Hydrologic Routing Reading: Applied Hydrology Sections 8.1, 8.2, 8.4 - PowerPoint Presentation

1. Hydrologic RoutingReading: Applied Hydrology Sections 8.1, 8.2, 8.4

2. 2Flow RoutingProcedure to determine the flow hydrograph at a point on a watershed from a known hydrograph upstreamAs the hydrograph travels, itattenuates gets delayedQtQtQtQt

3. 3Why route flows?Account for changes in flow hydrograph as a flood wave passes downstreamThis helps in Accounting for storagesStudying the attenuation of flood peaksQt

4. Watershed – Drainage area of a point on a streamConnecting rainfall input with streamflow outputRainfallStreamflow

5. Flood Control DamsDam 13AFlow with a Horizontal Water Surface

6. Floodplain Zones1% chance < 0.2% chanceMain zone of water flowFlow with a Sloping Water Surface

7. 7Types of flow routingLumped/hydrologicFlow is calculated as a function of time alone at a particular locationGoverned by continuity equation and flow/storage relationship Distributed/hydraulicFlow is calculated as a function of space and time throughout the systemGoverned by continuity and momentum equations

8. 8Hydrologic RoutingUpstream hydrographDownstream hydrographInput, output, and storage are related by continuity equation:DischargeInflowDischargeOutflowTransferFunctionQ and S are unknownStorage can be expressed as a function of I(t) or Q(t) or bothFor a linear reservoir, S=kQ

9. 9Lumped flow routingThree typesLevel pool method (Modified Puls)Storage is nonlinear function of QMuskingum methodStorage is linear function of I and QSeries of reservoir modelsStorage is linear function of Q and its time derivatives

10. 10S and Q relationships

11. 11Level pool routingProcedure for calculating outflow hydrograph Q(t) from a reservoir with horizontal water surface, given its inflow hydrograph I(t) and storage-outflow relationship

12. 12Level pool methodologyDischargeTimeStorageTimeInflowOutflowUnknownKnownNeed a function relatingStorage-outflow function

13. 13Level pool methodologyGiven Inflow hydrographQ and H relationshipStepsDevelop Q versus Q+ 2S/Dt relationship using Q/H relationshipCompute Q+ 2S/Dt using Use the relationship developed in step 1 to get Q

14. 14Ex. 8.2.1Given I(t)Given Q/HArea of the reservoir = 1 acre, and outlet diameter = 5ft

15. 15Ex. 8.2.1 Step 1Develop Q versus Q+ 2S/Dt relationship using Q/H relationship

16. 16Step 2Compute Q+ 2S/Dt usingAt time interval =1 (j=1), I1 = 0, and therefore Q1 = 0 as the reservoir is emptyWrite the continuity equation for the first time step, which can be used to compute Q2

17. 17Step 3Use the relationship between 2S/Dt + Q versus Q to compute Q Use the Table/graph created in Step 1 to compute Q What is the value of Q if 2S/Dt + Q = 60 ?So Q2 is 2.4 cfsRepeat steps 2 and 3 for j=2, 3, 4… to compute Q3, Q4, Q5…..

18. 18Ex. 8.2.1 results

19. 19Ex. 8.2.1 resultsInflowOutflowPeak outflow intersects with the receding limb of the inflow hydrographOutflow hydrograph

20. 20Q/H relationshipshttp://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.html Program for Routing Flow through an NRCS Reservoir

21. Hydrologic river routing (Muskingum Method)Wedge storage in reachAdvancingFloodWaveI > QRecedingFloodWaveQ > IK = travel time of peak through the reachX = weight on inflow versus outflow (0 ≤ X ≤ 0.5)X = 0  Reservoir, storage depends on outflow, no wedgeX = 0.0 - 0.3  Natural stream

22. 22Muskingum Method (Cont.)Recall:Combine:If I(t), K and X are known, Q(t) can be calculated using above equations

23. 23Muskingum - ExampleGiven:Inflow hydrographK = 2.3 hr, X = 0.15, Dt = 1 hour, Initial Q = 85 cfsFind:Outflow hydrograph using Muskingum routing method

24. 24Muskingum – Example (Cont.)C1 = 0.0631, C2 = 0.3442, C3 = 0.5927

25. HEC-HMS Model of Brushy CreekDam 7Walsh Dr

26. Watershed W1820

27. Junction J329W1820R580J329W1820J329

28. Reach R580