surface water. water sediment the lane diagram i. events during precipitation a. interception b....

Surface Water

WATER SEDIMENT

The Lane Diagram

http://csmres.jmu.edu/geollab/eaton/web/eaton_files/Courses/GEOL%20110/front%20page%20geol_110.htm

I. Events During PrecipitationA. InterceptionB. Stem Flow C. Depression StorageD. Hortonian Overland FlowE. InterflowF. Throughflow -> Return Flow

G. Baseflow

II. HydrographA. General

II. HydrographA. GeneralB. Storm Hydrograph


1. Shape and Distribution of “events”

direct ppt., runoff, baseflow, interflow


1. Shape and Distribution of 2. Hydrograph Separation


1. Shape and Distribution of 2. Hydrograph Separation

C. Predicting the rate of Baseflow Recession after a storm

Why care?

Predicting the rate of Baseflow Recession after a storm

An example problem….

Gaining and Losing Streams…..

III. Rainfall-Runoff Relationships

III. Rainfall-Runoff RelationshipsA. Time of Concentration

III. Rainfall-Runoff RelationshipsA. 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 = 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)

III. Rainfall-Runoff RelationshipsA. Time of Concentration

example problem

L = 13,385 ftH = 380 ft

tc = L 1.15

7700 H 0.38

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)

L = 31,385 ftH = 380 ft

Tc = 0.75 hrs, or 45 minutes

III. Rainfall-Runoff RelationshipsA. Time of ConcentrationB. Rational Equation


If the period of ppt exceeds the time of concentration, then the Rational Equation applies


Q=CIA


Q=CIAWhere Q=peak runoff rate (ft3/s)

C= runoff coeffic. I = ave ppt intensity (in/hr) A = drainage area (ac)

First: solve for timeof concentration (“Duration”);THEN: solve for rainfall intensity fora given X year storm.


example problem

The drainage basin that ultimately flows past the JMU football stadiumis 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 timeof concentration (“Duration”);THEN: solve for rainfall intensity fora given X year storm.

“45 minutes from previousexercise”


example problem

Q = ciA



example problem

Q = ciA

Q = (0.85)*(2.5 in/hr)*(90 acres)



example problem

Q = 191.3 ft3/s



example problem

Calculate the mean velocity if the cross sectional area of the channel is 40 ft2.



example problem

An industrial park with flat roofed buildings, parking lots, and very littleopen area has a drainage basin area of 90 acres. The 25 year floodhas an intensity of 2 in/hr. Find the peak discharge during the storm.


Discharge = Velocity x Area


example problem

An industrial park with flat roofed buildings, parking lots, and very littleopen area has a drainage basin area of 90 acres. The 25 year floodhas an intensity of 2 in/hr. Find the peak discharge during the storm.


Discharge = Velocity x Area191.3 ft3/s = 40ft2 * VV = 4.8 ft/s


Discharge = Velocity x Area191.3 ft3/s = 40ft2 * VV = 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 StreamflowA. Direct MeasurementsB. Indirect Measurements

III. Measurement of StreamflowA. Direct Measurements


1. Price /Gurley/Marsh-McBirney Current Meters


1. Price or Gurley Current Meter2. Weirs

Weirsrectangular

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 isincorrect (exponent is 3/2 asshown above)

WeirsV notch

Q=1.379 H 5/2

Where H = ht of backwater above weir crest (m)Q = m3/s

III. Measurement of StreamflowB. Indirect Measurements

III. Measurement of Streamflow B. Indirect Measurements

1.Manning Equation

V = R 2/3 S ½

n

WhereV = 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

nWhere

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

V = 1.49 * R 2/3 S ½

nWhere

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)

If using English units…..

Q = A R 2/3 S ½

nWhereQ = 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

If Q = V * A, then

V = R 2/3 S ½

n

Q = A R 2/3 S ½

NWhereQ = average flow dischargeA = area of channelr = hydraulic radiuss = 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/sDetermine Q in m3/s

B. Indirect Measurements1.Manning Equation2. SuperElevation Method3. Measurement of Cobbles

B. Indirect Measurements1.Manning Equation2. SuperElevation Method

B. Indirect Measurements1.Manning Equation2. SuperElevation Method

Q = A(Rc*g*cosS * tanΘ) ½

Q = discharge,A = average radial cross section in the bend,Rc= radius of curvatureS = slope of channel (degrees)Θ = angle between high water marks on opposite banks (degrees)

Example Problem:

B. Indirect Measurements3. Measurement of Cobbles


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

The “Costa Equation”


V = 0.18d 0.49

Where V =m/s, d=mm and 50 < d < 3200 mmMeasure the 5 largest boulders, intermediate axis

And hc = {V }1.5

{4.5*{(S + 0.001)0.17}}

Where V = velocity, in m/sS energy slope (decimal form) hc = competent flood depth (m)

Example Problem:Average of five largest boulders: 3.2m x 2.3m x 1.6 mAverage slope = 5.5 degrees

Find: average velocity and depth of flow

V. Hydraulic Geometry A. The relationships

Q = V*AQ = 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

“distance downstream trends”

