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1 Soils & Hydrology II 1 Soils & Hydrology II

1 Soils & Hydrology II - PowerPoint Presentation

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1 Soils & Hydrology II - PPT Presentation

Soil Water Precipitation and Evaporation Infiltration Streamflow and Groundwater Hydrologic Statistics and Hydraulics Erosion and Sedimentation Soils for Environmental Quality and Waste Disposal ID: 580966

flow runoff hydrograph soil runoff flow soil hydrograph water storage area peak number channel curve storm watershed soils increases

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Slide1

1

Soils & Hydrology II

Soil Water

Precipitation and Evaporation

Infiltration, Streamflow, and Groundwater

Hydrologic Statistics and Hydraulics

Erosion and Sedimentation

Soils for Environmental Quality and Waste Disposal

Issues in Water QualitySlide2

2

What is the significance of understanding streamflow?

Why are we concerned with how it relates to Landscapes?

Streamflow is important because it is related to:

Construction

of houses, bridges, spillways, and culverts

Surface Runoff

over landscapes, including flooding

The associated processes of

Erosion, Transport

, and

Deposition

.

Drinking

and

Irrigation

water supplies, especially during droughts

Recreational

activities, such as boating and fishing

Navigation

of commercial shipping and transportSlide3

3

Hydrograph:

Plots precipitation and runoff over time.

Runoff can be discharge, flow, or stageSlide4

4

Storm HydrographSlide5

5

Storm HydrographSlide6

6

Storm HydrographSlide7

7

Storm HydrographSlide8

8

Storm HydrographSlide9

9

Lag

TimeSlide10

10

Flow behavior for different streamsSlide11

11

Hydrograph BehaviorSlide12

12

Hydrograph Behavior:

Related to channel sizeSlide13

13

Hydrograph

for 1997 Homecoming Weekend

StormSlide14

14Slide15

15Slide16

16

Hydrograph Behavior:

Also related to channel patternsSlide17

17

Measurement Units

cfs: cubic feet per second

gpm: gallons per minute

mgd: million gallons per day

AF/day: Acre-Feet per day

cumec: cubic meters per second

Lps: liters per second

Lpm: liters per minute

1 cfs

2 AF/day

450 gpm

28.3 Lps

1 m

3

/s = 35.28 cfs

1 mgd

1.5 cfs

1 gpm = 3.785 LpmSlide18

18

WEIR: Used to provide accurate flow measurementsSlide19

19Slide20

20

Weir Types

Circular opening:

Q = c

r

2

h

1/2

Rectangular:

Q = c W h

3/2

Triangular:

Q = c h

5/2

where

Q is flow, cfs

c are weir coefficients

h is stage, ft

r is the pipe diameter, ft

W is the weir width, ftSlide21

21

Rectangular Weir

V-Notch (Triangular) Weir

Coweeta Hydrologic StationSlide22

22

Field Velocity Measurements

Flow Equation:

Q = v A

where

Q is the discharge, cfs

v is the water velocity, ft/s

A is the flow cross-sectional area, ft

2Slide23

23Slide24

24

Discharge MeasurementsSlide25

25

Manning's Equation

When flow velocity measurements are not available

v = (1.49/n) R

2/3

S

1/2

where

v is the water velocity, ft/s

n is the Manning's hydraulic roughness factor

R = A / P is the hydraulic radius, ft

A is the channel cross-sectional area, ft

2

P is the channel wetted perimeter, ft

S is the water energy slope, ft/ftSlide26

26

Hydrologic Statistics:

Trying to understand and predict streamflow

Peak Streamflow Prediction:

Our effort to predict catastrophic floods

Recurrence Intervals:

Used to assign probability to floods

100-yr flood:

A flood with a 1 chance in 100 years, or a flood with a probability of 1% in a year.Slide27

27

Return Period

T

r

= 1 / P

T

r

is the average recurrence interval, years

P is exceedence probability, 1/years

Recurrence Interval Formulas:

T

r

= (N+1) / m

Gringarten Formula: T

r

= (N+1-2a) / (m-a)

where

N is number of years of record,

a = 0.44 is a statistical coefficient

m is rank of flow (m=1 is biggest)Slide28

28

River Stage:

The elevation of the water surface

Flood Stage

The elevation when the river overtops the natural channel banks.Slide29

29

Bankfull Discharge

Q

bkf

= 150 A

0.63Slide30

30Slide31

31

Rating Curve

The relationship between river stage and dischargeSlide32

32Slide33

33Slide34

34

Peak Flows in Ungaged Streams

Q

n

= a A

x

P

n

where

A is the drainage area, and

P

n

is the n-year precipitation depth

Q

n

is the n-year flood flow

Q

2

= 182 A

0.622

Q

10

= 411 A

0.613

Q

25

= 552 A

0.610

Q

100

= 794 A

0.605Slide35

35

Channel flooding vs upland floodingSlide36

36

Curve Number Method

Most common method used in the U.S. for predicting stormflow peaks, volumes, and hydrographs.

