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2012 PE Review: 
IV-A: Hydrology and Hydraulics 
Michael C. Hirschi, PhD, PE, D.WRE 
Senior Engineer 
Waterborne Environmental, Inc. 
hirschim@waterborne-env.com 
also Professor Emeritus 
University of Illinois
Acknowledgements: 
Daniel Yoder, I-A, PE Review 2006 
Rafael (Rafa) Muñoz-Carpena, I-A, PE Review 2007-09 
Rod Huffman, PE Review coordinator
Session Topics 
• Hydrology 
• Hydraulics of Structures 
• Open Channel Flow
Hydrology 
• Hydrologic Cycle 
• Precipitation 
– Average over Area 
– Return Period 
• Abstractions from Rainfall 
• Runoff 
– Hydrographs 
– Determination methods
Hydraulics of Structures 
• Weir flow 
• Orifice flow 
• Pipe flow 
• Spillway flow 
– Stage-Discharge relationship
Open Channel Flow 
• Channel geometries 
– Triangular 
– Trapezoid 
– Parabolic 
• Manning’s equation 
– Manning roughness, “n” 
• Grass waterway design
A few comments 
• Material outlined is about 3 weeks or more in a 
3-semester hour class. I’m compressing at least 
6 hours of lecture and 3 laboratories into 2 
hours, so I will: 
– Review highlights and critical points 
– Do example problems 
• You need to: 
– Review and tab references 
– Do additional example problems, or at least 
thoroughly review examples in references
Hydrologic Cycle 
From Fangmeier et al. (2006)
Precipitation 
• Input to the Rainfall-Runoff process 
• Forms include: 
– Rainfall 
– Snow 
– Hail 
– Sleet 
• Measured directly 
• Varies temporally and areally
Rainfall Data 
• Daily 
• Hourly 
• 15-minute 
• Continuous 
• Reported as depth, which is really volume 
over a given area, over a period of time
Average Rainfall 
• Simple arithmetic average 
• Theissen Polygon
Example 1 
How do different calculation methods of rainfall average compare? 
Consider:
Raingage data 
• Gages (clockwise from upper left): 1.9”, 
2.1”, 1.8”, 1.9”, 2.1”, 2.2” 
Arithmetic average: 2.0”
Theissen Polygons 
• Areas closest to each raingage 
determined by perpendicular bisectors of 
each line between raingages. 
• Areas for each raingage, again clockwise 
from upper left: 65ac, 150ac, 55ac, 140ac, 
215 ac, 270ac 
• Figure is repeated with Theissen polygon 
construction added.
Why bisectors? 
• When perpendicular bisectors are 
constructed, they are, by definition, lines 
that are equidistant from the points at the 
ends of the lines they bisect. 
• So, the combination of the constructions 
delineate areas that are closest to a given 
point (raingage in this case)
Is the watershed average 
rainfall using the Theissen 
Polygon method most nearly: 
A. 2.0” 
B. 2.1” 
C. 2.2” 
D. 1.9”
Theissen calculation 
• Uses areal weighted average, so the sum of the 
products of area x depth divided by total area 
• Hint: If you measure the areas yourself, the 
denominator should be the sum of the areas, not 
the known watershed area 
• So, average Theissen rain: Answer B, 2.1” 
(65*1.9+150*2.1+55*1.8+140*1.9+215*2.1+270*2. 
2)/(65+150+55+140+215+270)=2.07”, which is 
best represented as 2.1” given most data is 2 
significant digits.
Any questions on Theissen Polygons?
Return Period (two descriptions) 
• A 10 year-24 hour rainfall volume is that 
depth of rainfall over a 24 hour period that 
is met or exceeded, on the long-term 
average, once every 10 years. 
• Another way to describe it is the 24 hour 
rainfall depth that has a 1 in 10 (10%) 
chance to be met or exceeded each year, 
on the long term average.
US 100yr-24hr Rainfall 
100yr-24hr data from TP-40 (Hershfield (1961) 
as referenced by Fangmeier et al. (2006)
Return Period Data 
• Constructed from historical rainfall data 
• Available in tabular form via website or 
state USDA-NRCS reports. 
