Surface Water
The Lane Diagram
WATER
SEDIMENT
I. Events During Precipitation
A. Interception
B. Stem Flow
C. Depression Storage
D. Hortonian Overland Flow
E. Interflow
F. Throughflow -> Return Flow
G. Baseflow
II. Hydrograph
A. General
II. Hydrograph
A. General
B. Storm Hydrograph
II. Hydrograph
A. General
B. Storm Hydrograph
1. Shape and Distribution of “events”
direct ppt., runoff, baseflow, interflow
II. Hydrograph
A. General
B. Storm Hydrograph
1. Shape and Distribution of
2. Hydrograph Separation
II. Hydrograph
A. General
B. Storm Hydrograph
1. Shape and Distribution of
2. Hydrograph Separation
C. Predicting the rate of Baseflow Recession after a storm
vs.
Why care?
Predicting the rate of Baseflow Recession after a storm
Predicting the rate of Baseflow Recession after a storm
An example problem….
Gaining and Losing Streams…..
III. Rainfall-Runoff Relationships
III. Rainfall-Runoff Relationships
A. Time of Concentration
III. Rainfall-Runoff Relationships
A. Time of Concentration
“The time required for overland flow and channel flow to reach
the basin outlet from the most distant part of the catchment”
III. Rainfall-Runoff Relationships
A. Time of Concentration
“The time required for overland flow and channel flow to reach
the basin outlet from the most distant part of the catchment”
tc = L 1.15
7700 H 0.38
III. Rainfall-Runoff Relationships
A. Time of Concentration
“The time required for overland flow and channel flow to reach
the basin outlet from the most distant part of the catchment”
tc = L 1.15
7700 H 0.38
tc = time of concentration (hr)
L = length of catchment (ft) along the mainstream from basin mouth to headwaters
(most distant ridge)
H = difference in elevation between basin outlet and headwaters (most distant ridge)
L = 13,385 ft
H = 380 ft
III. Rainfall-Runoff Relationships
A. Time of Concentration
tc = L 1.15
example problem
7700 H 0.38
L = 31,385 ft
H = 380 ft
Tc = 0.75 hrs, or 45 minutes
tc = (13,385) 1.15
7700 (380) 0.38
tc = time of concentration (hr)
L = length of catchment (ft) along the mainstream from basin mouth to headwaters
(most distant ridge)
H = difference in elevation (ft) between basin outlet and headwaters (most distant ridge)
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
If the period of ppt exceeds the time of concentration, then the
Rational Equation applies
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
Q=CIA
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
Q=CIA
Where Q=peak runoff rate (ft3/s)
C= runoff coeffic.
I = ave ppt intensity (in/hr)
A = drainage area (ac)
First: solve for time
of concentration (“Duration”);
THEN: solve for
rainfall intensity for
a given X year storm.
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
The drainage basin that ultimately flows past the JMU football stadium
is dominated by an industrial park with flat roofed buildings,
parking lots, shopping malls, and very little open area.
The drainage basin has an area of 90 acres.
Find the peak discharge during a storm that has a 25 year flood return interval.
First: solve for time
of concentration (“Duration”);
THEN: solve for
rainfall intensity for
a given X year storm.
“45 minutes from previous
exercise”
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
The drainage basin that ultimately flows past the JMU football stadium
is dominated by an industrial park with flat roofed buildings,
parking lots, shopping malls, and very little open area.
The drainage basin has an area of 90 acres.
Find the peak discharge during a storm that has a 25 year flood return interval.
Q = ciA
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
The drainage basin that ultimately flows past the JMU football stadium
is dominated by an industrial park with flat roofed buildings,
parking lots, shopping malls, and very little open area.
The drainage basin has an area of 90 acres.
Find the peak discharge during a storm that has a 25 year flood return interval.
Q = ciA
Q = (0.85)*(2.5 in/hr)*(90 acres)
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
The drainage basin that ultimately flows past the JMU football stadium
is dominated by an industrial park with flat roofed buildings,
parking lots, shopping malls, and very little open area.
The drainage basin has an area of 90 acres.
