Micro Design

System Capacity

D = gross application for what ever time period ( hrs, day or days) T= hours in time period used to decide “D” (max 22hrs/day) A= Acres irrigated

Wetted Area vs Area

Irrigated area vs. wetted area ET rates are computed and published assuming that the depth comes from the whole area, not just the canopy area or the wetted area. ET rate, expressed as depth per unit of time, is largely governed by the amount of energy available to convert liquid water into vapor (i.e. ET), and therefore does not depend upon the tree size (once the canopy exceeds 65% of the surface area), or the wetted area (as long as there is sufficient root mass to absorb the required water).

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 zones of 0.1 ac each. Design EU of 90% Q = 453*DA T D =.2” / system efficiency A = Area T = 22 hrs

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

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 zones, each of these zones 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.

Hours of irrigation per day to apply.2” (1 zone 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

WELL 10:15 a.m. zone 2 8:00 a.m. zone 1 2:45 p.m. zone 4 5:15 p.m. zone 5 7:30 p.m. zone 6 12:30 p.m. zone 3 9:45 p.m. zone 7 12:00 a.m. zone 8 2:15 a.m. zone 9 4:30 a.m. zone 10

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

WELL 8:00 a.m. Field 1 2 hrs and gpm 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!!!! 10:40 a.m. Field 2 2 hrs and gpm 1:20 p.m. Field 3 2 hrs and gpm 4:00 p.m. Field 4 2 hrs and gpm 6:40 p.m. Field 5 2 hrs and gpm 9:20 p.m. Field 6 2 hrs and gpm 12:00 a.m. Field 7 2 hrs and gpm 2:40 a.m. Field 8 2 hrs and gpm 5:20 a.m. Field 8 2 hrs and gpm

Adjust flow rate or set time If T a 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.

Practice problem – Set time and Q s

Pressure flow relationship (P a ) Where: q a = average emitter flow rate (gph) P a = average pressure (psi) x = emitter exponent K = flow constant

EU is related to Friction loss

Emission Uniformity

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

Design objective Limit the pressure differential to maintain the desired EU and flow variation The pressure differential is affected by ◦ Lateral length and diameter  Economics longer and Larger ◦ Manifold location ◦ slope

Four Cases Effects of slope

Lateral Flow flat slope

Lateral Flow 2% downhill slope

Lateral flow 2% uphill slope

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

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

Solution

Practice problem - allowable loss

Start with average lateral

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)

Lateral flow varied slope

Manifold location Location 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

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

Manifold placement zz

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

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 )

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)

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

Multiple outlet factor Christiansen's equation for computing the reduction coefficient (F) for pipes with multiple, equally spaced outlets where the first outlet is S l from the mainline is: F = Reduction factor N= number of outlets M= exponent depends on which friction equation is used

Caveats For pipes that have no flow past the last outlet (sprinkler) Cannot be directly applied to the estimation of friction losses only partway down the lateral pipe. Assumes that each outlet has a constant discharge, Equations are for use with laterals having nearly constant discharge per outlet, such as for hand lines, wheel-lines, solid set (fixed), and linear-move systems or drip laterals. The value of F approaches 0.36 when N > 35, which is often the case with sprinkler or drip laterals.

Multiple outlet factors Number of outlets F F >

Adjust length for barb and other minor losses

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

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

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

Example Given: lateral 1 diameter 0.50”, qave=1.5gpm, Barb diameter 0.10” lateral 2 diameter 0.50”, q ave =1.5gpm, k d =.10 Both laterals are 300’ long and emitter spacing is 4 ft Find: equivalent length for lateral 1 and h total for lateral 2

Solution Lateral 1Lateral 2

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

Practice problem Lateral length

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

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

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

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

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

Calculate final EU

Pressure Tanks

Practice problem pressure difference and EU

Block Hydraulics Avg lateral inlet pressure Becomes the avg outlet pressure for manifold

Manifold hydraulics

Allowable manifold loss Allowable pressure loss – lateral losses Calculate losses using Hazen-Williams etc. Find minimum and maximum outlet pressures for manifold use this to calculate maximum and minimum lateral flow rate Calculate block EU and flow variation

Block EU

Flow variation