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### 2011 PE Review:IV-A: Hydrology and Hydraulics

Michael C. Hirschi, PhD, PE, D.WRE

Professor and Assistant Dean

University of Illinois

mch@illinois.edu

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

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.

- rainfall using the Theissen
- Polygon method most nearly:
- 2.0”
- 2.1”
- 2.2”
- 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.

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?

- 4.5”
- 3.9”
- 4.1”
- Cannot estimate from available maps

Example

- Answer is C. From map, 10yr-24hr rainfall in NE Illinois is just over 4”, use 4.1” to be conservative.

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.

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”.

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?

A: 2.5 runoff-inches

B: 53 acre-inches

C: 630 acre-ft

D: 53 acre-ft

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

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 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

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

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)

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?
- 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

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)?

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

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:

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.

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.

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.

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)?

Resistance to flow

Kentucky bluegrass has a C resistance when unmowed

and a D resistance when cut to 2” height

Solution

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.

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