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Hydrologic Analysis Dr. Bedient CEVE 101 Fall 2013.

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Presentation on theme: "Hydrologic Analysis Dr. Bedient CEVE 101 Fall 2013."— Presentation transcript:

1 Hydrologic Analysis Dr. Bedient CEVE 101 Fall 2013

2 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

3 Typical Graphs for Hydrologic Analysis Hydrograph Hyetograph Cummulative Rainfall

4 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

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6 Hurricane Ike (September 13) and FAS2 Prediction Made by Nick FANG at the Phil Bedient Water Resources Research Group at Rice University.

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8 Typical Hydrographs  Rising limb  Crest segment  Recessive curve  Falling limb  Base Flow

9 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.

10 Hyetograph  Graph showing rainfall intensity vs time (in./hr)  Can be calculated if given cumulative rainfall (P) or gross rainfall (I)

11 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)

12 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.

13 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

14 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.

15 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

16 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)

17 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.

18 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.

19 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.

20 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

21 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)

22 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

23 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

24 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

25 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

26 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

27  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

28  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

29 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

30 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.

31 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

32 Manning’s variables for different cross- sections

33 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

34 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


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