 m 3 /s during 1 hour Hydro generation as a function of the discharge is denoted H(Q) General about hydropower

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10 What do we want? Want to optimize the operation of our stations in the hydro system. Want to use linear equations so that we can use LP when solving the optimization problem. Want to model the characteristics of hydro systems.

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11 Power generation Production equivalent: Measured in MWh/HE Marginal production equivalent: Measured in MWh/HE Definitions:

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12 Power generation Relative efficiency: Measured in percent Relative efficiency shows how much energy can be extracted from each m 3 water compared to the maximal possible. Definitions:

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13 Power generation Power generation in hydropower station H(Q) Q Relative efficiency in hydropower station Q (Q)

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14 Power generation H(Q) Q Power generation curves are not linear! –Approximated by piece-wise linear curves (segments) –Break points at best local efficiency points –Each segment has constant marginal production equivalent (the slope of the curve)

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15 Power generation Power generation in power station i: H i (Q) QiQi i, 1 i, 2 i, 3 i, 4 Segment 1Segment 2Segment 3Segment 4 Q i,1 Q i,2 Q i,3 Q i, 4

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16 Power generation Total discharge in power station i hour t: Q i,j,t = Discharge in station i, segment j, during hour t n i = Number of segments in station i Total production in station i hour t : i,j = Marginal production equivalent for station i, segment j

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17 Q: How make sure that discharge in segment 2 doesn’t occur before maximum discharge in segment 1 is reached? A: If the discharge will first occur in segment 1, then in segment 2, etc. H(Q) Q Q1Q1 Q2Q2 Q3Q3 Q4Q4 11 22 33 44 Power generation

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18 Power generation Low discharges often have bad efficiency, but in the linear model discharges up to the first break point are assumed to have best production equivalent! Want to avoid low discharges! Can be solved by introducing binary variables Forbidden discharges:

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19 Power generation Production equivalent according to figure Low efficiency for discharges lower than 50 HE Want to have discharges larger than 50 HE or no discharge at all Create a linear model 50 100150200250300 0.1 0.2 0.3 0.4 i (u) uiui MWh/HE HE Forbidden discharges: QiQi i(Qi)i(Qi)

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20 Power generation 50 100150200250300 0.1 0.2 0.3 0.4 i (u) uiui MWh/HE HE Q Ki,t Define 2 segments: 1.Origin 2.Discharges between 50 and 300 HE, Q Ki,t Assume constant marginal production equivalent in the interval 50-300 HE, Ki. Can e.g. chosen as the average value of i (Q i ) in the interval. Forbidden discharges: QiQi i(Qi)i(Qi)

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21 Power generation The total discharge and production can now be calculated as: Define binary variable representing which segment discharge occurs during the hour Forbidden discharges:

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22 Power generation Variable limits: Must define constraints to assure that the discharge in the second segment is 0 when z i,t = 0 Forbidden discharges:

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23 Planning problem Short term hydropower planning problem: maximizeincome during period - production costs during period + assets after the end of the period regarding hydrological coupling legal agreements physical limitations

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24 Income & cost The production costs can be neglected for hydropower production! They are very small. Income during the period can consist of: Sale of power on spot market Sale of power bilateral to customer t = price hour t For power station i:

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25 Assets after the end of period Assets after the end of the planning period consists of the value of stored water in the reservoirs. Depends on: –Future power price –How much power that you can expect to produce with the stored water Value of stored water at reservoir i: e = expected future power price M i,T = reservoir contents in reservoir i at the end of the planning period M i = set of indices for all station downstream station i (including station i ) j = expected prod. equivalent for station j

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26 Hydrological coupling The hydropower stations in a river are not independent, they are a part of a system The operation of one station affects the other stations in the river To operate the river system in an efficient way, the whole system needs to be considered

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27 Hydrological coupling Balance equation for a reservoir: new reservoir contents = old reservoir contents + water flowing into reservoir – water flowing out from reservoir Discharge from stations directly upstream Spillage from stations directly upstream Inflow from surrounding area DischargeSpillage

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28 Hydrological coupling Hydrological coupling equation for station i: Discharge & spillage in station i during hour t Spillage from stations directly upstream coming down to station i during hour t Contents in reservoir i at the end of hour t Contents in reservoir i at the end of hour t-1 Discharge from stations directly upstream coming down to station i during hour t Inflow to reservoir i during hour t

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29 Hydrological coupling Take some time for the water to flow from one station to the next, so called delay time: ji = Delay time between station j and the closest station downstream i ji is a complicated function of water flow, reservoir levels, etc. Assume constant delay time and let the delay time between station j and i be h j hours and m j minutes

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30 Hydrological coupling t t-h j hjhj mjmj t-h j -1 Discharge & spillage this hour will arrive hour t Weighted mean value of discharge and spillage h j +1 and h j hours earlier t+1

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31 Legal agreements & physical limitations Physical limitations & legal agreements limits the operation of the power stations Results in variable limits in the optimization problem:

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32 Legal agreements & physical limitations Contracts with costumers H i,t = production in station i hour t D t = contracted load hour t Can also be ”=” depending on the problem

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33 Planning problem - example Two hydropower stations (1 & 2) located after each other in a river. All produced power is sold on the power exchange Plan the operation for the 6 next-coming hours Known: –Price forecast for the 6 hours: λ(t), t=1,...,6 –Stored water can be used for generation of power that can be sold for the price λ f –The reservoirs are half full at the beginning of the planning period 1 2 Example:

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34 Planning problem - example Hydropower data: –Installed capacity: –Maximum discharge: –Maximum reservoir contents: –Local hydro inflow: Assume constant efficiency, i.e. constant production equivalent. Installed production capacity is reached at maximum discharge Example

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35 Planning problem - example Solution: The problem: maximizeincome from selling power on power exchange + value of stored water s.t.hydrological coupling is fulfilled Variables: Discharge in station i during hour t: Q i (t), i = 1,2, t = 1,...,6 Spillage in station i during hour t: s i (t), i = 1,2, t = 1,...,6 Reservoir contents, station i, at the end of hour t: M i (t), i = 1,2, t = 1,...,6 Example

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36 Planning problem - example Constant production equivalent Max. discharge gives installed capacity Value of sold power = Value of stored water = Objective function: Thus: Example

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37 Planning problem - example Hydrological constraints for the stations: Initial reservoir contents: Variable limits: Example

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38 General about hydropower Operation planning of a system of hydropower plants in a river Planning per hour Planning period 12-24 hours In this course we use linear or MILP models linear or MILP programming problem

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39 PROBLEM 12 - Symbols for Short-term Hydro Power Planning Problems State whether the following symbols denote optimization variables or parameters. a) γ i = expected future production equivalent for water stored in reservoir i b) M i, 0 = contents of reservoir i at the beginning of the planning period c) M i, t = contents of reservoir i at the end of hour t

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40 PROBLEM 12 - Symbols for Short-term Hydro Power Planning Problems State whether the following symbols denote optimization variables or parameters. d) Q i, t = discharge in power plant i during hour t e) S i, t = spillage from reservoir i during hour t f) V i, t = local inflow to reservoir i during hour t

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41 PROBLEM 13 - Value of Stored Water Which expression can be used for the value of stored water in a 24 hour planning of three hydro power plants in line along a river?*

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42 PROBLEM 14 - Hydrological Constraints Which expression can be used for the hydrological constraint of the third hydro power plant along a river?*

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