M = 0.5

M = 0.1

M = 0.4

Q W d V10.1 33.3 0.71 0.4311.3 31.5 0.67 0.54

15 38 0.79 0.528.9 38 0.94 0.8156.8 40.7 1.1 1.27106 29.1 2.14 1.7119 44.5 1.95 1.37125 42.5 1.58 1.86132 42 1.9 1.66133 30 2.08 2.13181 43 1.9 2.22201 43 2.04 2.29312 55 2.24 2.54494 70 6.04 1.17503 66 3.47 2.2629 73 3.7 2.33674 71 4.55 2.09

1100 72.5 4.97 3.061740 75 5.56 4.172930 215 3.38 4.04

Distance Downstream

y = 14.904x0.2375

y = 0.2754x0.395

y = 0.2448x0.3669

0

50

100

150

200

250

0 500 1000 1500 2000 2500 3000 3500

Q - Discharge

wid

th, d

epth

an

d v

elo

citi

es

y = 0.2754x0.395

y = 0.2448x0.3669

y = 14.904x0.2375

0.1

1

10

100

1000

1 10 100 1000 10000

Discharge (cfs)

wid

th,

dep

th a

nd

vel

oci

ty (

ft,s

)

W

D

V

y = 0.2375x + 1.1733

R2 = 0.6746

y = 0.395x - 0.5601

R2 = 0.875

y = 0.3669x - 0.6112

R2 = 0.8359

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3 3.5 4

Q Discharge

wid

th, d

epth

an

d v

elo

citi

es

LOG TRANSFORM PLOT

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


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 .


For multiple years: q = 1- ( 1-1/RI)n

where q = probability of flood with RI with a specified number of years n


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 duringa 100 year storm at the Harrison Gaging Station nearGrottoes, 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.

http://www.usgs.gov/



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

Year Q (cfs) Gage Ht (ft)Magnitude RI (yrs)

9/6/1996 28900 15.57 1 73.011/4/1985 28100 15.47 2 36.510/15/194

2 23100 17.2 3 24.39/19/2003 22000 14.41 4 18.36/21/1972 21300 15.25 5 14.6

1924-05-00 21000 16.6 6 12.29/6/1979 16200 13.47 7 10.4

10/5/1972 15300 13.24 8 9.13/18/1936 12600 13.07 9 8.13/19/1975 12400 12.2 10 7.39/28/2004 12300 12.26 11 6.68/16/1940 12100 12.91 12 6.14/17/2011 11900 12.15 13 5.64/26/1937 11700 13 14 5.29/18/1945 11300 12.8 15 4.98/20/1969 11100 12.72 16 4.61/25/2010 11100 11.9 17 4.311/29/200

5 10900 11.84 18 4.13/19/1983 10300 11.44 19 3.89/20/1928 10100 11.9 20 3.72/17/1998 10000 11.59 21 3.54/22/1992 9840 11.8 22 3.35/30/1971 9460 11.93 23 3.210/17/193

2 8700 11.5 24 3.012/1/1934 8340 11.3 25 2.99/19/1944 8340 11.33 26 2.810/9/1976 8250 10.62 27 2.72/14/1984 8250 10.6 28 2.64/17/1987 8120 11.08 29 2.56/18/1949 7980 11.06 30 2.41/26/1978 7800 10.38 31 2.4

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 streamwill be equaled or exceeded.”

B . Flow Duration Curves“shows the percentage of time that a given flow of a streamwill be equaled or exceeded.”

P = * m n+1

* (100)

B . Flow Duration Curves“shows the percentage of time that a given flow of a streamwill be equaled or exceeded.”

P = * m * n+1

(100)

VI. Flood Frequency B. Flow Duration

Example Problem: Determine the discharge that can be expected80% of the time at the Harrison Gaging Station nearGrottoes, 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.

http://www.usgs.gov/


Example Problem: Determine the discharge that can be expected80% of the time at the Harrison Gaging Station nearGrottoes, 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 yield for a full day?

How much water would this value of discharge yield for a full day?

81 ft3 * 3600 s * 24 hr = 6,998,400 ft3 of water in one day s 1 hr 1 d

V. Sediment TransportA. Shear Stress

τc = critical boundary shear stress

hc = minimal water depth required for flowρw = water density (assume 1.00 g/cm3)

g = gravitational acceleration (981 cm/s2)S = slope (decimal e.g., meters per meters)

V. Sediment TransportA. 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 Transport1. Shear Stress2. 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 flowρs, 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 Transport1. Shear Stress

2. Shields Equation

τc = hc ρw g S

τc = τ*c

(ρs – ρw)gD50τc = critical boundary shear stress





hc =(ρs – ρw) τ*c D50

ρwSOR

τc = hc ρw g S

τc = τ*c

(ρs – ρw)gD50τc = critical boundary shear stress





hc =(ρs – ρw) τ*c D50

ρwSOR

Problem: A gravel bed stream of slope 2 m per 1 km has a median grain sizeof 60 mm. Caculate: 1) the critical shear stress required for bedload mobilization; 2) The critical water depth to initiate motion

surface water. water sediment the lane diagram i. events during precipitation a. interception b....

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