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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

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.

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.

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 Front

warm air

cold air

cold air

warm air

Occluding Front

Occluded Front

warm air

warm air

cool air

cold air

cool air

cold air

Relationship between precipitation and climate

- Arid (desert): less than 10”/year
- Semi-arid (Mediterranean): 10 to 20”/yr
- Humid: 20 to 60”/yr
- Moist: more than 60”/yr

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

- 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

- 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)

Average Pan Evaporation

- 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.

- AET is less than PET
- If no moisture in soil, then plants run out of water
- Plant responds by
- 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
- PET is the pan evaporation
- Cp is a pan coefficient
- Approximately equal to 0.7, or 70 percent
- A very rough approximation

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

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.

- 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

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
- 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: 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. 300 acre-inches of rain

b. 1,089,000 cubic feet of rain

c. 30,861 cubic meters of rain

d. 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. A continental maritime meeting a tropical polar

b. A continental polar meeting a continental maritime

c. A tropical maritime meeting a continental polar

d. A tropical continental meeting a polar continental

4. Is canopy interception more like evaporation or transpiration? Explain your answer!!

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