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Screen and Bypass Design Bryan Nordlund, P.E. National Marine Fisheries Service Lacey, Washington Note: this presentation represents the views of the presenter,

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Presentation on theme: "Screen and Bypass Design Bryan Nordlund, P.E. National Marine Fisheries Service Lacey, Washington Note: this presentation represents the views of the presenter,"— Presentation transcript:

1 Screen and Bypass Design Bryan Nordlund, P.E. National Marine Fisheries Service Lacey, Washington Note: this presentation represents the views of the presenter, and in most cases, is based on fishway design experience in working for NMFS. Special thanks to Larry Swenson for the assistance with slide content.

2 Hydrology and Hydraulics

3 Hydrology relates to the science dealing with the occurrence, circulation and distribution of water on the earth's lands and in the atmosphere. Hydraulics refers to fluids in motion. Hydrology doesn't make fish barriers (unless streamflow gets too low or too high), but hydraulics can create a barrier.

4 Determining Fishway Design Flows

5 Determining Fishway Design flows 1. Locate daily average streamflow records (USGS, BOR, other) and import into Excel. 2. Determine Passage Season by discussion with agency fish biologists. 3. Truncate daily flow records outside of Passage Season. 4. Sort remaining records by highest to lowest flow, keeping date associated with flow record. 5. Fishway Design Flow range is the stream flow range where all criteria should be achieved with design. 6. NMFS Design Flow Ranges: 95% - 5% exceedence flows (90% of Passage Season flows)

6 Example: Determining Fishway Design flows For example, if the truncated flow records contain 2000 records: The 5% exceedence flow (Q5) is the streamflow exceeded 5% of the days in the passage season. The 95% exceedence flow (Q95)is the streamflow exceeded 95% of the days in the passage season. Q5 is the 100 th highest flow record of the sorted data set. (0.05 x 2000 = 100) Q95 is the 1900 th highest (or 100 th lowest) flow record of the sorted data set. (0.95 x 2000 = 1900)

7 Suppose assessment using previously demonstrated method yields: Q5 = 11,250 cfs and Q95 = 210 cfs. Using a tailwater rating curve, the water surface elevations for an bypass outfall location can be determined. Example: Using 5% and 95% exceedence flow range in fish passage design

8 Some Dam Tailwater Rating Curve River Flow, in CFS Tailwater Water Surface Elevation, in Feet Q5 = 11,250 cfs WSE = 628.8 ft Q95 = 210 cfs WSE = 624.4 ft

9 More About Design Flow guidelines Fishway design flow should consider specific migration timing information for all species and life stages intended to pass. This will contract, expand or shift the design flow range. Providing optimal passage for 90% of the passage season does not mean that 10% of the run is not passed.

10 In some rivers, passage may be impaired by extreme flow events. Note: Flow in lower ladder is flowing UP the ladder Bonneville Dam – May31, 1948 985,000 CFS

11 More About Design Flow guidelines (continued) Passage of the entire run is expected to occur as streamflow conditions improve. Passage facilities can provide passage beyond the design flow range even if the facility is not within design criteria.

12 Data needs for determining screen and bypass flows Rating curves (flow vs.water surface elevation) and flow records for point of diversion, canal (if applicable), and bypass outfall. The greater the data range - the better, but often you will need to work with what you have. Maximum and minimum diverted flow. Canal cross sections, at least at the proposed screen site.

13 Hydraulic Calculations in Fishway Design

14 Hydraulics - Objective: Given: hydrology, biological criteria, and the design criteria -- 1.Determine: size and hydraulic capacity of key fishway components 2.Calculate: Flow rates for a)Weirs b)Orifices c)Open Channels

15 Properties of Water

16 Calculating Discharge (Q) LJ S NOAA Fisheries 4-28-03 16

17 Velocity Head When water moves from point A to point B, velocity head, is calculated by : (equation 1) h v = V 2 / 2g, where h v is velocity head differential from A and B V is water velocity between A and B g is the gravitational constant 32.2 feet per second squared.

18 Velocity Head Why does Velocity Head matter? Because if velocity is fast enough, the water surface will decrease downstream.

19 Example: velocity head as flow approaches a weir If water velocity at point B is 4 fps, and at point A is nearly zero, then velocity head at point B is calculated as: hv = V2 / 2g (equation 1) = (4 fps)2 /(2 x 32.2 ft/s2 ) = 16/64.4 (do units check?) = 0.25 feet A velocity head of 0.25 feet means that the water surface will drop 0.25 feet from A to B, assuming that velocity at A is nearly zero.

