Micro Design

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

( ) ( )

( )

453 ( )in ac

hr

D ASystem Q gpm

T

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

**100

*(1 )r

gR

ET TGross Application F

EU L

Wetted Area vs Area

Irrigated area vs. wetted areaET 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 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 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 ac4.5 gpm

= 2 hrs and 15 minutes

WELL

10:1

5 a

.m.

zon

e 2

8:0

0 a

.m.

zon

e 1

2:4

5 p

.m.

zon

e 4

5:1

5 p

.m.

zon

e 5

7:3

0 p

.m.

zon

e 6

12:3

0 p

.m.

zon

e 3

9:4

5 p

.m.

zon

e 7

12:0

0 a

.m.

zon

e 8

2:1

5 a

.m.

zon

e 9

4:3

0 a

.m.

zon

e 1

0

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.

Practice problem – Set time and Qs

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

EU is related to Friction loss

x

a

n

a

n

x

a

n

a

n

q

q

P

P

P

P

q

q

1

Emission Uniformity

1 1.27 100

1 1.27 100

n

a

x

n

a

qCVEU

qn

or

PCV

Pn

Lateral Line DesignImportant lateral characteristics

◦Flow rate◦Location and spacing of manifolds◦Inlet pressure◦Pressure difference

Design objectiveLimit 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

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 Ranges from 2 to 3 but generally considered to be 2.5

DPs =allowable pressure loss for subunit

Pa = average emitter pressure

Pn = minimum emitter pressure

Example

Given: CV=0.03, 3 emitters per plant, qa = .43gph Pa=15 psi, EU=92, x=0.57

Find: qn, Pn, and P

Solution

1 1 1

.57

0.031 1.27 100 92 1 1.27 100

0.433

92 0.430.404

.031 1.27 100

3

0.404 15 13.474

0.43

2.5 2.5 15 13.47 3

n n

a

n

x xn n n

n aa a a

a n

q qCVEU

qe

q

P q qP P

P q q

P P P

.83psi

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)

6060aea

el

qnq

S

lq

Determine optimum lateral lengthEU SlopeBased on friction loss

◦ limited to ½ the allowable pressure difference (ΔPs)

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 factorChristiansen's equation for computing

the reduction coefficient (F) for pipes with multiple, equally spaced outlets where the first outlet is Sl from the mainline is:

F = Reduction factorN= number of outletsM= exponent depends on which friction equation is used

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

86.1711.0 ie DBf

OrOr use

equation

Where Fe= equivalent length of lateral, ft)K = 0.711 for English units)B = Barb diameter, inD = Lateral diameter, in

Adjusted length

e

ee

S

fSLL

L’ = adjusted lateral length (ft)

L = lateral length (ft)

Se = emitter spacing (ft)

fe = barb loss (ft)

Barb lossMore companies are giving a Kd

factor now days

2

2f e d ft f f e

Vh K h h eh

g

Example

Given: lateral 1 diameter 0.50”, qave=1.5gpm,

Barb diameter 0.10” lateral 2 diameter 0.50”,

qave=1.5gpm, kd=.10

Both laterals are 300’ long and emitter spacing is 4 ft

Find: equivalent length for lateral 1 and htotal for lateral 2

Solution

1.86

1.75

4.75

0.711*.1 .5 0.258

4 .258300 319.4

4

1.519.4 .36 0.00133 0.51

.5

ef

L

hf ft

fth

fth

gV

fpsV

et

e

h

675.75*009.0

009.009.0*1.0

09.02

45.2

45.20.5

1.5409.0

2

2

Lateral 1 Lateral 2

Procedure

Step 1 - Select a length calculate the friction loss

Step 2 – adjust length to achieve desired pressure difference ( 0.5DHs)

75.2

a

bab L

Lhfhf

36.0

af

bfab h

hLL

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 < 0 and –S > friction slope

75.211 zhzEh f

fhEh

fhElh

fhElh )(

Last condition

57.157.057.1

36.00.1

ff

ff

ff h

EFh

h

EFEorh

h

E

h

Eh

S < 0 and –S < Friction slope

Which ever is greater

Inlet pressureEstimate with the following equation

Single Lateral

Paired Lateral

Better to use computer program

30.433

4 2f

l a

h ElP h

3.753.750.75 1 2 1 0.4332l a fp

EP h h z z z

Find minimum lateral pressure

Where S > 0 or S=0

Where S < 0 and –S < Friction slope

Where S < 0 and –S > friction slope

n lP P P

n lP P

n lP P P

Calculate final EU

1 1.27 100

1 1.27 100

n

a

x

n

a

qCVEU

qn

or

PCV

Pn

Practice problem pressure difference and EU

Block HydraulicsAvg lateral inlet pressure

Becomes the avg outlet pressure for manifold

Manifold hydraulics

Allowable manifold lossAllowable pressure loss – lateral

lossesCalculate 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

𝑞𝑣𝑎𝑟=𝑞𝑚𝑎𝑥−𝑞𝑚𝑖𝑛

𝑞𝑚𝑎𝑥