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Micro Design

Micro Design. System Capacity. Crop Water Needs Example. Calculate capacity required for a proposed 1 ac. Micro irrigation system on Vegetables. Using drip tape with a flow of 0.45 gpm /100’ and 12” emitter spacing, 200 ft rows, 5 ft row spacing, and 10 fields of 0.1 ac each. Q = 453*DA

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Micro Design

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  1. Micro Design

  2. System Capacity

  3. Crop Water Needs Example • Calculate capacity required for a proposed 1 ac. Micro irrigation system on Vegetables. Using drip tape with a flow of 0.45 gpm/100’ and 12” emitter spacing, 200 ft rows, 5 ft row spacing, and 10 fields of 0.1 ac each. Q = 453*DA T D = .2” / system efficiency A = Area T = 22 hrs

  4. Crop Water Needs Example Answer Q = 453*DA T = 453 x (.2”/.9) x 1 ac 22 hrs = 4.5 gpm

  5. Each field will have a capacity of 4.5 gpm. • 200 ft rows with a 0.45 gpm/100’ drip tape flow will give you 0.9 gpm per row. • At 5 ft row spacing, 1 ac will have approximately 10 fields, each of these fields will have 5 rows, 200 ft long. • 5 rows times 0.9 gpm/row is 4.5 gpm per field. • Minimum water requirement is 4.5 gpm for 1 ac, we only need to run 1 field at a time to meet the crop water demand.

  6. Hours of irrigation per day to apply .2”(1 field of 0.1 ac each) T = 453*DA Q = 453 x (.2”/.9) x 0.1 ac 4.5 gpm = 2 hrs and 15 minutes

  7. 8:00 a.m. field 1 2 hrs and 15 min @ 4.5 gpm 10:15 a.m. field 2 2 hrs and 15 min @ 4.5 gpm 12:30 p.m. field 3 2 hrs and 15 min @ 4.5 gpm 2:45 p.m. field 4 2 hrs and 15 min @ 4.5 gpm 5:15 p.m. field 5 2 hrs and 15 min @ 4.5 gpm 7:30 p.m. field 6 2 hrs and 15 min @ 4.5 gpm 9:45 p.m. field 7 2 hrs and 15 min @ 4.5 gpm WELL 12:00 a.m. field 8 2 hrs and 15 min @ 4.5 gpm 2:15 a.m. field 9 2 hrs and 15 min @ 4.5 gpm 4:30 a.m. field 10 2 hrs and 15 min @ 4.5 gpm

  8. When you get to the field you discover that the well only produces 2.7 gpm. So @ .9 gpm/row water 3 rows and have15 sets

  9. Hours of irrigation per day to apply .2”(given a well capacity of 2.7 gpm and fields of 3 rows each) T = 453*DA Q = 453 x (.2”/.9) x 0.07 ac 2.7 gpm = 2 hrs and 40 minutes

  10. 8:00 a.m. Field 1 2 hrs and 40 min @ 2.7 gpm 10:40 a.m. Field 2 2 hrs and 40 min @ 2.7 gpm 1:20 p.m. Field 3 2 hrs and 40 min @ 2.7 gpm 4:00 p.m. Field 4 2 hrs and 40 min @ 2.7 gpm 6:40 p.m. Field 5 2 hrs and 40 min @ 2.7 gpm 9:20 p.m. Field 6 2 hrs and 40 min @ 2.7 gpm 12:00 a.m. Field 7 2 hrs and 40 min @ 2.7 gpm 2:40 a.m. Field 8 2 hrs and 40 min @ 2.7 gpm 5:20 a.m. Field 9 2 hrs and 40 min @ 2.7 gpm WELL At 6:00 a.m. the pump has been running for 22 hrs and at 8:00 a.m. we need to go back to Field 1 but we haven’t irrigated all the fields!!!!

  11. Watering strategies • Select emitter based on water required • Calculate set time

  12. Adjust flow rate or set time • If Ta is greater than 22 hr/day (even for a single-station system), increase the emitter discharge • If the increased discharge exceeds the recommended range or requires too much pressure, either larger emitters or more emitters per plant are required.

  13. Select the number of stations • If Ta ≈ 22 h/d, use a one-station system (N = l), select Ta < 22 hr/day, and adjust qa accordingly. • If Ta <11 h/d, use N = 2, select a Ta <11, and adjust qa accordingly. • If 12 < Ta < 18, it may be desirable to use another emitter or a different number of emitters per plant to enable operating closer to 90 percent of the time and thereby reduce investment costs.

  14. Pressure flow relationship (Pa) Where: qa= average emitter flow rate (gph) Pa = average pressure (psi) x = emitter exponent K = flow constant

  15. Start with average lateral

  16. Standard requires • Pipe sizes for mains, submains, and laterals shall maintain subunit (zone) emission uniformity (EU) within recommended limits • Systems shall be designed to provide discharge to any applicator in an irrigation subunit or zone operated simultaneously such that they will not exceed a total variation of 20 percent of the design discharge rate.

  17. Design objective • Limit the pressure differential to maintain the desired EU and flow variation • What effects the pressure differential • Lateral length and diameter • Manifold location • slope

  18. Four Cases

  19. Allowable pressure loss (subunit) This applies to both the lateral and subunit. Most of the friction loss occurs in the first 40% of the lateral or manifold DPs =allowable pressure loss for subunit Pa = average emitter pressure Pn = minimum emitter pressure

  20. Emission Uniformity

  21. Q is related to Pressure

  22. Hydraulics • Limited lateral losses to 0.5DPs • Equation for estimating • Darcy-Weisbach (best) • Hazen-Williams • Watters-Keller (easiest, used in NRCS manuals)

  23. Hazen-Williams equation hf =friction loss (ft) F = multiple outlet factor Q = flow rate (gpm) C = friction coefficient D = inside diameter of the pipe (in) L = pipe length (ft)

  24. Watters-Keller equation hf = friction loss (ft) K = constant (.00133 for pipe < 5” .00100 for > 5”) F = multiple outlet factor L = pipe length (ft) Q = flow rate (gpm) D = inside pipe diameter (in)

  25. Multiple outlet factors

  26. Adjust length for barb and other minor losses

  27. Adjusted length L’ = adjusted lateral length (ft) L = lateral length (ft) Se = emitter spacing (ft) fe = barb loss (ft)

  28. Start with average lateral

  29. Pressure Tanks

  30. Practice problem

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