Irrigation ManagementIrrigation Managementg gg gModule 6Module 6

IrrigationIrrigation SystemsSystemsIrrigation Irrigation SystemsSystemsandand

I i tiI i ti M tM tIrrigation Irrigation Management Management ininin in

Xeric Xeric GardensGardens

The Water CycleThe Water Cycle ETET

EvaporationEvaporation

yy

RunoffRunoff InfiltrationInfiltration

ratio

nra

tion

Root Root AbsorptionAbsorptionStorage inStorage in Tr

ansp

irTr

ansp

ir

Soil Surface AbsorptionAbsorptionStorage in Storage in

root zoneroot zone

TT

Deep DrainageDeep Drainage

Surface

Desirable PathwayDesirable Pathwayyy

Undesirable losses Undesirable losses

Goal of Efficient IrrigationGoal of Efficient Irrigation

Provide enough Provide enough supplemental water (above supplemental water (above precipitation) toprecipitation) to satisfy the plant’s requirements forsatisfy the plant’s requirements forprecipitation) to precipitation) to satisfy the plant s requirements for satisfy the plant s requirements for adequate growth and/or desired qualityadequate growth and/or desired quality..Prevent water wastePrevent water waste viavia runoff excessiverunoff excessivePrevent water waste Prevent water waste via via runoff, excessive runoff, excessive evaporation, evaporation, or or deep drainage below the plant’s root deep drainage below the plant’s root zonezonezone.zone.

XeriscapeXeriscape GardenGardenXeriscapeXeriscape Garden Garden Efficient Efficient IIrrigation Management rrigation Management g gg gSuccess depends on knowledge of three essential Success depends on knowledge of three essential

ttcomponents: components:

The output and efficiency of the irrigationThe output and efficiency of the irrigation systemsystemThe output and efficiency of the irrigation The output and efficiency of the irrigation system system (including individual emitters).(including individual emitters).

The waterThe water--holding characteristics of the soil. holding characteristics of the soil.

The estimated water required by each plant forThe estimated water required by each plant forThe estimated water required by each plant for The estimated water required by each plant for ‘acceptable’ growth and quality. ‘acceptable’ growth and quality.

The Irrigation SystemThe Irrigation Systemg yg y

To schedule irrigations effectively you must know theTo schedule irrigations effectively you must know theTo schedule irrigations effectively you must know the To schedule irrigations effectively you must know the output (water flow rate or water application rate) of output (water flow rate or water application rate) of the irrigation systemthe irrigation systemthe irrigation systemthe irrigation system..

The flow rate is usually measured in volume per unit time The flow rate is usually measured in volume per unit time ((i e gallons per minute [i e gallons per minute [gpmgpm] in sprinkler systems and gallons per hour] in sprinkler systems and gallons per hour((i.e. gallons per minute [i.e. gallons per minute [gpmgpm] in sprinkler systems and gallons per hour ] in sprinkler systems and gallons per hour [[gphgph] in drip systems).] in drip systems).

The application rate is usually measured in depth per unit The application rate is usually measured in depth per unit pp y p ppp y p ptime time (i.e. inches per hour [in/hr(i.e. inches per hour [in/hr]).]).

One can be converted to the other if the wetted area is One can be converted to the other if the wetted area is known known (measured).(measured).

Irrigation System Output Irrigation System Output (Flow Rate)(Flow Rate)

The maximum flow rateThe maximum flow rateDetermined by direct measurement from a Determined by direct measurement from a spigot spigot using a bucketusing a bucket and stopwatchand stopwatchusing a bucket using a bucket and stopwatch and stopwatch

Procedure:Procedure:Using a watch, time Using a watch, time how long it takes to how long it takes to fill a container having fill a container having a known volume. a known volume.

Example:Example: Suppose it takes 30 seconds to fill a Suppose it takes 30 seconds to fill a pp55--gallon gallon bucket.bucket.

Flow Flow rate = 5 gallons/0.5 minute = 10 rate = 5 gallons/0.5 minute = 10 gpmgpm..

Determining Determining the flow the flow rate of rate of an existing an existing irrigation system using a flow meter orirrigation system using a flow meter orirrigation system using a flow meter or irrigation system using a flow meter or installed water meterinstalled water meter

Most water meters show total Most water meters show total gallonsgallons..To determine the flow rate of To determine the flow rate of your irrigation systemyour irrigation system::

Insure only irrigation system is Insure only irrigation system is runningrunningrunningrunning..Record an initial flow meter Record an initial flow meter reading. reading. Run system for a timed periodRun system for a timed periodRun system for a timed period Run system for a timed period (i.e. 10, 15, 30 minutes for a (i.e. 10, 15, 30 minutes for a sprinkler system; a few hours for a sprinkler system; a few hours for a drip systemdrip system))RecordRecord the final meter readingthe final meter readingRecord Record the final meter readingthe final meter reading..Subtract the initial reading from Subtract the initial reading from the final reading.the final reading.DivideDivide the result bythe result by the time inthe time inDivide Divide the result by the result by the time in the time in minutes.minutes.

ExampleExample (Flow Meter)(Flow Meter)

Suppose the meter reading was 24,000 gallons Suppose the meter reading was 24,000 gallons before before turning on turning on the irrigation system and was the irrigation system and was 24,600 gallons after running the irrigation system 24,600 gallons after running the irrigation system for 30 minutesfor 30 minutesfor 30 minutes. for 30 minutes.

Total gallons used = 24,600 Total gallons used = 24,600 –– 24,000 = 600. 24,000 = 600.

