Download presentation

Presentation is loading. Please wait.

Published byAndy Densmore Modified over 2 years ago

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

2
0:10

3
Watershed water balance 1:45

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

5
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

6
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

7
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

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

9
Draw a Control Volume 6:50

10
Some Numbers Inflow, Q in = 2.5 gpm Outflow, Q out = 1 gpm (note, this is a cheat. The outflow flow rate increases as the chamber fills) Chamber dimensions: 8” tall, 24 in 2 cross sectional area 1 cubic in = 0.00433 gals 7:00

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

12
Feel free to ask someone else if you get stuck (there are different approaches) Inflow, Q in = 2.5 gpm Outflow, Q out = 1 gpm (note, this is a cheat. The outflow flow rate increases as the chamber fills) Chamber dimensions: 8” tall, 24 in 2 cross sectional area 1 cubic in = 0.00433 gals Qin – Qout = dV/dt

13
The Answer Initial volume = 0 Final volume = 24 in 2 * 8” = 192 in 3 Convert to gallons: 192 in 3 * (0.00433 gal/1 in 3 ) = 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

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

15
Draw Control Volume and Specify Inflows and Outflows

16
runoff evaporation precipitation Groundwater exchange Dam release Water supply diversions (demands) Effluent (returns) 14:50

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

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

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

20
What Is Unimpaired Inflow? 25:35

21
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

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

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

24
Storage-Area-Elevation Table Storage = volume of water Surface area elevation Mean sea level Storage (af) Area (acres) Elevation (ft) 05.832 118137843 9930178053 11957227254 34055497860 32:30

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

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

27
Fun with units I = ET + Q out + D + Storage end of day – Storage beginning of day Net demands, D Unimpaired inflow, I Net Evapotransporation, ET Q out 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

28
Net demands, D Unimpaired inflow, I Net Evapotransporation, ET Q out 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) 05.832 118137843 9930178053 11957227254 34055497860 I = ET + Q out + D + Storage end of day – Storage beginning of day 39:30

29
Stage = 54’Stage = 53’ 50 mgd 100 cfs 0.3” I = ET + Q out + D + Storage end of day – Storage beginning 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

Similar presentations

OK

CEE 3430, Engineering Hydrology David Tarboton. Overview Handouts –Syllabus –Schedule –Student Information Sheet –Homework 1 Web: https://usu.instructure.com/courses/392804https://usu.instructure.com/courses/392804.

CEE 3430, Engineering Hydrology David Tarboton. Overview Handouts –Syllabus –Schedule –Student Information Sheet –Homework 1 Web: https://usu.instructure.com/courses/392804https://usu.instructure.com/courses/392804.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on airport automation Ppt online viewers Ppt on national education policy 1986 ford Ppt on resistance temperature detector failures Ppt on ministry of corporate affairs ontario Ppt on travelling salesman problem branch and bound Ppt on classical economics time Ppt on accounting standards 10 Ppt on cross-sectional study research questions Seeley's anatomy and physiology ppt on cells