Your SlideShare is downloading. ×
  • Like
Irrigation Systems and Irrigation Management in Xeric Gardens - New Mexico State University
Upcoming SlideShare
Loading in...5

Thanks for flagging this SlideShare!

Oops! An error has occurred.


Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Irrigation Systems and Irrigation Management in Xeric Gardens - New Mexico State University


Irrigation Systems and Irrigation Management in Xeric Gardens - New Mexico State University

Irrigation Systems and Irrigation Management in Xeric Gardens - New Mexico State University

Published in Education
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads


Total Views
On SlideShare
From Embeds
Number of Embeds



Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

    No notes for slide


  • 1. Irrigation Management g gModule 6
  • 2. Irrigation Systems andIrrigation ManagementI i ti M t in Xeric Gardens
  • 3. The Water Cycle y ET Evaporation ration Runoff Infiltration TranspirSoil RootSurface Storage in Absorption T root zone Deep Drainage Desirable Pathway y Undesirable losses
  • 4. Goal of Efficient Irrigation Provide enough supplemental water (above precipitation) to satisfy the plant’s requirements for plant s adequate growth and/or desired quality. quality. Prevent water waste via runoff excessive runoff, evaporation, or deep drainage below the plant’s root zone. zone
  • 5. Xeriscape GardenEfficient Irrigation Management g gSuccess depends on knowledge of three essentialcomponents: t The output and efficiency of the irrigation system (including individual emitters). The water-holding characteristics of the soil. water- The estimated water required by each plant for ‘acceptable’ growth and quality.
  • 6. The Irrigation System g y To schedule irrigations effectively you must know the output (water flow rate or water application rate) of the irrigation system. system. The flow rate is usually measured in volume per unit time (i.e. gallons per minute [gpm] in sprinkler systems and gallons per hour ie [gpm] [gph] in drip systems). gph] The application rate is usually measured in depth p unit pp y p per time (i.e. inches per hour [in/hr]). [in/hr]). One can be converted to the other if the wetted area is known (measured).
  • 7. Irrigation System Output (Flow Rate)
  • 8. The maximum flow rateDetermined by direct measurement from a spigotusing a bucket and stopwatchProcedure:Using a watch, timehow long it takes tofill a container havinga known volume. Example: Suppose it takes 30 seconds to fill a p 5-gallon bucket. Flow rate = 5 gallons/0.5 minute = 10 gpm. gpm.
  • 9. Determining the flow rate of an existingirrigation system using a flow meter orinstalled water meter Most water meters show total gallons. gallons. To determine the flow rate of your irrigation system: system: Insure only irrigation system is running. running. Record an initial flow meter reading. Run system for a timed period (i.e. 10, 15, 30 minutes for a sprinkler system; a few hours for a drip system) system) Record the final meter reading. reading. Subtract the initial reading from the final reading. Divide the result by the time in minutes.
  • 10. Example (Flow Meter) Suppose the meter reading was 24,000 gallons before turning on the irrigation system and was 24,600 gallons after running the irrigation system for 30 minutes minutes. Total gallons used = 24,600 – 24,000 = 600. Flow rate in gpm = 600/30 = 20 gpm This might be a typical flow rate for a small 8-sprinkler g yp p irrigation zone (2.5 gpm from each sprinkler).
  • 11. Drip Systems p y Because of the very low flow rates of drip systems, you may h have t run the system for 2 to 3 hours to get to th t f t h t t accurate flow rates with a water meter. meter. Flow rates can be expressed in gallons per hour (gph) (gph) instead of gallons per minute (gpm). (gpm). Example with a flow (or water) meter: Water meter reading before irrigation start-up = 24,000 gallons start- Water meter reading after running the system for 2 hours = 24,300 gallons. Total gallons used = 24,300 – 24,000 = 300 Flow rate in gph = 300/2 = 150 gph. gph. • This might be a typical flow rate of a drip system irrigating a moderate sized (100 plant) xeriscape garden Smaller plants may be irrigated garden. with only 1 emitter while larger plants may be irrigated with 2 or 3 emitters.