Find the peak discharge during a storm that has a 25 year flood return interval.
Q = 191.3 ft3/s
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
The drainage basin that ultimately flows past the JMU football stadium
is dominated by an industrial park with flat roofed buildings,
parking lots, shopping malls, and very little open area.
The drainage basin has an area of 90 acres.
Find the peak discharge during a storm that has a 25 year flood return interval.
Calculate the mean velocity if the cross sectional area
of the channel is 40 ft2.
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
An industrial park with flat roofed buildings, parking lots, and very little
open area has a drainage basin area of 90 acres. The 25 year flood
has an intensity of 2 in/hr. Find the peak discharge during the storm.
Calculate the mean velocity if the cross sectional area
of the channel is 40 ft2.
Discharge = Velocity x Area
III. Rainfall-Runoff Relationships
A. Time of Concentration
B. Rational Equation
example problem
An industrial park with flat roofed buildings, parking lots, and very little
open area has a drainage basin area of 90 acres. The 25 year flood
has an intensity of 2 in/hr. Find the peak discharge during the storm.
Calculate the mean velocity if the cross sectional area
of the channel is 40 ft2.
Discharge = Velocity x Area
191.3 ft3/s = 40ft2 * V
V = 4.8 ft/s
Calculate the mean velocity if the cross sectional area
of the channel is 40 ft2.
Discharge = Velocity x Area
191.3 ft3/s = 40ft2 * V
V = 4.8 ft/s or 146.3 cm/s
If the channel is made of fine sand, will it remain stable?
Hjulstrom Diagram
146.3 cm/s
0.10-0.25 mm
(fine sand) size range
III. Measurement of Streamflow
III. Measurement of Streamflow
A. Direct Measurements
B. Indirect Measurements
III. Measurement of Streamflow
A. Direct Measurements
III. Measurement of Streamflow
A. Direct Measurements
1. Price /Gurley/Marsh-McBirney Current
Meters
III. Measurement of Streamflow
A. Direct Measurements
1. Price or Gurley Current Meter
2. Weirs
Weirs
rectangular
Q = 1.84 (L – 0.2H)H 3/2
Where L = length of weir crest (m),
H = ht of backwater above weir crest (m),
Q = m3/s
*note: eq. 2.16B in Fetter is
incorrect (exponent is 3/2 as
shown above)
Weirs
V notch
Q=1.379 H 5/2
Where H = ht of backwater above weir crest (m)
Q = m3/s
III. Measurement of Streamflow
B. Indirect Measurements
III. Measurement of Streamflow
B. Indirect Measurements
1.Manning Equation
V = R 2/3 S
n
½
Where
V = average flow velocity (m/s)
R = hydraulic radius (m)
S = channel slope (unitless)
n = Manning roughness coefficient
1.Manning Equation
V = R 2/3 S ½
n
Where
V = average flow velocity (m/s)
R = hydraulic radius (m)
S = channel slope (unitless)
n = Manning roughness coefficient
R = A/P
A = Area (m2)
P = Wetted Perimeter (m)
1.Manning Equation
If using English units…..
V = 1.49 * R 2/3 S ½
n
Where
V = average flow velocity (ft/s)
R = hydraulic radius (ft)
S = channel slope (unitless)
n = Manning roughness coefficient
R = A/P
A = Area (ft2)
P = Wetted Perimeter (ft)
V = R 2/3 S ½
n
If Q = V * A, then
Q = A R 2/3 S ½
n
Where
Q = average flow discharge (m3/s)
A = area of channel (m2)
R = hydraulic radius (m)
S = channel slope (unitless)
n = Manning roughness coefficient
R = A/P
A = Area
P = Wetted Perimeter
Q = A R 2/3 S ½
N
Where
Q = average flow discharge
A = area of channel
r = hydraulic radius
s = channel slope (unitless)
n = Manning roughness coefficient
R = A/P
A = Area
P = Wetted Perimeter
Example Problem:
A flood that occurred in a mountain stream comprised of cobbles, pebbles,
and few boulders creates a high water mark of 3 meters above the bottom of the
channel, and temporarily expands the channel width to 6 m. The slope of the
water surface is 100 meters of drop per 1 km of distance.