• Available as national maps (similar to 
previous slide) in several references such 
as Haan, Barfield & Hayes (1994).
Example 
A reservoir is to be designed to contain the 
runoff from a 10yr-24hr rainfall event in 
Northeastern Illinois. What rainfall 
volume is to be considered? 
A. 4.5” 
B. 3.9” 
C. 4.1” 
D. Cannot estimate from available maps
10yr-24hr map from Haan, Barfield & Hayes (1994)
Example 
• Answer is C. From map, 10yr-24hr rainfall 
in NE Illinois is just over 4”, use 4.1” to be 
conservative.
Questions on precipitation?
Abstractions from Rainfall 
• Abstractions from rainfall are “losses” from 
rainfall that do not show up as storm water 
runoff: 
– Interception 
– Evapotranspiration 
– Storage 
• In bank 
• On surface 
– Infiltration
Runoff by other names… 
• “Effective” rainfall 
• Rainfall “excess”
Runoff 
If rainfall rate exceeds the soil infiltration 
capacity, ponding begins, and any soil 
surface roughness creates storage on the 
surface. After at least some of those 
depressions are filled with water, runoff 
begins. Additional rain continues to fill 
depressional storage and runoff rate 
increases as more of the hill slope and 
subsequently the watershed contributes 
runoff.
Rainfall/Runoff process
Time of Concentration, tc 
The time from the beginning of runoff to the 
time at which the entire watershed is 
contributing runoff that reaches the 
watershed outlet is called the Time of 
Concentration. It is also described as the 
“travel time from the hydraulically most 
remote point in a watershed to the outlet”.
Curve Number method
CN Method, continued
Time of Concentration, tc
CN values
Runoff Volume determination
Runoff Example 
In a previous problem, a design rain event in NE 
Illinois was determined to be 4.1”. Assuming the 
watershed in question was a completed 300 ac 
residential area with an average lot size of ½ ac, 
all on Hydrologic Group C soils, what is the 
needed pond volume, if all runoff is to be 
retained? 
A: 2.5 runoff-inches 
B: 53 acre-inches 
C: 630 acre-ft 
D: 53 acre-ft
Runoff Example, continued
Runoff Volume determination
Answer to Runoff Example 
The answer is D, 53 acre-ft. From the table, 
the CN for Hyd group C soil with ½-ac lot 
is 80. Using the graph with a 4.1” rainfall, 
runoff depth is 2.1”. Volume is then 
300ac*2.1in = 630 ac-in, divided by 12 is 
53 ac-ft.
Additional example 
You discover that the subdivision is actually 100 
acres of ½ ac lots on C soils, 100 acres of ½ ac 
lots on D soils, 50 acres of ¼ ac lots on B soils 
and 50 acres of townhouses on A soils. What 
CN value would you use? 
A: 79 
B: 85 
C: 80 
D: 75
Add’l Runoff Example, cont.
Answer 
The correct answer is C, 80. Use an area-weighted 
average, similar to Theissen 
method. The respective CN values for ½ 
ac on C, ½ ac on D, ¼ ac on B and 
townhouses on A are 80, 85, 75 & 77. 
The area-weighted CN is then 
(80*100+85*100+75*50+77*50)/300 = 
80.33, which is more appropriately 80.
Peak Discharge 
The CN method also provides for Peak 
Discharge estimation, using graphs or 
tables. Required information includes 
average watershed slope, watershed flow 
path length, CN, and rainfall depth. The 
graphical method from the EFM is:
Peak Runoff Discharge
Peak Discharge Example 
Same residential watershed that produced 
2.1” of runoff from a 4.1” rainfall. Flow 
length is 2500’, slope is 2%. CN is 80, so 
S is 2.5”. Ia = 0.2*S = 0.5”. Ia/P = 
0.5/4.1=0.122. 