  • 12. Drip Emitter Flow R tD i E itt Fl Rates Drip emitters are designed to have a specific flow rate and many are marked (or color-coded) to designate that flow rate in color- gallons per hour (i.e. 0.5, 1.0, 2.0, 5.0 gph, etc.). gph, To estimate the flow rate for the entire system j y just add up the p flow rates of the individual emitters. If you’re uncertain (or, if you just want to verify) you can measure the t t f th output from each emitter using a stop watch (or watch with a h itt i t t h second hand) and a measuring cup.
  • 13. Example Place a measuring cup below the normally positioned emitter and mark the time. After 1 minute, remove the cup and record the volume. Suppose you measured ¼ cup or 2 ounces. Flow rate = 2 ounces/minute or 120 ounces/hour. There are 128 ounces in a gallon, so… so… The flow rate is about 1 gallon per hour (120 oz./128 oz. = 0.94 gph) gph)
  • 14. Troubleshooting for Leaks g In buried irrigation systems, on sandy soils, it may be difficult to visually detect a leak in the system. If you know the output of each of your sprinklers, or emitters (available from the dealer or manufacturer if the model number is known), you can add these up to get an estimate of what the total flow rate should be (at a given pressure). If the calculated flow rate using your meter readings is much more than the estimated theoretical flow rate there might be rate, an underground leak.
  • 15. Precipitation Rate(Sprinklers) The average precipitation rate of a sprinkler system can be… y Calculated from the flow rate (Q) and wetted area (A) or (A), or… Directly measured using catch cans.
  • 16. Sprinkler Precipitation Rate Calculated from flow rate (Q) and wetted area (A). ( ) Equation: PR = 96.3 x Q/A • Where: PR = water application rate in inches/hour • Q = flow rate in gallons/minute (gpm) (gpm) • A = area in square feet (width x length, feet) Example: E l • Suppose measured flow rate = 20 gpm • Suppose wetted area = 50’ x 100’ = 5000 sq. ft. pp q • Then PR = 96.3 x 20/5000 = 0.39 inches/hour
  • 17. Converting precipitation rate to flow rate Consider: It takes 0 623 gallons of water to cover 0.623 1 square foot to a depth of 1 inch, so: so: To T convert Precipitation Rate to Flow Rate: t P i it ti R t t Fl R t FR = (PR x A x 0.623)/60 • Where; FR = flow rate in gallons per minute (gpm) (gpm) PR = precipitation rate in inches per hour A = wetted area in square feet Example from previous slide: • S Suppose the wetted area was 50 f h d 0 feet by 100 f b feet (A = 5,000 sq. f ) 000 ft.) • The precipitation rate (PR) was 0.39 inches per hour. Solution: Solution: FR = (0 39 x 5000 x 0 623)/60 = 20 gpm (0 (0.39 0.39 0.623)/60
  • 18. Sprinkler System p yDirect Measurement of Precipitation Rate Set out a grid of straight-sided straight- catch cans (i.e. tuna cans, soup cans, cans coffee cans etc ) cans, etc.) Run the system for a timed period and then measure the water depth in straight-sided straight- cans with a ruler.
  • 19. Sprinkler System p yCalculating average precipitation rate To calculate the average precipitation rate: Find the total of all measurements and divide by the number of measurements. Example: Suppose you set out 10 cans and y ran the sprinkler pp y you p system for 30 minutes. Assume your measurements from the cans were: 0.25, 0.30, were: 0.30, 0.35, 0.25, 0.20, 0.35, 0.40, 0.20, and 0.30 inches. The average depth would be: 0.29 inches (the sum of the above measurements [2.9]) divided by 10 ( / ) (2.9/10) (2.9/10 The irrigation rate would be equal to 0.29 inches divided by 30 minutes (0 29/30) or 0 0097 inches per (0.29/30) 0.0097 minute or 0.58 inches per hour (0.0097 x 60).
  • 20. Drip S Systems With drip systems PR = FR/0.623/(D x D x 0.785) Where: • PR = precipitation rate in inches • FR = emitter(s) flow rate in gallons/hour • D = wetted diameter or plant canopy diameter Example: Suppose the emitter flow rate ( ) = 1 gph pp (FR) gp Suppose the plant diameter (D) = 3 ft Then; PR = 1/0.623/(3 x 3 x 0.785) = 0.23 in/hr
