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

HydrographHyetograph

Cummulative Rainfall

HydrographsHydrograph: 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.

Typical Hydrographs

Rising limb Crest segment Recessive curve Falling limb Base Flow

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)

0 0

30 0.2

60 0.1

90 0.3

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 0 + 0 (I0) = 0 0+ 0.2 (I1)= 0.2 0.2 + 0.1(I2) = 0.3 0.3+ 0.3 (I3)= 0.6

The resulting table will beTime (min) Cumulative Rainfall

(P) (in)

0 0

30 0.2

60 0.3

90 0.6

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= 9027.78 ac-in

(1.008 cfs-hr = 1 ac-in)

Time (hr) Q (cfs) Volume (cfs-hr)

0-2 100 200

2-4 300 600

4-6 500 1000

6-8 700 1400

8-10 650 1300

10-12 600 1200

12-14 500 1000

14-16 400 800

16-18 300 600

18-20 200 400

20-22 150 300

22-24 100 200

24-26 50 100

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

2600-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 = 9027.78 ac-in Step 2 = Divide ac-in by acres

9027.78 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 timeLag time: (L or Tp) time from the center of mass of rainfall to the peak of the hydrograph

Time of RiseTime of Rise: (Tr) the time from the start of rainfall excess to the peak of the hydrograph

Time of ConcentrationTime of Concentration: (Tc) the time from the end of the net rainfall to the inflection point of the hydrograph

Time BaseTime 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 Qn = storm hydrograph ordinate

Pi = rainfall excess

Uj = 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. 4th 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*U P2*U P3*U P4*U P5*U Q

0 0.5

1 0.5 1

2 0.5 1 1.5

3 0.5 1 1.5 0

4 0.5 1 1.5 0 0.5

5 0.5 1 1.5 0 0.5

6 0.5 1 1.5 0 0.5

7 0.5 1 1.5 0 0.5

8 0.5 1 1.5 0 0.5

9 0.5 1 1.5 0 0.5

10 1 1.5 0 0.5

11 1.5 0 0.5

12 0 0.5

13 0.5

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*U P2*U P3*U P4*U P5*U Q

0 0.5*0 = 0

1 0.5*100 = 50 1*0 = 0

2 0.5*320 = 160

1*100 = 100

1.5*0 = 0

3 0.5*450 = 225

1*320 = 320

1.5*100 = 150

0*0 = 0

4 0.5*370 = 185

1*450 = 450

1.5*320 = 480

0*100 = 0 0.5*0 = 0

5 0.5*250 = 125

1 *370 = 370

1.5*450 = 675

0*320 = 0 0.5*100 = 50

6 0.5*160 = 80 1*250 = 250

1.5*370 = 555

0*450 = 0 0.5*320 = 160

7 0.5*90 = 45 1*160 = 160

1.5*250 = 375

0*370 = 0 0.5*450 = 225

8 0.5*40 = 20 1*90 = 90 1.5*160 = 240

0*250 = 0 0.5*370 = 185

9 0.5*0 = 0 1*40 = 40 1.5*90 = 135 0*160 = 0 0.5*250 = 125

10 1*0 = 0 1.5*40 = 60 0*90 = 0 0.5*160 = 80

11 1.5*0 = 0 0*40 = 0 0.5*90 = 45

12 0*0 = 0 0.5*40 = 20

13 0.5*0 = 0

Storm Hydrograph from the Unit HydrographP = [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*U P2*U P3*U P4*U P5*U Q

0 0.5*0 = 0 0

1 0.5*100 = 50 1*0 = 0 50

2 0.5*320 = 160

1*100 = 100

1.5*0 = 0 160+100+0=260

3 0.5*450 = 225

1*320 = 320

1.5*100 = 150

0*0 = 0 225+320+150+0 = 695

4 0.5*370 = 185

1*450 = 450

1.5*320 = 480

0*100 = 0 0.5*0 = 0 1115

5 0.5*250 = 125

1 *370 = 370

1.5*450 = 675

0*320 = 0 0.5*100 = 50

1220

6 0.5*160 = 80 1*250 = 250

1.5*370 = 555

0*450 = 0 0.5*320 = 160

1045

7 0.5*90 = 45 1*160 = 160

1.5*250 = 375

0*370 = 0 0.5*450 = 225

805

8 0.5*40 = 20 1*90 = 90 1.5*160 = 240

0*250 = 0 0.5*370 = 185

535

9 0.5*0 = 0 1*40 = 40 1.5*90 = 135 0*160 = 0 0.5*250 = 125

300

10 1*0 = 0 1.5*40 = 60 0*90 = 0 0.5*160 = 80

140

11 1.5*0 = 0 0*40 = 0 0.5*90 = 45

45

12 0*0 = 0 0.5*40 = 20

20

13 0.5*0 = 0 0

Storm Hydrograph from the Unit HydrographP = [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 Pn= [0.5, 1.0, 1.5, 0.0, 0.5] in

Un= [0, 100, 320, 450, 370, 250, 160, 90, 40, 0] cfsTime (hr)

P1Un P2Un P3Un P4Un P5Un Qn

0 0 0

1 50 0 50

2 160 100 0 260

3 225 320 150 0 695

4 185 450 480 0 0 1115

5 125 370 675 0 50 1220

6 80 250 555 0 160 1045

7 45 160 375 0 225 805

8 20 90 240 0 185 535

9 0 40 135 0 125 300

10 0 60 0 80 140

11 0 0 45 45

12 0 20 20

13 0 0

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, cfsn = Manning’s Roughness Coefficient (ranges from 0.015 - 0.15)S = Slope of channel in longitudinal directionR = 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

Q=

1.49nAR

23 S

A AA

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 = 0.020 and S = .0002 for the concrete-lined channel.

From Bedient et al. Hydrology and Floodplain analysis 4th Ed. Example 7-3

Example of calculating flow Given

y = 25 ftn = 0.02S = 0.002b = 75 ft

Manning’s Equation is used to compute Q and from the geometry provided: =

A = = 4375 ft^2 P = = 281.16 ft

Then all the values are plugged in to give Q = 28,730 cfs

Q=

1.49nAR

23 S