2011 PE Review:IV-A: Hydrology and Hydraulics Michael C. Hirschi, PhD, PE, D.WRE Professor and Assistant Dean University of Illinois firstname.lastname@example.org
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.
Is the watershed average • 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
10yr, 1yr rainfall in NE IL From: http://www.erh.noaa.gov/er/hq/Tp40s.htm
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