  • 21. The Soil Once you’ve determined the output of your irrigation system, you need to get an idea of your soil’s water holding characteristics characteristics.
  • 22. Soil Water-Holding Characteristics Water-
  • 23. Soil h ld water lik a sponge!S il holds t like ! Available Water Saturated Soil Field Capacity Wilting Point
  • 24. Soil Texture The amount of water a particular soil will hold depends on it’s texture (the percentage of sand, p p g , silt, and clay particles). Sand particles: 0.05 mm to 2.0 mm Silt particles: 0.002 mm to 0.05 mm Clay particles: less than 0.002 mm
  • 25. Soil Water Holding Capacityis Related to Soil TextureClayey Soil Texture Sandy Soil Texture Small particles Larger particles Low water intake rate High water intake rate High water holding capacity Low water holding capacity (heavy soil) (light il) (li ht soil)
  • 26. Determining Soil Texture g Fill a quart canning jar with about 4 inches of tamped soil. Add water up to about 2 inches below rim of jar. Seal with lid so it won’t leak. Shake continuously for about 10 -15 minutes. Set on flat surface and do not disturb. After about 48 hours, the soil should settle out in the water into three distinct layers. Measure the total soil depth and the depths of each distinct layer. Sand will be the bottom layer (it settles out first), silt the middle layer, and clay the top layer (it settles out last). last) Calculate the percentages of sand, silt, and clay (see next page for example)
  • 27. Example Suppose you measured 4 5 inches of total soil 4.5 depth, 2.5 inches of sand (bottom layer), 1.5 inch of silt (middle layer) and 0.5 inch of clay (top layer). y layer). The texture would be 56% sand (2.5 ÷ 4.5 x 100), 33% silt (2.5 (1.5 ÷ 4.5 x 100), and 11% clay (0.5 ÷ 4.5 x 100). To define the soil type see the soil texture pyramid – next slide.
  • 28. Soil Texture Pyramid y 2. Draw another 1. Draw a li 1 D line line from upper right straight across to lower left from from % clay scale the % silt scale ( (example = 11%). p %) (example = 33%). 33%) Intersection: sandy loam soil (58% sand)
  • 29. Water Holding Capacity g p y Once the soil texture has been determined, the approximate available water holding capacity can be estimated. p y
  • 30. Approximate Available Soil Moisture pp in Various Textured Soils Soil Texture In./In. In./Ft. (available moisture) (available moisture)Coarse sand and gravel .04 0.5Sands .07 0.8Loamy sands .09 1.1Sandy loams .13 1.5Fine sandy loams .16 Our example 1.9Loams and silt loams .20 2.4Clay loams & silty clay .18 2.1loamsSilty clays and clays .16 1.9
  • 31. Feel Test for Soil Type and Moisture yp % Water Sandy *Loamy Clayey Of Field Capacity (coarse) (average) (fine) (fi ) 0-25% Dry,loose, slips Powdery, maybe Dry, cracked, not between fingers forming a slightly easily reduced to solid crust powder 25- 25-50% Seems dry, does Brittle but sticks Fairly plastic, not form a ball together when forms a ball when when squeezed h d squeezed d squeezedd 50- 50-75% Forms a loose ball Forms a plastic Forms a ball, can when squeezed ball, sticky when be stretched but easily falls squeezed between the apart thumb and index, smooth to touch 75- 75-100% Forms coherent Forms a very Same as above ball, not smooth plastic ball, easily smoothed *Loam soil is best.
  • 32. Soil Intake Rate Due to the larger particle size and hence larger pore spaces, a sandy soil will absorb water faster than th a clayey soil. l il
  • 33. Soil Intake Rates Basic Intake Rates Coarse sand 0.75 – 1.00 in/hr Fine sands 0.5 – 0.75 in/hr Fine sandy loams 0.35 0.50 0 35 – 0 50 in/hr Silt loams 0.25 – 0.40 in/hr Clay loams 0.10 – 0.30 in/hr
  • 34. Water Percolationi Diffin Different S il T Soil Textures (Drip Irrigation)
  • 35. Summary Because a clay soil holds more water than a sandy soil, a greater volume of water must be applied to a cla soil ol me ater m st clay to penetrate to an equal depth. However, However because of the lower intake rate of a clay soil soil, the water application rate (precipitation rate) must be less than on the sandy soil to prevent runoff or excessive y p puddling. Once an equal water penetration depth is reached on both soils, soils, the required irrigation frequency will be less on the clay soil since it holds more plant available water than the sandy soil. y