Tc = 2500^0.8*(1000/80-9)^0.7/1140/2^0.5 
=0.8hr
Peak Runoff Discharge
Example solution 
From graph, with Ia/P of 0.122 and Tc of 
0.8hr, unit peak discharge is 0.57 cfs/ac/in 
or qp = 0.57*300*2.1 = 360 cfs
Rational Method 
The Rational Equation is: 
Qp = CiA 
where: 
C is a coefficient 
i is rainfall intensity of duration tc 
A is area in acres 
C is approximately 0.4, A is 300ac, i is 2” in 30min, so 4iph, 
peak rate is then 0.4*300*4 = 480 cfs
Questions on runoff?
Hydraulics of Structures 
Flow through structures is important given 
that such structures control the rate of 
flow. Sizing of such structures is then 
important to allow flow to pass while 
protecting downstream areas from the 
effects of too high a flow rate. Structures 
may also be used for measurement of 
water flow. Each type of structure will 
produce different types of flow depending 
upon size and flow rate passing through it.
Weirs 
• Sharp-crested 
• Broad-crested
Weir Equation 
(from EFH-Ch03 Hydraulics)
Sharp-Crested Weir 
(from EFH-Ch03 Hydraulics)
More complex weirs 
(from Haan et al., 1994)
Example 
• You are measuring flow using a 90° V-notch 
weir. H is measured as 0.53’ at 
2.5’ upstream of the weir. What is the 
flow rate? 
A. 230 gpm 
B. 0.51 cfs 
C. 0.51 gpm 
D. A & B
Answer 
• The answer is D. The equation from Haan 
et al (1994) is:
Answer, continued 
• Q = 2.5*H^2.5, where Q is in cfs and H is 
in feet 
• Q=2.5*(0.53)^2.5=0.511 cfs or 0.51 cfs 
• Q=0.51 cfs*60sec/min*7.48gal/cf=230 
gpm 
• Note: Both answers contain 2 significant 
figures
Orifice Flow 
• Submerged vs Free Outlet 
• Shapes affecting C
Submerged Orifice
Free Discharge Orifice
Orifice Coefficients
Example 
• Markers Mark distillery just moved a 3’ diameter 
barrel of their bourbon over their charcoal filter 
bed to drain the bourbon into the system to be 
bottled. The bung plug is removed 
instantaneously, allowing barrel strength 
bourbon to flow freely from the 2” diameter 
bung, which can be considered a sharp-edged 
orifice. What is the initial flow rate (assuming 
same specific gravity as water, which is an 
incorrect assumption)?
Answers 
A: 0.5 cfs 
B: 83 gpm 
C: 26.6 gpm 
D: 200 L/hr
Solution 
Q=0.61*A*(2*g)^0.5*h^0.5 
=0.61*(π*1”^2)*(2*32.2f/s/s)^0.5*3’^0.5 
=0.61*3.1415/144*(64.4)^0.5*3^0.5 
=0.185 cfs 
Q=83 gpm (answer B)
Pipe flow 
When considering pipe flow in a structure, 
Bernoulli’s equation is used: 
Frictional losses are multiples of the velocity head (V2/2g) 
and are additive.
Head loss under pipe flow 
• Entrance loss (Ke) 
• Bend loss (Kb) 
• Pipe friction loss (Kc) 
• Each coefficient is documented in references 
Considering the Bernoulli equation for a spillway, 
the pressure at entrance and exit is atmospheric, 
the elevation difference is the water surface elevation 
difference between upstream and downstream, 
and the remaining term is the velocity head plus losses
Consider the following
Pipe flow
Spillway considerations 
A given spillway may have several 
discharge relationships (weir, orifice, pipe) 
depending upon the head (stage). The 
stage discharge curve then becomes a 
combination curve, with the type of 
relationship allowing the highest flow at a 
given head in control. 
Consider a drop inlet control structure:
Stage-Discharge Curve
Example 
An 18” CMP with an 18” vertical riser is used 
as the principal spillway for a pond. The 
pipe is 50’ long with one 90° bend. The 
top of the inlet is 10’ above the bottom of 
the outlet. Develop the stage-discharge 
relationship assuming a free outfall.
Weir flow 
Basic equation: 
Given 18” riser, length of weir is 2πr, or 4.7’, so
Orifice flow 
Basic orifice equation: 
Given 18” riser and assuming C’ of 0.6,
Pipe flow 
Basic pipe flow equation: 
After looking up each parameter:
Stage-Discharge Relationship
Questions on spillway hydraulics?
