15.4.2: Flushing the Injection and Irrigation System

At the end of a chemigation event it is important that the chemical injection pump and the injection line tubing and associated valves and the irrigation pipeline be flushed free of chemicals using fresh water. Flushing the injection system reduces the chance that chemical precipitates will foul the equipment components during future chemigation events. The irrigation pipeline system including the laterals should be flushed to prevent unexpected exposure of field workers to chemicals remaining in the line and to properly distribute the chemical in the field. The time required to flush a mainline is equal to the pipeline length divided by the mean velocity according to the following equation:

\(T_m=\dfrac{0.0408(ID)^2 \times L_m}{Q} \) (15.3)

where: T_m = flushing time of a mainline (min),

ID = inside diameter of the pipe (in),

L_m = the length of the mainline (ft), and

Q = irrigation system flow rate (gpm).

Equation 15.3 can also be used for manifolds and submains by solving it for each segment of the manifold. Flushing time of a lateral is different than the mainline. DeTar (1983) provides the following equation for calculation of flushing times in laterals that have outlets with equal discharge:

\(T_L=\dfrac{0.0408(ID)^2 \times S}{q}\left[0.577215+\ln\left(\dfrac{L_l}{S}\right)\right] \) (15.4)

where: T_L= flushing time in a lateral (min),

q = outlet discharge (gpm),

L_l = lateral length (ft), and

S = outlet spacing (ft).

The ratio Ll/S is the number of outlets, N , on the lateral and can be substituted into Equation 15.4 accordingly. Equation 15.4 is valid for laterals that have 10 outlets or more. Equation 15.4 can also be used for manifolds that have more than 10 laterals attached but the flow rate of each lateral must be the same. Figure 15.13 was developed using Equation 15.4.

Example 15.2

A 78-acre field is irrigated with a subsurface drip irrigation system. The flow rate of the system is 500 gpm. From the water source the water is conveyed 975 feet in an 8-inch mainline (ID = 7.762 in). The water then flows into a 4-inch manifold (ID 4.280 in). The manifold is 325 feet long (one side of the inlet tee) and has 65 laterals that are 5 ft apart. The flow rate in each lateral is 3.81 gp

Find: Total amount of time to flush the chemical out of the mainline and manifold.

Solution

The mainline flush time can be calculated with Eq. 15.3:

\(T_m=\dfrac{0.0408(ID)^2 \times L_m}{Q}=\dfrac{0.0408(7.762)^2 \times 925}{500}=5\text{ min} \)

The manifold flush time can be calculated with Eq. 15.4:

\(T_L=\dfrac{0.0408(ID)^2 \times S}{q}\left[0.577215+\ln\left(\dfrac{L_l}{S}\right)\right] \)

\(T_L=\dfrac{0.0408(4.280)^2 \times 5}{3.81}\left[0.577215+\ln\left(\dfrac{325}{5}\right)\right]=5\text{ min} \)

So, the total time required to flush the mainline and manifold is 10 min. This is also the time it will take to move chemical from the injection device to the last lateral on the manifold assuming the system has already been primed with water.

Figure 15.13. Travel or flush times in irrigation laterals, (a) aluminum sprinkler laterals, (b) drip irrigation tubing.

a)

b)

In Example 15.3 the three hours of flushing time is more than adequate since only 85 minutes total is necessary. The main point is that the time allowed for flushing must be equal to or greater that the total flushing time that was calculated. The travel time of chemical from the injection device to the most distant emitter equals the total flushing time. So the flushing time is important for the proper amount of chemical to be distributed in each zone. In Example 15.3 each zone will receive 4 hours of injected chemical but each emitter will only have discharged the correct amount of chemical when flushing is complete. It is important that the irrigation system be primed with water prior to injecting chemicals so that the priming time is not part of the injection time. Example 15.3 illustrates that there is flexibility in managing chemigation with set-type systems. The solution given in Example 15.3 is only one of many acceptable management approaches.

Example 15.3

For the same field in Example 15.2, a farmer plans to apply 15 lbs/ac of nitrogen using fertigation with injection of 28% UAN nitrogen solution into a subsurface drip irrigation system.