Flow Flow rate in rate in gpmgpm = 600/30 = 20 = 600/30 = 20 gpmgpmThis This might be a typical flow rate for a small might be a typical flow rate for a small 88--sprinkler sprinkler g ypg yp ppirrigation zone (2.5 irrigation zone (2.5 gpmgpm from each sprinkler).from each sprinkler).

Drip SystemsDrip Systemsp yp yBecause of the very low flow rates of drip systems, you Because of the very low flow rates of drip systems, you

h t th t f 2 t 3 h t th t th t f 2 t 3 h t tmay have to run the system for 2 to 3 hours to get may have to run the system for 2 to 3 hours to get accurate flow rates with a water meteraccurate flow rates with a water meter..

Flow rates can be expressed in gallons per hour (Flow rates can be expressed in gallons per hour (gphgph) ) instead of gallons per minute (instead of gallons per minute (gpmgpm).).

Example with a flow (or water) meter:Example with a flow (or water) meter:Water meter reading before irrigation startWater meter reading before irrigation start--up = 24,000 gallonsup = 24,000 gallonsWater meter reading after running the system forWater meter reading after running the system forWater meter reading after running the system for Water meter reading after running the system for 2 hours = 24,300 gallons. 2 hours = 24,300 gallons. Total gallons used = 24,300 Total gallons used = 24,300 –– 24,000 = 30024,000 = 300Flow rate in Flow rate in gphgph = 300/2 = 150 = 300/2 = 150 gphgph..

•• This might be a typical flow rate of a drip system irrigating a moderate This might be a typical flow rate of a drip system irrigating a moderate sized (100 plant)sized (100 plant) xeriscapexeriscape garden Smaller plants may be irrigatedgarden Smaller plants may be irrigatedsized (100 plant) sized (100 plant) xeriscapexeriscape garden. Smaller plants may be irrigated garden. Smaller plants may be irrigated with only 1 emitter while larger plants may be irrigated with 2 or 3 with only 1 emitter while larger plants may be irrigated with 2 or 3 emitters. emitters.

D i E itt Fl R tD i E itt Fl R tDrip Emitter Flow RatesDrip Emitter Flow Rates

Drip emitters are designed to have a specific flow rate and Drip emitters are designed to have a specific flow rate and many are markedmany are marked (or color(or color--coded)coded) to designate that flow rate into designate that flow rate inmany are marked many are marked (or color(or color coded)coded) to designate that flow rate in to designate that flow rate in gallons per hour gallons per hour (i.e. 0.5, 1.0, 2.0, 5.0 (i.e. 0.5, 1.0, 2.0, 5.0 gphgph, etc.). , etc.).

To estimate the flow rate for the entire system just add up the To estimate the flow rate for the entire system just add up the y j py j pflow rates of the individual emitters. flow rates of the individual emitters.

If you’re uncertain If you’re uncertain (or, if you just want to verify) (or, if you just want to verify) you can measure you can measure th t t f h itt i t t hth t t f h itt i t t hthe output from each emitter using a stop watch the output from each emitter using a stop watch (or watch with a (or watch with a second hand) second hand) and a measuring cup. and a measuring cup.

ExampleExampleExampleExample

Place a measuring cup below the normally positionedPlace a measuring cup below the normally positionedPlace a measuring cup below the normally positioned Place a measuring cup below the normally positioned emitter and mark the time.emitter and mark the time.After After 1 minute, remove the cup and record the volume1 minute, remove the cup and record the volume..

Suppose Suppose you measured ¼ cup or 2 ounces.you measured ¼ cup or 2 ounces.Flow Flow rate = 2 ounces/minute or rate = 2 ounces/minute or 120 120 ounces/hour. ounces/hour. There There are 128 ounces in a gallon, soare 128 ounces in a gallon, so……

The The flow rate is about 1 gallon per hour flow rate is about 1 gallon per hour (120 oz./128 oz. (120 oz./128 oz. = 0.94 = 0.94 gphgph) )

Troubleshooting for LeaksTroubleshooting for Leaksgg

In buried irrigation systems, on sandy soils, it may be difficultIn buried irrigation systems, on sandy soils, it may be difficultIn buried irrigation systems, on sandy soils, it may be difficult In buried irrigation systems, on sandy soils, it may be difficult to visually detect a leak in the system.to visually detect a leak in the system.

If you know the output of each of your sprinklers, or emittersIf you know the output of each of your sprinklers, or emittersIf you know the output of each of your sprinklers, or emitters If you know the output of each of your sprinklers, or emitters (available from the dealer or manufacturer if the model number is (available from the dealer or manufacturer if the model number is known)known), you can add these up to get an estimate of what the , you can add these up to get an estimate of what the total flow rate should betotal flow rate should be (at a given pressure)(at a given pressure)total flow rate should be total flow rate should be (at a given pressure)(at a given pressure)..

If the calculated flow rate using your meter readings is much If the calculated flow rate using your meter readings is much more than the estimated theoretical flow rate there might bemore than the estimated theoretical flow rate there might bemore than the estimated theoretical flow rate, there might be more than the estimated theoretical flow rate, there might be an underground leak. an underground leak.

Precipitation Precipitation RateRate(Sprinklers)(Sprinklers)

The average precipitation rate of a sprinkler The average precipitation rate of a sprinkler system can be…system can be…yy

Calculated from the flow rate (Q) and Calculated from the flow rate (Q) and wetted area (A) orwetted area (A) orwetted area (A), or…wetted area (A), or…

Directly measured using catch cans.Directly measured using catch cans.