Open Channel Flow 
Flow through open channels is another 
important area to consider and review. 
Velocity and flow rate are usually 
calculated using Manning’s equation, 
which considers flow geometry, channel 
roughness and slope.
Manning’s Equation 
Where: 
V= flow velocity in fps 
Rh = Hydraulic Radius in ft 
S = Energy gradeline slope in ft/ft (=bed slope for normal flow) 
n = Manning coefficient 
1.49 = conversion from SI to English units 
Hydraulic radius is the flow area divided by the wetted perimeter.
Open Channel Flow – Channel Geometry
Manning “n” values
Example 
What is the flow rate for a rectangular 
finished (clean) concrete channel with a 
base width of 8’, channel slope of 0.5%, 
with a “normal” water depth of 2’? 
A: 140 cfs 
B: 8.5 cfs 
C: 100 cfs 
D: 200 cfs
Solution 
n is 0.015, Rh is 8*2 sq.ft./(2+8+2) ft, S is 0.005 ft/ft, so 
V = 8.5 ft/sec 
Q = V*A= 8.5 ft/sec*16 sq.ft. = 140 cfs
Vegetated Waterway Design 
The design of a vegetated waterway is an 
iterative process, considering both 
capacity when the grass is unmowed and 
hence higher resistance to flow and 
stability when recently mowed and more 
susceptible to bed scour at high flow 
velocities. Fortunately, the EFM has tables 
of suitable channel dimensions.
Design steps from EFH:
Example 
A subdivision produces a peak runoff rate of 
60 cfs from a 10yr-24hr rainfall. A 
vegetated waterway with an average slope 
of 3% is to be planted with Kentucky 
bluegrass. The soil at the waterway site is 
easily eroded. The waterway will be 
constructed with a parabolic shape. What 
top width and depth are required (ignoring 
freeboard)?
Choices 
A: 20’, 2’ 
B: 18.5’, 1.1’ 
C: 15’, 1.5’ 
D: 12’, 0.6’
Permissible Velocity 
Kentucky bluegrass on a 3% slope easily eroded soil 
can handle up to 5 fps.
Resistance to flow 
Kentucky bluegrass has a C resistance when unmowed 
and a D resistance when cut to 2” height
EFM table 
Reading the chart for 60cfs, V1 of 5fps, a 
top width of 18.5’ with a depth of 1.1’ is 
suitable, so answer B.
Questions on open channel flow or 
vegetated waterways? 
Questions about anything in the 
whole presentation?
Thanks!

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2012 PE Review: Hydrology and Hydraulics Highlights

  • 1. 2012 PE Review: IV-A: Hydrology and Hydraulics Michael C. Hirschi, PhD, PE, D.WRE Senior Engineer Waterborne Environmental, Inc. hirschim@waterborne-env.com also Professor Emeritus University of Illinois
  • 2. Acknowledgements: Daniel Yoder, I-A, PE Review 2006 Rafael (Rafa) Muñoz-Carpena, I-A, PE Review 2007-09 Rod Huffman, PE Review coordinator
  • 3. Session Topics • Hydrology • Hydraulics of Structures • Open Channel Flow
  • 4. Hydrology • Hydrologic Cycle • Precipitation – Average over Area – Return Period • Abstractions from Rainfall • Runoff – Hydrographs – Determination methods
  • 5. Hydraulics of Structures • Weir flow • Orifice flow • Pipe flow • Spillway flow – Stage-Discharge relationship
  • 6. Open Channel Flow • Channel geometries – Triangular – Trapezoid – Parabolic • Manning’s equation – Manning roughness, “n” • Grass waterway design
  • 7. A few comments • Material outlined is about 3 weeks or more in a 3-semester hour class. I’m compressing at least 6 hours of lecture and 3 laboratories into 2 hours, so I will: – Review highlights and critical points – Do example problems • You need to: – Review and tab references – Do additional example problems, or at least thoroughly review examples in references
  • 8. Hydrologic Cycle From Fangmeier et al. (2006)
  • 9. Precipitation • Input to the Rainfall-Runoff process • Forms include: – Rainfall – Snow – Hail – Sleet • Measured directly • Varies temporally and areally
  • 10. Rainfall Data • Daily • Hourly • 15-minute • Continuous • Reported as depth, which is really volume over a given area, over a period of time
  • 11. Average Rainfall • Simple arithmetic average • Theissen Polygon
  • 12. Example 1 How do different calculation methods of rainfall average compare? Consider:
  • 13. Raingage data • Gages (clockwise from upper left): 1.9”, 2.1”, 1.8”, 1.9”, 2.1”, 2.2” Arithmetic average: 2.0”
  • 14. Theissen Polygons • Areas closest to each raingage determined by perpendicular bisectors of each line between raingages. • Areas for each raingage, again clockwise from upper left: 65ac, 150ac, 55ac, 140ac, 215 ac, 270ac • Figure is repeated with Theissen polygon construction added.
