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Basic Hydraulics of Flow (Pipe flow, Trench flow, Detention time) PowerPoint Presentation
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Basic Hydraulics of Flow (Pipe flow, Trench flow, Detention time)

Basic Hydraulics of Flow (Pipe flow, Trench flow, Detention time)

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Basic Hydraulics of Flow (Pipe flow, Trench flow, Detention time)

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  1. Basic Hydraulics of Flow (Pipe flow, Trench flow, Detention time) Math for Water Technology MTH 082 Lecture 4 Hydraulics Chapter 7 (pgs. 311-319-341)

  2. RULES FOR FLOW RATES • DRAW AND LABEL DIAGRAM • CONVERT AREA or VELOCITY DIMENSIONS • 3 .SOLVE EACH FORMULA INDIVIDUALLY (Velocity and Area) • 4. ISOLATE THE FlOW PARAMETERS NECESSARY • Q= (Velocity) (Area formula first) • 5. USE YOUR UNITS TO GUIDE YOU • 6. SOLVE THE PROBLEM • 7.CARRY OUT FINAL FLOW RATE CONVERSIONS

  3. Types of flow rate? • Instantaneous flow rate- Flow rate at a particular moment in time. Use cross sectional area and velocity in a pipe or channel • Average flow rate -Average of instantaneous flow rates over time. Records of time and flow

  4. How do we measure flow rate? Water Meter

  5. How do we measure flow rate? Differential Pressure Metering Devices • Most common (~50%) units in use today. • Measure pressure drop across the meter which is proportional to the square of the flow rate.

  6. How do we measure flow rate? • Weirs

  7. How do we measure flow rate? • Parshall flumes for open channels

  8. How do we measure flow rate? • Orifice meters for closed conduit • An orifice is simply a flat piece of metal with a specific-sized hole bored in it. Most orifices are of the concentric type, but eccentric, conical (quadrant), and segmental designs are also available.

  9. How do we measure flow rate? • Venturi meters for closed conduit • Venturi tubes have the advantage of being able to handle large flow volumes at low pressure drops. A venturi tube is essentially a section of pipe with a tapered entrance and a straight throat.

  10. Factors that influence flow rate? • Fluid dynamics:typically involves calculation of various properties of the fluid, such as velocity, pressure, density, and temperature as functions of space and time. • Viscosityis commonly perceived as "thickness", or resistance to pouring. Viscosity describes a fluids internal resistance to flow and may be thought of as a measure of fluid friction. Water is "thin", having a lower viscosity, while vegetable oil is "thick" having a higher viscosity. Low viscosity =fast moving; high viscosity slow moving • DensityForces that arise due to fluids of different densities acting differently under gravity. • Friction of the liquid in contact with the pipe. Friction=slower motion

  11. How do we select a flow meter? • What is the fluid being measured (air, water,etc…)? • Do you require rate measurement and/or totalization from the flow meter? • If the liquid is not water, what viscosity is the liquid? • Is the fluid clean? • Do you require a local display on the flow meter or do you need an electronic signal output? • What is the minimum and maximum flowrate for the flow meter? • What is the minimum and maximum process pressure? • What is the minimum and maximum process temperature? • Is the fluid chemically compatible with the flowmeter wetted parts? • If this is a process application, what is the size of the pipe?

  12. How do we quantify flow rate? • Because the pipe's cross-sectional area is known and remains constant, the average velocity is an indication of the flow rate. Q = V x A where Q = liquid flow through the pipe/channel(length(ft3)/time) V = average velocity of the flow (length (ft)/time) A = cross-sectional area of the pipe/channel(lengthft2) Units must match!!! ft3/min, ft3/d, etc. **MAKE AREA or VELOCITY CONVERSIONS FIRST!** **MAKEFLOW RATE CONVERSIONS LAST!!!!******

  13. Open Channel Flow Rate (ft3/time) A= W X L =ft2 W=width (ft) V= ft/time L=depth (ft) V=velocity (ft/time) Q (flow rate) = V X A =ft3/time Flow = 7.48 gal or 3.06 X 10-6 acre feetor1 mgd 1ft3 1 gal 1,000,000 gal Time = 24 hrs or 1440 min or 86,400 sec 1 day1 day 1 day

  14. Circular pipe Flowing Full (ft3/time) A= 0.785 (diameter (ft))2 = ft2 D=diameter (ft) V= ft/time V=velocity (ft/time) Q (flow rate) = V X A =ft3/time When pipe is flowing full you can use the full cross sectional area (0.785)

  15. Pipe Flowing Full An 8 in transmission main has a flow of 2.4 fps. What is the gpm flow rate through the pipe? 1. Label Figure! D= 8 in=0.67 ft V= 2.4 fps Solve for Q! 2. Is it flowing full? YES.. I can use 0.785 in equat. 3. Formula: Q= A * V 4. Substitute and Solve Q= (0.785)(0.67 ft)(0.67 ft)(2.4 fps) Q =0.85 cfs 5. Convert: (0.85 ft3/sec)(60 sec/ 1 min)(7.48 gal/ft3) = 380 gpm

