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Agriculture / Bases of Drip Irrigation Planning

In this text we will show practical steps in designing a drip irrigation system. To properly read and PoljoKapPoKap Ilustracija Cedi understand the article the reader is advised to be sure to read and learn the material from a simple article about the pump evaluation for manual watering (in preparation).

Due to the extensive topic on irrigation, in the article you will not find a replacement for textbooks. However, the purpose of the article is to establish precise sequence of steps of practical design, without detailed consideration of calculations, unit conversions and other minor procedures. Starting from this information an interested reader is invited to choose an example case of his own and calculate parameters of the selected irrigation system by the help of formulas and data in this article or in the literature.

In this way, prior to any investment, the reader can estimate the cost of his project. In any case, consultations and exchange of experience with the contractor or equipment distributor is recommended and required prior to procurement and deployment of equipment.

General remarks

In hydrodynamics and especially in irrigation systems and pumps, the pressure is generally and almost always expressed in meters instead in pascals or bars or similar units. If pressure is expressed in meters, this means that at this point we have the pressure equal to the hydrostatic pressure of water column of the same height. The relationship between the pressure in bars and the pressure in meters is 1:10. For example: a dropper that has a nominal pressure of 1 bar at its entrance, will have an equivalent hydraulic pressure of 10m high water column.

Suffix tr, such as in h tr means friction, ie friction losses expressed in equivalent hidrost. pressure (meters). Suffixes GR, SR, LAT, DROP, 0 in the variables mean the main distribution, secondary distribution, lateral, dropper, nominal value, respectively.

In all following formulas, the units are in International system of units (SI), unless indicated otherwise.

Network topology

The system can have a simple structure consisting only of main line and laterals. Alternatively there is two-level hierarchy system composed of main line, submains and laterals. For the largest property there is three-level hierarchy structure consisting of main line, submains, manifolds and laterals. Which of those hierarchical schemes of irrigation will be applied depends on the size of the property and also the maximum length of laterals that we can use.poljoKapPoKapMreza2

Now let's see the specific sequence of design steps.

Step 1:
topology selection and selection of manufacturers / distributors of equipment

agriculture KapPoKapIlustracija1 In the first place you should familiarize yourself, if you have done that already, with the plant watering needs. Secondly, in this step, outlining the property and on it you specify the irrigation topology: main lines, laterals, and if necessary submains and manifolds. The choice of topology should provide means for independent activation or deactivation of desired network segments, and of course provide watering of the whole property around the plans, according to the plants needs.

Pay attention to the maximum length of laterals that you use - all over 100m may later prove to be excessive, and that you will precisely check in the next step. Afterwords if  this length of laterals turns out to be too high in paragraph 2.3, we will have to return to this procedural step and redefine our topology so the length of laterals is acceptable.

Also you select the topology in which the laterals will spread only horizontally. With the help of the textbooks as we said, you should familiarize yourself how the exact plant should be watered in order tobe able to customize the design according to this knowledge.

It's not bad to remind you that the irrigation pipes nominal diameter is the outer diameter, not internal. The internal diameter is calculated from the thickness of the wall and outer diameter, or is read from the table provided by manufacturer.

Step 2: calculation of lateral

agriculture KapPoKapIlustracija2 Here we first deal with the required number of emitters (droppers). Then we find the required flow through the lateral and end up checking the uniformity of flow. Here are the exact sequence of design steps.

2.1 Number of emitters and distance between them

Number of emitters on lateral and the spacing between adjacent emitters are determined primarily based on the needs of plants and soil type in question. We shall not talk here about detalis of those, because they can be found in the literature and on the Internet.

2.2 Flow per lateral

Based on the emitter number and their nominal flow Q kap0, by multiplying we obtain the Q LAT, ie. nominal lateral f flow through the lateral.