  • 15. Why bisectors? • When perpendicular bisectors are constructed, they are, by definition, lines that are equidistant from the points at the ends of the lines they bisect. • So, the combination of the constructions delineate areas that are closest to a given point (raingage in this case)
  • 16. Is the watershed average rainfall using the Theissen Polygon method most nearly: A. 2.0” B. 2.1” C. 2.2” D. 1.9”
  • 17. Theissen calculation • Uses areal weighted average, so the sum of the products of area x depth divided by total area • Hint: If you measure the areas yourself, the denominator should be the sum of the areas, not the known watershed area • So, average Theissen rain: Answer B, 2.1” (65*1.9+150*2.1+55*1.8+140*1.9+215*2.1+270*2. 2)/(65+150+55+140+215+270)=2.07”, which is best represented as 2.1” given most data is 2 significant digits.
  • 18. Any questions on Theissen Polygons?
  • 19. Return Period (two descriptions) • A 10 year-24 hour rainfall volume is that depth of rainfall over a 24 hour period that is met or exceeded, on the long-term average, once every 10 years. • Another way to describe it is the 24 hour rainfall depth that has a 1 in 10 (10%) chance to be met or exceeded each year, on the long term average.
  • 20. US 100yr-24hr Rainfall 100yr-24hr data from TP-40 (Hershfield (1961) as referenced by Fangmeier et al. (2006)
  • 21. Return Period Data • Constructed from historical rainfall data • Available in tabular form via website or state USDA-NRCS reports. • Available as national maps (similar to previous slide) in several references such as Haan, Barfield & Hayes (1994).
  • 22. Example A reservoir is to be designed to contain the runoff from a 10yr-24hr rainfall event in Northeastern Illinois. What rainfall volume is to be considered? A. 4.5” B. 3.9” C. 4.1” D. Cannot estimate from available maps
  • 23. 10yr-24hr map from Haan, Barfield & Hayes (1994)
  • 24. Example • Answer is C. From map, 10yr-24hr rainfall in NE Illinois is just over 4”, use 4.1” to be conservative.
  • 26. Abstractions from Rainfall • Abstractions from rainfall are “losses” from rainfall that do not show up as storm water runoff: – Interception – Evapotranspiration – Storage • In bank • On surface – Infiltration
  • 27. Runoff by other names… • “Effective” rainfall • Rainfall “excess”
  • 28. Runoff If rainfall rate exceeds the soil infiltration capacity, ponding begins, and any soil surface roughness creates storage on the surface. After at least some of those depressions are filled with water, runoff begins. Additional rain continues to fill depressional storage and runoff rate increases as more of the hill slope and subsequently the watershed contributes runoff.
  • 30. Time of Concentration, tc The time from the beginning of runoff to the time at which the entire watershed is contributing runoff that reaches the watershed outlet is called the Time of Concentration. It is also described as the “travel time from the hydraulically most remote point in a watershed to the outlet”.