  16. Pipe Not Flowing Full An 8 in transmission main has a flow of 3.4 fps. What is the gpm flow rate through the pipe if the water is flowing at a depth of 5 inches? 1. Label Figure! D= 8 in=0.67 ft H2O depth= 5 in=0.41 ft V= 3.4 fps Solve for Q! 2. Is it flowing full? NO.. I need d/D ratio 3. d/D= 5”/8”=0.63 (ratio) =_0.5212_ from table 4. Formula: Q= A * V 5. Substitute and Solve Q= (0.5212)(0.67 ft)(0.67 ft)(3.4 fps) Q =0.8 cfs 6. Convert: (0.8 ft3/sec)(60 sec/ 1 min)(7.48 gal/ft3) = 359 gpm

  17. Example 1. Circular pipe Flowing Full (ft3/time)A 15 in diameter pipe is flowing full.What is the gallons per minute flow ratein the pipe if the velocity is 110 ft/min. Area (pipe)= 0.785 (diameter)2 A= 0.785 (1.25 ft)2 =1.23 ft2 V= 110 ft/min D=diameter (15 inches) Convert! (15in)(1ft/12in) D=1.25 ft Q= ?gpm V=110 (ft/min) Q (flow rate) = V X A 110 ft/min X 1.23 ft2 = 134.92ft3/min Q (flow rate) = 134.92ft3/min (7.48 gal/ft3)= 1,009 gpm

  18. D=diameter (1.5ft) D= 18 inches 8 hrs V = 2 ft/sec A full 18” raw sewage line has broken and has been leaking raw sewage into Arcade creek for 8 hours. What is the gpm flow rate through a pipe-- assume a velocity of 2 ft/sec? DRAW: • Given: • Formula: • Solve: Diameter= 18 in,1.5 ft; depth= 18”or 1.5ft,1 ft, V=2 ft/sec= Q? Q = V X A Area (pipe)= 0.785 (diameter)2 Area (pipe)= 0.785 (diameter)2 A= 0.785 (1.5 ft)2 =1.76 ft2 Q= V X A Q= 2 ft/sec X 1.76 ft2 = 3.56 ft3/sec 3.56 ft37.48 gal60sec = 1585 gpm sec 1ft3 1min • 1585 gpm • 507 gpm • 1057 gpm • 202929 gpm

  19. D=diameter (1.5ft) D= 18 inches V = 2 ft/sec 8 hrs How many gallons of raw sewage was released to Arcade Creek after 8 hrs? • 95100 gallons • 760,800 gallons • 1057 gpm • 202929 gpm 1585 gallons X 60 min X 8 hrs = 760,800 gal min 1 hr

  20. Example 2. Channel Flowing Full (ft3/time)What is the MGD flow ratethrough a channel that is 3ft wide with water flowing to a depth of 16 in.at a velocity of 2 ft/sec? Area (rect)= L X W A= (1.33 ft) 3 ft =3.99 ft2 V= 2 ft/sec Q= ?MGD W= 3ft L= (16 inches) Convert! (16in)(1ft/12in) D=1.33 ft V= 2 ft/sec Depth (L) =16 in Q (flow rate) = V X A 2 ft/sec X 3.99 ft2 = 7.98ft3/sec Q (flow rate)=(7.98ft3/sec) (60sec/1min) (7.48 gal/ft3) (1,440 min/day)= 5,157,251 gpd 5,157,251 gpd = 5.16 MGD

  21. Example 4. Water depth in channel (ft)A channel is 3 ft wide.If the flow in the channel is 7.5 MGDand the velocity of the flow is 185 ft/min, what is the depth (in feet) of water in the channel? Area (rect)= L X W A= (L=(?ft)) (3ft) W= 3 ft Depth (L)=? ft V= 185 ft/min V= 185 ft/min Q= 7.5 MGD =7,500,000 gpd Q=7,500,000 gpd(7.48ft3/1gal/)(1day/1440min) Q=696.3 ft3/min Q= V X A where A= L X W Q (flow rate) = V X (L X W) = L= Q÷V(W) L=696.3 ft3/min÷185 ft/min(3 ft) L= 1.25 ft

  22. Example 5. Velocity =rate (length/time)A float is placed in a channel. It takes 2.5 min to travel 300 ft.What is the flow velocity in feet per minute in the channel? V= rate (length/time) = 300 ft 2.5 min V= 120 ft/min V= 300 ft 2.5min 300 ft 2.5min

  23. Example 7. Velocity in a pipe flowing full (length/time)A 305 mm diameter pipe flowing full is carrying 35 L/sec. What is the velocity of the water (m/sec) through the pipe? Area (pipe)= 0.785 (.305m)2 A= 0.785 (.305 m)2 =.0703 m2 Q= 35 L/sec= 35L/sec (1m3/1000L) Q=.035 m3/sec D=diameter (305 mm) Convert! (305mm)(1m/1000mm) D=.305 m Q=30 L/sec V= ?(m/sec) Velocity??? Q= V X A where V= Q/A V=Q÷A V=.035 m3/sec÷(.0703 m2) V= .49 m/sec

