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 Irrigation Notes The Basics of Matched Precipitation Precipitation rate is the speed at which water is being applied to a specific area. When design- ing an irrigation system, it is important to ensure that the precipitation is even over each area or zone of coverage. To accomplish this, examine the amount of water falling into each individual area from all sprinklers supplying that area. Then, either select the appropri- ate nozzles, or zone together sprinklers with the same pre- cipitation rate. In other words, the criteria to consider are flow rates and arcs of coverage. The following il- lustration depicts three different sprinkler heads with matched precipitation rates. 360 = 4 GPM 180 = 2 GPM 90 = 1 GPM At first glance, it may not be apparent that the rates are matched. However, in each case, one gallon per minute (GPM) is applied to each quar- ter circle (assuming the radius of throw is the same for each head), and thus precipitation is matched. Let’s look at the mathematics of this example. If the radius of  the full-circle head is 30 feet, which is typical for a mid-range sprinkler, using the formula for finding the area of a circle (πR 2 ), we calculate its total area of cov- erage is 2,827 square feet: 3.1415 x 30 2  = 2827. Then, us- ing the precipitation-rate formula (GPM x 96.3 ÷ area), we find the average precipitation rate over the full circle is 0.136 inches per hour: 4 x 96.3 ÷ 2827 = 0.136. klers 1, 2, 4, and 5 – a quarter circle head, two half-circle heads, and one full-circle head, respectively. Area A receives 100% of the discharge from the quarter-circle head, 50% of each of the two half-circle heads, and 25% of the output of the full-circle head. If each head is supplying 4 GPM, the precipitation rate over Area A is computed as follows: 1. Find the total GPM for area A. 100% of Sprinkler 1 = 4 GPM 50% of Sprinkler 2 = 2 GPM 50% of Sprinkler 4 = 2 GPM 25% of Sprinkler 5 = 1 GPM 9 GPM 2. Apply the precipitation-rate formula. Avg. precip. over Area A = 9 GPM x 96.3 ÷ (30 ft. x 30 ft.)= 0.963 inches per hour. Next, Area B receives 50% of the discharge of the two half- circle heads (Sprinklers 2 and 3), and 25% of the discharge of Sprinklers 5 and 6, which are full-circle heads (note that the size of Area D is the equivalent of Area B). 1.If each head is supplying 4 GPM, the total GPM for Area B is: 50% of Sprinkler 2 = 2 GPM 50% of Sprinkler 3 = 2 GPM 25% of Sprinkler 5 = 1 GPM 25% of Sprinkler 6 = 1 GPM  6 GPM 2. The precipitation rate is: Avg. precip. over Area B = 6 GPM x 96.3 ÷ (30 ft. x 30 ft.)= 0.642 inches per hour. The area covered by the half- circle is exactly half that of the full-circle head. The discharge rate is also half that of the full- circle head or 2 GPM. There- fore, the precipitation rate over the half circle area is 0.136 inches per hour: 2 x 96.3 ÷ 1414 = 0.136. Similarly, the area covered by the quarter-circl e sprinkler is half that of the half-circle head. With a 1 GPM nozzle, the precipitation rate over the quarter-circle is 0.136 inches per hour as well. If, however, nozzles with the same flow rate (say 4 GPM) were used on all three heads, the precipitation rates would be 0.136, 0.273 and .0545 inches per hour, respectively. In this situation, the corners of an irri- gated area covered by quarter- circle heads (0.545 in./hr.), would flood long before enough water could be applied to the areas covered by the full-circle sprinklers (0.136 in./hr.). To make this clearer, let’s ex- amine a nine-head section of an irrigation system, assuming that 4 GPM nozzles are installed on all sprinklers, which are thirty feet apart on a square grid. The area covered by these nine sprinklers can be considered as four subsections, labeled A, B, C, and D, as shown. 5 4 3 2 9 8 7 1 6 A B C D Looking first at Area A, you can see it is watered by Sprin-

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  • gf eThe Basics oPrecipitation rate is the speed

    at which water is being appliedto a specific area. When design-ing an irrigation system, it isimportant to ensure that theprecipitation is even over eacharea or zone of coverage. Toaccomplish this, examine theamount of water falling intoeach individual area from allsprinklers supplying that area.Then, either select the appropri-ate nozzles, or zone togethersprinklers with the same pre-cipitation rate.

