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

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 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 (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
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 X 10-6 acre feet or 1 mgd
1ft gal ,000,000 gal
Time = 24 hrs or min or 86,400 sec
1 day day day

14. Circular pipe Flowing Full (ft3/time)
A= (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 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 2. Is it flowing full? NO.. I need d/D ratio
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. Area (pipe)= 0.785 (diameter)2 A= 0.785 (1.25 ft)2 =1.23 ft2
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)= (diameter)2
A= (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 = ft3/min
Q (flow rate) = ft3/min (7.48 gal/ft3)= 1,009 gpm

18. DRAW: Given: Formula: Solve:
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)= (diameter)2
A= (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 ft gal sec = gpm
sec ft min
D=diameter (1.5ft)
D= 18 inches
V = 2 ft/sec
8 hrs
1585 gpm
507 gpm
1057 gpm
gpm

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

20. Area (rect)= L X W A= (1.33 ft) 3 ft =3.99 ft2 V= 2 ft/sec Q= ?MGD
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)
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. Area (rect)= L X W A= (L=(?ft)) (3ft) V= 185 ft/min
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
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= ft3/min÷185 ft/min (3 ft)
L= 1.25 ft

22. V= rate (length/time) = 300 ft 2.5 min V= 120 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

23. Area (pipe)= 0.785 (.305m)2 A= 0.785 (.305 m)2 =.0703 m2 V= ?(m/sec)
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)= (.305m)2
A= (.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. Area (rect)= L X W A= (4 ft) (2.3ft) =9.2 ft2
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
V= 500ft/3min
V=166.6 ft/min
D=2.3 ft
W= 4 ft
Q=? ft3/min
Q (flow rate) = V X A
166.6 ft/min X 9.2 ft2 = 1533ft3/min

25. 12.56 gpm 5637 gpm 452 gpm 2851gpm DRAW: Given: Formula: Solve:
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
Area (pipe)= (diameter)2
Area (pipe)= (diameter)2
A= (2 ft)2 =1.6 ft2
Q= V X A
Q= 4 ft/sec X 1.6 ft2 = 6.35 ft3/sec
6.35 ft gal sec = gpm
sec ft min
D=diameter (2 ft)
D= 24 inches
depth=(1 ft)
d= 12 inches
12.56 gpm
5637 gpm
452 gpm
2851gpm

26. Math for Water Technology
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 )

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
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
Volume = (30 ft)(30ft)(15 ft) = ft3
10597 ft3 (7.48 gal/1ft3) = gal
DT= volume/flow = gal/700 gpm = 113 minutes
113 minutes minutes or 1 hr = 53 minutes or 1 hour and 53 minutes
1hr. 34min.
1hr. 53min.
1hr. 47min.
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= (Diameter)(Diameter)(depth)
Volume = (80 ft)(80ft)(12.2 ft) = ft3
61292 ft3 (7.48 gal/1ft3) = gal (1MG/1,000,000 gal) = 0.46 MG
DT= volume/flow = MG /5 MGD = .09 days (24 hr/1d) = 2.2 hrs
2.2 hrs.
1.68 hrs.
2.4 hrs.
1.74 hrs.

31. A 10 MG reservoir has a peak of 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
10 MG
2.8 MGD
3.5 hrs
85.7 hrs
0.28 hrs
4 hrs

32. 0.83 min .52 min 121 min 250 min DRAW: Given: Formula: Solve:
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
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= .502 MG/6 MGD
Detention Time =0.83 day
0.83 day (24 h/1day)(60 min/1 hr)= min
1.8 miles (5280 ft/1mile)= 9504 ft
36 in=3 ft
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. Strongly Agree Agree Disagree Strongly Disagree
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