basics of reservoir operations
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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Basics of Reservoir
Operations
Computer Aided Negotiations
Fall 2008
Megan Wiley Rivera
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0:10
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Watershed water balance1:45
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Water Budgets—Conservation of Mass
• Mass is not created or destroyed
• What goes in – what comes out = change in what’s inside
2:40
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Apply conservation of mass to an atm interaction
• Starting balance: $2000
• Deposit a check for $50
• Take out $30
• Ending balance: $2020
ATM
$2000 in account
$50
$30
What goes in – what comes out = change in what’s inside (final balance – initial balance)
$50 – $30 = $2020 - $2000
3:30
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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)
4:05
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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
5:15
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Try it for a Britta Filter
• How long can you leave your Britta pitcher filling in the sink before it starts overflowing?
6:20
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Draw a Control Volume6:50
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Some Numbers
• 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
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The Equation
• 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
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Feel free to ask someone else if you get stuck (there are different approaches)
• 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
Qin – Qout = dV/dt
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The Answer
• 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
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Now Let’s Apply It to a Reservoir
lakeriver
dam
town
13:00
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Draw Control Volume and Specify Inflows and Outflows
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Draw Control Volume and Specify Inflows and Outflows
runoff
evaporationprecipitation
Groundwater exchangeDam release
Water supply diversions (demands)
Effluent (returns)
14:50
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An Aside: Identifying Consumptive Uses (water removed from the basin)
runoff
evaporationirrigation
evaporation
infiltration
Groundwater exchange
16:15
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An Aside: Identifying Consumptive Uses (water removed from the basin)
runoff
evaporation
Water supply diversions (demands)
Effluent (returns)
Other consumptive uses (e.g. manufacturing)
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Back to Conservation of Mass
runoff
evaporationprecipitation
Groundwater exchangeDam release
Water supply diversions (demands)
Effluent (returns)
Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout
24:45
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What Is Unimpaired Inflow?25:35
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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)
26:25
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Use the equation to calculate unimpaired inflows (daily average)
runoff
evaporationprecipitation
Groundwater exchangeDam release
Water supply diversions (demands)
Effluent (returns)
Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout
32:20
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Use the equation to calculate unimpaired inflows (daily average)
Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout
measured
measured
measuredBeginning and end of day stages
Measured/modeled/estimated from meteorological info
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Storage-Area-Elevation Table
Storage = volume of water
Surface area
elevation
Mean sea level
Storage (af)
Area (acres)
Elevation (ft)
0 5.8 32
1181 378 43
9930 1780 53
11957 2272 54
34055 4978 60
32:30
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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
Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout Beginning and end of day stages
Work with your partner again
32:40
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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
Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout Beginning and end of day stages
35:20
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Fun with units
• I = ET + Qout + D + Storageend of day – Storagebeginning of day
Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout Beginning and end of day stages
Stage = 54’ Stage = 53’
50 mgd
100 cfs
0.3”
Do calculations first in af/day and then mgd
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Net demands, D
Unimpaired inflow, I
Net Evapotransporation, ET
Qout Beginning and end of day stages
Stage = 54’ Stage = 53’
50 mgd
100 cfs
0.3”
Do calculations first in af/day and then mgd
Storage (af) Area (acres) Elevation (ft)
0 5.8 32
1181 378 43
9930 1780 53
11957 2272 54
34055 4978 60
• I = ET + Qout + D + Storageend of day – Storagebeginning of day
39:30
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Stage = 54’ Stage = 53’
50 mgd
100 cfs
0.3”
• I = ET + Qout + D + Storageend of day – Storagebeginning of day
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