Sprinkler Precipitation RateSprinkler Precipitation Rate

Calculated from flow rate (Q) and wetted area (A).Calculated from flow rate (Q) and wetted area (A).(Q) ( )(Q) ( )Equation: PR = 96.3 x Q/AEquation: PR = 96.3 x Q/A

•• Where: PR = water application rate in inches/hourWhere: PR = water application rate in inches/hour•• Q = flow rate in gallons/minute (Q = flow rate in gallons/minute (gpmgpm))•• A = area in square feet (width x length, feet)A = area in square feet (width x length, feet)

E lE lExample:Example:•• Suppose measured flow rate = 20 Suppose measured flow rate = 20 gpmgpm•• Suppose wetted area = 50’ x 100’ = 5000 sq. ft.Suppose wetted area = 50’ x 100’ = 5000 sq. ft.pp qpp q•• Then PR = 96.3 x 20/5000 = 0.39 inches/hourThen PR = 96.3 x 20/5000 = 0.39 inches/hour

Converting precipitation rate to flow rateConverting precipitation rate to flow rateConverting precipitation rate to flow rateConverting precipitation rate to flow rate

Consider: It takes 0 623 gallons of water to coverConsider: It takes 0 623 gallons of water to coverConsider: It takes 0.623 gallons of water to cover Consider: It takes 0.623 gallons of water to cover 1 1 square foot to a depth of 1 inch, sosquare foot to a depth of 1 inch, so::

T t P i it ti R t t Fl R tT t P i it ti R t t Fl R tTo convert Precipitation Rate to Flow Rate:To convert Precipitation Rate to Flow Rate:FR = (PR x A x 0.623)/60FR = (PR x A x 0.623)/60

•• Where;Where;Where;Where;FR = flow rate in gallons per minute (FR = flow rate in gallons per minute (gpmgpm))PR = precipitation rate in inches per hourPR = precipitation rate in inches per hourA = wetted area in square feetA = wetted area in square feetA = wetted area in square feetA = wetted area in square feet

Example from previous slide: Example from previous slide: S h dS h d 00 f bf b 100100 f (Af (A 000000 f )f )•• Suppose the wetted area was Suppose the wetted area was 50 50 feet by feet by 100 100 feet (A = feet (A = 5,000 5,000 sq. ft.) sq. ft.)

•• The precipitation rate (PR) was The precipitation rate (PR) was 0.39 0.39 inches per hour.inches per hour.

SolutionSolution: FR = (: FR = (0 390 39 xx 50005000 x 0 623)/60 =x 0 623)/60 = 2020 gpmgpmSolutionSolution: FR = (: FR = (0.39 0.39 x x 5000 5000 x 0.623)/60 = x 0.623)/60 = 20 20 gpmgpm

Sprinkler SystemSprinkler Systemp yp yDirect Measurement of Precipitation RateDirect Measurement of Precipitation Rate

Set out a grid of Set out a grid of straightstraight--sided sided catch cans catch cans (i.e. tuna cans, soup (i.e. tuna cans, soup cans coffee cans etc )cans coffee cans etc )cans, coffee cans, etc.)cans, coffee cans, etc.)

Run the system for a timed Run the system for a timed period and then measure theperiod and then measure theperiod and then measure the period and then measure the water depth in straightwater depth in straight--sided sided cans with a ruler.cans with a ruler.

Sprinkler SystemSprinkler Systemp yp yCalculating Calculating average precipitation average precipitation raterate

To calculate To calculate the average precipitation rate:the average precipitation rate:Find the total Find the total of all measurements of all measurements and divide by the and divide by the number number of measurements.of measurements.

Example:Example:Suppose Suppose you set out 10 cans and you ran the sprinkler you set out 10 cans and you ran the sprinkler pppp y y py y psystem for 30 minutes. system for 30 minutes. Assume your measurements Assume your measurements from the cans werefrom the cans were: 0.25, 0.30, : 0.25, 0.30, 0.30, 0.35, 0.25, 0.20, 0.35, 0.40, 0.20, and 0.30 inches. 0.30, 0.35, 0.25, 0.20, 0.35, 0.40, 0.20, and 0.30 inches.

The average depth The average depth would would be: be: 0.29 inches (the 0.29 inches (the sum of the above sum of the above measurements [2.9]) measurements [2.9])

( /( / ))divided by 10 (2.9/10divided by 10 (2.9/10))

The The irrigation rate would be equal to 0.29 inches irrigation rate would be equal to 0.29 inches divided by 30 minutes (0 29/30) or 0 0097 inches perdivided by 30 minutes (0 29/30) or 0 0097 inches perdivided by 30 minutes (0.29/30) or 0.0097 inches per divided by 30 minutes (0.29/30) or 0.0097 inches per minute or 0.58 inches per hour (0.0097 x 60). minute or 0.58 inches per hour (0.0097 x 60).

SSDrip SystemsDrip Systems

With drip systemsWith drip systemsPR = FR/0.623/(D x D x 0.785)PR = FR/0.623/(D x D x 0.785)PR FR/0.623/(D x D x 0.785)PR FR/0.623/(D x D x 0.785)Where:Where:

•• PR = precipitation rate in inchesPR = precipitation rate in inches•• FR = emitter(s) flow rate in gallons/hourFR = emitter(s) flow rate in gallons/hour•• D = wetted diameter or plant canopy diameterD = wetted diameter or plant canopy diameter

Example: Example: Suppose the emitter flow rate (FR) = 1 Suppose the emitter flow rate (FR) = 1 gphgphpp ( )pp ( ) gpgpSuppose the plant diameter (D) = 3 ftSuppose the plant diameter (D) = 3 ftThen; PR = 1/0.623/(3 x 3 x 0.785) = 0.23 in/hrThen; PR = 1/0.623/(3 x 3 x 0.785) = 0.23 in/hr

The SoilThe SoilThe SoilThe Soil

Once you’ve determined Once you’ve determined the output of your the output of your irrigation system, you need to get an idea of your irrigation system, you need to get an idea of your soil’s water holding characteristicssoil’s water holding characteristicssoil’s water holding characteristics. soil’s water holding characteristics.