  24. W= 4 ft V= 500ft/3min V=166.6 ft/min D=2.3 ft Example 8. Flow rate in channel flowing full (ft3/time)A channel is 4 ft wide with water flowing to a depth of 2.3 ft.If a float placed in the water takes 3 min to travel a distance of 500 ft, what is the ft3/min flow rate in the channel? Area (rect)= L X W A= (4 ft) (2.3ft) =9.2 ft2 V= 500 ft/3min=166.6 ft/min Q=? ft3/min Q (flow rate) = V X A 166.6 ft/min X 9.2 ft2 = 1533ft3/min

  25. What is the gpm flow rate through a pipe that is 24 inch wide with water flowing to a depth of 12 in. at a velocity of 4 ft/sec? DRAW: • Given: • Formula: • Solve: Diameter= 24 in,2 ft; depth= 12”,1 ft, V=4 ft/sec= Q? Q = V X A d/D= 12/24=0.5=table 0.3927 Area (pipe)= 0.3927 (diameter)2 Area (pipe)= 0.3927(diameter)2 A= 0.3927 (2 ft)2 =1.6 ft2 Q= V X A Q= 4 ft/sec X 1.6 ft2 = 6.35 ft3/sec 6.35 ft37.48 gal60sec = 2851 gpm sec 1ft3 1min D=diameter (2 ft) D= 24 inches • 12.56 gpm • 5637 gpm • 452 gpm • 2851gpm depth=(1 ft) d= 12 inches

  26. Detention Time Math for Water Technology MTH 082 Lecture 4 Mathematics Ch 22 (pgs. 193-196) “how long a drop of water or suspended particle remains in a tank or chamber”

  27. What is detention time? Detention time (DT) = volume of tank = MG flow rate MGD

  28. The time it takes for a unit volume of water to pass entirely through a sedimentation basin is called • Detention time • Hydraulic loading rate • Overflow time • Weir loading rate

  29. What is the average detention time in a water tank given the following: diameter = 30' depth = 15' flow = 700 gpm • 1hr. 34min. • 1hr. 53min. • 1hr. 47min. • 2 hrs. 3 min. Volume = 0.785 (30 ft)(30ft)(15 ft) = 10597 ft3 10597 ft3 (7.48 gal/1ft3) = 79269 gal DT= volume/flow = 79269 gal/700 gpm = 113 minutes 113 minutes - 60 minutes or 1 hr = 53 minutes or 1 hour and 53 minutes

  30. What is the average detention time in a water tank given the following: diameter = 80' depth = 12.2' flow = 5 MGD • 2.2 hrs. • 1.68 hrs. • 2.4 hrs. • 1.74 hrs. V= 0.785 (Diameter)(Diameter)(depth) Volume = 0.785 (80 ft)(80ft)(12.2 ft) = 61292 ft3 61292 ft3 (7.48 gal/1ft3) = 458470 gal (1MG/1,000,000 gal) = 0.46 MG DT= volume/flow = 0.46 MG /5 MGD = .09 days (24 hr/1d) = 2.2 hrs

  31. 10 MG 2.8 MGD A 10 MG reservoir has a peak of 2.8 MGD. What is the detention time in the tank in hours? DRAW: • Given: • Formula: • Solve: tank= 10 MG, Flow rate 2.8 mgd DT= volume of tank/flow rate DT=VT/FR Time = 10 MG/2.8MGD Time= 3.5 day 3.5 day (24 h/1day)=85.7 hrs • 3.5 hrs • 85.7 hrs • 0.28 hrs • 4 hrs

  32. 36 in=3 ft 1.8 miles (5280 ft/1mile)= 9504 ft A 36 in transmission main is used for chlorine contact time. If the peak hourly flow is 6 MGD, and the main is 1.8 miles long, what is the contact time in minutes? DRAW: • Given: • Formula: • Solve: tank= 10 MG, Flow rate 2.8 mgd Volume = π*r2*h DT= volume of tank/flow rate DT=VT/FR Volume = π*r2*h V= π *(1.5ft)2*(9504 ft) V=6718 ft3 Convert to gallons V=6718 ft3(1 gal/7.48 ft3) V=502,505 gal or .502 MG DT=VT/FR DT= .502 MG/6 MGD Detention Time =0.83 day 0.83 day (24 h/1day)(60 min/1 hr)= 120.6 min • 0.83 min • .52 min • 121 min • 250 min

  33. What did you learn? • How is flow measured? • What equation is used to determine flow rate? • What are the units for flow rate, velocity? • What is detention time?

  34. Today’s objective: to become proficient with the concept of basic hydraulic calculations used in the waterworks industry applications of the fundamental flow equation, Q = A X V, and hydraulic detention time has been met. • Strongly Agree • Agree • Disagree • Strongly Disagree