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Chapter 10
Precipitation and Evaporation
Causes of Air Movement
• Solar energy does not heat Planet Earth
uniformly
• Hotter air near the equator rises
• Colder air near the poles sinks
General Circulation of the
Atmosphere
Lapse Rates:
The 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: P V = n R T
• 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.
Lapse Rates
Typical Lapse Rates
• Dry air:
– 1°C / 100 m
– 5.5°F / 1000 ft
• Wet air:
– 0.50°C / 100 m
– 2.7°F / 1000 ft
• Average conditions
– 0.65°C / 100 m
– 3.5°F / 1000 ft
Lapse Rate Examples
• Question:
– 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?
• Answer:
– 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.
• Question:
– 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?
• Answer:
– 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.
• Question:
– 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?
• Answer:
– Use the wet adiabatic lapse rate
– This gives 32°F + 5.4°F = 37.4°F
• Question:
– You now have a great job in Arizona.
– Unfortunately, it is often 110°F in Tucson (at
2,000 feet of elevation) 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 day?
• Answer:
– What lapse rate should I use?
– What is the resulting change in temperature?
• Question:
– During the winter, find the temperature at the
ski lodge on Mt. Lemmon if the temperature in
Tucson is 50°F on an average day.
• Answer:
– ??
• Question:
– Why is the wet lapse rate less than the dry rate?
• Answer:
– 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.
Relationship between Saturation
Vapor Pressure and Temperature
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
• The equation is:
– es = 6.11 exp {17.3 T / (T+237.3) }
– where
• es is the saturation vapor pressure, in millibar
• T is the air temperature, in °C
Example
• The air temperature is
– T = 30°C
• The saturated (maximum) vapor pressure at this
temperature is
– es = 42.6 millibars.
• The actual vapor pressure is:
– ea = 17.1 millibars.
• The relative humidity is:
– RH = ea / es = 17.1 / 42.6 = 40 %
• The dewpoint temperature is where RH = 100%, or
when ea = es, which is at 15°C.
Why Does it Rain?
• Air is forced to rise - for 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).
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.
Types of Air Masses
• Fronts occur at the boundary of air masses
• The types of air masses are:
Warm
Cold
Wet
Maritime Tropical
Maritime Polar
Dry
Continental Tropical
Continental Polar
Frontal Storm
Cold Front
Warm Front
warm air
cold air
warm air
Occluding Front
Occluded Front
warm air
warm air
cold air
cold air
cool air
cool air
cold air
Convective Storm
Orographic Precipitation
Rain Shadow
Hurricanes
Cyclones
Typhoons
Effect of Area on the
Maximum Precipitation
Precipitation variation in the US
Relationship between
precipitation and climate
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Arid (desert): less than 10”/year
Semi-arid (Mediterranean): 10 to 20”/yr
Humid: 20 to 60”/yr
Moist: more than 60”/yr
Seasonal Distribution of
Temperature and Rainfall
Precipitation Intensity
Return Period
• Return Period = 1 / Probability
– Tr = 1 / P
• if P = .01 = 1%, then Tr = 100 years
• if P = .10 = 10%, then Tr = 10 years
– A 100-yr flood has a 1% probability each year
– The median flood has a 50% probability, or a
return period of 2 years.
– An average flood happens every 2.5 years or so.
Evapotranspiration
• Evaporation:
– Loss of water to the atmosphere by abiotic
processes from soil and water surfaces
• Transpiration:
– Loss of water to the atmosphere by biotic
processes (pumping of water through roots to
leaves through stomata)
• Canopy Interception:
– Loss of water to the atmosphere from plant
surfaces
Evapotranspiration
• Composed of:
– Evaporation from soil and water surfaces
• Large if soil is moist and there is no mulch or leaf cover!
– Transpiration through plant tissue
• Plant Factors: Leaf area, root depth, plant type
• Soil Factors: Available water
– Interception on plant surfaces
• Evaporation from plant surfaces
• About 10-20% of precip for hardwoods, more in pines
Evaporation
• Proportional to the Wind Speed
• Proportional to the Vapor Pressure Deficit
– The VPD is how dry it is
– A large deficit means the air is very 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)
Evaporation Pan
Average Pan Evaporation
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Seattle, WA = 30”/yr
Massachussetts = 35”/yr
Minnnesota = 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.
• Actual Evapotranspiration, AET
– The actual amount of evapotranspiration loss per
time for given area
– It depends on the crop (or vegetation), stage of
growth, soil moisture, and climatic variables.
– AET is usually less than PET unless soil is moist
and the crop is nearly mature.
Actual Evapotranspiration
• 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 !
Predicting AET
• AET = Cp ·PET
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PET is the pan evaporation
Cp is a pan coefficient
Approximately equal to 0.7, or 70 percent
A very rough approximation
Weighing Lysimeter
Streamflow Depletion
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.
Water Budget Components
Paired Watershed Studies
Energy Budget Approach
Effect of Sun Angle
Effect of Shelterbelt Harvesting
on Incident Solar Radiation
Effect of Forest Harvesting on
Stream Temperatures
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 = FC - WP is the Maximum Available Water
• F =  - WP 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)
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Depth of rooting zone: Ds = 20 cm
Bulk density: BD = 1.70 g/cm3
Field capacity: FC = 0.20
Wilting point: WP = 0.08
Crop factor: Kc = 0.8
Soil factor: Ks = F / S
AET = Kc Ks PET = 0.8 (F / S) PET
PET: given in table
Maximum water content: S = [FC - WP] ·BD ·Ds
– S = [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
Chapter 10 Quiz
1. If three inches of rain falls on 100 acres, this is equal to:
a.
b.
c.
d.
300 acre-inches of rain
1,089,000 cubic feet of rain
30,861 cubic meters of rain
25 acre-feet of rain
2. Name the four precipitation mechanisms
3. Which of the following combinations of air masses will
produce the maximum precipitation:
a.
b.
c.
d.
A continental maritime meeting a tropical polar
A continental polar meeting a continental maritime
A tropical maritime meeting a continental polar
A tropical continental meeting a polar continental
4. Is canopy interception more like evaporation or transpiration?
Explain your answer!!