Irrigation Water Management

Irrigation Systems

2-TERMS & UNITS

Index
Units Conversion
Volume = Depth x Area
Amount = Rate x Time
Application Rate (in/hr) = Flow Rate (gpm) x 96.3/Area (sq-ft)
Pressure/Tension Units
Salinity Units
     Concentration
     Conductivity
     Osmotic Stress

Units Conversion

All units conversions are based on the concept that multiplying by 1 doesn't change any quantity:

37 x 1 = 37     -->     no change

Also, 1 = any amount ÷ same amount

 Example 1

2.5 days = how many hours?

A conversion factor between days and hours can be built from the equation

1 day = 24 hours

So, 24 hours ÷ 1 day = 1, because 24 hours and 1 day are the same quantity of time. The conversion factor 24 hours/day is just a convenient form of 1, and we can multiply (or divide, if necessary) by this factor without changing things.

           24 hours
2.5 days x --------
            1 day

Remember that units can cancel, just like numbers cancel in fractions. In the above equation, the "days" of 2.5 days and the "day" of 1 day cancel, leaving the "hours" of 24 hours as the only remaining unit.

           24 hours
2.5 days x -------- = 2.5 x 24 hours = 60 hours
            1 day

 Example 2

If water use is 250 gallons per person per day, a family of 4 uses

250 gal/person/day x 4 persons = 1,000 gallons/day

How many acre-feet per year is this?

1,000 gal/day x [1 cu-ft]/[7.48 gal] = 133.69 cu-ft/day

133.69 cu-ft/day x [1 acre]/[43,560 sq-ft] = 0.00307 ac-ft/day

The sq-ft cancel into the cu-ft, leaving feet, and feet x acres gives acre-feet, or ac-ft. There is nothing to cancel the "day", so the final unit is ac-ft/day.

0.00307 ac-ft/day x 365 days/year = 1.12 ac-ft/yr

The "days" cancel each other, leaving the unit as ac-ft/yr.

Note: This is the basis of the statement sometimes found in newspapers that one acre-foot of water is enough to serve a family of 4 for one year.

 Example 3

How many acres does one football field cover?

If you leave out the end zones, a football field is 100 yards long by 53 yards wide.

100 yd x 53 yd x 3 ft/yd x 3 ft/yd x 1 acre/[43,560 sq-ft] = 1.1 acres

Note: Approximately, one acre-foot of water will cover a football field with water 1 foot deep (another explanation often encountered in the newspapers).

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Volume = Depth x Area

 Example 4

How deep will 1 cu-ft of water spread out over 4 sq-ft of area?

Depth = Volume/Area = [1 cu-ft]/[4 sq-ft] = 0.25 ft = 3 inches

 Example 5

Water has been turned into a 4 acre reservoir until it stands 3 inches deep. How much water has been added?

3 inches x 1 ft/12 inches = 0.25 ft

Volume = Area x Depth = 4 acre x 0.25 ft = 1 ac-ft

Water depth can be a physical depth, as in the 2 previous examples. Other situations where water depths correspond to physical depths are
     standing water
     rainfall (3 inches of rain)
     irrigation amount (a 4 inch irrigation application)
     rain gage with straight sides
          1 inch rain => 1 inch standing water in bottom of gage

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Volume = Rate x Time

This expression can be used for many quantities, so long as the units for Volume, Rate and Time correspond. If the Time is in minutes and the Amount is in gallons, then the Rate must be in gallons/minute (gpm).

 Example 6

A pump delievers 6 gpm to a field for 10 hours. How much water is applied to the field?

10 hours X 60 min/hr = 600 min

Volume Applied = Flow Rate X Time
Volumen Applied = 6 gpm X 600 min = 3600 gal

 Example 7

A water source delivers 10 gpm (cubic feet per second) for 8 hours. How many acre feet of water are delivered?

1 cfs = 1 ac-in/hr (approximately)

10 cfs = 10 ac-in/hr

Amount = Rate x Time = 10 ac-in/hr x 8 hr = 80 ac-in

80 ac-in x 1 ft/12 in = 6.67 ac-ft

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Application Rate (in/hr) = Flow Rate (gpm) x 96.3/Area (sq-ft)

Though not as basic a relationship as Volume = Rate x Time, the following is so useful that it is worth treating as a basic relationship:

This is a units conversion formula with all of the units conversions already worked out for you. For example, the conversion of gallons to cubic feet, so it can be divided by square feet, and the conversion of the resulting feet to inches is all bound into the "96.3" figure. Also in there is the conversion of the minutes in gpm to hours in in/hr.

This expression is set up to assess individual sprinklers. For example,

 Example 8

Sprinklers with a flow rate of 3.5 gpm are used in a system with a 30 ft x 40 ft sprinkler spacing. What is the sprinkler application rate (in/hr)?