Determine V in m/s
Determine Q in m3/s
B. Indirect Measurements
1.Manning Equation
2. SuperElevation Method
3. Measurement of Cobbles
B. Indirect Measurements
1.Manning Equation
2. SuperElevation Method
B. Indirect Measurements
1.Manning Equation
2. SuperElevation Method
Q = A(Rc*g*cosS * tanΘ) ½
Q = discharge,
A = average radial cross section in the bend,
Rc= radius of curvature
S = slope of channel (degrees)
Θ = angle between high water marks on opposite banks (degrees)
Example Problem:
B. Indirect Measurements
3. Measurement of Cobbles
B. Indirect Measurements
3. Measurement of Cobbles
The “Costa Equation”
V = 0.18d 0.49
Where V =m/s, d=mm
where 50 < d < 3200 mm
Measure the 5 largest boulders, intermediate axis, take the average
B. Indirect Measurements
3. Measurement of Cobbles
V = 0.18d 0.49
Where V =m/s, d=mm and 50 < d < 3200 mm
Measure the 5 largest boulders, intermediate axis
And hc = {V
}1.5
{4.5*{(S + 0.001)0.17}}
Where V = velocity, in m/s
S energy slope (decimal form)
hc = competent flood depth (m)
Example Problem:
Average of five largest boulders: 3.2m x 2.3m x 1.6 m
Average slope = 5.5 degrees
Find: average velocity and depth of flow
V. Hydraulic Geometry
A. The relationships
Q = V*A
Q=V*w*d
w = aQb
d = cQ f
v = kQ m
V. Hydraulic Geometry
A. The relationships
B. “at a station”
C. “distance downstream”
A. Hydraulic Geometry
“at a station trends”
M = 0.26
M = 0.4
M = 0.34
A. Hydraulic Geometry
M = 0.5
M = 0.4
“distance downstream trends”
M = 0.1
Distance Downstream
Q
10.1
11.3
15
28.9
56.8
106
119
125
132
133
181
201
312
494
503
629
674
1100
1740
2930
W
33.3
31.5
38
38
40.7
29.1
44.5
42.5
42
30
43
43
55
70
66
73
71
72.5
75
215
d
V
0.71
0.67
0.79
0.94
1.1
2.14
1.95
1.58
1.9
2.08
1.9
2.04
2.24
6.04
3.47
3.7
4.55
4.97
5.56
3.38
0.43
0.54
0.5
0.81
1.27
1.7
1.37
1.86
1.66
2.13
2.22
2.29
2.54
1.17
2.2
2.33
2.09
3.06
4.17
4.04
250
width, depth and velocities
200
150
y = 14.904x0.2375
100
50
y = 0.2754x0.395
y = 0.2448x0.3669
0
0
500
1000
1500
2000
Q - Discharge
2500
3000
3500
1000
100
0.2375
width, depth and velocity (ft,s)
y = 14.904x
W
10
y = 0.2754x
0.395
y = 0.2448x0.3669
D
1
V
0.1
1
10
100
Discharge (cfs)
1000
10000
2.5
LOG TRANSFORM PLOT
y = 0.2375x + 1.1733
R2 = 0.6746
2
width, depth and velocities
1.5
y = 0.395x - 0.5601
R2 = 0.875
1
0.5
y = 0.3669x - 0.6112
R2 = 0.8359
0
-0.5
-1
0
0.5
1
1.5
2
Q Discharge
2.5
3
3.5
4
VI. Flood Frequency
A. Flood Frequency Analysis
B. Flow Duration Curves
VI. Flood Frequency
A. Flood Frequency Analysis
Flood recurrence interval (R.I.)
use Weibull Method - calculates the R.I.
by taking the average time between 2 floods of equal or
greater magnitude.
RI = (n + 1) / m
where n is number of years on record,
m is magnitude of given flood
VI. Flood Frequency
A. Flood Frequency Analysis
What does this mean???
the curve estimates the magnitude of a flood that can be
expected within a specified period of time
The probability that a flow of a given magnitude will occur during any
year is P = 1/RI.