2.3 Checking the maximum permissible length of laterals

In most cases, the laterals manufacterer presents table where you will find the recommended maximum allowable length of laterals. This length should ensure relatively uniform flow through all branches of the lateral drain. In the event that these tables are not available, this problem can be resolved by the variable L in the system of equations:

{V = Q / S}      (S = d 2 π / 4)

poljo KapPoKapIlustracijaHW{{h_{tr}} = f \cdot 6,82 \cdot \frac{{{{(v/C)}^{1,852}}}}{{{d^{1,17}}}} \cdot L}  Hazen Williams formula (HW), apply when Re≥2000

{h_{tr}} = f \cdot \frac{{64}}{{{R_e}}} \cdot \frac{L}{d} \cdot \frac{{{v^2}}}{{2g}}   D'Arcey Weisbach formula (DW), apply when Re<2000

{{R_e} = 930 \cdot d[mm] \cdot v}   Reynolds number

f = \frac{1}{{m + 1}} + \frac{1}{{2n}} + \frac{{\sqrt {m - 1} }}{{6{n^2}}}     Christiansen's factor, m=1,852 for HW formula, m=2 for DW

{h_{tr}} = 2{p_{kap}}\left[ {bar} \right]     htr is the result of HW or DW forumla, depending on Reynolds number
All variables with no units shown are expressed in SI-system. Here: v = velocity through lateral, Q = the water flow, S = surface area of ​​the inner section of the pipe or laterals, f = Christiansen's coefficient, C = constant in the formula mostly depending on material, d = internal diameter of lateral, L = length of lateral, and pkap is the nominal pressure of emitter when it achieves its nominal flow, as stated by the manufacturer.

The equation for v is the generally valid equation for fluid flow through any surface. Formulae Hazen Williams and Darcy Vajsbah apply alternatively depending on the value of the Reynolds number. Both formulas calculate the pressure loss htr that exists along pipe. The constant C in the formula HV normally is in the range between 120 and 150 for pipes that are commonly used in irrigation (eg 140 for okiten ie PE hoses, 150 for PVC pipe). Christiansen's factor is a correction koefficient in HW or DW formula. Last formula used above is modeling the uniform flow through the lateral.

On the Internet and in the literature you have a lot of material with detailed descriptions of these formulas, so in this article I will continue to deal with them as it is and this is quite sufficient for solving the system of equations for the variable L.


2.4 The choice of pipe and losses

Diameter of offered laterals is selected from the chosen manufacturer data. The pressure loss in lateral, htr LAT is calculated with formulas from step 2.3. In choosing the diameter we will start from smaller values of course. If later in the section 2.6 we learn this does not fit the requirement of uniformity, we have to return to this procedural step and increase the diameter of lateral in order to reduce the losses.

2.5 The pressure at the entrance of lateral

At the lateral entrance there will be equivalent hidrost. pressure hLAT=htrLAT+hkap0, where hkap0 is the nominal equivalent hydraulic pressure of emitter.

2.6 Checking the uniformity of flow in lateral

This requirement of uniformity is expressed by equation: htrLAT≤0,2*hLAT.

Step 3: Design of submain line

Firstly we determine the number of laterals per submain distribution line. So we deduce the flow through submain line and based on that calculation result we can determine the diameter of the pipe and losses. Finally make sure that the flow through submain line is uniform: if everything is OK we continue design to the next step, if not go back to increase the diameter of the pipe, until you meet the uniformity of flow principle.

All these calculations we shall perform only on one segment of the submain.

3.1 The number of laterals on submain line agriculture KapPoKapIlustracijaGrananje

Here, based on the need for watering the plants, which we met in Step 1, choose a number of laterals on the submain pipe.

3.2 Flow through submain pipe

This we deduce directly out of the previous data by multiplication.

3.3 Selection of pipe dimensions and calculation of losses

Here also we work as in paragraph 2.4 for laterals.

3.4 Checking speed

v ≤ 1.7m / sec. If this is not fulfilled, then return to point 3.3 with a larger diameter pipe.

3.5 The pressure at the entrance of submain line

The pressure at the entrance of submain is hSR=htrSR+hLAT.

3.6 Checking the uniformity of flow through the submain distribution

This uniformity is checked as if in 2.6: 20% of the submain pressure: htrSR≤0,2hSR

poljo KapPoKapIlustracijaBigPipe

Step 4: calculation of main line

Here also all the calculations are done only on one segment of the main lines. Design steps resemble those of step #3.

4.1 Flow through the segment of main distribution

This information is obtained by multiplying values from some of the previous steps.

4.2 The choice of pipes and calculation of losses

Similarly as in Step 3.

4.3 Checking speed

Similarly as in Step 3.

4.4 Pressure at the entrance of main line

The pressure at the entrance of the main line is thus hGR=htrGR+hSR

4.5 Practical assessment for acceptable losses in the main line

At this point we will adopt practical assessment: hGR<15м. On the bases of your experience you can later adopt some other criteria. Needless to say - that if this condition is not met, you can go back and relax appropriate parameters order to meet this requirement.