App Rate (in/hr) = Flow Rate (gpm) x 96.3 / Area (sq-ft)

App Rate (in/hr) = 3.5 gpm x 96.3 / (30 ft x 40 ft)

App Rate (in/hr) = 0.28 in/hr

It can also be used to assess entire systems or fields if we remember to convert acres to square feet:

 Example 9

A center pivot sprinkler system irrigates a field of 130 acres, and is to be designed to meet a peak application rate of 0.35 in/day. Assuming 24 hours/day operation during the peak period, what flow rate (gpm) should the system have?

First, convert the 0.35 in/day to inches per hour:

0.35 in/day x 1 day/24 hours = 0.014583 in/hr

Next, convert 130 acres to square feet:

130 acres x 43,560 sq-ft/acre = 5,662,800 square feet

Next, re-write the formula to solve for Flow Rate (gpm):

Application Rate (in/hr) = Flow Rate (gpm) x 96.3 / Area (sq-ft)

Flow Rate (gpm) = App Rate (in/hr) x Area (sq-ft) / 96.3

Flow Rate (gpm) = 0.01458 in/hr x 5,662,800 sq-ft / 96.3

Flow Rate (gpm) = 857 gpm

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Pressure/Tension Units

 Example 10

What is the pressure created by one foot of water. One cubic foot of water weighs 62.4 pounds. Think of that cubic foot of water as a perfect cube, with dimensions of one foot in each direction. Particularly,
* the water stands 1 foot high, and
* the weight of the water presses down on 1 sq-ft of area

1 sq-ft x 12 in/ft x 12 in/ft = 144 sq-in

So, 62.4 pounds are pressing down on 144 sq-in. This is a "Pressure" of

62.4 lb ÷ 144 sq-in = 0.433 lb/sq-in = 0.433 psi

Note:  psi = pounds per square inch

 Example 11

How high does water have to stand to generate a pressure of 1 psi? Consider a column of water of height H (ft) pressing down on a 1 sq-ft area (144 sq-in). The total volume of the water will be

volume = depth x area = H x 1 sq-ft = H cu-ft

each cu-ft weighs 62.4 pounds, so

weight = H cu-ft x 62.4 lb/cu-ft = 62.4 H pounds

pressure = weight/area = 62.4 H lb/144 sq-in

Our problem asks about a pressure of 1 psi, so set

1 psi = 62.4 H lb/144 sq-in = (62.4 H/144) psi = 0.433 H psi

H = 1 psi ÷ 0.433 psi = 2.31

So 2.31 feet of water = 1 psi of pressure

Other Units for Pressure or Tension

Tension is (suction) is just the reverse of pressure. Pressure is a PUSH - tension is a PULL. A TENSION of +155 cb = a PRESSURE of -155 cb.

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Salinity Units

There are 2 approaches to measuring salinity: concentration and conductivity.

Concentration

is the amount of salt per unit of water

 Example 12

Show that 1 milligram per Liter (1 mg/L) equals 1 part per million (1 ppm).

1 Liter (L) = 1,000 milliliters (ml)

1 ml of water has a mass of 1 gram (g)

1 ml = 1 g = 1,000 milligrams (mg)

1 L = 1,000 ml x 1,000 mg/ml = 1,000,000 mg (1 million mg)

1 milligram of salt in 1 Liter of water gives a concentration of

1 mg/1 million mg = 1 part per million = 1 ppm

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Conductivity

is the ability to pass electric current  -->  the reverse of resistance

Resistance is measured in Ohms, so conductivity uses Mhos (ohms spelled backwards) in its unit. In irrigation, salinity is usually much smaller than a mho, so the unit millimho (1/1000 of a mho) is used. For technical reasons, the geometry of the measurement is important, so the actual unit used is the

mmho/cm = millimho per centimeter

More recently, the metric system has changed the unit to dS/m (deciSeimens per meter), but fortunately, it turns out that numerically these units are the same

1 dS/m = 1 mmho/cm

Electrical conductivity is abbreviated "EC", so one might say "The EC of the irrigation water is 1.3 dS/m."

There is a general relationship between concentration and conductivity. The exact relationship depends on the chemistry of the water - the actual ions making up the salts in the water, but usually,

1 dS/m = 600-800 ppm

For the following example, we will use the approximation

1 dS/m = 700 ppm

 Example 13

The EC of the irrigation water is 1.3 dS/m. What is the concentration of salts in the irrigation water?