Soil WaterSoil Water--Holding CharacteristicsHolding Characteristics

S il h ld t lik !S il h ld t lik !Soil holds water like a sponge!Soil holds water like a sponge!

Available WaterAvailable Water

Saturated SoilSaturated SoilSaturated SoilSaturated Soil Field CapacityField CapacityField CapacityField Capacity Wilting PointWilting PointWilting PointWilting Point

Soil TextureSoil TextureSoil TextureSoil Texture

The amount of water a particular soil will hold The amount of water a particular soil will hold depends on it’s texture depends on it’s texture ((the percentage of sand, the percentage of sand, pp (( p g ,p g ,silt, and clay particles). silt, and clay particles).

Sand particles: 0.05 mm to 2.0Sand particles: 0.05 mm to 2.0 mmmmSand particles: 0.05 mm to 2.0 Sand particles: 0.05 mm to 2.0 mmmm

Silt particles: 0.002 mm to 0.05 Silt particles: 0.002 mm to 0.05 mmmm

Clay particles: less than 0.002 mmClay particles: less than 0.002 mm

Soil Water Holding CapacitySoil Water Holding CapacitySoil Water Holding Capacity Soil Water Holding Capacity is is Related to Soil TextureRelated to Soil Texture

Clayey Soil TextureClayey Soil Texture Sandy Soil TextureSandy Soil Texture

Small particlesLow water intake rateHigh water holding capacity

Larger particlesHigh water intake rateLow water holding capacity(li ht il)(heavy soil) (light soil)

Determining Soil Texture Determining Soil Texture ggFill a quart canning jar with about Fill a quart canning jar with about 4 4 inches of inches of tamped soil. tamped soil. Add water up to about 2 inches below rim of jar.Add water up to about 2 inches below rim of jar.Seal Seal with lid so it won’t leak.with lid so it won’t leak.Shake Shake continuously for about 10 continuously for about 10 --15 15 minutes.minutes.Set Set on on flat surface and flat surface and do not disturb.do not disturb.After After about 48 about 48 hours, the soil should settle out in the water hours, the soil should settle out in the water into three distinct layers. into three distinct layers. Measure Measure the total soil depth and the depths of each distinct the total soil depth and the depths of each distinct layer. layer. SandSand will be the bottom layerwill be the bottom layer (it settles out first)(it settles out first) silt the middlesilt the middleSand Sand will be the bottom layer will be the bottom layer (it settles out first)(it settles out first), silt the middle , silt the middle layer, and clay the top layer layer, and clay the top layer (it settles (it settles out lastout last)). . Calculate the percentages of sand, silt, and clay Calculate the percentages of sand, silt, and clay (see next page for example) (see next page for example)

ExampleExampleExampleExample

Suppose you measuredSuppose you measured 4 54 5 inches of totalinches of total soilsoilSuppose you measured Suppose you measured 4.5 4.5 inches of total inches of total soil soil depth, 2.5 depth, 2.5 inches of sandinches of sand (bottom layer)(bottom layer), 1.5 inch of silt , 1.5 inch of silt ((middle layer) middle layer) and 0.5 inch of clay and 0.5 inch of clay (top layer(top layer).).yy

The texture would be The texture would be 56% 56% sand (sand (2.5 2.5 ÷÷ 4.5 x 100), 33% 4.5 x 100), 33% silt silt ((1.51.5 ÷÷ 4.5 x 100),4.5 x 100), andand 11%11% clayclay ((0.50.5 ÷÷ 4.5 x 100).4.5 x 100).((1.5 1.5 4.5 x 100), 4.5 x 100), and and 11% 11% clay clay ((0.5 0.5 4.5 x 100).4.5 x 100).

To define the soil type see the soil texture pyramidTo define the soil type see the soil texture pyramid–– next slide. next slide.

Soil Texture PyramidSoil Texture Pyramidyy

1 D li2. Draw another

1. Draw a line straight across from % clay scale (example = 11%).

line from upper right to lower left from the % silt scale (example = 33%)( p %) (example = 33%).

Intersection: sandy loam soil (58% sand)

Water Holding Capacity Water Holding Capacity

Once the soil texture has been determined,Once the soil texture has been determined,

g p yg p y

Once the soil texture has been determined,Once the soil texture has been determined,the approximate available water holding the approximate available water holding capacity can be estimated.capacity can be estimated.p yp y

Approximate Available Soil Moisture Approximate Available Soil Moisture ppppin Various Textured Soilsin Various Textured Soils

Soil TextureSoil Texture In./In. In./In. (available moisture)(available moisture)

In./Ft. In./Ft. (available moisture)(available moisture)

Coarse sand and gravelCoarse sand and gravel .04 .04 0.50.5SandsSands .07.07 0.80.8Loamy sandsLoamy sands .09.09 1.11.1Sandy loamsSandy loams .13.13 1.51.5Fine sandy loamsFine sandy loams .16.16 1.91.9Loams and silt loamsLoams and silt loams .20.20 2.42.4

