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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) – PowerPoint PPT presentation

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


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.
  • Viscosity is 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
  • Density Forces that arise due to fluids of
    different densities acting differently under
    gravity.
  • Friction of the liquid in contact with the pipe.
    Frictionslower motion

11
How do we select a flow meter?
  1. What is the fluid being measured (air,
    water,etc)?
  2. Do you require rate measurement and/or
    totalization from the flow meter?
  3. If the liquid is not water, what viscosity is the
    liquid?
  4. Is the fluid clean?
  5. Do you require a local display on the flow meter
    or do you need an electronic signal output?
  6. What is the minimum and maximum flowrate for the
    flow meter?
  7. What is the minimum and maximum process pressure?
  8. What is the minimum and maximum process
    temperature?
  9. Is the fluid chemically compatible with the
    flowmeter wetted parts?
  10. 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 (length
ft2)
Units must match!!! ft3/min, ft3/d, etc.
MAKE AREA or VELOCITY CONVERSIONS
FIRST! MAKE FLOW RATE CONVERSIONS
LAST!!!!
13
Open Channel Flow Rate (ft3/time)
A W X L ft2
Wwidth (ft)
V ft/time
Ldepth (ft)
Vvelocity (ft/time)
Q (flow rate) V X A ft3/time
Flow 7.48 gal or 3.06 X 10-6 acre feet or 1
mgd 1ft3 1
gal 1,000,000 gal Time
24 hrs or 1440 min or 86,400 sec
1 day 1 day
1 day
14
Circular pipe Flowing Full (ft3/time)
A 0.785 (diameter (ft))2 ft2
Ddiameter (ft)
V ft/time
Vvelocity (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 in0.67 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?
D 8 in0.67 ft
1. Label Figure!
H2O depth 5 in0.41 ft
V 3.4 fps
Solve for Q!
2. Is it flowing full? NO.. I need d/D ratio
3. d/D 5/80.63 (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 rate in 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
Ddiameter (15 inches) Convert!
(15in)(1ft/12in) D1.25 ft
Q ?gpm
V110 (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
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 18or 1.5ft,1 ft,
V2 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 ft3 7.48 gal
60sec 1585 gpm sec 1ft3
1min
  1. 1585 gpm
  2. 507 gpm
  3. 1057 gpm
  4. 202929 gpm

19
How many gallons of raw sewage was released to
Arcade Creek after 8 hrs?
1585 gallons X 60 min X 8 hrs 760,800 gal
min 1 hr
  1. 95100 gallons
  2. 760,800 gallons
  3. 1057 gpm
  4. 202929 gpm

20
Example 2. Channel Flowing Full (ft3/time) What
is the MGD flow rate through 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) D1.33 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
MGD and 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 Q7,500,000
gpd(7.48ft3/1gal/)(1day/1440min) Q696.3 ft3/min
Q V X A where A L X W Q (flow rate) V X (L
X W) L QV(W) L 696.3 ft3/min185
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
Ddiameter (305 mm) Convert! (305mm)(1m/1000mm) D
.305 m
Q30 L/sec
V ?(m/sec)
Velocity???
Q V X A where V Q/A V QA V .035
m3/sec(.0703 m2) V .49 m/sec
24
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/3min166.6 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, V4
ft/sec Q? Q V X A d/D 12/240.5table
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 ft3 7.48 gal 60sec
2851 gpm sec 1ft3
1min
Ddiameter (2 ft) D 24 inches
depth(1 ft) d 12 inches
  1. 12.56 gpm
  2. 5637 gpm
  3. 452 gpm
  4. 2851gpm

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

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

Tank Detention Time
Flash mixing basin 30-60 sec
Flocculation basin 20-60 min
Sedimentation basin 1-12 h
28
The time it takes for a unit volume of water to
pass entirely through a sedimentation basin is
called
  1. Detention time
  2. Hydraulic loading rate
  3. Overflow time
  4. Weir loading rate

29
What is the average detention time in a water
tank given the following diameter 30' depth
15' flow 700 gpm
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
  1. 1hr. 34min.
  2. 1hr. 53min.
  3. 1hr. 47min.
  4. 2 hrs. 3 min.

30
What is the average detention time in a water
tank given the following diameter 80' depth
12.2' flow 5 MGD
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
  1. 2.2 hrs.
  2. 1.68 hrs.
  3. 2.4 hrs.
  4. 1.74 hrs.

31
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 DTVT/FR Time 10
MG/2.8MGD Time 3.5 day 3.5 day (24 h/1day)85.7
hrs
  1. 3.5 hrs
  2. 85.7 hrs
  3. 0.28 hrs
  4. 4 hrs

32
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
pr2h DT volume of tank/flow rate DTVT/FR Volum
e pr2h V p (1.5ft)2(9504 ft) V6718
ft3 Convert to gallons V6718 ft3(1 gal/7.48
ft3) V502,505 gal or .502 MG DTVT/FR DT .502
MG/6 MGD Detention Time 0.83 day 0.83 day (24
h/1day)(60 min/1 hr) 120.6 min
  1. 0.83 min
  2. .52 min
  3. 121 min
  4. 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
Todays 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.
  1. Strongly Agree
  2. Agree
  3. Disagree
  4. Strongly Disagree
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