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# Basics of Reservoir Operations - PowerPoint PPT Presentation

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

Computer Aided Negotiations

Fall 2008

Megan Wiley Rivera

Watershed water balance

Water Budgets—Conservation of Mass

• Mass is not created or destroyed

• What goes in – what comes out = change in what’s inside

\$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

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)

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

Try it for a Britta Filter

• How long can you leave your Britta pitcher filling in the sink before it starts overflowing?

Draw a Control Volume

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

Qin – Qout = dV/dt

Work with a partner to figure it out

8:40

The Equation

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

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

town

dam

lake

river

Now Let’s Apply It to a Reservoir

Draw Control Volume and Specify Inflows and Outflows

evaporation

Effluent (returns)

precipitation

Water supply diversions (demands)

Dam release

runoff

Groundwater exchange

An Aside: Identifying Consumptive Uses (water removed from the basin)

evaporation

evaporation

irrigation

infiltration

runoff

Groundwater exchange

An Aside: Identifying Consumptive Uses (water removed from the basin)

evaporation

Effluent (returns)

Water supply diversions (demands)

Other consumptive uses (e.g. manufacturing)

runoff

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

What Is Unimpaired Inflow?

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)

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

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

32:30 average)

Storage-Area-Elevation Table

Surface area

elevation

Storage = volume of water

Mean sea level

32:40 average)

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

35:20 average)

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

Net Evapotransporation, ET average)

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

39:30 average)

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

0.3” average)

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