Step 5: minor losses

Minor losses occur in pipe components such as valves, fittings and everything else that is not the streight pipe. There are multiple ways of their calculation, but by far the simplest method is the method of equivalent pipe length. Fitting loss is expressed as Le/d, where d is the inner diameter, and Le is the length of an imaginary pipe by which is the fitting replaced. The equivalent pipe produces the same losses as the fitting itself. Sometime this method is applied in the form of a nomogram:

poljo KapPoKapGubiciNomogram

Another way is to specify the coefficient of losses, by which the losses are calculated according to the formula htr=k * v2/(2g), where v is the speed of the water flow, and g=9,81m/s2. One of the simpler tables for the application of this method is presented below.

poljo KapPoKapGubici

Which method you will use depends on your choice. On the Internet, search the pressure loss in fittings to get the complete table.

Step 6: Calculating parameters of pump

First we start from the pump flow, then from the previous steps we determine the equivalent hydrostatic pressure of the pump needed to be achieved.

agriculture KapPoKapIlustracijaPumpa

Then we choose the efficiency of the pump and thus arrive at the required pump electrical power.

6.1 Determination of pump flow

We just read the flow in the main line.

6.2 Determination of the equivalent hydrostatic pressure in network

I'm just rewriting hGR from step 4.3, so easy, right?

6.3 Establish the height difference hpum0 at the site of the pump

That's the difference in height between the whole pipelin network and the intake of the pump.

6.4 Calculation of the minor losses hpom el

Losses on supporting elements like valves, couplings, elbows, couplings etc. are very numerous, generally. In a separate literature you can find the coefficients for the fittings and how each of them contribute to the velue of equivalent hidr. pressure. This equivalent hydrostatic pressure is to be treated the same way as you treated the loss as a result of friction. If irrigating say ten hectares until you find the precise parameters, take for example 10m as losses in minor elements, at least as a rough measure.

6.5 Equivalent hydrostatic pressure of the pump

hpum=hGR+hpum0+hpom el

6.6 Power of the pump

This you already know from the previous article on irrigation pump (yet to be added in this english version).  For a quick and easy calculation assume the multiplication of efficiency factor and transmission factor E * η equals to 50% for the pump with an electric motor, and for the pump with a generator that is 40%.=

The important case: elevation

What is skipped in the literature is the explanation of elevation.

agriculture KapPoKapElevacija

In the example below you can see a side view of the tube elevated helev after the pressure gauge. The fluid flows from left to right. What is the pressure level at the place of valve, due to the increase in height?

Between the gauge and the valve there are two differences: there will be the effect of friction of fluid in the pipe and there will be the potential energy difference. The friction loss is calculated by means of DW or HW formulas.

About the potential energy: it exists on the basis of energy consumption that some element of water needs to reach the plac of valve starting from the place of pressure gauge. This also means that we will shall take helev into account when calculating the resistance that the pump must overcome. If the pipe has a negative elevation, helev will count as a negative value, ie. it will be a contribution to the work of the pump.

It is important to understand that this discussion on the elevation so far is confined to the main line, submain line and manifold. In the case of lateral something like this should not be done. The lateral on elevated ground should be installed at an angle of about -1% (down), not more and not less. In this way the emitters flow will be uniform.

Do your own example

Now you are armed with an arsenal of real physical-mathematical weapons to resolve your own drop irrigation system. The question is if you will know how to use this arsenal: your actual knowledge will depend on how meticulously you read this article and whether you know how to expand your knowledge if something is missing in your mental picture. If you lack imagination, here is a task which can be solved for practice. agriculture KapPoKapZadatak

On the picture is a sketch of a simple rectangular property with already selected distribution topology and with the well in the middle. Let us suppose that property is an orchard whose seedlings are spaced 1,1m x 1,1m. On each tree we need 4 emitters. Every emitter has capacity 4L/H and nominal pressure of 1 bar. All you have to do is to specify the required pump and pipe diameters of main line, submain line and laterals. If you get electrical power of pump somewhere in the range between 8 and 11 kW you have probably a good planning. If not, well start again until you get it right!

How would you set the distribution network up on one side of a hill?


 Drip Irrigation, Moshe Sne, 2005

 Drip Irrigation Design Fundamentals & Guidelines, IRWA Team Lebanon, 2004

 Handbook on Pressurized Irrigation Techniques, FAO, 2007

 The handbook of Technical Irrigation Information, Hunter, 2002

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