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

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**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)**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**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**How do we measure flow rate?**Water Meter**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.**How do we measure flow rate?**• Weirs**How do we measure flow rate?**• Parshall flumes for open channels**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.**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.**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**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?**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!!!!********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**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)**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**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**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**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**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**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**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**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**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**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**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**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”**What is detention time?**Detention time (DT) = volume of tank = MG flow rate MGD**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**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**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**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**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**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?**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