Useful for designing ditches, culverts, detention ponds, and water quality treatment facilities. Slide37

37

P = Precipitation, usually rainfall

Heavy precipitation causes more runoff than light precipitation

S = Storage Capacity

Soils with high storage produce less runoff than soils with little storage.

F = Current Storage

Dry soils produce less runoff than wet soilsSlide38

38

r = Runoff Ratio => how much of the rain runs off?

r = Q / P

r = 0 means that little runs off

r = 1 means that everything runs off

r = Q / P = F / S

r = 0 means that the bucket is empty

r = 1 means that the bucket is full

F = P - Q or r = Q / P = (P - Q) / S

the soil fills up as it rains

Solving for Q yields:

Q = P

2

/ (P + S)Slide39

39

S is maximum available soil moisture

S = (1000 / CN) - 10

CN = 100 means S = 0 inches

CN = 50 means S = 10 inches

F is actual soil moisture content

F / S = 1 means that F = S, the soil is full

F / S = 0 means that F = 0, the soil is empty

Land Use CN S, inches

Wooded areas 25 - 83 2 - 30

Cropland 62 - 71 4 - 14

Landscaped areas 72 - 92 0.8 - 4

Roads 92 - 98 0.2 - 0.8Slide40

40

Curve Number Procedure

First we subtract the initial abstraction, I

a

, from the observed precipitation, P

Adjusted Rainfall: P

a

= P - I

a

No runoff is produced until rainfall exceeds the initial abstraction.

I

a

accounts for interception and the water needed to wet the organic layer and the soil surface.

The initial abstraction is usually taken to be equal to 20% of the maximum soil moisture storage, S, => I

a

= S / 5

The runoff depth, Q, is calculated from the adjusted rainfall, P

a

, and the maximum soil moisture storage, S, using:

Q = P

a

2

/ (P

a

+ S)

or use the graph and the curve number

We get the maximum soil moisture storage, S, from the Curve Number, CN:

S = 1000 / CN - 10

CN = 1000 / (S + 10)

We get the Curve Number from a Table. Slide41

41

Example

A typical curve number for forest lands is CN = 70, so the maximum soil storage is: S = 1000 / 70 - 10 = 4.29"

A typical curve number for a landscaped lawn is 86, and so

S = 1000 / 86 - 10 = 1.63”

A curve number for a paved road is 98, so S = 0.20”

Why isn’t the storage equal to zero for a paved surface?

The roughness, cracks, and puddles on a paved surface allow for a small amount of storage.

The Curve Number method predicts that I

a

= S / 5 = 0.04 inches of rain must fall before a paved surface produces runoff.

For a CN = 66, how much rain must fall before any runoff occurs?

Determine the maximum potential storage, S = 1000 / 66 - 10 = 5.15"

Determine the initial abstraction, I

a

= S

/ 5

= 5.15” / 5 = 1.03"

It must rain 1.03 inches before runoff begins.

If it rains 3 inches, what is the total runoff volume?

Determine the effective rainfall, P

a

= P - I

a

= 3" - 1.03" = 1.97"

Determine the total runoff volume, Q = 1.97

2

/ (1.97 + 5.15) = 0.545"Slide42

42

Unit HydrographsSlide43

43

Unit HydrographSlide44

44

Unit Area HydrographsSlide45

45

Unit Hydrograph Example

A unit hydrograph has been developed for a watershed

The peak flow rate is 67 L/s for 1 mm of runoff and an area of 100 ha

What is the peak flow rate for this same watershed if a storm produces 3 mm of runoff?

The unit hydrograph method assumes that the hydrograph can be scaled linearly by the amount of runoff and by the basin area.

In this case, the watershed area does not change, but the amount of runoff is three times greater than the unit runoff.

Therefore, the peak flow rate for this storm is three times greater than it is for the unit runoff hydrograph, or 3 x 67 L/s = 201 L/s.

What would be the peak flow rate for a nearby 50-ha watershed for a 5-mm storm?

