Transcript Lecture Powerpoint

Soils & Hydrology II
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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 Quality
arid: < 10”/year
semi-arid: 10 - 20”/yr
humid: 20 - 60”/yr
moist: > 60”/yr
Causes of Air Movement
- Solar energy doesn’t heat Planet Earth uniformly
- Air rises near the Equator and Polar Fronts
- Air sinks near the Horse Latitudes and Poles
These result in “Hadley cells”
Sensible Heat: heat used to raise temperature
Latent Heat of Fusion: heat used to melt ice
Latent Heat of Vaporization: heat used to evaporate water
Absolute Humidity: mass of water vapor in a unit volume of air (mg/L)
Relative Humidity: ratio of actual vapor pressure to saturation vapor pressure
– Vapor Pressure: partial pressure of water vapor (mb)
– Saturation Vapor Pressure: maximum vapor pressure (mb)
Dewpoint Temperature: temperature at which the air is saturated, RH=100%
• Dewpoint Temperature
– The temperature to which a parcel of air with a given vapor pressure has to be cooled
in order to reach saturation.
– Air warmer than the dewpoint, RH < 100%
– Air cooler than dewpoint, RH > 100%, causes clouds and rain!
• Saturation Vapor Pressure
– The pressure is like a tea kettle, the more water in the air, the higher the pressure
– The saturation vapor pressure is the maximum amount of water that can be held
– Warm air holds more water than cold air
• Example:
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Actual vapor pressure = 17.1 mb
Air temp = 30°C
Saturated vapor pressure = 42.6 mb
Relative Humidity = 17.1 / 42.6 = 40%
Dewpoint = 15°C
mb = millibar, a unit of pressure
Lapse Rate: Change in temperature with altitude
• Dry air lapse rate:
– Holds for a clear, cloudless day.
– Air cools because the pressure drops
with altitude
– This can be blamed on the ideal gas law:
PV=nRT
• Wet air lapse rate:
– Holds for cloudy conditions.
– Wet air does not cool as quickly as dry
air because water vapor gives off heat as
it condenses, just like water absorbs
heat when it evaporates.
• Average, or environmental, lapse rate:
– The actual change in temperature with
altitude.
– The average rate is more typical for
partly cloudy conditions.
Dry air: 1°C / 100 m
5.5°F / 1000 ft
Wet air: 0.50°C / 100 m
2.7°F / 1000 ft
Average: 0.65°C / 100 m
3.5°F / 1000 ft
Lapse Rate Examples
• If the temperature in Athens is 40°F on a relatively dry fall day, what is the
likely temperature at an elevation 2000 feet higher in the Georgia
mountains?
– Use the dry adiabatic lapse rate of 5.5°F/1000 feet.
– For an elevation that is 2000 ft higher, this gives a temperature that is 11°F
cooler, or 29°F.
• If the temperature in Athens is 90°F on a humid summer day, what is the
likely temperature at 2000 feet in the Georgia mountains?
– Use the wet adiabatic lapse rate of 2.7°F per 1000 ft.
– This gives a temperature that is 5.4°F cooler, or 84.6°F.
• What does the temperature in Athens have to be on a rainy day for there to
be snow falling at an elevation of 2000 feet in the Georgia mountains?
– Use the wet adiabatic lapse rate
– This gives 32°F + 5.4°F = 37.4°F
• You now have a great job in Tucson, Arizona (elev = 2000’).
– Unfortunately, it’s often 110°F during the summer.
– You see Mt. Lemmon, which rises to 9,000 feet right outside of town.
– What is the temperature at the summit on a clear, dry, summer day?
– During the winter, find the temperature at the ski lodge if the temperature in
Tucson is 50°F.
• Why is the wet lapse rate less than the dry rate?
– As wet air rises, the atmosphere becomes saturated and the relative
humidity reaches 100%.
– To cool further requires that the atmosphere release some of it's moisture as
precipitation - rain if the air is above freezing, snow or ice if it’s below
freezing.
– The condensation of water releases heat - just as evaporation cools.