    In other words, the criteria toconsider are flow rates and arcsof coverage. The following il-lustration depicts three differentsprinkler heads with matchedprecipitation rates.

    360 = 4 GPM180 = 2 GPM90 = 1 GPMAt first glance, it may not be

    apparent that the rates arematched. However, in eachcase, one gallon per minute(GPM) is applied to each quar-ter circle (assuming the radiusof throw is the same for eachhead), and thus precipitation ismatched.

    Lets look at the mathematicsof this example. If the radius ofthe full-circle head is 30 feet,which is typical for a mid-rangesprinkler, using the formula forfinding the area of a circle (R2),

    we calculate its total area of cov-erage is 2,827 square feet:3.1415 x 302 = 2827. Then, us-ing the precipitation-rate formula(GPM x 96.3 area), we find theaverage precipitation rate overthe full circle is 0.136 inches perhour: 4 x 96.3 2827 = 0.136.Irri

    Matched PrThe area covered by the half-

    circle is exactly half that of thefull-circle head. The dischargerate is also half that of the full-circle head or 2 GPM. There-fore, the precipitation rate overthe half circle area is 0.136inches per hour: 2 x 96.3 1414 = 0.136.

    Similarly, the area coveredby the quarter-circle sprinkleris half that of the half-circlehead. With a 1 GPM nozzle,the precipitation rate over thequarter-circle is 0.136 inchesper hour as well.

    If, however, nozzles with thesame flow rate (say 4 GPM)were used on all three heads, theprecipitation rates would be0.136, 0.273 and .0545 inchesper hour, respectively. In thissituation, the corners of an irri-gated area covered by quarter-circle heads (0.545 in./hr.),would flood long before enoughwater could be applied to theareas covered by the full-circlesprinklers (0.136 in./hr.).

    To make this clearer, lets ex-amine a nine-head section of anirrigation system, assuming that4 GPM nozzles are installed onall sprinklers, which are thirtyfeet apart on a square grid. Thearea covered by these ninesprinklers can be considered asfour subsections, labeled A, B,C, and D, as shown.

    98754

    321

    6

    A B

    CD

    Looking first at Area A, youcan see it is watered by Sprin-ation Notes

    cipitationklers 1, 2, 4, and 5 a quartercircle head, two half-circleheads, and one full-circle head,respectively.

    Area A receives 100% of thedischarge from the quarter-circlehead, 50% of each of the twohalf-circle heads, and 25% ofthe output of the full-circle head.If each head is supplying 4GPM, the precipitation rate overArea A is computed as follows:

    1.Find the total GPM for area A.

    100% of Sprinkler 1 = 4 GPM

    50% of Sprinkler 2 = 2 GPM

    50% of Sprinkler 4 = 2 GPM

    25% of Sprinkler 5 = 1 GPM

    9 GPM

    2.Apply the precipitation-rateformula.

    Avg. precip. over Area A =

    9 GPM x 96.3 (30 ft. x 30 ft.)=

    0.963 inches per hour.

    Next, Area B receives 50% ofthe discharge of the two half-circle heads (Sprinklers 2 and 3),and 25% of the discharge ofSprinklers 5 and 6, which arefull-circle heads (note that thesize of Area D is the equivalentof Area B).

    1.If each head is supplying 4GPM, the total GPM forArea B is:

    50% of Sprinkler 2 = 2 GPM

    50% of Sprinkler 3 = 2 GPM

    25% of Sprinkler 5 = 1 GPM25% of Sprinkler 6 = 1 GPM

    6 GPM

    2. The precipitation rate is:

    Avg. precip. over Area B =

    6 GPM x 96.3 (30 ft. x 30 ft.)=

    0.642 inches per hour.

