Presentation on theme: "Hydrologic Analysis Dr. Bedient CEVE 101 Fall 2013."— Presentation transcript:
1Hydrologic AnalysisDr. Bedient CEVE 101 Fall 2013
2ReviewA watershed- a basic unit used in most hydrologic calculations relating to the water balance or computation of rainfall- runoffWatershed Response-how the watershed response to rainfall. Several characteristics will effect responseDrainage AreaChannel SlopeSoil TypesLand UseLand CoverMain channel and tributary characteristics- channel morphologyThe shape, slope and character of the floodplainThere is a way to plot the flow from the outlet over a space of time. This graphical representation is called a hydrograph
3Typical Graphs for Hydrologic Analysis HydrographHyetographCummulative Rainfall
4Hydrographs Hydrograph: continuous plot of instantaneous discharge Flow rate (cfs or cms) vs. timeWatershed factors of importance:Size and shape of drainage areaSlope of the land surface and the main channelSoil types and distribution in watershedMeteorological Factors that influence the shape and volume of runoff:Rainfall intensity and patternAreal distribution of rainfall over the basinSize and duration of the storm event
9The Watershed Response Hydrograph As rain falls over a watershed area, a certain portion will infiltrate the soil. Some water will evaporate to atmosphere.Rainfall that does not infiltrate or evaporate is available as overland flow and runs off to the nearest stream.Smaller tributaries or streams then begin to flow and contribute their load to the main channel at confluences.As accumulation continues, the streamflow rises to a maximum (peak flow) and a flood wave moves downstream through the main channel.The flow eventually recedes or subsides as all areas drain out.
10Hyetograph Graph showing rainfall intensity vs time (in./hr) Can be calculated if given cumulative rainfall (P) or gross rainfall (I)
11Example of plotting hyetographs and rainfall For the rainfall given below, plot cumulative rainfall (P) and gross rainfall hyetograph with ∆t = 30 min.Time (min)Rainfall (I) (in)300.2600.1900.3
12Example of plotting hyetograph and rainfall We are given gross rainfall, to find cumulative rainfall you need to take rainfall from each time step and you add it to the previous time step’s result so0 + 0 (I0) = 0(I1)= 0.2(I2) = 0.3(I3)= 0.6The resulting table will beTime (min)Cumulative Rainfall (P) (in)300.2600.3900.6To go from P to I you take your rainfall at a time step and subtract the previous time step-try it and see if you get the same results.
13Example of plotting hyetograph and rainfall To plot rainfall intensity (in./hr) you take your gross rainfall (I) and divide it by your time step (30 min = 0.5 hr)0 / 0.5 = 0 in/hr0.2 / 0.5 = 0.4 in/hr0.1 / 0.5 = 0.2 in/hr0.3 / 0.5 = 0.6 in/hr
14Determining volume of runoff The volume of runoff from a watershed is equal to the area under the hydrograph.In graph form, an approximation can be made by which estimates the volume as a bar graph. Each individual bar is then added to give volume.In table form, this is done by multiplying flow (Q) by the time step.This is summarized in the next example.
15Example of determine total volume of runoff Given the hydrograph, determine the total volume of runoff for a 2600 acres-basin.Adapted from Bedient et al, Hydrology and floodplain analysis, 4th ed. Example 2-1
16Example of determine total volume of runoff Step 1: We can determine the volume by creating a bar graph to estimate volume as shown in the next figure:Step 2 = Sum the bar graphsSo total volume = 9100 cfs-hr= ac-in(1.008 cfs-hr = 1 ac-in)Time (hr)Q (cfs)Volume (cfs-hr)0-21002002-43006004-650010006-870014008-10650130010-12120012-1414-1640080016-1818-2020-2215022-2424-2650
17Finding the volume left to infiltration Depending on the soil types, runoff will not begin until the soil is completely saturated.There is a way to find how much water was infiltrated into the soil during a rain event.If given an amount of rainfall and the amount of direct runoff. Infiltration is equal to the difference between rainfall and direct runoff (evaporation is ignored)This is summarized in the next example.
18Example of finding the volume left to infiltration. Given our previous acre basin with a runoff of cfs-hr. Determine the amount of volume left to infiltration knowing that there was 4.0 in. of rainfall.
19Example of finding the volume left to infiltration. Step 1: Convert cfs-hr to ac-in9100 cfs = ac-inStep 2 = Divide ac-in by acresac-in / 2600 ac = 3.47 = 3.5 inches- amount of direct runoffStep 3 = Subtract direct runoff from rainfall4.0 – 3.5 = 0.5 in was left to infiltration.