Our example

Clay loams & silty clay Clay loams & silty clay loamsloams

.18.18 2.12.1

SiltySilty clays and claysclays and clays .16.16 1.91.9

Feel Test for Soil Type and MoistureFeel Test for Soil Type and Moistureypyp% Water% Water Sandy Sandy

(coarse)(coarse)*Loamy *Loamy

(average)(average)ClayeyClayey(fi )(fi )(coarse)(coarse) (average)(average) (fine)(fine)

00--25%25% Dry,loose, slips Dry,loose, slips between fingersbetween fingers

Powdery, maybe Powdery, maybe forming a slightly forming a slightly solid crustsolid crust

Dry, cracked, not Dry, cracked, not easily reduced to easily reduced to powderpowder

Of Field Capacity

solid crustsolid crust powderpowder

2525--50%50% Seems dry, does Seems dry, does not form a ball not form a ball

h dh d

Brittle but sticks Brittle but sticks together when together when

dd

Fairly plastic, Fairly plastic, forms a ball when forms a ball when

ddwhen squeezedwhen squeezed squeezedsqueezed squeezedsqueezed

5050--75%75% Forms a loose ball Forms a loose ball when squeezed when squeezed

Forms a plastic Forms a plastic ball, sticky when ball, sticky when

Forms a ball, can Forms a ball, can be stretched be stretched

but easily falls but easily falls apartapart

squeezedsqueezed between the between the thumb and index, thumb and index, smooth to touchsmooth to touch

7575--100%100% Forms coherent Forms coherent ball, not smoothball, not smooth

Forms a very Forms a very plastic ball, easily plastic ball, easily smoothedsmoothed

Same as aboveSame as above

*Loam soil is best.*Loam soil is best.

Soil Intake RateSoil Intake Rate

Due to the larger particle size and hence largerDue to the larger particle size and hence largerDue to the larger particle size and hence larger Due to the larger particle size and hence larger pore spaces, a sandy soil will absorb water faster pore spaces, a sandy soil will absorb water faster th l ilth l ilthan a clayey soil.than a clayey soil.

Soil Intake RatesSoil Intake RatesSoil Intake RatesSoil Intake Rates

Basic Intake RatesBasic Intake Rates

Coarse sandCoarse sand 0.75 0.75 –– 1.00 in/hr1.00 in/hrFine sandsFine sands 0.5 0.5 –– 0.75 in/hr0.75 in/hrFine sandy loamsFine sandy loams 0 350 35 0 50 in/hr0 50 in/hrFine sandy loamsFine sandy loams 0.35 0.35 –– 0.50 in/hr0.50 in/hrSilt loamsSilt loams 0.25 0.25 –– 0.40 in/hr0.40 in/hrClay loamsClay loams 0.10 0.10 –– 0.30 in/hr0.30 in/hr

Water Water PercolationPercolationii Diff S ilDiff S il TTin in Different Soil Different Soil Textures Textures (Drip Irrigation)(Drip Irrigation)

SummarySummary

Because a clay soil holds more water than a sandy soil, Because a clay soil holds more water than a sandy soil, aa greater ol me of ater m st be applied to a cla soilgreater ol me of ater m st be applied to a cla soila a greater volume of water must be applied to a clay soil greater volume of water must be applied to a clay soil to to penetrate to an equal depthpenetrate to an equal depth..

However because of the lower intake rate of a clay soilHowever because of the lower intake rate of a clay soilHowever, because of the lower intake rate of a clay soil, However, because of the lower intake rate of a clay soil, the the water application rate (precipitation rate) water application rate (precipitation rate) must be must be less less than than on the sandy soil to prevent runoff or excessive on the sandy soil to prevent runoff or excessive y py ppuddlingpuddling. .

Once an equal water penetration depth is reached on both Once an equal water penetration depth is reached on both soilssoils, , the the required irrigation frequency will be less on the required irrigation frequency will be less on the clay soil since it holds more plant available water than the clay soil since it holds more plant available water than the sandy soil. sandy soil. yy

NotesNotes

Most native, xeric plants prefer a sandy soil with good drainage. Most native, xeric plants prefer a sandy soil with good drainage.

Root rot may occur in heavy clay soils that hold too much waterRoot rot may occur in heavy clay soils that hold too much water..

If irrigated properlyIf irrigated properly using drip irrigation deepusing drip irrigation deep drainage anddrainage andIf irrigated properly, If irrigated properly, using drip irrigation, deep using drip irrigation, deep drainage and drainage and runoff will not occur in a runoff will not occur in a XeriscapeXeriscape™™. .

Determining Soil Chemistry & FertilityDetermining Soil Chemistry & Fertilityg y yg y y

SampleSample

Examine Sampling Kit can be obtained Sampling Kit can be obtained from NMSU Cooperativefrom NMSU Cooperativefrom NMSU Cooperative from NMSU Cooperative Extension Extension Services throughout Services throughout the state.the state.

Ex. NMSU CES San Juan Ex. NMSU CES San Juan County 213County 213--A S. Oliver Ave. A S. Oliver Ave. AztecAztec, , NM 87410NM 87410PhonePhone: : 505505--334334--94969496

I Fi ld S il M i tI Fi ld S il M i tIn Field Soil MoistureIn Field Soil Moisture

During the year, periodically examine During the year, periodically examine the top foot of your soil the top foot of your soil near near the base of your plants using a long screwdriver or the base of your plants using a long screwdriver or i il di il dsimilar rod. similar rod.