  • 36. Notes Most native, xeric plants prefer a sandy soil with good drainage. Root rot may occur in heavy clay soils that hold too much water. water. If irrigated properly using drip irrigation deep drainage and properly, irrigation, runoff will not occur in a Xeriscape™.
  • 37. Determining Soil Chemistry & Fertility g y y SampleSampling Kit can be obtained Examinefrom NMSU CooperativeExtension Services throughoutthe state.Ex. NMSU CES San JuanCounty 213-A S. Oliver Ave. 213-Aztec,Aztec, NM 87410Phone: 505-334-Phone: 505-334-9496
  • 38. In Fi ld Soil MoistureI Field S il M i t During the year, periodically examine the top foot of your soil near the base of your plants using a long screwdriver or similar rod. i il d By experiencing what the soil feels like when it’s wet and when it s dry you ll 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 more effectively.
  • 39. Plant Water Requirements Saving S i water i the l d in h landscape i more a f is function of careful i f f l irrigation management than of plant selection. selection. Many xeric (drought-tolerant) plants can (and will) use just (drought- as much water as non-xeric plants, if this water is available non- to them. them. The difference is – xeric plants can survive and prosper under low- low-water conditions that would be detrimental to (or kill) species not adapted to arid conditions.
  • 40. Additional CAdditi l Considerations id ti With landscape plants, economic yield or plant size is not of p primary concern. y concern. The primary goal is to provide just enough water, fertilizer, etc. to result in an acceptable plant specimen for a quality landscape landscape.
  • 41. Evapotranspiration (ET) Evaporation – loss of water directly from soil and plant surfaces. surfaces. Transpiration – loss of water from plant internal tissues through the leaf stomates, during photosynthesis. stomates,
  • 42. ET is related to… Atmospheric demands (weather) (weather) Air temperature Solar radiation Relative humidity Wind Generally, ET increases with higher temperature, greater solar radiation higher wind and lower humidity radiation, wind, humidity.
  • 43. ET is also related to… Size of the plant. plant. Live leaf area (canopy) of the plant. plant. Microhabitat (i e microclimate) and other factors (i.e. factors.
  • 44. Estimating ETfor Irrigation Scheduling For many years, irrigation managers have used climate-based climate- reference (or potential) ETR values, along with correction factors (crop-coefficients or KC values) to help them efficiently schedule crop- irrigations on agricultural crops and turfgrass. turfgrass. Due to a lack of research, the technique has received limited use for scheduling irrigations on landscape plants, particularly xeriscapes. xeriscapes. i
  • 45. How th system worksH the t k A reference ET (ETR) is calculated from weather data (ET collected from an approved, standard weather station. See: http://weather nmsu edu. See: ETR is then multiplied by a correction factor (crop coefficient) to estimate the ET of a crop based on the plants size, canopy area, maturity, etc. t it t Generally, for alfalfa the crop coefficient (KC) is equal to (K 1 during most of the year. year. Cool Season Turf: 0.8, Warm season turf: 0.6 The canopy area for alfalfa and grass is usually considered py g y to be 100% of the (irrigated) area. area. ™ The canopy area for a well-developed, mature Xeriscape well- may be 60% or less of the total (irrigated) area area.
  • 46. Automated Weather Station – Calculation of ETO Wind Speed Wind Direction Solar Radiation Air Temperature Precipitation Data Logger D t L Relative H midit Relati e Humidity Solar Charger 12V Battery
