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
Basics of Reservoir
Operations
Computer Aided Negotiations
Fall 2008
Megan Wiley Rivera
0:10
Watershed water balance1:45
Water Budgets—Conservation of Mass
• Mass is not created or destroyed
• What goes in – what comes out = change in what’s inside
2:40
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
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
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
Try it for a Britta Filter
• How long can you leave your Britta pitcher filling in the sink before it starts overflowing?
6:20
Draw a Control Volume6:50
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
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
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
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
Now Let’s Apply It to a Reservoir
lakeriver
dam
town
13:00
Draw Control Volume and Specify Inflows and Outflows
Draw Control Volume and Specify Inflows and Outflows
runoff
evaporationprecipitation
Groundwater exchangeDam release
Water supply diversions (demands)
Effluent (returns)
14:50
An Aside: Identifying Consumptive Uses (water removed from the basin)
runoff
evaporationirrigation
evaporation
infiltration
Groundwater exchange
16:15
An Aside: Identifying Consumptive Uses (water removed from the basin)
runoff
evaporation
Water supply diversions (demands)
Effluent (returns)
Other consumptive uses (e.g. manufacturing)
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
What Is Unimpaired Inflow?25:35
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
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
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
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
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
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
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
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
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