micro design
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
Micro Design
BASIC SYSTEM DATA
(Refer to NRCS Standard 441- Irrigation System, Micro Irrigation, for design requirements)
Total area irrigated ……..…………….... (acres)
Available water supply flow rate …….... (gpm) Available flow per acre: (gpm/acre)
System design flow rate …………….... (gpm) @ operating pressure of (psi)
# Zones Planned …………….... Area irrigated by each zone: (acres)
# Zones irrigated concurrently……….... Application rate per zone: (inches/hr)
Lateral line material: _________________________________ Inside diameter: (inches)
Drip tape/line material: ________________________________
Inside diameter: (inches)
Drip tape/line spacing: ………………….. (inches); Drip tape/line depth:
(inches)
Flushing velocity: ………………………. (ft/s); Flushing end pressure: (psi)
Flushing flow rate: (gpm)
Describe Emitter (make, model, etc.):
Type (circle one): laminar laminar/turbulent turbulent Emitter spacing: (inches)
Emitter path width:……………………..... (inches); Emitter path height: (inches)
Describe Filter system (type, model, capacity in gpm):
Pressure loss across filter: ….………. (psi); Head required at filter: psi
Time required for backwash: …………. (hours); Backwash flow rate: gpm
Describe Sand Separator (type, model, capacity in gpm):
Describe Chemigation Valve (type, model, location):
Describe Check Valve (type, model, location):
Zone Number: Type of drip tape/line:
Emitter data (model, type,
etc.)
Spacing: (inches)
Design manifold inlet pressure downstream of zone control valve: (psi)
Emitter discharge = q = Kd Hx (gal/hr) Kd = x =
Manufacturer’s Coefficient of Variation, (Cv):
Average emitter design discharge, qave: (gal/hr) @ line pressure of (psi)
Maximum emitter discharge, qmax: (gal/hr) @ line pressure of (psi)
Location of maximum discharge emitter:
Minimum emitter discharge, qmin: (gal/hr) @ line pressure of (psi)
Location of maximum discharge emitter:
Flow Variation = _________ % Emission Uniformity, (EU), = __________ %
System Capacity
**100
*(1 )r
gR
ET TGross Application F
EU L
dayplantgalf
SSFF rpg
dgp //**
623.0)/(
100( )11.6 ( / )
( )ft tapeQ gpm
Application Rate in hrLateral spacing in
( ) ( )
( )
453 ( )in ac
hr
D ASystem Q gpm
T
Crop Water Needs ExampleCalculate 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 TD = .2” / system
efficiencyA = AreaT = 22 hrs
Crop Water Needs Example Answer
Q = 453*DA T
= 453 x (.2”/.9) x 1 ac22 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 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.
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 ac4.5 gpm
= 2 hrs and 15 minutes
WELL
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
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
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
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 ac2.7 gpm
= 2 hrs and 40 minutes
WELL
8:00 a.m. Field 1
2 hrs and 40 min @ 2.7 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 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
Watering strategies
Select emitter based on water required
Calculate set time
adgp
a qe
FT )/(
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.
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.
Pressure flow relationship (Pa)
1
xa
a
qP
K
Where:
qa= average emitter flow rate (gph)
Pa = average pressure (psi)
x = emitter exponentK = flow constant
Start with average lateral
Standard requiresPipe 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.
Design objectiveLimit the pressure differential to
maintain the desired EU and flow variation
What effects the pressure differential◦Lateral length and diameter◦Manifold location◦slope
Four Cases
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
nas PPP 5.2
DPs =allowable pressure loss for subunit
Pa = average emitter pressure
Pn = minimum emitter pressure
Emission Uniformity
1 1.27 100n
a
qCVEU
qn
Q is related to Pressure
1
1 1.27 100
xx
n n n n
a a a a
x
n
a
q P P q
q P P q
So
PCV
Pn
HydraulicsLimited lateral losses to 0.5DPs
Equation for estimating◦Darcy-Weisbach (best)◦Hazen-Williams◦Watters-Keller (easiest, used in NRCS
manuals)
LDC
QFhf 87.4
852.1
5.10
C factor Pipe diameter (in)
130 ≤ 1
140 < 3
150 ≥ 3
130 Lay 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
75.4
75.1
D
QKFLh f
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)
Multiple outlet factors
Number of outlets
FNumber of
outlets
F
1.851 1.752 1.851 1.752
12345678
1.000.640.540.490.460.440.430.42
1.000.650.550.500.470.450.440.43
910-1112-1516-2021-3031-70>70
0.410.400.390.380.370.360.36
0.420.410.400.390.380.370.36
Adjust length for barb and other minor losses
Adjusted length
e
ee
S
fSLL
L’ = adjusted lateral length (ft)
L = lateral length (ft)
Se = emitter spacing (ft)
fe = barb loss (ft)
Start with average lateral
Pressure Tanks
Practice problem