Peak Flow: Q

p

= Q

o

(A / A

o

) (R / R

o

)

where

Q

p

is the peak flow rate and Q

o

is for a reference watershed,

A is the area of watershed and A

p

is the area of reference watershed.

Q = (67 L/s) (50 ha / 100 ha) (5 mm / 1 mm) = 168 L/s

In this case, the peak runoff rate was scaled by both the watershed area and the runoff amount. Slide46

46

Flood RoutingSlide47

47

Streamflow and Land Management

BMPs improve soil and water quality

Most of our attention is placed on preventing pollution, decreasing stormwater, and improving low flows.

Forestry

Forest streams have less stormflow and total flow, but more baseflow

Forest litter (O-Horizon) increases infiltration

Forest canopies intercept more precipitation (higher Leaf-Area Indices, LAI)

Forest have higher evapotranspiration rates

Forest soils dry faster, have higher total storageSlide48

48

Forest Management

Harvesting

High-lead yarding on steep slopes reduces soil compaction

Soft tires reduces soil compaction

Water is filtered using vegetated stream buffers (SMZs)

Water temperatures also affected by buffers

Roads

Road runoff can be dispersed onto planar and convex slopes

Broad-based dips can prevent road erosion

Site Preparation

Burning a site increases soil erosion and reduces infiltration

Leaving mulch on soils increases infiltration

Piling mulch concentrates nutrients into local "hot spots"

Distributing mulch returns nutrients to soils

Some herbicides cause nitrate increase in streamsSlide49

49

Forestry Compliance in Georgia with Water Quality Protection Standards (1991-2004)Slide50

50

Agricultural Land Management

Overland flow is a main concern in agriculture

increases soil erosion, nutrients, and fecal coliform

increases herbicides, pesticides, rodenticides, fungicides

Plowing

exposes the soil surface to rainfall (and wind) forces

mulching + no-till reduces runoff and increases infiltration

terracing and contour plowing also helps

Pastures (livestock grazing)

increases soil compaction

reduces vegetative plant cover

increases bank erosion

rotate cattle between pastures and fence streams

Urban Land Management

Urban lands have more impervious surfaces

More runoff, less infiltration, recharge, and baseflow

Very high peak discharges, pollutant loads

Less soil storage, channels are straightened and piped, no floodplains

Baseflows are generally lower, except for irrigation water (lawns & septic)Slide51

51

Benefits of Riparian Buffers

Bank Stability:

The roots of streambank trees help hold the banks together.

When streambank trees are removed, streambanks often collapse, initiating a cycle of sedimentation and erosion in the channel.

A buffer needs to be at least 15 feet wide to maintain bank stability.

Pollutant Filtration:

As dispersed overland sheet flow enters a forested streamside buffer, it encounters organic matter and hydraulic roughness created by the leaf litter, twigs, sticks, and plant roots.

The organic matter adsorbs some chemicals, and the hydraulic roughness slows down the flow.

The drop in flow velocity allows clay and silt particles to settle out, along with other chemicals adsorbed to the particles.

Depending on the gradient and length of adjacent slopes, a buffer needs to be 30-60 feet wide to provide adequate filtration. Slide52

52

Denitrification:

Shallow groundwater moving through the root zones of floodplains is subject to significant denitrificiation.

Removal of floodplain vegetation reduces floodplain denitrification

Shade:

Along small and mid-size streams, riparian trees provide significant shade over the channel, thus reducing the amount of solar radiation reaching the channel so summer stream temperatures are lower and potential dissolved oxygen levels are higher.

Buffers need to be at least 30 feet wide to provide good shade and microclimate control, but benefits increase up to 100 feet.

Organic Debris Recruitment:

River ecosystems are founded upon the leaves, conifer needles, and twigs that fall into the channel.

An important function of riparian trees is providing coarse organic matter to the stream system.

Buffers only need to encompass half the crown diameter of full-grown trees to provide this function. Slide53

53

Large Woody Debris Recruitment:

Large woody debris plays many important ecological functions in stream channels.

It helps scour pools, a favored habitat for many fish.

It creates substrate for macroinvertebrate and algae growth, and it forms cover for fish.

It also traps and sorts sediment, creating more habitat complexity.

Woody debris comes from broken limbs and fallen trees.

The width of a riparian buffer should be equal to half a mature tree height to provide good woody debris recruitment.

Wildlife Habitat:

Many organisms, most prominently certain species of amphibians and birds use both aquatic and terrestrial habitat in close proximity.

Maintaining a healthy forested riparian corridor creates important wildlife habitat.

The habitat benefits of riparian buffers increase out to 300 feet.