EX: a 50 year flood has a 1/50th chance, or 2 percent chance
of occurring in any given year .
VI. Flood Frequency
A. Flood Frequency Analysis
For multiple years:
q = 1- ( 1-1/RI)n
where q = probability of flood with RI with a
specified number of years n
VI. Flood Frequency
A. Flood Frequency Analysis
For multiple years:
q = 1- ( 1-1/RI)n
where q = probability of flood with RI with a
specified number of years n
EX: a 50 year flood has an 86% chance of occurring over 100 years
Example Problem: Determine the water height during
a 100 year storm at the Harrison Gaging Station near
Grottoes, Virginia.
VI. Flood Frequency
A. Flood Frequency Analysis
Example Problem: Determine the water height during
a 100 year storm at the Harrison Gaging Station near
Grottoes, Virginia.
Method:
•Access data at www.usgs.gov; select water tab
•Select “water watch” under ‘streams, lakes, rivers’ option
•Choose the current stream flow map,
your state and the respective station location
•Open station page by clicking on the station number
•Select “surface water - peak streamflow” option
•Choose ‘tab separated file’ format
•Highlight, copy, and paste (special) your data to Excel
for analysis.
VI. Flood Frequency
A. Flood Frequency Analysis
Example Problem: Determine the water height during
a 100 year storm at the Harrison Gaging Station near
Grottoes, Virginia.
Method (continued):
•Clean up data so that only ‘Year’ , ‘Q’, and ‘Gage Ht.’ are present
•Sort data based on Q in descending order
•Add magnitude (m) ranking (highest = 1)
•Add RI formula, where RI = (n+1)/m
•Create graph depicting RI vs. Q
•Create graph of Q vs. Gage Ht.
•Determine Gage Height with respect to the given RI
Magnitud
Year
Q (cfs)
Gage Ht (ft) e
RI (yrs)
9/6/1996
28900
15.57
1
73.0
11/4/1985
28100
15.47
2
36.5
10/15/1942
23100
17.2
3
24.3
9/19/2003
22000
14.41
4
18.3
6/21/1972
21300
15.25
5
14.6
1924-05-00
21000
16.6
6
12.2
9/6/1979
16200
13.47
7
10.4
10/5/1972
15300
13.24
8
9.1
3/18/1936
12600
13.07
9
8.1
3/19/1975
12400
12.2
10
7.3
9/28/2004
12300
12.26
11
6.6
8/16/1940
12100
12.91
12
6.1
4/17/2011
11900
12.15
13
5.6
4/26/1937
11700
13
14
5.2
9/18/1945
11300
12.8
15
4.9
8/20/1969
11100
12.72
16
4.6
1/25/2010
11100
11.9
17
4.3
11/29/2005
10900
11.84
18
4.1
3/19/1983
10300
11.44
19
3.8
9/20/1928
10100
11.9
20
3.7
2/17/1998
10000
11.59
21
3.5
4/22/1992
9840
11.8
22
3.3
5/30/1971
9460
11.93
23
3.2
10/17/1932
8700
11.5
24
3.0
12/1/1934
8340
11.3
25
2.9
9/19/1944
8340
11.33
26
2.8
10/9/1976
8250
10.62
27
2.7
2/14/1984
8250
10.6
28
2.6
4/17/1987
8120
11.08
29
2.5
6/18/1949
7980
11.06
30
2.4
1/26/1978
7800
10.38
31
2.4
35000
y = 7030.7ln(x) + 671.64
30000
Discharge (cfs)
25000
20000
15000
10000
5000
0
1.0
10.0
Recurrence Interval (yrs)
100.0
20
18
y = 3.4469ln(x) - 19.854
16
14
Gage Height (ft)
12
10
8
6
4
2
0
0
5000
10000
15000
20000
Discharge (cfs)
25000
30000
35000
VI. Flood Frequency
B . Flow Duration Curves
VI. Flood Frequency
B. Flow Duration Curves
“shows the percentage of time that a given flow of a stream
will be equaled or exceeded.”
