Epa Net Sprinkler System Example

2Mar201207:21:04-EPANETSprin 2Mar201207:21: 04-EPANETSprinklerSystemExample.sm klerSystemExample.sm

Spri Sp rink nkle ler r __ __irri ir riga gati tion on  __  __syst em __analysis anal ysis__usin g __EPAN ET __2.0 By Gilberto Gilberto E. Urroz, Urroz, March 2012 Sprinkle Sprinklers rs are are commonly commonly used to irrigate irrigate house house yards, yards, parks, or agricult agricultural ural plots. plots. The network network mapped below represent represents s a sprinkle sprinkler r irrigation irrigation system system for a small small park. park.

The pipes pipes in this network network have the followin following g length, diameters, and and Hazen-Williams coefficients: ------------------- -------------------Pipe Pipe L(ft L(ft) ) D(in D(in) ) Pipe ipe L(ft) (ft) D(in) (in) ------------------- -------------------P1 10 00 4 P8 40 0 2 P2 30 0 4 P9 40 0 2 P3 40 0 2 P1 0 50 0 4 P4 40 0 2 P1 1 40 0 2 P5 50 0 4 P1 2 40 0 2 P6 40 0 2 P1 3 50 0 4 P7 40 0 2 P1 4 50 0 4 ------------------- -------------------The pump pump curve curve is define defined d by the follow following ing curve: ---------------Q(cfs) Q(cfs) hP(ft) hP(ft) ---------------0. 00 17 0 0. 67 13 5 1. 00 10 0 ----------------

All nodes nodes are at zero zero elevat elevation ion, , while while the reserv reservoir oir R1 R1 has a total total head (water surface elevation) of 10 ft. The sprinkle sprinkler r heads located at juncti junctions ons J6, J7, J8, J9, J10, J10, J11, J11, J12, J12, J13, J13, J14, J14, and J16 have have emitte emitter r coefficients of 0.04 cfs/(psi)^0.5, except except for for those those at J13 and J14, J14, whose whose emitter emitter coeffici coefficients ents are are 0.05 cfs/(psi)^0.5 Sprinkler heads heads are represented by junction junctions, s, some some of which which are terminal terminal junction junctions s (e.g., (e.g., J7, J8, J10, J12, J13, J13, J14, and J15). J15). The dischar discharge, ge, Q, produced produced by a sprinkler sprinkler head is related related to the local local pressure, pressure, p, by: Q

C

E

p

In entering entering data data for this network, network, we select CFS (cubic feet feet per second) second) as the default default unit of discharg discharge, e, and H-W (Hazen-Wi (Hazen-William lliams) s) as the friction friction loss loss equation equation to use. The emitter emitter coefficien coefficients ts are are entered entered in the node properti properties es in in the proper proper units, i.e., i.e., in this case, case, in CFS/(ft) CFS/(ft)^0.5 ^0.5. . Irrigati Irrigation on systems systems are typically typically operated operated under steady-s steady-state tate condit conditions ions for a given period period of time. time. Thus, Thus, for the present present case a steady-st steady-state ate solution solution suffic suffices. es. The figure figure below shows the hydraulic hydraulic grade grade line elevatio elevations ns at at nodes nodes and flow discharges discharges in the pipes pipes for the steady steady state state solution. solution.

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2Mar201207:21:04-EPANETSprinklerSystemExample.sm

Notice that the suction side of the pump, J1, shows a negative pressure, as expected, whereas the discharge side, J2, shows a positive pressure. Since all elevations are set at the same level (zero), the pressure difference across the pump is related to the pump head as follows: p p J2 J1 hP γ where γ is the specific weight of water. Taking γ this case, is:

hP

27.09 psi 62.4

24.4483 psi lbf  ft

62.4 lbf/ft^3, the pump head, in

, i.e.,

hP 118.9345 ft , while the

3

pump supplies a total of 0.83 cfs to the irrigation system network. The pressures at the sprinklers vary from 1.25 psi (J10, J12) to 9.08 psi (J7,J8). The following figure shows the node demands, which, in this case, basically represent the sprinkler discharges. The figure also shows the discharge directions in the pipes.

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In most cases, sprinkler irrigation system discharges are given in GPM (gallons per minute) rather than in CFS (cubic feet per second). For practice, you should repeat this exercise using GPS for the default discharge units. The network properties are basically the same except for the emitter coefficient values which should be given in gpm/(psi)^0.5. Thus, the values to use are: 3 ft s

cfs

0.04

cfs

 psi

1 2

17.9532

gpm

gpm

 psi

1 2

0.05

gal min

cfs

 psi

1 2

22.4416

gpm

 psi

1 2

Also, the pump curve needs to have the discharges, Q, converted from CFS to GPM: 0.67 cfs

300.7169 gpm

1.00 cfs

448.8311 gpm

Thus, the resulting pump curve is: ---------------Q(gpm) hP(ft) ---------------0.0 170 300.7 135 448.8 100 ---------------Booster __pump Since all the sprinklers are set at a zero elevation and the source reservoir, R1, is at a higher elevation, theoretically a pump is not needed to supply water to the system. However, in this case, an elevation of 10 ft will produce much smaller pressures at the sprinkler heads if the pump were not present. The following figure shows the pressures and flows for the case in which the pump is removed:

The pump, in this case, is referred to as a booster pump because it "boosts" the discharge delivered to the system.

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Calcul ati ng __prinkl er' __emitter s __coefficient Manufacturers provide data detailing the discharge characteristics of sprinkler heads. For example, the figure below shows performance data for different nozzle sizes fora particu model.

The emitter coefficient can be calculated from the discharge and pressure data listed in the table above. For example, for the 1/8" nozzle size, the p and Q data ar

p

25 30 35 (psi) 40 45 50

Q

The following code calculates the emitter coefficients: 2.26 2.48 0.452 n length p 2.68 (gpm) 0.4528 for k 1 .. n 2.86 gpm 0.453 Q k 3.03 C C E 0.4522 ps E 3.20 p k k 0.4517 0.4525

An average value for the emitter coefficient for this case is n C C

E_ave

k= 1 n

E

k =>

C

E_ave

0.4524

gpm psi

or

0.4524

gpm  psi

0.001

cfs  psi

 NOTE: _   _   _   _Modeling  _  _free-discharging  _  _pipelines  _  _in  _  _EPANET Consider the following example shown in page 103 of the Spring 2012 CEE 35100 Reader:

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The solution for Q found using SMath Studio was Q = 1.38 cfs. To solve such a system in EPANET, point (2) should be represented as a node with a very large emitter coefficient. Since the pressure at that point should be zero (or very close to zero in the EPANET solution), then a very large emitter coefficient will ensure that the demand at that node has a reasonable value. In addition, to account for the velocity head at the free-discharging node (2), a loss coefficient of 1.0 must be included in the pipe connecting reservoir (1) with outlet (2). In the problem statement it is indicated that minor losses (in this case, reservoir entrance losses only) are to be ignored. However, the loss coefficient of 1.0 is necessary for a complete solutio In setting up the EPANET model we created the following map (here showing the pressures and the flow discharge after running the mode

We use a total head of 60 ft for R1, and an elevation of 55 ft for J1. Pipe P1 has a length of 100 ft, a diameter of 6 inches, a Hazen-Williams coefficient of 110, and a (minor) loss coefficient of 1.0. Node J1 uses a emitter coefficient of 1000 cfs/psi^0.5. After running the program you get a warning that negative pressures were detected in the system. Ignore this message and check the final result, Q = 1.39 cfs, very close to the value found using SMath Studio (Q = 1.385 cf

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