By experiencing what the soil feels like when it’s wet and when By experiencing what the soil feels like when it’s wet and when it’s dry you’ll be able to estimate how much water each plantit’s dry you’ll be able to estimate how much water each plantit s dry, you ll be able to estimate how much water each plant it s dry, you ll be able to estimate how much water each plant is using and you’ll be able to manage your irrigations much is using and you’ll be able to manage your irrigations much more effectively. more effectively.

Plant Water RequirementsPlant Water RequirementsPlant Water RequirementsPlant Water Requirements

S i i h l d i f i f f lS i i h l d i f i f f lSaving water in the landscape is more a function of careful Saving water in the landscape is more a function of careful irrigation irrigation management than of plant selectionmanagement than of plant selection..

Many xeric (droughtMany xeric (drought--tolerant) plants can (and will) use just tolerant) plants can (and will) use just as as much water as nonmuch water as non--xeric xeric plants, plants, if this water is available if this water is available to to themthem..

The difference is The difference is –– xeric plants can survive and prosper under xeric plants can survive and prosper under lowlow--waterwater conditions that would be detrimental to (or kill)conditions that would be detrimental to (or kill)lowlow water water conditions that would be detrimental to (or kill) conditions that would be detrimental to (or kill) species species not not adapted to arid conditions. adapted to arid conditions.

Additi l C id tiAdditi l C id tiAdditional ConsiderationsAdditional Considerations

With landscape plants, economic With landscape plants, economic yield yield or plant size is not of or plant size is not of primary primary concernconcern..p yp y

The primary goal is to provide just enough water, fertilizer, etc. The primary goal is to provide just enough water, fertilizer, etc. toto result in an acceptable plant specimen for a quality landscaperesult in an acceptable plant specimen for a quality landscapeto to result in an acceptable plant specimen for a quality landscape. result in an acceptable plant specimen for a quality landscape.

EvapotranspirationEvapotranspiration (ET)(ET)

Evaporation Evaporation –– loss of water directly from soil and plant surfacesloss of water directly from soil and plant surfaces..

Transpiration Transpiration –– loss of water from plant internal tissues through loss of water from plant internal tissues through the the leaf leaf stomatesstomates, during , during photosynthesis.photosynthesis.

ET is related to…ET is related to…Atmospheric demands Atmospheric demands (weather(weather))

Air Air temperaturetemperature

Solar Solar radiationradiation

Relative Relative humidityhumidity

WindWindWindWind

Generally, ET increases with higher Generally, ET increases with higher temperature, temperature, greatergreater solar radiation higher windsolar radiation higher wind andand lower humiditylower humiditygreater greater solar radiation, higher wind, solar radiation, higher wind, and and lower humidity. lower humidity.

ET is also related to…ET is also related to…

Size of the plantSize of the plant..

Live leaf area (canopy) of the plantLive leaf area (canopy) of the plant..

Microhabitat (i e microclimate)Microhabitat (i e microclimate) and other factorsand other factorsMicrohabitat (i.e. microclimate) Microhabitat (i.e. microclimate) and other factors. and other factors.

Estimating ETEstimating ETEstimating ET Estimating ET for for Irrigation SchedulingIrrigation Scheduling

For many years, irrigation managers have used climateFor many years, irrigation managers have used climate--based based referencereference (or potential)(or potential) ETETRR values, along with correction factorsvalues, along with correction factorsreference reference (or potential) (or potential) ETETRR values, along with correction factors values, along with correction factors ((cropcrop--coefficients or Kcoefficients or KCC values)values) to help them efficiently schedule to help them efficiently schedule irrigations on agricultural crops and irrigations on agricultural crops and turfgrassturfgrass..

Due to a lack of research, the technique has received limited Due to a lack of research, the technique has received limited use for scheduling irrigations on landscape use for scheduling irrigations on landscape plants, particularly plants, particularly

iixeriscapesxeriscapes. .

H th tH th t kkHow the system How the system worksworks

A reference ET (A reference ET (ETETRR) ) is calculated from weather data is calculated from weather data collected collected from from an approved, standard weather station. an approved, standard weather station. SeeSee:: http://weather nmsu eduhttp://weather nmsu eduSeeSee: : http://weather.nmsu.eduhttp://weather.nmsu.edu..ETETRR is then multiplied by a correction factor is then multiplied by a correction factor (crop coefficient) (crop coefficient) to to estimate the ET of a crop based on the plants size, canopy estimate the ET of a crop based on the plants size, canopy

t it tt it tarea, maturity, etc. area, maturity, etc. Generally, for alfalfa the crop coefficient (Generally, for alfalfa the crop coefficient (KKCC) ) is equal to is equal to 11 duringduring mostmost of the yearof the year..1 1 during during most most of the yearof the year..

Cool Season Turf: 0.8, Warm season turf: Cool Season Turf: 0.8, Warm season turf: 0.60.6

The canopy area The canopy area for alfalfa and grass is usually considered for alfalfa and grass is usually considered pypy g yg yto be 100% of the to be 100% of the (irrigated)(irrigated) areaarea..The The canopy area for a wellcanopy area for a well--developed, mature developed, mature XeriscapeXeriscape™™

may be 60% or less of the totalmay be 60% or less of the total (irrigated)(irrigated) areaareamay be 60% or less of the total may be 60% or less of the total (irrigated) (irrigated) area. area.

