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 rainfallrunoff
Watershed Response-how the watershed
response to rainfall. Several
characteristics will effect response

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Drainage Area
Channel Slope
Soil Types
Land Use
Land Cover
Main channel and tributary characteristicschannel 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
7
6
Flow (cfs)

5
4
3
2
1
0
0
1
2
3
4
5
6
Time (hr)
7
8
9
10
Typical Graphs for Hydrologic Analysis
Hydrograph
4
700
2
0
600
1
4
Cummulative
Rainfall
400
Cumulative Mass Curve
300
100
0
0
5
10
Time (hr)
15
20
3
2.5
2
1.5
1
0.5
0
16:05
16:20
16:35
16:50
17:05
17:20
17:35
17:50
18:05
18:20
18:35
18:50
19:05
19:20
19:35
19:50
20:05
20:20
20:35
20:50
200
Flow (cfs)
Flow (cfs)
500
2
3
Time (hr)
I (in./hr)
Hyetograph
800
Time
Hydrographs
Hydrograph: continuous plot of instantaneous discharge
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Watershed factors of importance:



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Flow rate (cfs or cms) vs. time
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
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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)
4
2
0
1
2
3
Time (hr)
4
I (in./hr)

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 be

Time (min)
Cumulative Rainfall (P) (in)
0
0
30
0.2
60
0.3
90
0.6
Cumulative Rainfall (P) (in)
Rainfall (P) (in.)

0.8
0.6
0.4
0.2
0
0
20
40
60
Time (Min.)
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.
80
100
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)
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
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
0.7
0.6
Rainfall (P) (in.)

0.5
0.4
0.3
0.2
0.1
0
0
30
60
Time (Min.)
90
Determining volume of runoff
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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
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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 2600acre 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


Step 2 = Divide ac-in by acres


9100 cfs = 9027.78 ac-in
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

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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

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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
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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
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Lag time: (L or Tp) time from
the center of mass of rainfall to
the peak of the hydrograph
Time of Rise: (Tr) the time
from the start of rainfall excess
to the peak of the hydrograph
Time of Concentration: (Tc)
the time from the end of the net
rainfall to the inflection point of
the hydrograph
Time Base: (Tb) the total
duration of the DRO
hydrograph
Developing a Storm Hydrograph from a 1 hr Unit
Hydrograph-Unit Hydrograph Convolution
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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
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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
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
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
1
1.5
0
0.5
1.5
0
0.5
0
0.5
10
11
12
13
P2*U
P3*U
P4*U
P5*U
0.5
Q
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*U
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
1*0 = 0
1.5*40 = 60
0*90 = 0
0.5*160 = 80
1.5*0 = 0
0*40 = 0
0.5*90 = 45
0*0 = 0
0.5*40 = 20
10
11
12
13
P2*U
P3*U
P4*U
P5*U
0.5*0 = 0
Q
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

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
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
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
1*0 = 0
1.5*40 = 60
0*90 = 0
0.5*160 = 80
140
1.5*0 = 0
0*40 = 0
0.5*90 = 45
45
0*0 = 0
0.5*40 = 20
20
0.5*0 = 0
0
10
11
12
13
P2*U
P3*U
P4*U
P5*U
Q
0
50
160+100+0=260
225+320+150+0 = 695
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] cfs
Time
(hr)
P1 Un
0
0
1
50
0
2
160
100
0
3
225
320
150
0
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
0
60
0
80
140
0
0
45
45
0
20
20
0
0
10
11
12
13
P 2 Un
P 3 Un
P 4 Un
P 5 Un
Qn
0
50
Storm Hydrograph - Qn
260
695
Outflow (cfs)

1400
1200
1000
800
600
400
200
0
0
1
2
3
4
5
6 7 8
Time (hr)
9
10 11 12 13
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
2
1.49
Q=
AR 3 S
n
A
P = Wetted Perimeter
A
Pipe P = Circum.
A
Natural Channel
Q = Flowrate, cfs
n = Manning’s Roughness Coefficient (ranges from 0.015 - 0.15)
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
Manning’s variables for different crosssections
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 ft
n = 0.02
S = 0.002
b = 75 ft

Manning’s Equation is used to compute Q and from the
2
1.49
geometry provided:
=
3
Q=


n
AR
S
A=
= 4375 ft^2 P =
= 281.16 ft
Then all the values are plugged in to give Q = 28,730 cfs