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Published byVincent Peres Modified over 2 years ago

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Lateral Design

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

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Typical Layouts

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More layouts

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SDI

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Lateral installation

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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

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

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

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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.

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Start with average lateral

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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

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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

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Emission Uniformity

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EU is related to Friction loss

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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

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Solution

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Practice problem

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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)

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Slope and topography

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Four Cases

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Lateral Flow flat slope

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Lateral Flow 2% downhill slope

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Lateral flow 2% uphill slope

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Lateral flow varied slope

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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

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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

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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

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

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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 )

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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)

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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)

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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

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Adjust length for barb and other minor losses

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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

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

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

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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

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Solution Lateral 1Lateral 2

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Practice problem

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

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

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

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

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

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

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Handout Pressure summary

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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

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Practice problem

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Class design problem

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3.4 Equal friction method This method of sizing is used for supply, exhaust and return air duct system and employs the same friction lose per foot of length.

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