– This release of heat warms the air slightly, so the air does not cool as fast as
dry air would.
Why Does it Rain?
• Air is forced to rise (reasons described below!)
• Rising air cools because the ideal gas law says that the
temperature falls when the air pressure decreases.
• The air cools at the dry lapse rate until it reaches its dewpoint.
• Once the air reaches its dewpoint, the relative humidity reaches
100%, and clouds form.
• As the air continues to rise, the air cools at the wet lapse rate,
causing precipitation to form because the colder air can not hold
the excess moisture.
• The condensing water generates heat, causing the air to warm
slightly, so that the wet air lapse rate is less than the dry rate.
• The excess heat generated by the condensing water causes the
air to rise faster (because warmer air rises through colder air).
Raingages
Thiessen Polygons
• Used to estimate watershed precipitation
• Individual raingages are assigned the area
closest to them
• The area is found by:
– drawing lines between gages
– bisecting the lines and drawing
perpendiculars
– the volume of runoff is the depth for the
gage times the area.
Types of Precipitation Events
• Frontal
– when a cold air mass collides with a warm air mass.
– At least one of the air masses must be maritime.
• Convective
– when moist, warm (maritime tropical) air heats near the ground surface, it
warms, rises, cools, and releases its moisture as rain, hail, etc.
• Orographic
– when moist (maritime) air is forced upward over mountains, it cools,
releasing its moisture as rain or snow.
• Cyclones (hurricanes)
– when a self-sustaining (non frontal) low pressure system develops in the
tropics.
• Mesoscale Convective Complex
– Mid-latitude storm complex covers large area, but does not persist
• Lake Effect Storms
– Downwind of warm lake, lake evaporation increases rain and snow
Types of Air Masses
• Fronts occur at the boundary of air masses
• The types of air masses are:
Dry = Continental, c
Wet = Maritime, m
Warm = Tropical, T
cT = Continental Tropical
mT = Maritime Tropical
Cold = Polar, P
cP = Continental Polar
mP = Maritime Polar
Frontal Storms
Cold Front
Warm Front
Occluded Front
Convective Storm
Orographic Precipitation
Rain Shadow
Hurricanes
Cyclones
Typhoons
Mesoscale Convective Complex
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Occurs over a large area
Persists for many hours, then dies away
Not long-lived like a front or hurricane
Associated with heavy rains and flooding
Affects Midwest and sometimes Georgia
Lake Effect Storms
Seasonal Distribution of Temperature and Rainfall
Return Period, Tr = 1 / P
• A 100-yr flood has a 1% probability each year
• P = 1/ Tr = 1/100 = 0.01 = 1%,
• A 10-yr flood has a 10% probability
• P = 1/ Tr = 1/10 = 0.10 = 10%
• A median flood has a 50% probability, Tr = 2 years
• Tr = 1/P = 1 / 0.5 = 2 yrs
• An average flood happens every 2.5 years or so.
Precipitation Intensity
Effect of Area on the
Maximum Precipitation
Evapotranspiration
• Evaporation from soil & water surfaces:
– Loss of water to the atmosphere by abiotic processes
– Large if soil is moist and there’s no mulch or leaf cover!
• Transpiration through plant tissue:
– Loss of water to the atmosphere by biotic processes
• Plant Factors: Leaf area, root depth, plant type. Pumping of water
through roots to leaves through stomata
• Soil Factors: Plant Available Water
• Interception:
– Precipitation falling on plant surfaces that then evaporates
– About 10-20% of total precip for hardwoods, more in
pines
• Evaporation, a function of:
– wind speed
– vapor pressure deficit (VPD)
• The VPD is how dry it is, a large deficit means the air is dry
• VPD = es - ea = es ( 1 - RH )
– RH = ea / es is the relative humidity
– ea is the actual vapor pressure
– es is saturated (maximum) vapor pressure, f(temp)
Seattle, WA = 30”/yr
Massachussetts = 35”/yr
Minnesota = 30 to 45”/yr
Pennsylvania = 40”/yr
Rocky Mountains = 45”/yr
North Georgia = 55”/yr
Los Angeles, CA = 60”/yr
East Texas = 70 to 80”/yr
Tucson, Arizona = 95”/yr
West Texas = 100 to 120”/yr
Imperial Valley, CA = 120”/yr
• Potential Evapotranspiration, PET
– The maximum possible transpiration by plants with unlimited soil
moisture.
