11.6: Sprinkler System Design
- Page ID
- 44635
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Detailed design of sprinkler lateral systems is beyond the scope here; however, some general relationships are needed to manage systems properly. We have considered the hydraulics of sprinkler laterals and the pressure variation along the lateral. Two important considerations that are still needed are: how to select sprinkler nozzles to satisfy capacity requirements of the system, and how many laterals are required for the field in a moved lateral system.
Sprinkler systems must apply enough water to satisfy crop water requirements and to account for inefficiencies and nonuniformities in the irrigation system and the field. From Chapter 5, the net system capacity requirement and the application efficiency for the system can be estimated. These quantities are used to compute the gross capacity, Qc (gallons per minute per acre, gpm/ac). The problem is to determine how that capacity is used to arrange sprinkler laterals, and to select the appropriate nozzles and pressure for the sprinkler system. For a moved lateral system with multiple laterals each having the same length, the minimum discharge required from each sprinkler on the lateral can be determined from:
\(q_s=\left(\dfrac{Q_cS_LS_m}{43560}\right)\left(\dfrac{N_s}{N_l}\right)\left(\dfrac{T_s}{T_0}\right)\left(\dfrac{I_i}{I_i-T_d}\right)\) (11.14)
where: Qc = gross system capacity requirement (gpm/ac),
Ns = number of sets required to irrigate the field,
Nl = number of laterals used to irrigate the field,
To = time of actual operation per set (hr),
Ts = set time (hr),
Ii = irrigation interval (days), and
Td = downtime for system (days); other parameters are as already defined.
The time of operation is the actual time that water is applied during the total set time. For example, a lateral may only operate 10 hours out of a 12-hour set. This provides time to drain and move the lateral. The irrigation interval is the amount of time between successive irrigations of the field. The downtime is the time required to maintain the engine, system, etc., to prepare the laterals for the next irrigation, and for any harvesting, farming, or other operations.
The number of sets in the field is determined by:
\(N_s=\dfrac{W_f}{S_m}\) (11.15)
where Wf is the width of the field as shown in Figure 11.12. It is often the case that more than one lateral is needed to irrigate a field.
The nozzle size(s) needed for the sprinklers on the lateral can now be determined using Tables 11.1 and 11.2. The spacing criteria must be considered as the nozzle size(s) is determined. The total flow required for the lateral is the product of the number of sprinklers on the lateral and the required discharge for each sprinkler:
\(Q_i=\dfrac{q_s L}{S_L} \) (11.16)
where: Qi = inflow to lateral (gpm) and
L = length of lateral.
The inflow to the lateral must be less than the maximum allowable inflow determined in the Section 11.5. If the inflow is excessive, more laterals are generally required with longer set times or shorter lengths.
Considering the operation of the lateral system requires a balance of management factors including the time of operation, the sprinkler and lateral spacing, the number of laterals required, and the application efficiency. Pressure and flow limitations must also be considered for proper operation. Often, a trial-and-error procedure is needed to balance all factors, and tradeoffs frequently are required. The landowner’s and/or irrigation manager’s preferences for operation should be incorporated into a management plan for the system.
The layout of laterals on sloping fields can be crucial. It is generally best to run the mainline up and down the hillslope while positioning the lateral so that it is relatively level. If the lateral must run up and down the hill, it is best to run the lateral downslope if possible. The prevailing wind direction and speed during the irrigation season should also be considered.
Figure 11.12. Field layout for a moved-lateral sprinkler system
Compute the minimum sprinkler discharge required for the system described below.
Given: A square field (1,200 feet x1,200 feet) is irrigated with a portable set-move (moved lateral) sprinkler system. The gross system capacity has been determined to be 6.0 gpm/ac. The spacing of sprinklers is 40 feet along the lateral and 50 feet between lateral sets. The system operates for 10 hours out of a 12-hour set. The field must be irrigated at least once every 10 days and 2 days are needed to move laterals to the beginning side and for equipment maintenance.
Find: Compute the minimum sprinkler discharge required for the system.
Solution
The number of sets in the field are Ns = Wf/Sm = 1,200 ft/50 ft = 24 sets. (Eq. 11.15) With 12-hour set times, 2 sets can be irrigated daily so 12 days of continual irrigation would be required with one lateral. We only have 8 days available to irrigate since 2 out of 10 days are used for downtime. Therefore, 2 laterals will be needed (Nl = 2). Each lateral must irrigate 12 sets taking 6 days. Thus, the irrigation interval could be 8 days rather than 10.
Then using Equation 11.14:
\(q_s=\left(\dfrac{Q_cS_LS_m}{43,560}\right)\times\left(\dfrac{N_s}{N_l}\right)\times\left(\dfrac{T_s}{T_0}\right)\times\left(\dfrac{I_i}{I_i-T_d}\right)\)
\(q_s=\left(\dfrac{(60\text{ gpm/ac})(40\text{ ft})(50\text{ ft})}{43,560\text{ ft}^2\text{/ac}}\right)\left(\dfrac{24\text{ sets}}{2\text{ laterals}}\right)\left(\dfrac{12\text{ hr}}{10\text{ hr}}\right)\left(\dfrac{8\text{ days}}{8-2\text{ days}}\right)\)
\(q_s=\dfrac{6.0\text{ gpm}}{\text{ac}}\times\dfrac{0.55\text{ ac}}{\text{sprinkler}} \times1.2\times1.33=5.3\text{ gpm/sprinkler}\)