Center Pivot Irrigation Systems Peter Waller Muluneh Yitayew Chapter

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Abstract A center pivot irrigation system is a movable pipe structure that rotates around a central pivot point connected to a water supply. Center pivot irrigation systems are the most popular sprinkler irrigation systems in the world because of their high efficiency, high uniformity, ability to irrigate uneven terrain, and low capital, maintenance, and management costs. The history of center pivot irrigation systems began in Nebraska in the 1950s, and there are now hundreds of thousands of center pivot irrigation systems in the world. Center pivots are “perhaps the most significant mechanical innovation in agriculture since the replacement of draft animals by the tractor” (Splitter, Scientific American). The systems move through the field by electrically powered tractor wheels. Sprinkler flow rates increase toward the outer end of the pivot because the end of the pivot travels faster. The primary design constraint is the prevention of runoff at the end of the pivot, where application rates are highest. This chapter covers center pivot pipeline and mainline design, selection of sprinklers, and optimization of the design with respect to yield, energy requirement, components, and economics.

Keywords Center pivots Sprinklers Spray heads Rotator heads Infiltration rate Application rate Sprinkler selection Simulation Pipe head loss Optimization This is a preview of subscription content,

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Questions 1. Discuss the different types of sprinkler nozzles and systems used on center pivots. Discuss the strengths and weakness of the systems.

2. What is the difference between a linear move and a center pivot irrigation system?

3. Calculate the percent evaporation from sprinkler droplets for the parameters in Example 12.1 except that relative humidity is 50 %. If the application depth is 25 mm to a mature corn crop from overhead impact sprinklers, then what is the total depth of evaporation + canopy interception loss?

4. Calculate the percent evaporation from overhead sprinklers for the parameters in Example 12.1 except that relative humidity is 60 %. What is the depth of evaporation for an application depth of 25 mm to a mature corn crop from overhead impact sprinklers? Consider canopy interception and droplet evaporation?

5. Calculate the flow rate of a center pivot that has a length of 350 m, and gross application depth 15 mm/day. The pivot operates for 21 hours/day.

6. Calculate the maximum application rate for the parameters in question 12.5. The sprinkler wetted diameter is 4 m and the percent evaporation is 14 %. Then, show that the maximum application rate is the same if the pivot rotates three times per day (7 hour rotation). Show calculation and explain why the maximum application rate is the same in both cases.

7. Using the parameters in question 12.6, calculate the depth of runoff and maximum application rate during each pass for an intake family 3 soil with 2 mm surface storage. Include the 0.6985 initial infiltration depth. Calculate for three revolutions per day. The percent evaporation is 14 %. Use Chapter 12 Center pivot program and hand calculations.

8. Repeat Question 12.7, but change to one rotation per day.

9. Calculate the flow rate and runoff at the middle sprinkler (175 m from pivot point) for the pivot described in questions 12.6–12.9 for a single 21 hour rotation per day. Sprinkler spacing is 2 m at the middle of the center pivot. Fraction evaporation is 14 %. Sprinkler wetted diameter is 3.5 m. The soil intake family is 1. Surface storage is 4 mm.

10. Derive the Christensen’s F factor in Eq. 12.12 by assuming that a center pivot has four sprinklers (1/4, 1/2, 3/4, and full length down the pivot).

11. Low-pressure sprinkler nozzles at Paradise Cattle Company have a wetted diameter (Dw) of 2 m. The average rotation period of the center pivot is 8 hours. The pivot is 400 m radius and the flow rate is 95 L/sec. 5 % of water is lost to evaporation. Calculate the daily gross application rate, and plot the instantaneous application rate as a function of time at the 400 m. The Paradise Cattle Company pivots used in this and the next problem are described in the Center pivot pump and pipe network design section.

12. Plot the instantaneous application rate vs. time for at 200 m for the same parameters as in question 12.11.

13. Calculate the pressure loss in a center pivot that has a length of 350 m, and gross application depth applied during each pass is 5 mm. Time of rotation is 8 hours. Use 198 mm pipe. There is no down time.

14. A sprinkler has a flow rate of 5 GPM at 20 PSI. What is the flow rate at 25 PSI?

15. Use the Chap. 12 Center pivot model to find the optimal water application depth for CV values of 0.1 and 0.3.

16. Is it worth adding pressure regulators for the elevations shown in Fig. 12.23 ? Regulators cost is $5.00 per. This is the same as question 12.15, but add the regulators.

17. A center pivot irrigation system requires 200 kPa sprinkler pressure. There are five pivots each with a flow rate of 110 L/sec. Pressure regulators are used. Sprinklers are 1 m above the land surface. There is a 4 m pressure loss in the pivot pipeline, and 10 m head loss in the pipe network between the pumps and the worst case pivot. The maximum elevation of the land surface is 20 m higher than the reservoir. Make all other necessary assumptions. How many pumps are required? At what pressure and flow rate should the pumps operate?

© Springer International Publishing Switzerland 2016

Authors and Affiliations Peter Waller Muluneh Yitayew 1. Agricultural and Biosystems Engineering University of Arizona Tucson USA