  • C8 9Finally, Area C is affected byfour full-circle heads (Sprinklers5, 6, 8, and 9).

    Each contributes 25% of itsoutput to this area, so each headcontributes 1 GPM to this area,for a total of 4 GPM. Thus, theprecipitation rate for Area C is:

    Avg. precip. over Area C =

    4 GPM x 96.3 (30 ft. x 30 ft.)=

    0.428 inches per hour.

    These precipitation rates arequite different if you are go-ing to use nozzles with thesame flow rate on quarter-circle, half-circle and full-circleheads, it will be necessary tozone each type of head sepa-rately. And further, you willhave to run the heads for differ-ent periods of time to achievematched precipitation, or areasof flooding or stress will result.(See Hunters LIT-088 for in-formation on scheduling.)

    Now lets look at the samenine-head section of a systemset up for matched precipita-tion. Assume that 4 GPMnozzles are installed in the full-circle heads, 2 GPM nozzles inthe half-circle heads, and 1GPM nozzles are in the quar-ter-circle heads. Remember,when examining precipitationrates, we are looking at theamount of water coming fromeach sprinkler into the area inbetween the sprinklers.

    Lets start with Area A again.

    A

    4 5Hunt1940

    21Sprinklers 1, 2, 4, and 5 are aquarter-circle head, two half-circle heads, and one full-circlehead, respectively. The totalGPM for the area is:

    100% of Sprinkler 1 (1 GPM) = 1 GPM

    50% of Sprinkler 2 (2 GPM) = 1 GPM

    50% of Sprinkler 4 (2 GPM) = 1 GPM

    25% of Sprinkler 5 (4 GPM) = 1 GPM

    4 GPM

    and the precipitation rate is:

    Avg. precip. over Area A =

    4 GPM x 96.3 (30 ft. x 30 ft.)=

    0.428 inches per hour.

    Area B is affected by Sprin-klers 2, 3, 5, and 6.

    B

    5 6

    32These are two half-circle

    heads with 2 GPM nozzles,and two full-circle heads with4 GPM nozzles. The totalGPM is:

    50% of Sprinkler 2 (2 GPM) = 1 GPM

    50% of Sprinkler 3 (2 GPM) = 1 GPM

    25% of Sprinkler 5 (4 GPM) = 1 GPM

    25% of Sprinkler 6 (4 GPM) = 1 GPM

    4 GPM

    and the precipitation rate is:

    Avg. precip. over Area B (and D) =

    4 GPM x 96.3 (30 ft. x 30 ft.)=

    0.428 inches per hour.

    Finally, Area C is unchanged.

    With four full-circle heads(Sprinklers 5, 6, 8, and 9) eacher Industries Incorporated The IrrigationDiamond St. San Marcos, California 92069

    contributing 25% of its 4 GPMoutput to this area, the total forthe area is 4 GPM.

    Avg. precip. over Area C =

    4 GPM x 96.3 (30 ft. x 30 ft.)=

    0.428 inches per hour.

    Thus, the four areas will bereceiving the same amount ofwater per hour, and the differ-ent patterns (quarter-circle,half-circle, and full-circle) canbe zoned together. If the sprin-klers are assigned to more thanone zone, the zones can be setto run the same amount oftime, depending on the needs ofthe landscape.

    As you can see, it is impor-tant that sprinklers are nozzledand zoned in a manner thatprovides even coverage. Design-ing and installing systems withmatched precipitation in mindwill reduce flooding and stressproblems, as well as save water.

    For additional information,refer to the following Hunterpublications: LIT-087, Sprin-kler Spacing and LIT-088,Scheduling Irrigation.

    65 Innovators P/N 700274U.S.A. TEL: (1) 760-744-5240 FAX: (1) 760-744-7461 LIT-084.PDF 5/97