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Water and Data: what are the links?

Explore how collecting and evaluating data on water can help predict water availability, manage water systems, and understand human behavior. Includes examples and calculations.

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Water and Data: what are the links?

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  1. TIM 155 – Summer 2016 Week 2 Water and Data: what are the links? M. Loik

  2. Collecting and evaluating data on water help us: • Predict how much water will be available • Predict how much demand there will be • Operate water systems better Many following images from Prof. Michael Loik

  3. Kinds of data collected: • Location: Political, Socio-economic, health • Quantity: Water amounts and flow rates • Quality: Temperature, salinity, biological concentrations • Human behavior: uses, responses to prices, willingness to invest in alternatives • Operational: pump performance, treatment equipment • Financial:cost, project management • Scientific: atmospheric, membrane performance

  4. Population Affected by Drought: 37,014,848

  5. What Happened to the Rain (until this winter)?

  6. Storm track shifts north…

  7. Winter 2015 Warm & Dry in San Francisco Cold & Snowy in Boston

  8. Winter 2016 An El Nino Year

  9. Atmospheric Rivers

  10. Atmospheric Rivers

  11. Let’s start with water conversion questions

  12. Example: A farmer sells 50 af/year of her water supply to a nearby city. Assuming each family in the city uses 500 gallons per day, how many families will that water supply?

  13. Example: A farmer sells 50 af/year of her water supply to a nearby city. Assuming each family in the city uses 500 gallons per day, how many families will that water supply? Approach: (1) make all the units the same (e.g., gallons, days). (2) figure out the units requested in the answer. (families).

  14. Quick math note: A/B / A/(B*C) = C

  15. Quick math note: A/B / A/(B*C) = C Why? First cancel the As and Bs. Then multiply by (C/1 / C/1) …this cancels the denominator and leaves C alone in the numerator. In our example, C is going to be “families”

  16. Example: A farmer sells 50 af/year of her water supply to a nearby city. Assuming each family in the city uses 500 gallons per day, how many families will that water supply? (1) 50 af / year x (1 year / 365 days) = 0.137 af / day (2) 0.137 af / day x (1,000,000 gal / 3.07 af ) = 44,621 gallons / day 44,621 gallons / day / 500 gallons / family.day = 89 families

  17. Example 2: A stream is flowing at 50 cfs. Assume no instream flows are required so a city can take all the water if it chooses. Will there be enough water to service a city that uses 10 mgd? Show your work…

  18. cfs = cubic feet per second Example 2: A stream is flowing at 50 cfs. Assume no instream flows are required so a city can take all the water if it chooses. Will there be enough water to service a city that uses 10 mgd? Show your work… mgd = million gallons per day

  19. cfs = cubic feet per second 1 cfs is a measurement of the flow of water. It measures one cubic foot of water (7.48 gallons) taking one second to flow past a given point. This is a common measurement of flow in a river or canal - e.g., 50 cfs, 1,000 cfs.

  20. Example 2: A stream is flowing at 50 cfs. Assume no instream flows are required so a city can take all the water if it chooses. Will there be enough water to service a city that uses 10 mgd? Show your work… Approach - change cfs to mgd and compare the two quantities

  21. Example 2: A stream is flowing at 50 cfs. Assume no instream flows are required so a city can take all the water if it chooses. Will there be enough water to service a city that uses 10 mgd? Show your work… 50 ft3/second x 60 sec/minute x 60 min/hour x 24 hr/day = 4.32 million ft3/day 4.32 million ft3/day x 7.48 gallons/1 ft3 = 32.3 million gallons/day (or mgd)

  22. Example 2: A stream is flowing at 50 cfs. Assume no instream flows are required so a city can take all the water if it chooses. Will there be enough water to service a city that uses 10 mgd? Show your work… 50 ft3/second x 60 sec/minute x 60 min/hour x 24 hr/day = 4.32 million ft3/day 4.32 million ft3/day x 7.48 gallons/1 ft3 = 32.3 million gallons/day (or mgd) YES - this is more than enough water to supply the city.

  23. Another conversion question…. The minimum streamflow required to protect American River fisheries is 500 cfs from September through December, and 250 cfs January through August. Assuming that each month has 30 days, how many acre-feet at a minimum are reserved for American River fisheries each year?

  24. Another question… The southern California coastal plain (Los Angeles, San Diego, etc.) has the following water-delivery capacities (million acre-feet/year): Local water: 1.1 L.A. aqueduct 0.47 Colorado River Aqueduct 1.21 California Aqueduct 2.01 Total: 4.79 If per capita consumption in the region remains at 200 gallons per day, at what level of population would fully-utilized delivery capacity equal total consumption?

  25. Estimating pollution in a lake using a Pollutant Diffusion Model Adapted from Consider a Spherical Cow, by John Harte, Mill Valley, Ca.: University Science Books, 1988, p. 36.