  • 36. Runoff Example In a previous problem, a design rain event in NE Illinois was determined to be 4.1”. Assuming the watershed in question was a completed 300 ac residential area with an average lot size of ½ ac, all on Hydrologic Group C soils, what is the needed pond volume, if all runoff is to be retained? A: 2.5 runoff-inches B: 53 acre-inches C: 630 acre-ft D: 53 acre-ft
  • 39. Answer to Runoff Example The answer is D, 53 acre-ft. From the table, the CN for Hyd group C soil with ½-ac lot is 80. Using the graph with a 4.1” rainfall, runoff depth is 2.1”. Volume is then 300ac*2.1in = 630 ac-in, divided by 12 is 53 ac-ft.
  • 40. Additional example You discover that the subdivision is actually 100 acres of ½ ac lots on C soils, 100 acres of ½ ac lots on D soils, 50 acres of ¼ ac lots on B soils and 50 acres of townhouses on A soils. What CN value would you use? A: 79 B: 85 C: 80 D: 75
  • 42. Answer The correct answer is C, 80. Use an area-weighted average, similar to Theissen method. The respective CN values for ½ ac on C, ½ ac on D, ¼ ac on B and townhouses on A are 80, 85, 75 & 77. The area-weighted CN is then (80*100+85*100+75*50+77*50)/300 = 80.33, which is more appropriately 80.
  • 43. Peak Discharge The CN method also provides for Peak Discharge estimation, using graphs or tables. Required information includes average watershed slope, watershed flow path length, CN, and rainfall depth. The graphical method from the EFM is:
  • 45. Peak Discharge Example Same residential watershed that produced 2.1” of runoff from a 4.1” rainfall. Flow length is 2500’, slope is 2%. CN is 80, so S is 2.5”. Ia = 0.2*S = 0.5”. Ia/P = 0.5/4.1=0.122. Tc = 2500^0.8*(1000/80-9)^0.7/1140/2^0.5 =0.8hr
  • 47. Example solution From graph, with Ia/P of 0.122 and Tc of 0.8hr, unit peak discharge is 0.57 cfs/ac/in or qp = 0.57*300*2.1 = 360 cfs
  • 48. Rational Method The Rational Equation is: Qp = CiA where: C is a coefficient i is rainfall intensity of duration tc A is area in acres C is approximately 0.4, A is 300ac, i is 2” in 30min, so 4iph, peak rate is then 0.4*300*4 = 480 cfs
  • 50. Hydraulics of Structures Flow through structures is important given that such structures control the rate of flow. Sizing of such structures is then important to allow flow to pass while protecting downstream areas from the effects of too high a flow rate. Structures may also be used for measurement of water flow. Each type of structure will produce different types of flow depending upon size and flow rate passing through it.
  • 51. Weirs • Sharp-crested • Broad-crested
  • 52. Weir Equation (from EFH-Ch03 Hydraulics)
  • 53. Sharp-Crested Weir (from EFH-Ch03 Hydraulics)
  • 54. More complex weirs (from Haan et al., 1994)
  • 55. Example • You are measuring flow using a 90° V-notch weir. H is measured as 0.53’ at 2.5’ upstream of the weir. What is the flow rate? A. 230 gpm B. 0.51 cfs C. 0.51 gpm D. A & B
  • 56. Answer • The answer is D. The equation from Haan et al (1994) is:
  • 57. Answer, continued • Q = 2.5*H^2.5, where Q is in cfs and H is in feet • Q=2.5*(0.53)^2.5=0.511 cfs or 0.51 cfs • Q=0.51 cfs*60sec/min*7.48gal/cf=230 gpm • Note: Both answers contain 2 significant figures
  • 58. Orifice Flow • Submerged vs Free Outlet • Shapes affecting C
  • 62. Example • Markers Mark distillery just moved a 3’ diameter barrel of their bourbon over their charcoal filter bed to drain the bourbon into the system to be bottled. The bung plug is removed instantaneously, allowing barrel strength bourbon to flow freely from the 2” diameter bung, which can be considered a sharp-edged orifice. What is the initial flow rate (assuming same specific gravity as water, which is an incorrect assumption)?
