Transcript Lecture 4 Hydraulics and detention time

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
1. DRAW AND LABEL DIAGRAM
2. 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.
• 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.
Friction=slower motion
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?
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=VxA
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!!!!******
Open Channel Flow Rate
(ft3/time)
W=width (ft)
A= W X L =ft2
V= ft/time
L=depth (ft)
V=velocity (ft/time)
Q (flow rate) = V X A =ft3/time
Flow = 7.48 gal
1ft3
or
Time = 24 hrs
or
1 day
3.06 X 10-6 acre feet or 1 mgd
1 gal
1,000,000 gal
1440 min or 86,400 sec
1 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
V= 3.4 fps H2O depth= 5 in=0.41 ft
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 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
Q= ?gpm
D=diameter (15 inches)
Convert! (15in)(1ft/12in)
D=1.25 ft
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
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:
Diameter= 18 in,1.5 ft; depth= 18”or 1.5ft,1 ft, V=2 ft/sec= Q?
D=diameter (1.5ft)
D= 18 inches
• Given:
=VXA
• Formula:Q
Area (pipe)= 0.785 (diameter)2
• Solve: Area (pipe)= 0.785 (diameter)2
A= 0.785 (1.5 ft)2 =1.76 ft2
8 hrs
V = 2 ft/sec
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
60%
30%
10%
gp
m
m
20
29
29
gp
57
10
50
7
gp
gp
m
m
0%
85
1585 gpm
507 gpm
1057 gpm
202929 gpm
15
1.
2.
3.
4.
How many gallons of raw sewage was released to Arcade Creek
after 8 hrs?
100%
D=diameter (1.5ft)
D= 18 inches
gp
m
m
29
29
20
57
10
ga
l
80
0
0%
76
0,
gp
lo
n
ns
llo
ga
0
10
95
0%
...
0%
V = 2 ft/sec
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
8 hrs
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
V= 2 ft/sec
W= 3ft
L= (16 inches)
Convert! (16in)(1ft/12in)
D=1.33 ft
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 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)
V= 185 ft/min
W= 3 ft
Depth (L)=? ft
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
V= 185 ft/min
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
Q=30 L/sec
V= ?(m/sec)
Q= V X A where
D=diameter (305 mm)
Convert!
(305mm)(1m/1000mm)
D=.305 m
Velocity???
V= Q/A
V= Q÷A
V= .035 m3/sec÷(.0703 m2)
V= .49 m/sec
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
W= 4 ft
D=2.3 ft
V= 500ft/3min
V=166.6 ft/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:
Diameter= 24 in,2 ft; depth= 12”,1 ft, V=4 ft/sec= Q?
• Given: Q = V X A
• Formula: d/D= 12/24=0.5=table 0.3927 2
Area (pipe)= 0.3927 (diameter)
• Solve:
2
D=diameter (2 ft)
D= 24 inches
depth=(1 ft)
d= 12 inches
Area (pipe)= 0.3927(diameter)
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
20%
m
51
gp
28
gp
m
45
2
gp
37
56
6
gp
m
m
0%
.5
12.56 gpm
5637 gpm
452 gpm
2851gpm
30%
12
1.
2.
3.
4.
50%
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)
What is detention time?
Detention time (DT) = volume of tank =
flow rate
Tank
Detention
Time
30-60 sec
Flash mixing
basin
Flocculation
20-60 min
basin
Sedimentation 1-12 h
basin
MG
MGD
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
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25% 25% 25% 25%
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
m
in
.
2
hr
s.
3
7m
in
.
1h
r.
4
3m
in
.
r.
5
1h
r.
3
1hr. 34min.
1hr. 53min.
1hr. 47min.
2 hrs. 3 min.
1h
1.
2.
3.
4.
4m
in
.
25% 25% 25% 25%
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
hr
s.
25%
1.
74
hr
s.
4
2.
1.
68
hr
s.
2
2.2 hrs.
1.68 hrs.
2.4 hrs.
1.74 hrs.
2.
1.
2.
3.
4.
hr
s.
DT= volume/flow = 0.46 MG /5 MGD = .09 days (24 hr/1d)25%= 2.2
hrs
25% 25%
A 10 MG reservoir has a peak of 2.8 MGD.
What is the detention time in the tank in
DRAW:
hours?
10 MG
DT= volume of tank/flow rate
DT=VT/FR
Time = 10 MG/2.8MGD
Time= 3.5 day
2.8 MGD
hr
s
25%
4
28
.7
hr
s
25%
hr
s
25%
hr
s
5
3.5 hrs
85.7 hrs
0.28 hrs
4 hrs
3.
1.
2.
3.
4.
25%
0.
3.5 day (24 h/1day)=85.7 hrs
85
• Given:
• Formula:
• Solve:
tank= 10 MG, Flow rate 2.8 mgd
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:
36 in=3 ft
1.8 miles (5280 ft/1mile)= 9504 ft
38%
25%
0
25
1
m
in
m
in
12
83
m
in
0%
m
in
0.83 min
DT=VT/FR
.52 min DT= .502 MG/6 MGD
Detention Time =0.83 day
121 min 0.83 day (24 h/1day)(60 min/1 hr)= 120.6 min
250 min
38%
2
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
0.
1.
2.
3.
4.
tank= 10 MG, Flow rate 2.8 mgd
.5
• Given:
• Formula:
• Solve:
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.
77%
Strongly Agree
Agree
Disagree
Strongly Disagree
15%
8%
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