Automated Weather Automated Weather StationStation –– Calculation of ETCalculation of ETOO

Wind DirectionWind Speed

Solar Radiation

D t L

Air Temperature

Relati e H miditData Logger Relative HumidityPrecipitationSolar Charger

12V Battery

XeriscapeXeriscape™™ Water RequirementsWater Requirements

Based on Based on observations observations of of differentially irrigated xeric differentially irrigated xeric l t t thl t t th A i lt l S i C t t F i tA i lt l S i C t t F i tplants at the plants at the Agricultural Science Center at Farmington, Agricultural Science Center at Farmington,

a a crop coefficient of between 0.2 and 0.4 should be crop coefficient of between 0.2 and 0.4 should be sufficientsufficient for most xeric plantsfor most xeric plants (compared to(compared to 1 0 for alfalfa1 0 for alfalfasufficient sufficient for most xeric plants for most xeric plants (compared to (compared to 1.0 for alfalfa 1.0 for alfalfa and 0.8 and 0.8 and 0.6 for and 0.6 for cool season and warm season turf, cool season and warm season turf, respectively).respectively).

AlsoAlso, the canopy area for a well, the canopy area for a well--developed, mature developed, mature XeriscapeXeriscape™™ is generally considered to be about 60% of is generally considered to be about 60% of pp g yg ythe the landscape landscape (compared to 100% for alfalfa and grass(compared to 100% for alfalfa and grass).).

Drip IrrigationDrip Irrigation SchedulingSchedulingDrip Irrigation Drip Irrigation SchedulingSchedulingConsiderations Considerations

Reference ET and irrigation recommendations are usually Reference ET and irrigation recommendations are usually provided in inches but drip emitter flow rates are expressed in provided in inches but drip emitter flow rates are expressed in gallonsgallonsgallonsgallons..It takes 0.623 gallons of water to cover 1 square foot to a It takes 0.623 gallons of water to cover 1 square foot to a depth depth of of 1 inch1 inch..With xeric landscape plants, the wetted area (or plant canopy With xeric landscape plants, the wetted area (or plant canopy area) is usually circulararea) is usually circular..Form la for con erting ET (or irrigation or precipitation) depthForm la for con erting ET (or irrigation or precipitation) depthFormula for converting ET (or irrigation or precipitation) depth Formula for converting ET (or irrigation or precipitation) depth in inches in inches to gallons for a landscape plant:to gallons for a landscape plant:

Gallons = Gallons = ET ET x 0.623 x Ax 0.623 x A•• Where:Where:

ET ET = = Irrigation Irrigation or or precipitation precipitation depth in inchesdepth in inchesA = wetted or plant canopy area in square feet A = wetted or plant canopy area in square feet

A f i l di t dA f i l di t d (d(d22) 0 785) 0 785•• Area of a circle = diameter squared Area of a circle = diameter squared (d(d22) x 0.785) x 0.785

SummarySummary

Determining the water requirement in gallons Determining the water requirement in gallons for a for a xeriscapexeriscape plant:plant:

Gallons = ETGallons = ET x Kx K x 0 623 x CAx 0 623 x CAGallons = ETGallons = ETR R x Kx KC C x 0.623 x CAx 0.623 x CAWhere:Where:

•• ETET = reference ET from weather station data (inches)= reference ET from weather station data (inches)•• ETETRR = reference ET from weather station data (inches)= reference ET from weather station data (inches)•• KKCC = correction factor (or crop coefficient)= correction factor (or crop coefficient)•• 0.623 = constant to convert inches to gallons0.623 = constant to convert inches to gallons•• CA = crop canopy area in square feetCA = crop canopy area in square feet

ClimateClimate--Based Irrigation Scheduling Based Irrigation Scheduling E l P bl fE l P bl f X iX i ™™ Pl tPl tExample Problem for a Example Problem for a XeriscapeXeriscape PlantPlant

Given: Given: The reference ET (ETR) for the week is 2.0 inches.The reference ET (ETR) for the week is 2.0 inches.The The plant plant to be irrigated has a canopy diameter to be irrigated has a canopy diameter (D) of 4 (D) of 4 feetfeet..The crop coefficient The crop coefficient (K(KCC) or ) or correction correction factor factor is 0.3. is 0.3. There is one 2 There is one 2 gphgph emitter emitter at the base of the plantat the base of the plant..

QuestionsQuestions::What is the plant’s canopy What is the plant’s canopy area (CA)? area (CA)?

AA DD DD 0 785 4 4 0 785 12 56 ft0 785 4 4 0 785 12 56 ft•• Area = Area = D D x x D D x 0.785 = 4 x 4 x 0.785 = 12.56 sq. ft. x 0.785 = 4 x 4 x 0.785 = 12.56 sq. ft. How many gallons should be provided How many gallons should be provided per week to per week to satisfy the satisfy the recommended irrigation recommended irrigation (or plant ET) depth?(or plant ET) depth?

•• Gallons = ETGallons = ETRR x x KKCC x 0.623 x A = x 0.623 x A = 2 x 0.3 2 x 0.3 x 0.623 x x 0.623 x 12.56 = 4.7 12.56 = 4.7 galsgals..

How long will the system need to be run to provide theHow long will the system need to be run to provide theHow long will the system need to be run to provide the How long will the system need to be run to provide the recommended precipitation depthrecommended precipitation depth..