20Unit HydrographsUnit Hydrograph: The unit hydrograph represents the basin response to 1 inch (1 cm) of uniform net rainfall for a specified duration.Works best for relatively small subareas (1-10 sq miles)AssumptionsRainfall excesses of equal duration are assumed to produce hydrographs with equivalent time basesRainfall distribution is assumed to be the same for all storms of equal durationDirect runoff ordinates for a storm of given duration are assumed directly proportional to rainfall excesses volumes2X the rainfall produces a doubling of hydrograph ordinates
21Unit Hydrographs In summary The hydrologic system is linear and time invariant.This means that complex storm hydrographs (the hydrographs we’ve looked at) can be produced by adding up individual unit hydrographs, adjusted for rainfall volumes and added and lagged in time.This is known as hydrograph convolution (see next example)
22Timing Parameters: UHLag time: (L or Tp) time from the center of mass of rainfall to the peak of the hydrographTime of Rise: (Tr) the time from the start of rainfall excess to the peak of the hydrographTime of Concentration: (Tc) the time from the end of the net rainfall to the inflection point of the hydrographTime Base: (Tb) the total duration of the DRO hydrograph
23Developing a Storm Hydrograph from a 1 hr Unit Hydrograph-Unit Hydrograph Convolution To find the storm hydrograph from the unit hydrograph it is necessary to have the rainfall ordinates for that given storm.The flow (U) ordinates of the Unit Hydrograph are also needed.Then, the (U) will be multiplied by P1 then by P2 until it has been multiplied by all the rainfall ordinates.Everytime you move to a different rainfall ordinate, you lag by one hour (since we have a 1-hr UH).Once everything has been multiplied and lagged. For each hour, the resulting P*U for that hour are added. This gives the flow of the storm (Q) for that hour.These steps are shown in the next example
24Unit Hydrograph Convolution Deriving hydrographs from multiperiod rainfall excessorWhereQn = storm hydrograph ordinatePi = rainfall excessUj = UH ordinatewhere j = n - i + 1
25Storm Hydrograph from the Unit Hydrograph Given the rainfall excess and the 1-hr UH derive the storm hydrograph for the watershed using hydrograph convolution. Compute the resulting hydrograph and assume no losses.P = [0.5, 1.0, 1.5, 0, 0.5] in.U = [0,100, 320, 450, 370, 250, 160, 90, 40, 0] cfsFrom Bedient et al. Hydrology and floodplain analysis. 4th Ed. Example 2-5
26Storm Hydrograph from the Unit Hydrograph P = [0.5, 1.0, 1.5, 0.0, 0.5] in.U = [0,100, 320, 450, 370, 250, 160, 90, 40, 0] cfsStep 1Want to multiply U by each P and lag by one hour everytime you move to the next P.Then for each hour you add your results for that following time, that will give you Q.Time (hr)P1*UP2*UP3*UP4*UP5*UQ0.5121.5345678910111213
27Storm Hydrograph from the Unit Hydrograph P = [0.5, 1.0, 1.5, 0.0, 0.5] in.U = [0,100, 320, 450, 370, 250, 160, 90, 40, 0] cfsStep 2Want to multiply U by each P and lag by one hour everytime you move to the next P.Then for each hour you add your results for that following time, that will give you Q.Time (hr)P1*UP2*UP3*UP4*UP5*UQ0.5*0 = 010.5*100 = 501*0 = 020.5*320 = 1601*100 = 1001.5*0 = 030.5*450 = 2251*320 = 3201.5*100 = 1500*0 = 040.5*370 = 1851*450 = 4501.5*320 = 4800*100 = 050.5*250 = 1251 *370 = 3701.5*450 = 6750*320 = 060.5*160 = 801*250 = 2501.5*370 = 5550*450 = 070.5*90 = 451*160 = 1601.5*250 = 3750*370 = 080.5*40 = 201*90 = 901.5*160 = 2400*250 = 091*40 = 401.5*90 = 1350*160 = 0101.5*40 = 600*90 = 0110*40 = 01213
28Storm Hydrograph from the Unit Hydrograph P = [0.5, 1.0, 1.5, 0.0, 0.5] in.U = [0,100, 320, 450, 370, 250, 160, 90, 40, 0] cfsStep 3Want to multiply U by each P and lag by one hour everytime you move to the next P.Then for each hour you add your results for that following time, that will give you Q.Time (hr)P1*UP2*UP3*UP4*UP5*UQ0.5*0 = 010.5*100 = 501*0 = 05020.5*320 = 1601*100 = 1001.5*0 = 0=26030.5*450 = 2251*320 = 3201.5*100 = 1500*0 = 0= 69540.5*370 = 1851*450 = 4501.5*320 = 4800*100 = 0111550.5*250 = 1251 *370 = 3701.5*450 = 6750*320 = 0122060.5*160 = 801*250 = 2501.5*370 = 5550*450 = 0104570.5*90 = 451*160 = 1601.5*250 = 3750*370 = 080580.5*40 = 201*90 = 901.5*160 = 2400*250 = 053591*40 = 401.5*90 = 1350*160 = 0300101.5*40 = 600*90 = 0140110*40 = 045122013
30Uniform Open-Channel Flow Uniform open channel flow is the hydraulic condition in which the water depth and the channel cross section do not change over some reach of the channel.The total energy change over the channel reach is exactly equal to the energy losses of boundary friction and turbulence.Strict uniform flow is rare in natural streams because of the constantly changing channel conditions.Often assumed in natural streams for engineering calculations.To calculate flow within a channel equations such as the Chezy eq. and Manning’s equation are used.
31Manning’s EquationAAAP = Wetted PerimeterPipe P = Circum.Natural ChannelQ = Flowrate, cfs n = Manning’s Roughness Coefficient (ranges from ) S = Slope of channel in longitudinal direction R = A/P, the hydraulic radius, where: A = Cross-sectional Area of Flow (area of trapezoid or flow area) P = Wetted Perimeter (perimeter in contact with water)Note: this equation is for U.S Customary units-for the metric units 1.49 is 1
32Manning’s variables for different cross-sections
33Example of calculating flow Brays Bayou can be represented as a single trapezoidal channel with a bottom width b of 75 ft and a side slope of 4:1 (horizontal:vertical) on average. If the normal bankfull depth is 25 ft at the Main St. bridge, compute the normal flow rate in cfs for this section. Assume that n = and S = for the concrete-lined channel.From Bedient et al. Hydrology and Floodplain analysis 4th Ed. Example 7-3
34Example of calculating flow Giveny = 25 ftn = 0.02S = 0.002b = 75 ftManning’s Equation is used to compute Q and from the geometry provided: =A = = 4375 ft^2 P = = ftThen all the values are plugged in to give Q = 28,730 cfs