This document provides an overview of topics to be covered in a 3-week professional engineering exam review session on hydrology and hydraulics. It will cover key aspects of hydrology including the hydrologic cycle, precipitation, runoff analysis using the Curve Number method, and peak discharge calculations. Hydraulics topics will include flow through common structures like weirs, orifices, and pipes. Example problems will be worked through for each major topic to illustrate concepts and calculations. Attendees are advised to review references and practice additional example problems.

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.

25. Questions on precipitation?

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.

29. Rainfall/Runoff process

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”.

31. Curve Number method

32. CN Method, continued

33. Time of Concentration, tc

34. CN values

35. Runoff Volume determination

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

37. Runoff Example, continued

38. Runoff Volume determination

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

41. Add’l Runoff Example, cont.

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:

44. Peak Runoff Discharge

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

46. Peak Runoff Discharge

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

49. Questions on runoff?

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

59. Submerged Orifice

60. Free Discharge Orifice

61. Orifice Coefficients

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

67. Consider the following

68. Pipe flow

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:

71. Stage-Discharge Curve

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:

76. Stage-Discharge Relationship

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

81. Manning “n” values

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.

85. Design steps from EFH:

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?

92. Thanks!