•• 4.7 gallons 4.7 gallons ÷÷ 2 2 gphgph = 2.35 hours or 2 hrs and 20 minutes= 2.35 hours or 2 hrs and 20 minutes

Average Daily Reference ET in inches per day (Average Daily Reference ET in inches per day (ETETRR) ) and plant ETand plant ET Estimates in Gallons for Xeric Plants atEstimates in Gallons for Xeric Plants atand plant ET and plant ET Estimates in Gallons for Xeric Plants at Estimates in Gallons for Xeric Plants at Different Different Canopy Areas Canopy Areas (CA) during (CA) during a Typical Season a Typical Season in Farmingtonin Farmington, NM , NM (K(KCC = 0.30)= 0.30)CC

MonthMonth MayMay JuneJune JulyJuly AugAug SeptSept

Days ofDays of 11 1515 1616 3131 11 3030 11 3131 11 3131 11 1515 1616 3030Days of Days of MonthMonth

11--1515 1616--3131 11--3030 11--3131 11--3131 11--1515 1616--3030

ETETRR per dayper day 0.350.35 0.400.40 0.420.42 0.380.38 0.280.28 0.250.25 0.220.22R R p yp y

Plant D in ft. Plant D in ft. (CA, ft(CA, ft22) )

Gallons of Water per Plant per Week* Gallons of Water per Plant per Week*

1 (0.8)1 (0.8)2 (3.1)2 (3.1)3 (7 1)3 (7 1)

0.40.41.41.43 23 2

0.40.41.61.63 73 7

0.40.41.71.73 93 9

0.40.41.51.53 53 5

0.30.31.11.12 62 6

0.30.31.01.02 32 3

0.20.20.90.92 02 03 (7.1)3 (7.1)

4 (12.6)4 (12.6)5 (19.6)5 (19.6)6 (28 3)6 (28 3)

3.23.25.75.78.98.9

12 912 9

3.73.76.66.610.210.214 714 7

3.93.96.96.9

10.710.715 515 5

3.53.56.26.29.79.7

14 014 0

2.62.64.64.67.17.1

10 310 3

2.32.34.14.16.46.49 29 2

2.02.03.63.65.65.68 18 16 (28.3)6 (28.3) 12.912.9 14.714.7 15.515.5 14.014.0 10.310.3 9.29.2 8.18.1

*Gallons/week/plant = *Gallons/week/plant = ETETRR x 0.3 x 0.62 x x 0.3 x 0.62 x CA x CA x 7 days7 days

More complete detailed More complete detailed list list at:at:pp

httphttp://cahe nmsu edu/aes/farm/documents/2007xericrevproceedingsforia pdf://cahe nmsu edu/aes/farm/documents/2007xericrevproceedingsforia pdfhttphttp://cahe.nmsu.edu/aes/farm/documents/2007xericrevproceedingsforia.pdf://cahe.nmsu.edu/aes/farm/documents/2007xericrevproceedingsforia.pdf

When irrigations are scheduled properly aWhen irrigations are scheduled properly aWhen irrigations are scheduled properly, a When irrigations are scheduled properly, a XeriscapeXeriscape™™ garden can save significant amounts of garden can save significant amounts of water compared towater compared to turfgrassturfgrass..water compared to water compared to turfgrassturfgrass..

Refer to next slide.Refer to next slide.AssumptionsAssumptions

•• Cool season turf KCool season turf KCC = 0.80; coverage = 100%= 0.80; coverage = 100%

W t f KW t f K 0 60 100%0 60 100%•• Warm season turf KWarm season turf KCC = 0.60; coverage = 100%= 0.60; coverage = 100%

•• XeriscapeXeriscape KKCC = 0.30; coverage = 60% = 0.30; coverage = 60%

Average Monthly Reference ET Average Monthly Reference ET (ET(ETRR) ) and and g yg y (( RR))Required Irrigation per 1000 Square Required Irrigation per 1000 Square Feet: Feet: Turf vs. Turf vs. XeriscapeXeriscape™ ™ ((XXSCAPESCAPE))

MonthMonth ETETRR CS TurfCS Turf WS TurfWS Turf XScapeXScape

inchesinches Gallons/1000 sq. ft.Gallons/1000 sq. ft.

MayMay 11 611 6 57815781 43364336 13011301MayMay 11.611.6 57815781 43364336 13011301

JuneJune 12.612.6 62806280 47104710 14131413

JulyJuly 11.811.8 58815881 44114411 13231323

AugAug 8 78 7 43364336 32523252 976976AugAug 8.78.7 43364336 32523252 976976

SeptSept 7.17.1 35393539 26542654 796796

TotalTotal 51.851.8 2581725817 1936319363 58095809

XeriscapeXeriscape™™ Irrigation SummaryIrrigation SummaryXeriscapeXeriscape Irrigation Summary Irrigation Summary

The irrigation requirement of xeric plants depends upon theThe irrigation requirement of xeric plants depends upon theThe irrigation requirement of xeric plants depends upon the The irrigation requirement of xeric plants depends upon the speciesspecies, the plant size, and the weather (including , the plant size, and the weather (including precipitationprecipitation).).

While many While many xeric plantsxeric plants, once established, will grow and , once established, will grow and exhibit acceptable quality without irrigation… exhibit acceptable quality without irrigation… p q y gp q y g

Most will benefit from between 4 and 10 gallons of Most will benefit from between 4 and 10 gallons of supplemental irrigation water per week (8 to 20 gallonssupplemental irrigation water per week (8 to 20 gallonssupplemental irrigation water per week (8 to 20 gallons supplemental irrigation water per week (8 to 20 gallons every 2 weeks) during every 2 weeks) during midmid--summersummer..

End of Module 6End of Module 6

IRRIGATION MANAGEMENTIRRIGATION MANAGEMENTEnd of Module 6End of Module 6