Water Resources Planning and Management River Basin Modeling Water Resources Planning and Management Daene C. McKinney

Water Resources Water at: What to do? Manipulate the hydrologic cycle Wrong place, wrong quantity, wrong time What to do? Manipulate the hydrologic cycle Build facilities? Remove facilities? Reoperate facilities? Reservoirs Canals Levees Other infrastructure

Scales Time Scales Water management plans Flood management Consider average conditions within discrete time periods Weekly, monthly or seasonal Over a long time horizon Year, decade, century Shortest time period No less than travel time from the upper basin to mouth For shorter time periods some kind of flow routing required Flood management Conditions over much shorter periods Hours, Days, Week

Processes Processes we need to describe: Precipitation Runoff Infiltration Percolation Evapotranspiration Chemical concentration Groundwater

Data Measurement Reservoir losses Missing data Data sources Flow conditions Natural Present Unregulated Regulated Future Reservoir losses Missing data Precipitation-runoff models Stochastic streamflow models Extending and filling in historic records

Hydrologic Frequency Analysis Flow duration curves Percent of time during which specified flow rates are equaled or exceeded at a given location Pr{ X > x }

Central Asia Syr Darya Naryn River

Naryn River Annual Flows Median flow Min. flow Glacier melt

Naryn River Annual Flows

Quantiles X is a continuous RV p-th quantile is xp Median: x50 equally likely to be above as below that value Examples Floodplain management - the 100-year flood x0.99 Water quality management minimum 7-day-average low flow expected once in 10 years 10th percentile of the distribution of the annual minima of the 7-day average flows p xp X FX(x)

Quantiles Observed values, sample of size n Order statistics (observations ordered by magnitude Sample estimates of quantiles can be obtained by using

Flow Duration Curve P(X>x)= Ranked Year Flow Rank 1-i/(n+1)= x i 1911 10817 1 0.99 6525 1912 11126 2 0.98 7478 1913 11503 3 0.97 8014 1914 11428 4 0.96 8161 1915 10233 5 0.95 8378 … 1997 10343 87 0.06 15062 1998 14511 88 0.05 15242 1999 14557 89 0.04 16504 2000 12614 90 0.03 16675 2001 12615 91 0.02 18754 2002 92=n 0.01 20725

Flow Duration Curve Flow duration curve - Discharge vs % of time flow is equaled or exceeded. Firm yield is flow that is equaled or exceeded 100% of the time

Increase Firm Yield - Add storage To increase the firm yield of a stream, impoundments are built. Need to develop the storage-yield relationship for a river

Simplified Methods Mass curve (Rippl) method Graphical estimate of storage required to supply given yield Constructed by summing inflows over period of record and plotting these versus time and comparing to demands Time interval includes “critical period” Time over which flows reached a minimum Causes the greatest drawdown of reservoir

Rippl method

Rippl Method Qt Rt Accumulated Inflows, Q Capacity K Accumulated Releases, R

Sequent Peak Method

Reservoirs Hoover Dam 158 m 35 km3 2,074 MW Toktogul Dam 140 m Grand Coulee Dam 100 m 11.8 km3 6,809 MW

Dams Masonry dams Embankment dams Spillways Arch dams Gravity dams rock-fill and earth-fill dams Spillways

Reservoir Qt Rt St K

Management of a Single Reservoir 2 common tasks of reservoir modeling: Determine coefficients of functions that describe reservoir characteristics Determine optimal mode of reservoir operation (storage volumes, elevations and releases) while satisfying downstream water demands

Reservoir Operation Compute optimal operation of reservoir given a series of inflows and downstream water demands where: St End storage period t, (L3); St-1 Beginning storage period t, (L3); Qt Inflow period t, (L3); Rt Release period t, (L3); Dt Demand, (L3); and K Capacity, (L3) Smin Dead storage, (L3)

Comparison of Average and Dry Conditions

Results Storage Input Release Demand t0 15000 t1 13723 426 1700 t2   t1 13723 426 1700 t2 12729 399 1388 t3 11762 523 1478 t4 11502 875 1109 t5 12894 2026 595 t6 15838 3626 637 t7 17503 2841 1126 t8 17838 1469 1092 t9 18119 821 511 t10 17839 600 869 t11 17239 458 1050 t12 16172 413 1476

Toktogul Power = 1,200 MW Height = 140 m Capacity = 19.5 km3

Toktogul on the Naryn River in Kyrgyzstan Lt At Evaporation Toktogul on the Naryn River in Kyrgyzstan Lt Losses from reservoir A Surface area of reservoir et ave. evaporation rate

Evaporation Lt At

Hydropower Production Hoover Dam 2,074 MW 158 m 35 km3 Toktogul Dam 1,200 MW 140 m 19.5 km3 Grand Coulee Dam 6,809 MW 100 m 11.8 km3

Reservoir with Power Plant earliest known dam - Jawa, Jordan - 9 m high x1 m wide x 50 m long, 3000 BC Hoover Dam

Reservoir with Power Plant Qt Rt St K Et Ht Qt

Power Production K Lt St Et Ht Qt Qt = Release (m3/period) qt = Flow (m3/sec) Pt = Power (kW) Et = Energy (kWh) Ht = Head (m) e = efficiency (%) timet = sec in period t Qt K St Et Ht Lt

Head vs Storage Relation Toktogul on the Naryn River in Kyrgyzstan

Model Lt St Et Qt Rt K Rt = Release (m3/period) Dt = Demand for water (m3/period)

K Qt Rt St Et Lt Toktogul Operation