Transcript Chapter 12

Chapter 12
Hydrologic Statistics and Hydraulics
Storm Hydrograph
• Stream Hydrographs:
– A plot of discharge (= flow rate) or stage (= water level)
versus time.
• Stormflow Hydrograph:
– A plot of discharge or stage before, during, and after a
specific storm.
• Rising Limb:
– The steep advance portion of the hydrograph that reflects
the onset of runoff
• Falling Limb:
– Flow that tapers off gradually following the peak.
• Peak Stormflow:
– Generally produced by surface runoff, either by partial
area contribution or Hortonian overland flow as well as
direct precipitation on the channels.
• Interflow:
– Flow that takes longer to reach the channel
– Dominates the falling limb of the hydrograph.
• Baseflow:
– The flow before and after the storm
– Generated principally by ground water discharge and
unsaturated interflow.
• Stormflow Volume:
– Total volume of streamflow associated with that storm
– It can be determined from the area under the hydrograph
when the hydrograph plots flow (not stage) vs time.
Hydrograph
for 1997
Homecoming
Weekend
Storm
Stream Hydrograph
Flow behavior for different streams
Hydrograph Behavior
Hydrograph Behavior:
Also related to
channel patterns
Streamflow
Variability
Measurement Units
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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
Useful Conversions
• 1 cfs 
– 2 AF/day
– 450 gpm
– 28.3 Lps
• 1 m3/s = 35.28 cfs
• 1 mgd  1.5 cfs
• 1 gpm = 3.785 Lpm
Weir Construction
Weir Types
Weir Equations
• Submerged Pipe:
– Q = c  r2 h1/2
• Rectangular Weir:
– Q = c W h3/2
• V-notch Weir:
– Q = c h5/2
• where
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–
–
–
–
Q is flow, cfs
c are weir coefficients
h is stage, ft
r is the pipe diameter, ft
W is the weir width, ft
Field Velocity Measurements
• Flow Equation:
– Q=vA
• where
– Q is the discharge, cfs
– v is the water velocity, ft/s
– A is the flow cross-sectional area, ft2
Discharge Measurements
Manning's Equation
• v = (1.49/n) R2/3 S1/2
• 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, ft2
• P is the channel wetted perimeter, ft
– S is the water energy slope, ft/ft
Manning’s Equation
• River Stage:
– The elevation of the water surface
• Flood Stage
– The elevation when the river overtops the natural
channel banks.
• Rating Curve
– The relationship between river stage and discharge
Rating Curve
• 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.
Return Period
• Tr = 1 / P
– Tr is the average recurrence interval, years
– P is exceedence probability, 1/years
• Recurrence Interval Formulas:
– Tr = (N+1) / m
– Gringarten Formula: Tr = (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)
Flood
Prediction
Peak Flows in Ungaged Streams
• Qn = a Ax Pn
• where
– A is the drainage area, and
– Pn is the n-year precipitation depth
– Qn is the n-year flood flow
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Q2 = 182 A0.622
Q10 = 411 A0.613
Q25 = 552 A0.610
Q100 = 794 A0.605
Bankfull Discharge
Qbkf = 150 A0.63
Curve Number Method
– Most common method used in the U.S. for
predicting stormflow peaks, volumes, and
hydrographs for precipitation events.
– It is useful for designing ditches, culverts,
detention ponds, and water quality treatment
facilities.
Curve Number Method
• 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 soils
• 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=F/S
• r = 0 means that the bucket is empty
• r = 1 means that the bucket is full
• F=P-Q
– the soil fills up as it rains
• Combining equations yields:
– Q = P (P - Q) / S
• Solving for Q yields:
– Q = P2 / (P + S)
• 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
Wooded areas
Cropland
Landscaped areas
Roads
CN
25 - 83
62 - 71
72 - 92
92 - 98
S, inches
2 - 30
4 - 14
0.8 - 4
0.2 - 0.8
Curve Number Procedure
– First we subtract the initial abstraction, Ia, from
the observed precipitation, P
• Adjusted Rainfall: Pa = P - Ia
– No runoff is produced until rainfall exceeds the
initial abstraction.
– Ia 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, => Ia = S / 5
– The runoff depth, Q, is calculated from the
adjusted rainfall, Pa , and the maximum soil
moisture storage, S, using:
• Q = Pa2 / (P_a + S)
• or by using 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.
Examples
• 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 Ia = S /
5 = 0.04 inches of rain must fall before a paved
surface produces runoff.
Another CN Example
• For a watershed with a curve number of 66, how
much rain must fall before any runoff occurs?
– Determine the maximum potential storage, S:
• S = 1000 / 66 - 10 = 5.15"
– Determine the initial abstraction, Ia
• Ia = 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, Pa
• Pa = P - Ia = 3" - 1.03" = 1.97"
– Determine the total runoff volume, Q
• Q = 1.972 / (1.97 + 5.15) = 0.545"
Unit Hydrographs
Unit Hydrograph
Flood Routing
Unit Area Hydrographs
Unit Hydrograph Example
• A unit hydrograph has been developed for a 100 hectare
watershed
• The peak flow rate for a storm that produces 1 mm of
runoff is 67 L/s
• 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: Qp = Qo (A / Ao ) (R / Ro )
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where
Qp is the peak flow rate,
Qo is peak flow for reference watershed,
A is the area of watershed,
Ap 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.
Forest Management
• Forest streams have less stormflow and total
flow
• 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 storage
BMPs improve soil and water quality
• 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 streams
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)
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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)
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.
• 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.
• 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.
Chapter 12 Quiz
1. A Curve Number of 95 is most likely typical of:
a. farmland b. forestland c. suburbs d. parking lot
2. Manning's equation is used to measure (circle any)
a. flow depth b. flow velocity c. stream discharge d. flow area
3.What is cross-sectional area, A, and discharge of a
stream, Q, if the average depth is 9”, the width is 20
feet, and the velocity is 27 ft/min (circle any)
a.
b.
c.
d.
A = 15 ft2
A = 150 in3
A = 1.5 ft
A = 180 in2
Q = 6.75 cfs
Q = 4,860 AF/yr
Q = 400 in/day
Q = 3000 gpm