B . Flow Duration Curves
“shows the percentage of time that a given flow of a stream
will be equaled or exceeded.”
P = * m * (100)
n+1
B . Flow Duration Curves
“shows the percentage of time that a given flow of a stream
will be equaled or exceeded.”
P = * m * (100)
n+1
VI. Flood Frequency
B. Flow Duration
Example Problem: Determine the discharge that can be expected
80% of the time at the Harrison Gaging Station near
Grottoes, Virginia.
Method:
•Access data at www.usgs.gov; select water tab
•Select “water watch” under ‘streams, lakes, rivers’ option
•Choose the current stream flow map,
your state and the respective station location
•Open station page by clicking on the station number
•Select “daily data” option,
•Then click ‘mean discharge’ option
•Choose the earliest date of record through ‘present’
•Choose ‘tab separated file’ format, and select ‘go’
•Highlight, copy, and paste (special) your data to Excel
for analysis.
VI. Flood Frequency
A. Flood Frequency Analysis
Example Problem: Determine the discharge that can be expected
80% of the time at the Harrison Gaging Station near
Grottoes, Virginia.
Method (continued):
•Clean up data so that only ‘Year’ , ‘Q’, and ‘Gage Ht.’ are present
•Sort data based on Q in descending order
•Add magnitude (m) ranking (highest = 1)
•Add P formula, where P = m/(n+1)
•Pick out desired probability value, and record the respective
discharge
P = * m * 100
n+1
How much water would this value of discharge y
for a full day?
How much water would this value of discharge y
for a full day?
81 ft3
s
* 3600 s * 24 hr = 6,998,400 ft3 of wate
1 hr
1d
V. Sediment Transport
A. Shear Stress
τc = critical boundary shear stress
hc = minimal water depth required for flow
ρw = water density (assume 1.00 g/cm3)
g = gravitational acceleration (981 cm/s2)
S = slope (decimal e.g., meters per meters)
V. Sediment Transport
A. Shear Stress
τc = hc ρw g S
τc = critical boundary shear stress (force per unit area) (g/cm-s2)
hc = minimal water depth required for flow (cm)
ρw = water density (assume 1.00 g/cm3)
g = gravitational acceleration (981 cm/s2)
S = slope (decimal, e.g., meters per meters)
V. Sediment Transport
1. Shear Stress
2. The Shields Equation
τc = hc ρw g S
τc
= τ*c
(ρs – ρw)gD50
τc = critical boundary shear stress
hc = minimal water depth required for flow
ρs, ρw = grain density (assume 2.65 g/cm3) and water density
g = gravitational acceleration (981 cm/s2)
D50 = median bed material grain size
τ*c = dimensionless critical shear stress (the Shields number)
0.03 for sand, 0.047 for gravel
V.
Sediment Transport
1. Shear Stress
τc = hc ρw g S
2. Shields Equation
τc
= τ*c
(ρs – ρw)gD50
OR
hc =(ρs – ρw) τ*c D50
ρwS
τc = critical boundary shear stress
hc = minimal water depth required for flow
ρs, ρw = grain density (assume 2.65 g/cm3) and water density
g = gravitational acceleration (981 cm/s2)
D50 = median bed material grain size
τ*c = dimensionless critical shear stress (the Shields number)
0.03 for sand, 0.047 for gravel
Problem: A gravel bed stream of slope 2 m per 1 km has a median grain size
of 60 mm. Caculate:
1) the critical shear stress required for bedload mobilization;
2) The critical water depth to initiate motion
τc = hc ρw g S
τc
= τ*c
(ρs – ρw)gD50
OR
hc =(ρs – ρw) τ*c D50
ρwS
τc = critical boundary shear stress
hc = minimal water depth required for flow
ρs, ρw = grain density (assume 2.65 g/cm3) and water density
g = gravitational acceleration (981 cm/s2)
D50 = median bed material grain size
τ*c = dimensionless critical shear stress (the Shields number)
0.03 for sand, 0.047 for gravel