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Basics of Reservoir Operations

Basics of Reservoir Operations. Computer Aided Negotiations Fall 2008 Megan Wiley Rivera. 0:10. 1:45. Watershed water balance. 2:40. Water Budgets—Conservation of Mass. Mass is not created or destroyed What goes in – what comes out = change in what’s inside. 3:30. $50. ATM $2000 in

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Basics of Reservoir Operations

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  1. Basics of Reservoir Operations Computer Aided Negotiations Fall 2008 Megan Wiley Rivera

  2. 0:10

  3. 1:45 Watershed water balance

  4. 2:40 Water Budgets—Conservation of Mass • Mass is not created or destroyed • What goes in – what comes out = change in what’s inside

  5. 3:30 $50 ATM $2000 in account $30 Apply conservation of mass to an atm interaction • Starting balance: $2000 • Deposit a check for $50 • Take out $30 • Ending balance: $2020 What goes in – what comes out = change in what’s inside (final balance – initial balance) $50 – $30 = $2020 - $2000

  6. 4:05 A dollar is easier to track than a unit of water • Water is “incompressible” • a unit volume of water is not created or destroyed • Must define boundaries to apply equation (control volume)

  7. 5:15 Time must be considered as well • Often times, inflows and outflows are measured as flow rates • The change in storage must therefore also be specified over some length of time

  8. 6:20 Try it for a Britta Filter • How long can you leave your Britta pitcher filling in the sink before it starts overflowing?

  9. 6:50 Draw a Control Volume

  10. Inflow, Qin = 2.5 gpm Outflow, Qout = 1 gpm (note, this is a cheat. The outflow flow rate increases as the chamber fills) Chamber dimensions: 8” tall, 24 in2 cross sectional area 1 cubic in = 0.00433 gals 7:00 Some Numbers

  11. What goes in – what comes out = change in what’s inside Qin – Qout = dV/dt Work with a partner to figure it out 8:40 The Equation

  12. Inflow, Qin = 2.5 gpm Outflow, Qout = 1 gpm (note, this is a cheat. The outflow flow rate increases as the chamber fills) Chamber dimensions: 8” tall, 24 in2 cross sectional area 1 cubic in = 0.00433 gals Feel free to ask someone else if you get stuck (there are different approaches) Qin – Qout = dV/dt

  13. Initial volume = 0 Final volume = 24 in2 * 8” = 192 in3 Convert to gallons: 192 in3 * (0.00433 gal/1 in3) = 0.83 gal Apply equation: 2.5 gpm – 1 gpm = 0.83 gal/x min Solve equation: x min = 0.83 gal/(1.5 gpm) = 0.55 min or about 30 second 12:30 The Answer

  14. 13:00 town dam lake river Now Let’s Apply It to a Reservoir

  15. Draw Control Volume and Specify Inflows and Outflows

  16. 14:50 Draw Control Volume and Specify Inflows and Outflows evaporation Effluent (returns) precipitation Water supply diversions (demands) Dam release runoff Groundwater exchange

  17. 16:15 An Aside: Identifying Consumptive Uses (water removed from the basin) evaporation evaporation irrigation infiltration runoff Groundwater exchange

  18. An Aside: Identifying Consumptive Uses (water removed from the basin) evaporation Effluent (returns) Water supply diversions (demands) Other consumptive uses (e.g. manufacturing) runoff

  19. 24:45 Back to Conservation of Mass Net demands, D Net Evapotransporation, ET evaporation Effluent (returns) precipitation Water supply diversions (demands) Dam release runoff Groundwater exchange Qout Unimpaired inflow, I

  20. 25:35 What Is Unimpaired Inflow?

  21. 26:25 Why Might We Want to Calculate It • If you want to model different operational scenarios, you need to know how much water is reaching the river via runoff (as opposed to upstream operations) • Also gives information about flow in the river without the presence of reservoirs (possible point of comparison)

  22. 32:20 Use the equation to calculate unimpaired inflows (daily average) Net demands, D Net Evapotransporation, ET evaporation Effluent (returns) precipitation Water supply diversions (demands) Dam release runoff Groundwater exchange Qout Unimpaired inflow, I

  23. Use the equation to calculate unimpaired inflows (daily average) Measured/modeled/estimated from meteorological info measured Net Evapotransporation, ET Net demands, D measured Qout Beginning and end of day stages measured Unimpaired inflow, I

  24. 32:30 Storage-Area-Elevation Table Surface area elevation Storage = volume of water Mean sea level

  25. 32:40 Net Evapotransporation, ET Net demands, D Qout Beginning and end of day stages Unimpaired inflow, I Use the equation to calculate unimpaired inflows (daily average) • What goes in – what comes out = change in what’s inside • Qin – Qout = dV/dt, or over the day: • Qin,daily ave – Qout, daily ave = Storageend of day – Storagebeginning of day • I – ET – Qout – D = Storageend of day – Storagebeginning of day • I = ET + Qout + D + Storageend of day – Storagebeginning of day Work with your partner again

  26. 35:20 Net Evapotransporation, ET Net demands, D Qout Beginning and end of day stages Unimpaired inflow, I Use the equation to calculate unimpaired inflows (daily average) • What goes in – what comes out = change in what’s inside • Qin – Qout = dV/dt, or over the day: • Qin,daily ave – Qout, daily ave = Storageend of day – Storagebeginning of day • I – ET – Qout – D = Storageend of day – Storagebeginning of day • I = ET + Qout + D + Storageend of day – Storagebeginning of day

  27. Net Evapotransporation, ET Net demands, D Qout Beginning and end of day stages Unimpaired inflow, I Fun with units • I = ET + Qout + D + Storageend of day – Storagebeginning of day 0.3” 50 mgd 100 cfs Stage = 54’ Stage = 53’ Do calculations first in af/day and then mgd

  28. 39:30 Net Evapotransporation, ET Net demands, D Qout Beginning and end of day stages Unimpaired inflow, I • I = ET + Qout + D + Storageend of day – Storagebeginning of day 0.3” 50 mgd 100 cfs Stage = 54’ Stage = 53’ Do calculations first in af/day and then mgd

  29. 0.3” 50 mgd • I = ET + Qout + D + Storageend of day – Storagebeginning of day 100 cfs Stage = 54’ Stage = 53’ ET: Multiply by average surface area for the day (see SAE) = 2026 acres 0.3” * 2026 acres = 0.025’ * 2026 acres = 50.7 af in one day Outflow: 100 cfs * 1.98 af/day / 1cfs = 198 af/day Demands: 50 mgd * 1 af/day / 3.069 mgd = 16 af/day I = 50.7 af/day + 198 af/day + 16 af/day + (11957 af – 9930 af)/day = 2292 af/day This is 747 mgd

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