2013 PE Review: Hydrology and Hydraulics. Michael C. Hirschi, PhD, PE, CPESC, D.WRE Senior Engineer Waterborne Environmental, Inc. email@example.com also Professor Emeritus University of Illinois . Acknowledgements: Daniel Yoder, I-A, PE Review 2006
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Michael C. Hirschi, PhD, PE, CPESC, D.WRE
Waterborne Environmental, Inc.
also Professor Emeritus
University of Illinois
Daniel Yoder, I-A, PE Review 2006
Rafael (Rafa) Muñoz-Carpena, I-A, PE Review 2007-09
Rod Huffman, past PE Review coordinator
From Fangmeier et al. (2006)
How do different calculation methods of rainfall average compare?
Arithmetic average: 2.0”
(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.
100yr-24hr data from TP-40 (Hershfield (1961)
as referenced by Fangmeier et al. (2006)
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?
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.
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”.
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
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.
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?
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.
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:
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
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
The Rational Equation is:
Qp = CiA
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
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.
(from EFH-Ch03 Hydraulics)
(from EFH-Ch03 Hydraulics)
B. 0.51 cfs
C. 0.51 gpm
D. A & B
A: 0.5 cfs
B: 83 gpm
C: 26.6 gpm
D: 200 L/hr
Q=83 gpm (answer B)
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.
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
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:
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.
Given 18” riser, length of weir is 2πr, or 4.7’, so
Basic orifice equation:
Given 18” riser and assuming C’ of 0.6,
Basic pipe flow equation:
After looking up each parameter:
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.
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.
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
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
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.
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)?
A: 20’, 2’
B: 18.5’, 1.1’
C: 15’, 1.5’
D: 12’, 0.6’
Kentucky bluegrass on a 3% slope easily eroded soil
can handle up to 5 fps.
Kentucky bluegrass has a C resistance when unmowed
and a D resistance when cut to 2” height
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 about anything in the