  • 47. Xeriscape™ Water Requirements Based on observations of differentially irrigated xeric plants at the A i lt l S i l t t th Agricultural Science C t at F Center t Farmington, i t a crop coefficient of between 0.2 and 0.4 should be sufficient for most xeric plants (compared to 1 0 for alfalfa 1.0 and 0.8 and 0.6 for cool season and warm season turf, respectively). Also, Also, the canopy area for a well-developed, mature well- Xeriscape™ is generally considered to be about 60% of p g y the landscape (compared to 100% for alfalfa and grass). grass).
  • 48. Drip Irrigation SchedulingConsiderations Reference ET and irrigation recommendations are usually provided in inches but drip emitter flow rates are expressed in gallons. gallons. It takes 0.623 gallons of water to cover 1 square foot to a depth of 1 inch. inch. With xeric landscape plants, the wetted area (or plant canopy area) is usually circular. circular. Formula Form la for converting ET (or irrigation or precipitation) depth con erting in inches to gallons for a landscape plant: Gallons = ET x 0.623 x A • Where: ET = Irrigation or precipitation depth in inches A = wetted or plant canopy area in square feet • A f i l diameter squared (d2) x 0.785 Area of a circle = di t d 0 785
  • 49. Summary Determining the water requirement in gallons for a xeriscape plant: Gallons = ETR x KC x 0 623 x CA 0.623 Where: • ETR = reference ET from weather station data (inches) • KC = correction factor (or crop coefficient) • 0.623 = constant to convert inches to gallons • CA = crop canopy area in square feet
  • 50. Climate-Climate-Based Irrigation Scheduling ™Example P blE l Problem f a X i for Xeriscape Plant Pl t Given: The reference ET (ETR) for the week is 2.0 inches. The plant to be irrigated has a canopy diameter (D) of 4 feet. feet. The crop coefficient (KC) or correction factor is 0.3. There is one 2 gph emitter at the base of the plant. plant. Questions: Questions: What is the plant’s canopy area (CA)? • A Area = D x D x 0 785 = 4 x 4 x 0 785 = 12 56 sq. ft 0.785 0.785 12.56 ft. How many gallons should be provided per week to satisfy the recommended irrigation (or plant ET) depth? • Gallons = ETR x KC x 0.623 x A = 2 x 0.3 x 0.623 x 12.56 = 4.7 gals. gals. How long will the system need to be run to provide the recommended precipitation depth. depth. • 4.7 gallons ÷ 2 gph = 2.35 hours or 2 hrs and 20 minutes
  • 51. Average Daily Reference ET in inches per day (ETR) (ETand plant ET Estimates in Gallons for Xeric Plants atDifferent Canopy Areas (CA) during a Typical Seasonin Farmington, NM (KC = 0.30) Farmington,Month May June July Aug SeptDays of 1-15 16- 16-31 1-30 1-31 1-31 1-15 16- 16-30 MonthETR per day p y 0.35 0.40 0.42 0.38 0.28 0.25 0.22Plant D in ft. Gallons of Water per Plant per Week* (CA, ft2)1 (0.8) 0.4 0.4 0.4 0.4 0.3 0.3 0.22 (3.1) 1.4 1.6 1.7 1.5 1.1 1.0 0.93 (7.1) (7 1) 3.2 32 3.7 37 3.9 39 3.5 35 2.6 26 2.3 23 2.0 204 (12.6) 5.7 6.6 6.9 6.2 4.6 4.1 3.65 (19.6) 8.9 10.2 10.7 9.7 7.1 6.4 5.66 (28 3) (28.3) 12.9 12 9 14.7 14 7 15.5 15 5 14.0 14 0 10.3 10 3 9.2 92 8.1 81*Gallons/week/plant = ETR x 0.3 x 0.62 x CA x 7 days
  • 52. More complete detailed list at: p http://cahe nmsu edu/aes/farm/documents/2007xericrevproceedingsforia pdf ://cahe
  • 53. When irrigations are scheduled properly a properly, ™Xeriscape garden can save significant amounts ofwater compared to turfgrass.Refer to next slide. Assumptions • Cool season turf KC = 0.80; coverage = 100% • Warm season turf KC = 0 60 coverage = 100% W t f 0.60; • Xeriscape KC = 0.30; coverage = 60%
  • 54. Average Monthly Reference ET (ETR) and g y (Required Irrigation per 1000 Square Feet:Turf vs. Xeriscape™ (XSCAPE) Month ETR CS Turf WS Turf XScape inches Gallons/1000 sq. ft. May 11.6 11 6 5781 4336 1301 June 12.6 6280 4710 1413 July 11.8 5881 4411 1323 Aug 8.7 87 4336 3252 976 Sept 7.1 3539 2654 796 Total 51.8 25817 19363 5809
  • 55. Xeriscape Irrigation Summary ™ The irrigation requirement of xeric plants depends upon the species, species, the plant size, and the weather (including precipitation). precipitation). While many xeric plants, once established, will grow and plants, exhibit acceptable q p quality without irrigation… y g Most will benefit from between 4 and 10 gallons of supplemental irrigation water per week (8 to 20 gallons every 2 weeks) during mid-summer. mid-summer.
  • 56. End of Module 6IRRIGATION MANAGEMENT