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![Page 1: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/1.jpg)
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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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 tocreate specialized models for short-term planning problems.
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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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General about short term planning
Optimization problem!
maximize income during period + future income – costs during period – future costs
regarding physical limitationseconomical restrictionslegal restrictions
General short term planning problem:
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General about short term planning• In this course: Hydro-thermal systems.• Quite different characteristics and modeling.
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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
![Page 7: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/7.jpg)
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• Hydropower station with low head
General about hydropower
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• Hydropower station with high head
General about hydropower
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• Variables in hydropower models:– Discharge, Q– Spillage, S– Reservoir contents, M
• All measured in hour equivalents, HE– Discharge, spillage: 1 HE = 1 m3/s during 1 h.– Reservoir contents: 1 HE = Volume
represented by the flow 1 m3/s during 1 hour
• Hydro generation as a function of the discharge is denoted H(Q)
General about hydropower
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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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Power generation
• Production equivalent:Measured in MWh/HE
• Marginal production equivalent:Measured in MWh/HE
Definitions:
( )( )
H QQ
Q
( )dH Q
dQ
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Power generation
• Relative efficiency:Measured in percent
• Relative efficiency shows how much energy can be extracted from each m3 water compared to the maximal possible.
max
( )( )
max max ( )Q
Q
Definitions:
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Power generation
Power generation in hydropower station
H(Q)
Q
Relative efficiency in hydropower station
Q
(Q)
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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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Power generation• Power generation in power station i:
Hi(Q)
Qi
i,1
i,2
i,3
i,4
Segment 1 Segment 2 Segment 3 Segment 4
Qi,1
Qi,2
Qi,3
Qi,4
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Power generation• Total discharge in power station i hour t:
, , ,1
in
i t i j tj
Q Q
Qi,j,t = Discharge in station i, segment j, during hour tni = Number of segments in station i
• Total production in station i hour t:
, , , ,1
in
i t i j i j tj
H Q
i,j = Marginal production equivalent for station i, segment j
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• 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.
,1 ,2 ,...ii i i n
H(Q)
Q
Q1Q2
Q3Q4
1
2
3
4
Power generation
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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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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
50 100 150 200 250 300
0.1
0.2
0.3
0.4
i(u)
ui
MWh/HE
HE
Forbidden discharges:
Qi
i(Qi)
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Power generation
50 100 150 200 250 300
0.1
0.2
0.3
0.4
i(u)
ui
MWh/HE
HE
QKi,t
• Define 2 segments:1. Origin2. Discharges between
50 and 300 HE, QKi,t
• Assume constant marginal production equivalent in the interval 50-300 HE, Ki. Can e.g. chosen as the average value of i(Qi) in the interval.
Forbidden discharges:
Qi
i(Qi)
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Power generation
• The total discharge and production can now be calculated as:
, , ,
, , , ,
50
50i t i t Ki t
i t Ki t i t Ki Ki t
Q z Q
H z Q
Define binary variable representing which segment discharge occurs during the hour
Forbidden discharges:
,
0, if 0 HE
1, if 50 HEi
i ti
Qz
Q
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Power generation
• Variable limits:
,
,
0
0,1
Ki t
i t
Q
z
, ,Ki t i tKiQ Q z
• Must define constraints to assure that the discharge in the second segment is 0 when zi,t = 0
Forbidden discharges:
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Planning problem
• Short term hydropower planning problem:
maximize income during period -production costs during
period +assets after the end of the
period
regarding hydrological couplinglegal agreementsphysical limitations
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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 marketSale of power bilateral to customer
,1
IncomeT
t i tt
H
t = price hour t
For power station i:
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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:
,( )i
i i e i T jj M
B M M
e = expected future power priceMi,T = reservoir contents in reservoir i at the end of the planning periodMi = set of indices for all station downstream station i (including station i)j = expected prod. equivalent for station j
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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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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
Discharge Spillage
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Hydrological coupling• Hydrological coupling equation for station i:
, ,, , 1 , , , , ,j i j i
i i
i t i t i t i t j t j t i tj Κ j Κ
M M Q S Q S V
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
![Page 29: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/29.jpg)
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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 hj hours and mj minutes
![Page 30: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/30.jpg)
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Hydrological coupling
tt-hj
hjmj
t-hj-1
Discharge & spillage this hour will arrive hour t
,, , 1 ,
60
60 60j i j j
j jj t j t h j k h
m mQ Q Q
Weighted mean value of discharge and spillage hj+1 and hj hours earlier
t+1
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Legal agreements & physical limitations
• Physical limitations & legal agreements limits the operation of the power stations
• Results in variable limits in the optimization problem:
,
,
,, ,
i t ii
ii i t
i Ti T i T
Q Q Q
M M M
M M M
![Page 32: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/32.jpg)
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Legal agreements & physical limitations
• Contracts with costumers
tIi
ti DH
,
Hi,t = production in station i hour tDt = contracted load hour t
Can also be ”=” depending on the problem
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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 λf
– The reservoirs are half full at the beginning ofthe planning period
1
2
Example:
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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
2,1, iH i
2,1, iQ i
2,1, ivi
2,1, iM i
Example
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Planning problem - example
Solution:
The problem:
maximize income from selling power on power exchange + value of stored water
s.t. hydrological coupling is fulfilled
Variables:
Discharge in station i during hour t: Qi(t), i = 1,2, t = 1,...,6 Spillage in station i during hour t: si(t), i = 1,2, t = 1,...,6 Reservoir contents, station i, at the end of hour t: Mi(t), i = 1,2, t = 1,...,6
Example
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Planning problem - example
• Constant production equivalent • Max. discharge gives installed capacity 2,1, i
Q
H
i
i
i
Value of sold power =
6
1
2
1
)(t i
iiQtlambda
Value of stored water = )6()6( 22121 MMlambdaf
Objective function:
Thus:
)6()6()( 22121
6
1
2
1
MMlambdaQtlambdaz tt i
ii
Example
![Page 37: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/37.jpg)
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Planning problem - example
Hydrological constraints for the stations:
6,...,1,)()()()()()1(
6,...,1,)()()()1(
2112222
11111
tvtstQtstQtMtM
tvtstQtMtM
Initial reservoir contents:
ii MM 5.0)1(
Variable limits:
6,...,1,2,1,)(0
6,...,1,2,1),(0
6,...,1,2,1,)(0
tiMtM
tits
tiQtQ
ii
i
ii
Example
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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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PROBLEM 12 - Symbols for Short-term Hydro Power Planning Problems
State whether the following symbols denoteoptimization variables or parameters.a) γi = expected future production equivalent for
water stored in reservoir ib) Mi, 0 = contents of reservoir i at the beginning
of the planning periodc) Mi, t = contents of reservoir i at the end of
hour t
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PROBLEM 12 - Symbols for Short-term Hydro Power Planning Problems
State whether the following symbols denoteoptimization variables or parameters.
d) Qi, t = discharge in power plant i during hour t
e) Si, t = spillage from reservoir i during hour t
f) Vi, t = local inflow to reservoir i during hour t
![Page 41: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/41.jpg)
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PROBLEM 13 - Value of Stored Water
Which expression can be used for the value ofstored water in a 24 hour planning of three hydropower plants in line along a river?*
![Page 42: 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.](https://reader036.fdocuments.us/reader036/viewer/2022081519/56649e635503460f94b5fb8c/html5/thumbnails/42.jpg)
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PROBLEM 14 - Hydrological Constraints
Which expression can be used for the hydrologicalconstraint of the third hydro power plantalong a river?*