20 Weir Flow – Free Discharge

21 Q=C w L ( H + 2 ) 3/2 VoVo 2g Sharp crested weir Where: C w = Weir Coefficient (handbook) L = Weir Length H= Head across weir V o 2 /2g = Velocity Head Equation 2:

22 Submerged Weir Flow

23 Submerged Weirs

24 Weir Flow – Submerged Discharge Q submerged = Q/Q 1 x Q unsubmerged

25 Example: Submerged Weir Then Q submerged = 0.85 x Q unsubmerged, with Q unsubmerged from equation 2 If H 1 =1 ft and H 2 =.33 ft, then H 2 /H 1 =.33 If H 2 /H 1 =.33, then Q/Q 1 = 0.85 (chart)

26 Orifice Flow – Free Discharge

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28 Free Discharge Orifice DHDH

29 Orifice Flow – Contraction (C c ), Velocity (C v ) and Discharge (C d ) Coefficients Thin Wall OrificesShort TubeBell Mouth

30 (Circular) Orifice Flow – Free Discharge Q = AV = C d A o (2g D H) 0.5 Where : C d is orifice discharge coefficient A o is the area of the orifice ΔH is the water surface drop through the orifice to impact point

31 Orifice Flow – Submerged Discharge

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36 (Circular) Orifice Flow – Submerged Discharge C d = C c C v Q = AV = C d A o (2g D H) 0.5

37 Example: Priest Rapids Fishway Orifices

38 The entire fishway flow passes through two 18” x 24” orifices with a 0.75 foot difference in water surface elevation. The forebay velocity is 0.1 ft/s. Calculate the orifice flow rate.

39 Example: calculation of orifice flow First, calculate the velocity head (equation 2): h v = 0.1 2 / (2 x 32.2) =.00016 ft Using equation 4: Q = 0.61 x A x [2g(H+ h v )] ½ Q = 0.61 x 18/12 ft x 24/12 ft x [2 x 32.2 x (9/12 + 0.00016) ft] ½ = 0.61 x 1.5 x 2 x 6.95 = 12.7 cfs, Or, Q = 25.4 cfs for both orifices Note that the calculated velocity head is negligible (slow forebay velocity) Note that the coefficient of 0.61 is only for a rectangular orifice. For further guidance on various orifice coefficients for a variety of shapes, see “Water Measurement Manual”, U.S. Bureau of Reclamation, Denver, Colorado, 1981.

40 Open Channel Flow – Manning’s Equation n = Manning’s Roughness Coefficient (find using Google) R h = Hydraulic radius in feet = flow area (A) ÷ flow perimeter (p) S o = Slope of channel in feet/feet Note: Flow perimeter is where flow contacts the channel sides (not the water surface)

41 Open Channel Flow – Hydraulic Variables

42 Manning’s Equation – Rectangular Channel

43 Flow ---> 250 feet Elev. 1.0’ Elev.0.0’ Elevation View - Concrete channel Cross section 3 feet 1 foot Ao = 3 x 1 = 3 square feet So = (1.0 – 0.0)/250 = 0.004 ft/ft Rh = 1 +1 +3 = 5 feet V= 1.49 0.015 N = 0.015 (smooth concrete) x 5 2/3 x 0.004 1/2 = 18.4 feet per second Flow (Q) = V x A = 18.4 x 3 = 55 cubic feet per second

44 Modeling Tools Computational Fluid Dynamics (CFD) Models Scaled Physical Models

45 Hydraulic Modeling

46 Numerical Modeling

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48 Handy Conversions 1 cubic feet per second = 448.8 gallons per minute 1 gallon per minute = 1440 gallons per day 1 cubic meter per second = 35.31 cubic feet per second 1 cubic foot per second= 2 acre-feet per day 1 acre-foot per day = 0.504 cubic feet per second 1 cubic feet = 7.48 gallons 1 cubic foot of water = 62.4 pounds 1 gallon of water = 8.34 pounds 1 foot per second = 0.3048 meters per second (degrees F – 32) x 5/9 = degrees Celsius 1 kilogram = 2.2 pounds 1 foot per second = 1.097 kilometers per hour = 0.682 miles per hour = 16.4 miles per day Or e-mail me at Bryan.Nordlund@noaa.gov for a handy conversions freeware

49 The End


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