– We usually take a percentage (e.g., 70%) of pan evaporation to
estimate PET.
• Actual Evapotranspiration, AET
– The actual amount of evapotranspiration loss per time for given area
– Depends on the type of plant, stage of growth, soil moisture, and
climatic variables.
• AET is less than PET
– If no moisture in soil, then plants run out of water
– Plant responds by
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wilting
twisting the petiole so leaves are perpendicular to sun
flutter to help dissipate heat
close their leaves
– Pan has plenty of water, soil doesn’t!
Weighing Lysimeter
AET Equation
• AET = Kc · Ks · PET
– AET is actual evapotranspiration
– PET is potential (max) evapotranspiration
• From evaporation pans or models
– Kc is a crop factor - changes with time
• See next slide
– Ks is a soil factor - changes with soil moisture
• Ks = F / S, where
– F is how much water in soil
– S is how much water the soil can hold
Variation of the Crop Factor
Soil Factor
• We use a very simple approach:
– Ks = F / S
• S is the Maximum Available Water
• F is the Actual Available Water
• This means:
– If F = S then Ks = 1 and AET = PET
– If F = 0 then Ks = 0 and AET = 0
• We calculate the soil storage using:
– F = P - (Q + AET)
Water Budget Procedure
• Find the initial water storage in the root zone
– set equal to the field capacity, F(1) = S
– appropriate in the spring after soaking rains
• Calculate the soil factor
– Ks = F / S
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Calculate the AET = Kc Ks PET
Subtract AET from the soil storage, F' = F - AET
If rainfall, then add, F'' = F' + P
Subtract drainage and runoff if soil is too wet
– if F'' > S, then Q = F'' - S, and F''' = S
• Carry over soil moisture to next day
– say from end of day 1 to beginning of day 2
– F(2) = F'''(1)
Depth of rooting zone:
Bulk density:
Field capacity:
Wilting point:
Crop factor:
Soil factor:
Ds = 20 cm
BD = 1.70 g/cm3
FC = 0.20
WP = 0.08
Kc = 0.8
Ks = F / S
AET = Kc Ks PET = 0.8 (F / S) PET
PET:
Given in table
Max water content: S = [FC - WP] ·BD ·Ds
= [0.20 - 0.08]·1.70·20 = 4.1 cm
Initial water content: F(1) = S
Precipitation:
Given in table
Irrigation Scheduling Procedure
– Find 25% of the maximum available water:
• F* = 0.25 S = 0.25 · 4.1 cm = 1.02 cm
– Irrigate when F falls below F*:
• F < 1.02 cm on Day 9.
– Determine how much water to add to bring rooting
depth back to FC:
• I = S - F = 4.10 - 0.97 = 3.13 cm
– Determine how long to irrigate:
• t = 3.13 cm / 1 cm/hr = 3.13 hours
Water Budget Approach
• ET = P - R - I
– P is precipitation
– R is runoff
– I is interception
• Mature Hardwoods: P = 150, R = 70, I = 18, ET = 62 cm/yr
• White Pines:
P = 150, R = 52, I = 36, ET = 62 cm/yr
Paired Watershed Studies:
1. Select two watersheds of approximate equal size, shape
and aspect.
2. Monitor streamflow for several years and find the
correlation between the two.
3. Hold one watershed as the control, and alter the second
watershed, in this case by converting to grass.
4. Monitor the change in streamflow and compare to what
would have happened if the watershed had not been
treated.
Paired Watershed Studies
Effect of Sun Angle
Effect of Shelterbelt Harvesting
Effect of Forest Harvesting on
Stream Temperatures
Streamflow Depletion
Energy Budget Approach