  • 63. Answers A: 0.5 cfs B: 83 gpm C: 26.6 gpm D: 200 L/hr
  • 64. Solution Q=0.61*A*(2*g)^0.5*h^0.5 =0.61*(π*1”^2)*(2*32.2f/s/s)^0.5*3’^0.5 =0.61*3.1415/144*(64.4)^0.5*3^0.5 =0.185 cfs Q=83 gpm (answer B)
  • 65. Pipe flow When considering pipe flow in a structure, Bernoulli’s equation is used: Frictional losses are multiples of the velocity head (V2/2g) and are additive.
  • 66. Head loss under pipe flow • Entrance loss (Ke) • Bend loss (Kb) • Pipe friction loss (Kc) • Each coefficient is documented in references Considering the Bernoulli equation for a spillway, the pressure at entrance and exit is atmospheric, the elevation difference is the water surface elevation difference between upstream and downstream, and the remaining term is the velocity head plus losses
  • 69. Spillway considerations A given spillway may have several discharge relationships (weir, orifice, pipe) depending upon the head (stage). The stage discharge curve then becomes a combination curve, with the type of relationship allowing the highest flow at a given head in control. Consider a drop inlet control structure:
  • 70.
  • 72. Example An 18” CMP with an 18” vertical riser is used as the principal spillway for a pond. The pipe is 50’ long with one 90° bend. The top of the inlet is 10’ above the bottom of the outlet. Develop the stage-discharge relationship assuming a free outfall.
  • 73. Weir flow Basic equation: Given 18” riser, length of weir is 2πr, or 4.7’, so
  • 74. Orifice flow Basic orifice equation: Given 18” riser and assuming C’ of 0.6,
  • 75. Pipe flow Basic pipe flow equation: After looking up each parameter:
  • 77. Questions on spillway hydraulics?
  • 78. Open Channel Flow Flow through open channels is another important area to consider and review. Velocity and flow rate are usually calculated using Manning’s equation, which considers flow geometry, channel roughness and slope.
  • 79. Manning’s Equation Where: V= flow velocity in fps Rh = Hydraulic Radius in ft S = Energy gradeline slope in ft/ft (=bed slope for normal flow) n = Manning coefficient 1.49 = conversion from SI to English units Hydraulic radius is the flow area divided by the wetted perimeter.
  • 80. Open Channel Flow – Channel Geometry
  • 82. Example What is the flow rate for a rectangular finished (clean) concrete channel with a base width of 8’, channel slope of 0.5%, with a “normal” water depth of 2’? A: 140 cfs B: 8.5 cfs C: 100 cfs D: 200 cfs
  • 83. Solution n is 0.015, Rh is 8*2 sq.ft./(2+8+2) ft, S is 0.005 ft/ft, so V = 8.5 ft/sec Q = V*A= 8.5 ft/sec*16 sq.ft. = 140 cfs
  • 84. Vegetated Waterway Design The design of a vegetated waterway is an iterative process, considering both capacity when the grass is unmowed and hence higher resistance to flow and stability when recently mowed and more susceptible to bed scour at high flow velocities. Fortunately, the EFM has tables of suitable channel dimensions.
  • 86. Example A subdivision produces a peak runoff rate of 60 cfs from a 10yr-24hr rainfall. A vegetated waterway with an average slope of 3% is to be planted with Kentucky bluegrass. The soil at the waterway site is easily eroded. The waterway will be constructed with a parabolic shape. What top width and depth are required (ignoring freeboard)?
  • 87. Choices A: 20’, 2’ B: 18.5’, 1.1’ C: 15’, 1.5’ D: 12’, 0.6’
  • 88. Permissible Velocity Kentucky bluegrass on a 3% slope easily eroded soil can handle up to 5 fps.
  • 89. Resistance to flow Kentucky bluegrass has a C resistance when unmowed and a D resistance when cut to 2” height
  • 90. EFM table Reading the chart for 60cfs, V1 of 5fps, a top width of 18.5’ with a depth of 1.1’ is suitable, so answer B.
  • 91. Questions on open channel flow or vegetated waterways? Questions about anything in the whole presentation?

Editor's Notes

  1. Time scale of individual process influences/dictates whether it is significant/used for a given situation.