1.3 dS/m x [700 ppm]/[1 dS/m] = [1.3 x 700] ppm = 910 ppm

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Osmotic Stress

One effect of salinity is to cause a stress on plants similar to the stress created by a lack of moisture in the soil. The salinity stress is called "osmotic" stress. The moisture stress can be measured by the moisture tension in the soil (bars, cb or psi), and the osmotic stress can be related to these same units. It turns out that the salinity resulting in an EC of 1 dS/m is approximately related to stress as follows

1 dS/m = 0.30 bar = 30 cb approximately

 Example 14

If the soil moisture tension is 50 cb and the EC of the soil water is 2 dS/m, what is the total (osmotic + moisture) tension experienced by the crop?

osmotic tension = 2 dS/m x [30 cb]/[1 dS/m] = [2 x 30] cb = 60 cb

moisture tension = 50 cb

total tension = osmotic + moisture tension = 60 cb + 50 cb = 110 cb

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Irrigation Water Management

Terms & Units Continued

Index
Efficiency
     Destination or Use
     Beneficial Use
          Evapotranspiration (ET)
          Leaching Requirement (LR)
          Other Beneficial Uses
     Non-Beneficial Uses
     Irrigation Efficiency (IE)
     Application Efficiency (AE)

Efficiency

Efficiency shows how effective a system is in converting a resource (input) to a result (output):

Input -->  SYSTEM  -->  Output

 Example 15

Input = 100  -->  SYSTEM  -->  Output = 80
                   |
                   + --> Loss = 20

Here, the efficiency is 80%

Input is also called "gross" (the larger number) and output is also called "net" (what's left over after the loss). In irrigation terms, what the pump delivers if the "gross" amount of water, and because there are always water losses, the water that is beneficial to the crop is the "net" amount.

Pump (applied)  <-->  Gross
Crop  <-->  Net

Efficiency is defined by

Eff = 100 x Output/Input = 100 x Net/Gross = Benefically Used Irrigation water / Irrigation Water Infiltrated

This relationship can also be expressed as

Net = Eff/100 x Gross

Gross = Net /[Eff/100]

 Example 16

Suppose the efficiency, as in the above example, is 80%, but that you really need 200 units net output. How much gross input is required to achieve this?

Gross = Net/[Eff/100] = 100 units/[80/100]

Gross = 200/0.80 = 250 units required input

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Destination or Use

A destination is any place irrigation water can go, it refers to what happens to the water. It may also be referred to as a water use. Irrigation water that is pumped to a field or that is turned out of a canal gate can have various destinations. Some irrigation water destinations are as follows:

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Beneficial Use

An irrigation water that serves an agronomic purpose, that benefits the crop being grown, is called a beneficial use. Note that for the special purpose of defining irrigation efficiency (or application efficiency), 'beneficial' is limit in scope to the crop being grown. Other uses might serve some useful societal purpose, but unless it serves the crop being grown, it is not called beneficial.

Suppose irrigation water runs of a field and collects in a nearby depression, providing water for wildflowers and habitat for birds and small animals. From the societal standpoint, this use of the water might support environmental goals, and therefore be considered beneficial in the general sense of the word. However, this water does no good for the crop in the field, and is therefore considered non-beneficial from an irrigation standpoint.

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Evapotranspiration (ET)

Probably the most significant beneficial use of irrigation water is for crop evapotranspiration (ET). ET is the sum of plant Transpiration (T) and Evaporation (E) from wet plant and soil surfaces.

Usually, irrigation water infiltrated into the soil is absorbed by the plant roots, taken into the plant, and transpires from the leaves as a result of the process of photosynthesis. Some water 'substitutes' for transpiration as it evaporates from soil or plant surfaces because it cools the surrounding air mass, and thereby reduces the transpiration that would otherwise take place.

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Leaching Requirement (LR)

The second most significant beneficial use of irrigation water is water that percolates through and below the root zone and washes away (leaches) salts that have accumulated in the root zone that may be harmful to the plant.

All irrigation water contains at least some amount of salt. As more and more irrigation water is added to the field, more and more salt is added as well. Plants take up essentially pure water, leaving the salts behind. If that salt is not washed away, it will eventually accumulate to the point that crop production suffers.

To get rid of this salt, more water is added to the soil than the plant itself needs, and the extra washes salts below the root zone. The amount of water necessary to do this is called the Leaching Requirement (LR). LR is usually expressed as a fraction: that fraction of the applied irrigation water that is required for leaching purposes.

If more irrigation water than LR goes below the root zone, it is called 'excess deep percolation' and is considered non-beneficial.

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Other Beneficial Uses

Other beneficial uses include water used to achieve some agronomic purpose or to support non-crop plants that achieve some agronomic purpose. Examples include

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Non-Beneficial Uses

Not all uses of irrigation water are beneficial

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Irrigation Efficiency (IE)

IW here means irrigation water. Note that we only count the IW applied and the uses of IW. If there is some rainfall, the rainfall amount is not considered part of the total, nor is any rainfall water that happens to go to a beneficial use (crop ET for example) counted in the numerator. The change in IW stored in the root zone can be significant over short time periods, but is usually assumed to be zero on an annual basis.