Given:

Irrigation systems characteristics:

Field width = 2600 ft
Field length = 1300 ft< br /> Field Area = 78 ac
Q = 500 gpm
Lateral spacing = 5 ft
Emitter spacing = 18 in
Discharge per emitter = 0.264 gal/hr
Lateral length = 1300 ft
Emitters per lateral = 866
Nominal lateral diameter = 7/8 in
ID = 0.859 in
Discharge per lateral = 3.18 gpm
Laterals per zone = 130
Number of irrigation zones in field = 4

Irrigation Management:

Irrigation interval = 2 d
Set time per zone = 9 h
Zone area = 19.5 ac

Fertigation Plan:

Nitrogen per acre = 15 lb/ac
28% UAN (3 lb N/gal)

Find: Time required to flush laterals (through the emitters, not the downstream flushing manifold) Injection rate in gal/h

Solution

Application of the DeTar Equation, Eq 15.4 (applicable only when flushing through the emitters)

\(S=1.5\text{ ft} \)

\(q=0.2\text{ gal/h}=0.0044\text{ gpm}\)

\(T_L=\dfrac{0.0408(ID)^2 \times S}{q}\left[0.577215+\ln\left(\dfrac{L_l}{S}\right)\right] \)

\(T_L=\dfrac{0.0408(0.859)^2 \times 1.5}{0.0044}\left[0.577215+\ln\left(\dfrac{1300}{1.5}\right)\right]=75\text{ min} \)

The same value can be obtained from Figure 15.13b.Thus, it will take 75 minutes to flush the laterals and it will take a total of 85 minutes to flush the mainline and manifold (Example 15.2) plus the laterals.

Assuming that an automatic controller and valves are used to change the water between zones the following approach is one alternative for managing this system.

Since the entire irrigation duration will take 36 hours one approach is to fertigate for the first 24 hours (four 6-hour sets) then terminate fertigation and flush each zone with fresh water for 3 hours giving a total of 12 additional irrigation hours. The injection rate would then be:

At 3 lbs N/gal, 15 lb/ac requires 5 gal/ac or solution to be injected.

Using Eq. 15.1:

\(q_i=\dfrac{G_p A_s}{t_i}=\dfrac{5\text{ gal/ac} \times 78\text{ ac}}{24\text{ h}}=16.2\text{ gal/h}\)

Thus the injection rate needed is 16 gal/h.

For systems like center pivots, the discharge on the lateral varies by distance along the lateral as discussed in Chapter 13. Buttermore and Eisenhauer (1989) developed the following flushing time equation for the center pivot lateral problem:

\(T=\dfrac{0.0102(ID)^2 (2L_l+S)}{Q}\ln\left(1+4\dfrac{L_l}{S}\right)\) (15.5)

The symbols L_l , S, and Q have the same meanings as in Equations 15.3 and 15.4. Equation 15.5 only applies to the case when an end gun is not operating and the diameter of the lateral is constant along its length. Buttermore and Eisenhauer (1989) presented equations that apply for the end gun condition and for the case where laterals have multiple diameters along their length. In general operating an end gun will result in shorter flush times than when not operating. The application of Equations 15.2 and 15.5 are illustrated in Example 15.4. In Example 15.4 Ri was based on the total area irrigated in the field and the total time it takes to irrigate the field and the calculated injection rate was based on this average irrigation rate. In this example if an end gun operated in the four corners the Ri would be about 5.45 ac/h when the end gun is off and 6.05 ac/h when it is on. Thus with constant injection rate and constant angular speed of the pivot chemical application rate errors would be at least 5 to 6 % with the chemical application being too high when the end gun is off and too low when it is on.

Example 15.4

A farmer plans to apply 1.5 pints per acre of an insecticide through a center pivot system.

Given: Field area = 126 ac

Q = 860 gpm

Pivot length = 1280 ft

Sprinkler spacing = 20 ft

Nominal lateral diameter = 6 in ID = 6.395 in

Time for 1 revolution = 22 h

Injection occurs at the pivot point (no mainline)

Find: Injection rate in gal/h

Time required to flush the center pivot lateral following the a

Solution

The injection rate can be calculated using Equation 15.2. We will assume that Ri is equal to the field area divided by the time to irrigate the field, i.e.,

\(R_i=\dfrac{126\text{ ac}}{22\text{ h}}=5.72\text{ ac/h}\)

\(q_i=G_pR_i=1.5\text{ pts/ac} \times 5.72 \text{ ac/h}=8.6\text{ pts/h}=1.07\text{ gal/h}\)

So the injection device should be calibrated to a flow of 1.1 gal/h.

Apply Equation 15.5 for determining flush time:

\(T=\dfrac{0.0102(ID)^2 (2L+S)}{Q}\ln\left(1+4\dfrac{L}{S}\right)\)

\(T=\dfrac{0.0102(6.395)^2 (2\times 1280+20)}{860}\ln\left[1+4\left(\dfrac{1280}{20}\right)\right]=6.9\text{ min}\)

The time to flush the system, assuming the end gun is not operating while flushing is about 7 min.

Assuming that chemical injection started when the center pivot began moving the center pivot should continue to irrigate for 7 more minutes after injection has stopped so that the chemical is flushed and the beginning of the revolution obtains the correct amount of chemical.