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Hydrologic Analysis Dr. Bedient CEVE 101 Fall 2013
Review A watershed- a basic unit used in most hydrologic calculations relating to the water balance or computation of rainfall- runoff Watershed Response-how the watershed response to rainfall. Several characteristics will effect response Drainage Area Channel Slope Soil Types Land Use Land Cover Main channel and tributary characteristics- channel morphology The shape, slope and character of the floodplain There is a way to plot the flow from the outlet over a space of time. This graphical representation is called a hydrograph
Typical Graphs for Hydrologic Analysis Hydrograph Hyetograph Cummulative Rainfall
Hydrographs Hydrograph: continuous plot of instantaneous discharge Flow rate (cfs or cms) vs. time Watershed factors of importance: Size and shape of drainage area Slope of the land surface and the main channel Soil types and distribution in watershed Meteorological Factors that influence the shape and volume of runoff: Rainfall intensity and pattern Areal distribution of rainfall over the basin Size and duration of the storm event
Hurricane Ike (September 13) and FAS2 Prediction Made by Nick FANG at the Phil Bedient Water Resources Research Group at Rice University.
The 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.
Hyetograph Graph showing rainfall intensity vs time (in./hr) Can be calculated if given cumulative rainfall (P) or gross rainfall (I)
Example 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)
Example 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 so (I 0 ) = 0 (I 1 )= 0.2 (I 2 ) = 0.3 (I 3 )= 0.6 The resulting table will be Time (min)Cumulative Rainfall (P) (in) To 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.
Example 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/hr 0.2 / 0.5 = 0.4 in/hr 0.1 / 0.5 = 0.2 in/hr 0.3 / 0.5 = 0.6 in/hr
Determining 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.
Example 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, 4 th ed. Example 2-1
Example 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 graphs So total volume = 9100 cfs-hr = ac-in (1.008 cfs-hr = 1 ac-in) Time (hr)Q (cfs)Volume (cfs-hr)
Finding 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.
Example of finding the volume left to infiltration. Given our previous acre basin with a runoff of 9100 cfs-hr. Determine the amount of volume left to infiltration knowing that there was 4.0 in. of rainfall.
Example of finding the volume left to infiltration. Step 1: Convert cfs-hr to ac-in 9100 cfs = ac-in Step 2 = Divide ac-in by acres ac-in / 2600 ac = 3.47 = 3.5 inches- amount of direct runoff Step 3 = Subtract direct runoff from rainfall 4.0 – 3.5 = 0.5 in was left to infiltration.
Unit Hydrographs Unit 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) Assumptions Rainfall excesses of equal duration are assumed to produce hydrographs with equivalent time bases Rainfall distribution is assumed to be the same for all storms of equal duration Direct runoff ordinates for a storm of given duration are assumed directly proportional to rainfall excesses volumes 2X the rainfall produces a doubling of hydrograph ordinates
Unit 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)
Timing Parameters: UH Lag time Lag time: (L or Tp) time from the center of mass of rainfall to the peak of the hydrograph Time of Rise Time of Rise: (Tr) the time from the start of rainfall excess to the peak of the hydrograph Time of Concentration Time of Concentration: (Tc) the time from the end of the net rainfall to the inflection point of the hydrograph Time Base Time Base: (Tb) the total duration of the DRO hydrograph
Developing 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
Unit Hydrograph Convolution Deriving hydrographs from multiperiod rainfall excess or Where Q n = storm hydrograph ordinate P i = rainfall excess U j = UH ordinate where j = n - i + 1
Storm 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] cfs From Bedient et al. Hydrology and floodplain analysis. 4 th Ed. Example 2-5
Storm Hydrograph from the Unit Hydrograph Want 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*UQ P = [0.5, 1.0, 1.5, 0.0, 0.5] in. U = [0,100, 320, 450, 370, 250, 160, 90, 40, 0] cfs Step 1
Want 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*UQ 00.5*0 = *100 = 501*0 = *320 = 1601*100 = *0 = *450 = 2251*320 = *100 = 1500*0 = *370 = 1851*450 = *320 = 4800*100 = 00.5*0 = *250 = 1251 *370 = *450 = 6750*320 = 00.5*100 = *160 = 801*250 = *370 = 5550*450 = 00.5*320 = *90 = 451*160 = *250 = 3750*370 = 00.5*450 = *40 = 201*90 = 901.5*160 = 2400*250 = 00.5*370 = *0 = 01*40 = 401.5*90 = 1350*160 = 00.5*250 = *0 = 01.5*40 = 600*90 = 00.5*160 = *0 = 00*40 = 00.5*90 = *0 = 00.5*40 = *0 = 0 Storm 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] cfs Step 2
Want 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*UQ 00.5*0 = *100 = 501*0 = *320 = 1601*100 = *0 = = *450 = 2251*320 = *100 = 1500*0 = = *370 = 1851*450 = *320 = 4800*100 = 00.5*0 = *250 = 1251 *370 = *450 = 6750*320 = 00.5*100 = *160 = 801*250 = *370 = 5550*450 = 00.5*320 = *90 = 451*160 = *250 = 3750*370 = 00.5*450 = *40 = 201*90 = 901.5*160 = 2400*250 = 00.5*370 = *0 = 01*40 = 401.5*90 = 1350*160 = 00.5*250 = *0 = 01.5*40 = 600*90 = 00.5*160 = *0 = 00*40 = 00.5*90 = *0 = 00.5*40 = *0 = 00 Storm 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] cfs Step 3
UH Convolution Example P n = [0.5, 1.0, 1.5, 0.0, 0.5] in U n = [0, 100, 320, 450, 370, 250, 160, 90, 40, 0] cfs Time (hr) P1UnP1Un P2UnP2Un P3UnP3Un P4UnP4Un P5UnP5Un QnQn
Uniform 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.
Manning’s Equation Q = 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) P = Wetted Perimeter Pipe P = Circum. Natural Channel A A A Note: this equation is for U.S Customary units-for the metric units 1.49 is 1
Manning’s variables for different cross- sections
Example 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 =.0002 for the concrete-lined channel. From Bedient et al. Hydrology and Floodplain analysis 4 th Ed. Example 7-3
Example of calculating flow Given y = 25 ft n = 0.02 S = b = 75 ft Manning’s Equation is used to compute Q and from the geometry provided: = A = = 4375 ft^2 P = = ft Then all the values are plugged in to give Q = 28,730 cfs