Water Resources Planning andManagement
Daene C. McKinney
System PerformanceIndicators
Reservoir Management
•Important task forwater managers aroundthe world.
•Models used to
–simulate or optimizereservoir performance
–design reservoirs orassociated facilities(spillways, etc.).
Operating Rules
•Allocate releases among purposes, reservoirs, and timeintervals
•In operation (as opposed to design), certain systemcomponents are fixed:
–Active and dead storage volume
–Power plant and stream channel capacities
–Reservoir head-capacity functions
–Levee heights and flood plain areas
–Monthly target outputs for irrigation, energy, water supply, etc
•Others are variable: Allocation of
–stored water among reservoirs
–stored and released water among purposes
–stored and released water among time intervals
Standard Operating Policy
Qt
X2t
K
St
Rt
X1t
Dt
Rt
Dt
Dt
Dt+K
St+ Qt
Release availablewater &
deficits occur
Release demand
spill excess
Sufficient waterto meet demands
Reservoir fillsand demand met
Release demand &
demand met
Demand
•Reservoir operating policy - releaseas function of storage volume andinflow
Rt = Rt(St,Qt)
Hedging Rule
•Reduce releases in times of drought (hedging) to save waterfor future releases in case of an extended period of lowinflows.
hedginghedging
DD
KK
Done?
No
System SimulationLoucks (Chapter 7, Section 9)
•Create network representation of system
•Need inflows for each period for each node
•For each period:
Perform mass balance calculations for each node
Determine releases from reservoirs
Allocate water to users
Start
t = 0
St = S0
St+1 = St +Qt -Rt
Stop
Yes
t = t + 1
Read Qt
File
Compute
Rt, Xit, i=1,…n
Data
Storage
Qt
X3t
K
St
R
X2t
X1t
Operating Policy
Allocation Policy
Example
•Using unregulated river for irrigation
•Proposed Reservoir
•Capacity: K = 40 million m3 (active)
•Demand: D = 30 40 45 million m3
•Winter instream flow: 5 mil. m3 min.
•45 year historic flow record available
•Evaluate system performance for a 20 yearperiod
•Simulate
•Two seasons/year, winter (1) summer(2)
•Continuity constraints
•Operating policy
Qt
X2t
K
St
R
X1t
Flow statistics
R2,t
Dt
Dt
Dt+K
S2,t+ Q2,t
K
Release availablewater
Releasedemand
Release demand
+ excess
Summer Operating Policy
Storage at beginning of summer
Performance Evaluation
•How well will the system perform?
–Define performance criteria
•Indices related to the ability to meet targets and theseriousness of missing targets
–Simulate the system to evaluate the criteria
–Interpret results
•Should design or policies be modified?
Performance Criteria - Reliability
•Reliability – Frequency with which demand was satisfied
–Define a deficit as:
•Then reliability is:
•where n is the total number of simulation periods
Performance Criteria - Resilience
•Resilience = probability that once the system is in a period ofdeficit, the next period is not a deficit.
•How quickly does system recover from failure?
Performance Criteria - Vulnerability
•Vulnerability = average magnitude of deficits
•How bad are the consequences of failure?
Simulate the System
System
Policies
Input
Output
x
g(x)
y
h(y)
Reservoir operating policy
Allocation policy
Hydrologic
time series
Model
output
Model
Uncertainty
•Deterministic process
–Inputs assumed known.
–Ignore variability
–Assume inputs are wellrepresented by average values.
–Over estimates benefits andunderestimates losses
•Stochastic process
–Explicitly account for variabilityand uncertainty
–Inputs are stochastic processes
–Historic record is one realizationof process.
FY(y)
Simulate the System
Policies
Simulate each
Input sequence
X
FX(x)
x
g(x)
y
h(y)
y
h(y)
Compute
statistics of
outputs
System
Generate multiple
input sequences
x
g(x)
Get multiple
output sequences
Reservoir operating policy
Allocation policy
Model
Distribution of inputs
The Simulation
•Simulate reservoir operation
–Perform 23 equally likely simulations
–Each simulation is 20 years long
–Each simulation uses a different sequence of inflows(realization)
Example – Realization 1
Rmin
0.5
K
4
Realization
1
Winter
Summer
Year
S1y
Q1y
S+Q
R1y
S2y
Q2y
S+Q
D2y
R2y
Deficit
1
0.000
4.740
4.740
0.740
4.000
1.805
5.805
3.000
3.000
0.000
2
2.805
2.918
5.723
1.723
4.000
1.499
5.499
3.200
3.200
0.000
3
2.299
2.747
5.045
1.045
4.000
1.548
5.548
3.400
3.400
0.000
4
2.148
2.819
4.966
0.966
4.000
1.753
5.753
3.600
3.600
0.000
5
2.153
3.871
6.023
2.023
4.000
2.229
6.229
3.800
3.800
0.000
6
2.429
3.585
6.015
2.015
4.000
2.235
6.235
4.000
4.000
0.000
7
2.235
4.736
6.971
2.971
4.000
2.984
6.984
4.100
4.100
0.000
8
2.884
3.275
6.159
2.159
4.000
2.212
6.212
4.200
4.200
0.000
9
2.012
3.188
5.200
1.200
4.000
2.666
6.666
4.300
4.300
0.000
10
2.366
3.401
5.767
1.767
4.000
1.240
5.240
4.300
4.300
0.000
11
0.940
3.811
4.750
0.750
4.000
2.371
6.371
4.400
4.400
0.000
12
1.971
3.435
5.407
1.407
4.000
2.421
6.421
4.400
4.400
0.000
13
2.021
2.460
4.481
0.500
3.981
1.317
5.298
4.400
4.400
0.000
14
0.898
2.377
3.275
0.500
2.775
1.896
4.671
4.400
4.400
0.000
15
0.271
3.692
3.963
0.500
3.463
1.831
5.293
4.500
4.500
0.000
16
0.793
3.302
4.095
0.500
3.595
1.300
4.895
4.500
4.500
0.000
17
0.395
2.548
2.944
0.500
2.444
2.047
4.491
4.500
4.491
-0.009
18
0.000
2.454
2.454
0.500
1.954
1.658
3.612
4.500
3.612
-0.888
19
0.000
3.139
3.139
0.500
2.639
2.768
5.407
4.500
4.500
0.000
20
0.907
2.910
3.816
0.500
3.316
1.445
4.762
4.500
4.500
0.000
Deficit
-0.897
Number
2.000
%
0.100
Results
Average failure frequency = 0.165
Average reliability = 1- 0.165 = 0.835 = 83.5%
Actual failure frequency [0, 0.40]
Actual Reliability [100%, 60%]