5.3.4: Application of Efficiency of the Low Quarter

It is important that all water “losses” during application be considered in an efficiency calculation. These losses shown in Figure 5.4 include:

• evaporation and drift,
• runoff,
• deep percolation due to nonuniform infiltration, and
• deep percolation due to excessive application.

Deep percolation occurs whenever infiltration exceeds the SWD. Excess infiltration can be caused by both the nonuniformity of application and excessive application. Non-uniformity of application is usually a result of a problem with the system for sprinkler and microirrigation, while excessive application is a result of system management. With surface irrigation, non-uniformity of application can also be a result of system management, e.g., if the flow rate in furrows is too low. Percolation caused by the nonuniformity occurs because the manager must decide how much of the field should be adequately irrigated. A common, albeit somewhat arbitrary, approach is to use the average low-quarter depth as the “management depth.” Managing according to the average low-quarter depth results in approximately 90% of the field being adequately irrigated and potentially about 10% of the field being under irrigated. Conservation of mass requires that the following water balance equation holds when conveyance losses (discussed later) are ignored:

d_g = d_z + d_r + d_ev

where: d_g = gross depth applied,

d_z = average depth infiltrated,

d_r = depth of runoff, and

d_ev = depth of evaporation and drift.

Rearranging Equation 5.6 results in:

d_z = d_g – d_r – d_ev

Note that Equation 5.7 accounts for above-ground losses, but the dz includes both water that will be stored in the root zone and deep percolation. Rearranging Equation 5.2 yields:

d_LQ = (DU)(d_z)

The effectiveness of d_LQ depends upon the quantity of infiltration relative to the SWD. The effective depth (d_e) is the irrigation water that remains in the root zone for plant use, accounting for SWD and assuming that any irrigation depth in excess of the d_LQ will be lost to deep percolation (i.e., assuming a 90% adequacy of irrigation). The de, a managed term, is the amount of water that will be used in irrigation scheduling; its utility will be illustrated in Chapter 6. Figure 5.7 illustrates the concept of de with four scenarios. In 5.7a, the infiltrated water is perfectly uniform (DU = 1.0) and equal to SWD. No deep percolation would occur in this scenario. In this case, d_LQ = d_z = d_e.

In Figure 5.7b, the infiltrated water is perfectly uniform, but, due to excessive application, infiltration exceeds SWD. In this case, d_LQ = d_z and d_e = SWD. The excessive application can be caused by irrigating too frequently or operating the system too long for the existing SWD. The interval between irrigations can be increased as long as SWD does not exceed the allowable depletion (AD)–a concept discussed in Chapter 6.

Nonuniform infiltration is illustrated in 5.7c. Here, the d_LQ = SWD = d_e. In this case, deep percolation is not due to excessive application caused by applying too much water or applying water too frequently but is due to the nonuniformity of the infiltration. The majority of the field (approximately 90%) experiences deep percolation because of the management decision to only allow about 10% of the field to be under irrigated.

Figure 5.7d illustrates the case where there are deep percolation losses due to both excess application and nonuniform infiltration. The figure illustrates the division of the two losses. In this case, d_e = SWD.

Figure 5.7. Distribution of infiltrated irrigation water and deep percolation under four scenarios

Figure 5.7 can be summarized by the following equations:

If \(d_{LQ} \leq SWD\), then \(d_e = d_{LQ}\) (5.9)

If \(d_{LQ} > SWD\), then \(d_e = SWD\) (5.10)

Finally, the concepts of uniformity (irrigation adequacy), \(d_{LQ}\), and \(d_e\) can be incorporated into the definition of application efficiency. The application efficiency of the low-quarter (\(E_{LQ}\)), discussed by Burt et al. (1997), is defined as:

\(E_{LQ} = 100\% \left( \frac{d_e}{d_a} \right)\) (5.11)

where: \(E_{LQ}\) = application efficiency of the low-quarter (%), and

\(d_a\) = depth applied from the original source.

Determination of the depth of water from the original source is straightforward except when runoff recovery is part of the system. Either Equation 3.1 or 3.3 can be used for the calculation of d_a. Without runoff recovery, d_a and d_g are equal; d_a is always equal to the volume of water taken from the original source, such as a well, divided by the total land area irrigated. Runoff recovery, discussed in detail in Chapter 10, is a common practice in surface irrigation. If conveyance losses are ignored, the relationship between d_a and d_g for a closed runoff recovery system (runoff water reapplied on the same field) is:

\(d_a = d_g - d_r R_i\)

\(d_a = d_g (1 - R_r R_i)\) (5.12a)

while, for an open runoff recovery system (runoff water reapplied on different field):

\(d_a = \frac{d_g}{1 + R_r R_i}\) (5.12b)

where: d_g = gross depth applied which includes the volume applied from the runoff recovery system,

d_r = depth of runoff,

R_r = runoff ratio (d_r / d_g), and

R_t = return ratio, the depth of water returned (reused) divided by the depth of runoff.

Example 5.2

In Example 5.1, the DU was 0.75 and d_z equaled 2.0 in. If d_a = 2.2 in, runoff is zero, and SWD = 1.6 in, determine the system’s E_LQ and d_ev.

_Given: d_z=2.0 in

d_a=2.2 in

d_r=0

SWD=0.75

DU=0.75

Find: d_ev

E_LQ

Solution:

Rearranging Equation 5.6

d_ev = d_g – d_z – d_r (Eq. 5.6)

d_ev = 22 in – 20 in – 0 = 0.2 in

Using Equations 5.8, 5.9, and 5.11, you will find that

d_LQ = (DU)(d_z) (Eq. 5.8)

d_LQ = (0.75)(2.0 in) = 1.5 in

Since d_LQ < SWD, d_e = 1.5 in, according to the criteria in Equation 5.9.

Since d_r = 0, d_a = d_g = 2.2 in

\(E_{LQ} = \left(\frac{d_e}{d_a}\right) \times 100\%\) (Equation 5.11)

Thus, \(E_{LQ} = \left(\frac{1.5 \text{ in}}{2.2 \text{ in}}\right) \times 100\% = 68\%\)

Example 5.3

Repeat Example 5.2 if SWD equaled 1.2 in.

Solution:

Now, \(d_{LQ} > SWD\), thus, Equation 5.10 applies and \(d_e = SWD = 1.2\) in

Thus, \(E_{LQ} = (1.2 \text{ in})/(2.2 \text{ in}) \times 100\% = 55\%\)