1. Lateral Design

2. Lateral Material/Types Drip tape Thin wall drip line Heavy wall drip line Polypipe with punch emitters Polypipe with sprays

3. Typical Layouts

6. More layouts

9. SDI

10. Lateral installation

11. SDI burial depth CropBurial DepthLine spacing Trees and grapes>16 inches (0.4m)As per row spacing Berries, Vines> 8 inches (0.2m)As per row spacing Row crops – corn, cotton ≥ 12 inches (0.3m) 24 – 80in (0.6 -2.03m) Raised beds – single row Tomatoes, melons 2-4 inches (0.05 – 0.1m)One line 4- 6 inches (0.1 - 0.15m) from center of bed Raised beds – double row Onions, peppers, strawberries 2-4 inches (0.05 – 0.1m)One line down center of bed Raised beds – double row > 30 inch (0.75m) bed width 3-6 inches (0.075 -0.15m)Two lines spaced ½ the bed width apart

12. Tape orientation ◦ One tape or more per bed ◦ Holes upward Tape thickness ◦ Trend toward thicker Tape materials ◦ Stretch vs. breakage

13. Lateral Line Design Important lateral characteristics ◦ Flow rate ◦ Location and spacing of manifolds ◦ Inlet pressure ◦ Pressure difference

14. 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.

15. Start with average lateral

16. Design objective Limit the pressure differential to maintain the desired EU and flow variation What effects the pressure differential ◦ Lateral length and diameter  Economics longer and smaller ◦ Manifold location ◦ slope

17. 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 Ranges from 2 to 3 but generally considered to be 2.5 D P s =allowable pressure loss for subunit P a = average emitter pressure P n = minimum emitter pressure

18. Emission Uniformity

19. EU is related to Friction loss

20. Example Given: CV=0.03, 3 emitters per plant, qa =.43gph P a =15 psi, EU=92, x=0.57 Find: q n, P n, and  P

21. Solution

22. Practice problem

23. Flow rate Where: l = Length of lateral, ft. (m). Se = spacing of emitters on the lateral, ft. (m). ne = number of emitters along the lateral. qa = average emitter flow rate, gph (L/h)

24. Slope and topography

25. Four Cases

26. Lateral Flow flat slope

27. Lateral Flow 2% downhill slope

28. Lateral flow 2% uphill slope

29. Lateral flow varied slope

30. Manifold spacing Spacing is a compromise between field geometry and lateral hydraulics Lateral length is based on allowable pressure - head difference. Have the same spacing throughout the field in all crops

31. Manifold Location More efficient to place in middle two laterals extend in opposite directions from a common inlet point on a manifold, they are referred to as a pair of laterals. Manifold placed to equalize flow rates on the uphill and downhill laterals

33. Manifold placement zz 0.00.51.00.85 0.10.561.20.89 0.20.601.40.92 0.30.651.60.94 0.40.691.80.96 0.50.722.00.98 0.60.752.20.99 0.70.782.41.00 0.80.812.61.00 0.90.832.751.00

34. Determine optimum lateral length EU Slope Based on friction loss ◦ limited to ½ the allowable pressure difference ( ΔP s )

35. Hydraulics Limited lateral losses to 0.5 D P s Equation for estimating ◦ Darcy-Weisbach(best) ◦ Hazen-Williams ◦ Watters-Keller ( easiest, used in NRCS manuals )

36. C factorPipe diameter (in) 130≤ 1 140< 3 150≥ 3 130Lay flat 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)

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

38. Multiple outlet factors Number of outlets F F 1.85 1 1.75 2 1.85 1 1.75 2 1234567812345678 1.00 0.64 0.54 0.49 0.46 0.44 0.43 0.42 1.00 0.65 0.55 0.50 0.47 0.45 0.44 0.43 9 10-11 12-15 16-20 21-30 31-70 >70 0.41 0.40 0.39 0.38 0.37 0.36 0.42 0.41 0.40 0.39 0.38 0.37 0.36

39. Adjust length for barb and other minor losses

40. Or Or use equation Where Fe= equivalent length of lateral, ft) K = 0.711 for English units) B = Barb diameter, in D = Lateral diameter, in

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

42. Barb loss More companies are giving a K d factor now days

43. Example Given: lateral 1 diameter 0.50”, qave=1.5gpm,Barb diameter 0.10” lateral 2 diameter 0.50”, q ave =1.5gpm, k=.25 Both laterals are 300’ long and emitter spacing is 4 ft Find: equivalent length for lateral 1 and h etotal for lateral 2

44. Solution Lateral 1Lateral 2

45. Practice problem

46. Procedure Step 1 - Select a length calculate the friction loss Step 2 – adjust length to achieve desired pressure difference ( 0.5 D H s )

47. Step 3 - adjust length to fit geometric conditions Step 4 - Calculate final friction loss Step 5 – Find inlet pressure Step 6 – Find minimum pressure

48. Next step is to determine Δh Paired Lateral Single Lateral – ◦ Slope conditions  S > 0  S = 0 ◦ Slope Conditions  S friction slope

49. Last condition S < 0 and –S < Friction slope Which ever is greater

50. Find minimum lateral pressure Where S > 0 or S=0 Where S < 0 and –S < Friction slope Where S friction slope

51. Inlet pressure Estimate with the following equation Single Lateral Paired Lateral Better to use computer program

52. Handout Pressure summary

53. Design Considerations Select emitter/flow rate Determine required operating pressure Calculate friction loss ◦ Quick estimate use multiple outlet factor ◦ Manufacture’s software ◦ Built spreadsheet Decide whether to use single or paired laterals Make adjustments Determine Δh and hose EU

54. Practice problem

55. Class design problem