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Application Efficiency (AE)

Technically, IE refers to an annual assessment, or a time period long enough that the ultimate fate of water stored in the root zone is known (was it used by the crop, or did it become excess deep percolation as the result of an ill-timed irrigation later in the season?).

For individual irrigation events, we use the term Application Efficiency (AE).

The 'target' here refers to the management goals for the irrigation, such as 'replenish the soil moisture deficit plus add x% over for leaching.'

 Example 17

Suppose a water source delivers 10 cfs to a 40 acre field for 24 hours. AE = 75%. What is the net benefit to the crop, expressed as an irrigation depth (inches)?

1 cfs = 1 ac-in/hr approximately

10 cfs = 10 ac-in/hr

amount = rate x time = 10 ac-in/hr x 24 hr = 240 ac-in

depth = volume/area = 240 ac-in/40 ac = 6 inches

So far, we are talking about water that has come out of the pump. So this is a Gross amount. To find the net amount, use the expression relating gross, net and efficiency

net = gross x AE/100

     = 6 in (gross) x 0.75 = 4.5 inches (net)

 Example 18

A water source delivers 10 cfs to a 40 acre field. AE = 75%. How long must the pump run to deliver a net benefit of 8 inches to the crop?

time = amount/rate

the rate is 10 cfs = 10 ac-in/hr. So all we must do is compute the amount, which in this case is the gross amount delivered by the pump.

gross = net/[AE/100] = 8 in/[0.75] = 10.7 inches gross

10.7 in x 40 acres = 428 ac-in gross

time = amount/rate = 428 ac-in/10 ac-in/hr = 42.8 hours

 Example 19

Sprinklers with a discharge of 3.5 gpm are used on a 30 ft x 40 ft spacing. In order to leave time to move the pipe, the irrigation set time is 11 hours. AE = 70%. If the crop ET (Et crop) is 0.3 in/day, how long can we go between irrigations?

time = amount/rate

rate = Et crop = 0.3 in.day

amount = net benefit from an irrigation

in/hr = [3.5 gpm x 96.3]/[30 x 40 sq-ft] = 0.28 in/hr

0.28 in/hr x 11 hr = 3.08 in (gross)

net = gross x AE/100 = 3.08 in x 0.70 = 2.163 in (net)

time = amount/rate = 2.163 in/0.3 in/day = 7.2 days

round DOWN to nearest day
     --> can go 7 days between irrigations

 Example 20

Sprinklers with a discharge of 5 gpm are used on a 40 ft x 50 ft spacing. In order to leave time to move the pipe, the irrigation set time is 11 hours. AE = 70%. At the time of irrigation, it will take 3.9 inches to refill the root zone (net benefit to the crop). How long should the sprinklers be run for this irrigation?

gross application rate = [5 gpm x 96.3]/[40x50 sq-ft] = 0.24 in/hr

net application rate = gross x AE/100
     = 0.24 in/hr x 0.70 = 0.168 in/hr

time = amount/rate = 3.9 in/0.168 in/hr = 23.2 hr

So irrigate for 23 hours, use 1 hour to move the pipe, then start the next day's irrigation set.

 Example 21

Over the course of the year, an irrigation system applies a gross amount of 48 inches of water. An evaluation determines that: 34 inches goes to satisfy crop ET; 8 inches runs off and is not collected for reuse; 5 inches passes through the root zone as deep percolation (total deep percolation), of which 3 inches is beneficial leaching for salt removal; and 1 inch is ET from weeds. What is the irrigation efficiency (%)?

IE(%) = 100 x [beneficial IW]/[Total IW]

     = 100 x [Et crop + Leaching]/[Total IW]

     = 100 x [34 + 3]/[48] = 100 x 37/48 = 77%

 Example 22

A useful concept for sizing pumps and irrigation systems is flow rate per unit area (such as gpm/acre). Suppose crop ET is Et crop = 0.3 in/day, and IE = 70%. What flow rate per unit area is required?

Et crop = 0.3 in/day (net)

gross = net/[IE/100] = [0.3 in/day]/[0.70] = 0.429 in/day

0.429 in/day x 1 day/24 hr = 0.01786 in/hr

Since we are interested in the flow rate for one acre, we'll use 43,560 sq-ft in the formula for in/hr

0.01786 in/hr = [Q gpm x 96.3]/[43,560 sq-ft]

Q gpm = [0.01786 x 43,560]/[96.3] = 8.08 gpm (for one acre)

Thus each acre of crop requires 8.08 gpm, or 8.08 gpm/acre

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