## System Capacity

System capacity is the maximum amount of water that an irrigation system can deliver on a continuous basis. Different units are used to describe system capacity. These are acre-feet (ac-ft, that is, the amount of water it will take to cover one acre, one foot deep), acre-inch (ac-in), gallons per minute (gal/min or gpm), and cubic feet per second (ft3/sec or cfs). The required pumping capacity of an irrigation system depends on the area to be irrigated (ac), the depth of water to apply (in), and the length of time that the irrigation system is operated (hr). Length of operation time refers to pumping time, not clock time. Pumping time is only the time water is flowing.

The amount of time per day that an irrigation system can operate depends on the type of system and the amount of maintenance it requires. A self-propelled unit may be able to run several days without stopping, whereas manual-move, tractor towed, and self-moved systems must be shut down at regular intervals. For systems other than center pivots and lateral moves, only a portion of a field is irrigated at one time and time is required to move the system from one portion of the field "set" to another "set." The term irrigation period is used to designate the number of days that a system can apply the water for one irrigation to a given area. Note that it is necessary for the irrigation period to be equal to or less than the irrigation interval. The required capacity of a system, in gallons per minute, can be determined by the following equation:

IRPx HPD

where RSC = Required system capacity (gal/min); 450 = Units conversion constant; A = Area irrigated (ac); DWA = Depth of water to apply per irrigation (in); IRP = Irrigation period (day); HPD = Time operating (hr/day).

Problem: Determine the required system capacity (gal/min) for the corn crop in the previous problem when the field area is 200 acres, and the system can operate for 18.0 hr per day for 7.7 days.

Solution:

424290 gal

138.6 min

For 200 acres of long-season corn grown on silt loam soil over compacted subsoil, irrigated with a system that is 70% efficient and limited to operating 18 hours per day for 7.7 days, the system must be able to deliver 3,040 gallons of water per minute.

Note: This is an example of an equation with a units conversion constant. The same problem can be solved using the units cancellation method.

min 60 min 18 hr 7.7 day 231 in3 1

144 in2 43560 ft2 200 ac 5833555200

1 ft2 1 ac 1 1920996

In some situations it might be necessary to use units of capacity other than gallons per minute. For example, water supplied from a large reservoir is often measured in acre-feet. In these cases, units cancellation and/or the appropriate conversion factors (Appendix I or II) can be used to convert the units.

Problem: What will the system capacity need to be in units of acre-feet/min?

Solution: Using units cancellation and system capacity from the previous problem:

ac-ft gal 1 ft3 1 ac

min min 4.48 gal 43,560 ft2

ac-ft

As noted earlier, system capacity is a function of four variables: area (ac); water flow rate (gal/min, ft3/min, ac-ft/min, etc.); depth of water applied or peak use (in); and time (min, hr, or days). This relationship is expressed mathematically as:

where D = Depth of water, either applied or peak use (in); A = Area irrigated (ac); Q = Water flow rate (cfs); T = Length of time water is applied (hr).

When any three of the variables are known, the remaining one can be calculated by rearranging the equation and substituting the values of the known variables. You must enter flow rate (Q) in cubic feet per second, depth in inches, and time in hours. The following discussion will illustrate several uses of this equation.

In the previous problem we determined the system capacity using units cancellation. If it is necessary to know how much water has been applied, the peak use does not accurately describe what we are solving for. When we want to know the depth of water that has been applied, D becomes the depth of water applied (DWA). This will work because the unit of measure is the same for both peak use and DWA (inches).

Problem: A producer spends 120 hr irrigating 90.0 acres. The pump discharges 1,350 gallons per minute. What average depth of water (in) is applied?

Solution: Because we want to know the amount of water applied, not the amount available to the plants, the efficiency factor is not used. Also, Q must be converted from gal/min to ft3/sec. Rearranging the equation, substituting depth of water to apply (DWA) for the depth (D), and including the conversion factor

1 ft3/sec = 2.25 x 10—3 gal/min1: DWA x A = Q x T Q x T

90 ac

An examination of this problem shows that the units do not cancel. However, when we enter the values of the variables with the units listed above we can obtain an answer very close to the true value. The symbol = means approximately equal. The exact solution using unit conversion/cancellation is:

1,350 37- x 231 — x __ . 3 ) x 120 hr x 60 — 12 —

90 ac x 43,560

In this case, the error in the approximate solution is:

3.98

Variations occur in the use of the equation for different types of irrigation systems. In situations where the limiting factor is the availability of water, the problem is to determine the maximum area that can be irrigated with the available water supply.

Problem: What is the largest size of lawn (ft2) that can be irrigated in 12 hr if a minimum of 0.5 inch of water is applied at each irrigation, the system is 90% efficient, and the water supply delivers 3.5 gal/min?

ft3 1 gal 1 rnin 1ft3 3 ft3 1 eal 2.25 x 10—3 ft3

Iff- = 1gal x ^m x - = 2.25 x 10—3 —then1gal = -

sec min 60.0 sec 7.40 gal sec mln 1 sec

Solution: Rearranging the equation, adding the efficiency factor, and converting the area to square feet:

1 ac

43,560 ft2

1 ac

43,560 ft2

0.90

168.5

0.90

If flood irrigation is used to water a field, assuming that the water flow rate is limited, it usually is necessary to determine the amount of time that the water should flow to cover the field at the desired depth.

Problem: How long will it take to apply 4 inches of water uniformly over 120 acres when the water is available at the rate of 20 cfs? (Assume 100% efficiency.)

Exact Solution:

DxA=QxT

ac ft3 sec min

sec min h

1,742,400 72,000

### 24.2 hr

During furrow irrigation it is important to know how long the water must run to apply the desired amount for each set of furrows. Three values are necessary to calculate time: the water flow rate for each furrow or for the entire set, the area of the furrow or the set, and the amount of water to be applied. The area is determined from the number of rows in the set, the row spacing, and the length of the row.

Problem: How much time is required to apply 3 inches of water to sixty 32-inch rows when the rows are one half mile long, and the system capacity is 30 gal/min/row?

Solution:

Number x Spacing (ft) x Lenght (ft) ft2

43,560 — ac in 1 ft ft 60 rows x 32-x -x 0.5 mile x 5280

row 12 in mile ft2

4.05

It will take 7.2 hr to apply 3.0 inches of water to the field.

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