  26. PRACTICE PROBLEM Factory Pollution outfall Stream inflow to lake Lake Stream outflow from lake

  27. Pollution Residence Time Question Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? Adapted from Consider a Spherical Cow, by John Harte, Mill Valley, Ca.: University Science Books, 1988, p. 36

  28. Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? In this problem, pollution builds up until inflow of pollution equals outflow - that’s the “steady state.” We want to know how much pollution is present at that time.

  29. Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? These are the assumptions that simplify the problem and make it easier to solve.

  30. Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? This is the data you will use for your calculations.

  31. “ACTUAL” SITUATION Factory Pollution outfall Stream inflow to lake Lake Stream outflow from lake

  32. “SIMPLIFIED,” DATA ADDED Factory Pollution outfall = 0.16 tonnes/day Stream inflow to Lake = outflow Lake Volume 4 x 107 m3 Stream outflow from lake = 8 x 104 m3/day Pollutant is evenly dispersed throughout the lake.

  33. Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? These are the two questions - You need to provide two separate answers!

  34. Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? “Steady state” means pollution outflow = inflow so the amount of pollution isn’t changing in the lake.

  35. Question: A stable but highly soluble pollutant is dumped into a lake at the rate of 0.16 tonnes per day. The lake volume is 4 x 107 m3, and the average water flow-through rate is 8 x 104 m3/day. Ignore evaporation from the lake and assume the pollutant is instantly uniformly mixed in the lake. What is the eventual steady-state stock of pollutant in the lake? What steady-state concentration will the pollutant reach in ppm (parts per million)? Measured in “parts per million” Measured in kilograms (because the units in the problem are all metric)

  36. parts per million= ppm For every million molecules in the lake, how many are pollution molecules? That number can be expressed as “parts per million.” If the answer is 2 ppm, that is equal to: .000002 = 2/1,000,000 = 2 ppm Scientists and water professionals say “two ppm” because it is quicker and easier than saying “two millionths” Note that to get to ppm, you need to cancel all the other units, such as volume (cubic meters) or mass (kilograms). There’s also ppt and ppb - what are these?

  37. Answer: To determine the steady-state stock of the pollutant, we first must calculate the residence time of the pollutant in the lake. Since the pollutant mixes instantly, its residence time will be equal to that of the water that passes through the lake.

  38. This was a simplifying assumption given in the problem. Answer: To determine the steady-state stock of the pollutant, we first must calculate the residence time of the pollutant in the lake. Since the pollutant mixes instantly, its residence time will be equal to that of the water that passes through the lake.

  39. Answer: To determine the steady-state stock of the pollutant, we first must calculate the residence time of the pollutant in the lake. Since the pollutant mixes instantly, its residence time will be equal to that of the water that passes through the lake. Residence time = Stock of water in the lake Flow of water in the lake for example: m3 = days m3 day

  40. Answer: To determine the steady-state stock of the pollutant, we first must calculate the residence time of the pollutant in the lake. Since the pollutant mixes instantly, its residence time will be equal to that of the water that passes through the lake. The problem gave us: stock of water in the lake = 4 x 107 m3 flow-through rate of water = 8 x 104 m3/day residence time of water in the lake = stock / flow = 500 days. (confirm this - do the math!)

  41. So the residence times in the lake of both the water and the pollutant are 500 days. To determine the steady-state stock of the pollutant in the lake, multiply the pollutant’s residence time by the rate at which it is being loaded: 500 days residence time x 0.16 tonnes per day = 80 tonnes of pollutant present in the lake in steady-state conditions.

  42. To determine the concentration of pollutant in the lake, we take the ratio of pollutant to water, using the same units - tonnes. First we convert m3 to tonnes:

  43. To determine the concentration of pollutant in the lake, we take the ratio of pollutant to water, using the same units - tonnes. First we convert m3 to tonnes: We want the lake volume (cubic meters) switched to mass units (tonnes) so that it can be compared to the tonnes of pollutant. Also, the units will cancel, which helps get us to ppm.

  44. To determine the concentration of pollutant in the lake, we take the ratio of pollutant to water, using the same units - tonnes. First we convert m3 to tonnes: 4 x 107 m3 x 1 tonne/1 m3 = 4 x 107 tonnes water in the lake.

  45. To determine the concentration of pollutant in the lake, we take the ratio of pollutant to water, using the same units - tonnes. First we convert m3 to tonnes: 4 x 107 m3 x 1 tonne/1 m3 = 4 x 107 tonnes water in the lake. equals one for water

  46. To determine the concentration of pollutant in the lake, we take the ratio of pollutant to water, using the same units - tonnes. First we convert m3 to tonnes: 4 x 107 m3 x 1 tonne/1 m3 = 4 x 107 tonnes water in the lake.

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