5.6: Determining System Capacity Requirements

Determining the C_n is difficult. Irrigation systems must supply enough water over prolonged periods to satisfy the difference between ET demands and rainfall. Water stored in the crop root zone can supply part of the crop demand. However, the volume of water that can be extracted from the soil should not exceed the amount that will induce crop water stress and likely yield loss. A careful accounting of the soil water status is required if stored soil water is used to supply crop water needs during periods when the crop ET demands are larger than the Cn plus any rainfall. Some irrigation designs have been developed to completely meet peak ET without reliance on either rain or stored soil water. Other techniques intentionally rely on stored soil water to meet peak crop requirements to reduce the required capacity, which decreases the initial cost of the irrigation system. The most conservative method is to provide enough capacity to meet the maximum expected or “peak” ET rate of the crop. In this case, rain and stored soil water are not considered in selecting the C_n. This design procedure relies on determining the distribution of crop ET during the year. The ET during the season varies from year to year (USDA-SCS, 1993). With the peak ET method, the maximum daily ET for each year is determined. Then the annual maximum daily ET rates are ranked and plotted. The Cn required to meet peak daily ET 70% of the time (i.e., in 7 of 10 yr) is normally taken as the acceptable capacity when using this method. A method to predict the daily peak period ET rate for general conditions was presented by the USDA-SCS (1970) as shown in Table 5.3. This relationship should only be used for general estimates and only if more localized peak data are not available.

Table 5.3. Peak daily crop ET rates as related to maximum monthly ET for the crop during the season and the net depth applied per irrigation (i.e., allowable depletion).

Allowable Depletion (in)	Maximum Monthly Crop Evapotranspiration (in/mo)
	5	6	6.5	7	7.5	8	8.5	9	9.5	10	10.5	11	11.5	12
	Peak Daily Evapotranspiration (ETd) in/d
1.0	.20	.24	.26	.28	.31	.33	.35	.37	.40	.42	.44	.46	.49	.51
1.5	.19	.23	.25	.27	.29	.32	.34	.36	.38	.41	.43	.45	.47	.50
2.0	.18	.23	.25	.27	.29	.31	.33	.35	.37	.39	.41	.44	.46	.48
2.5	.18	.22	.24	.26	.28	.30	.32	.34	.36	3.9	.41	.43	.45	.47
3.0	.18	.22	.24	.26	.28	.30	.32	.34	.36	.38	.40	.42	.44	.46
3.5	.18	.21	.23	.25	.27	.29	.31	.33	.35	.37	.39	.41	.44	.46
4.0	.17	.21	.23	.25	.27	.29	.31	.33	.35	.37	.39	.41	.43	.45
4.5	.17	.21	.23	.25	.27	.29	.31	.33	.35	.37	.39	.41	.43	.45
5.0	.17	.21	.23	.25	.27	.29	.30	.32	.34	.36	.38	.40	.42	.44
5.5	.17	.21	.22	.24	.26	.28	.30	.32	.34	.36	.38	.40	.42	.44
6.0	.17	.20	.22	.24	.26	.28	.30	.32	.34	.36	.38	.40	.41	.43

The peak ET method is based on selecting a C_n that can supply water at a rate equal to the peak ET for a period. However, it is unlikely that several periods with water requirements equal to the peak ET will occur consecutively. The crop water use during the combined time period can come from the irrigation supply or from rain and stored soil water. Therefore, the capacity could be reduced if rain is likely or if stored soil water can contribute part of the ET demand. Relying on soil water can reduce capacity requirements in two ways. First, the soil water can supply water for short periods of time when climatic demands exceed the capacity. The soil water used during the short period can be stored prior to its need or be replaced to some extent during the subsequent period when the ET demand decreases. When the C_n is less than the peak ET rate, there will be periods of shortage when crop water use must come from the soil or rain (Figure 5.9). However, during other periods, the capacity may exceed the ET and the water supplied during the surplus period can replenish some of the depleted soil water (Figure 5.9).

Figure 5.9. An example of the shortage and surplus periods for a system where the net system capacity is less than the average ET during a peak water use period.

The second way soil water can contribute to reduced capacity requirements is through allowable depletion (AD). This is the amount of water that can be depleted from the soil before crop stress occurs. The minimum capacity that maintains soil water above the AD during critical periods of the season can be used to design the irrigation system. An example of the effect of C_g on soil water mining and the magnitude of SWD during the season are shown in Figure 5.10. The positive bars in Figure 5.10 represent the amount of rainfall and ET during 10-d periods. After mid-May ET exceeds rain. The deficit bars represent the difference between ET and rain. The largest 10-d deficit occurs in midJuly. Without considering the use of soil water, the irrigation system would have to supply all of the deficit in that period. The peak 10-d irrigation requirement would be 3.3 in per 10 d (or 6.24 gpm/ac). For the 130-ac field shown in Figure 5.10, the C_n for the peak 10-d period would be 810 gpm, and, using an 85% E_LQ, the C_g requirement would be approximately 950 gpm. The amount of water that a 500 gpm capacity system, with an 85% E_LQ and assuming no D_t, can supply is also shown in Figure 5.10. The C_n for this system is:

\(C_n = 500 \text{ gpm} \times \frac{1 \text{ ac-in/hr}}{450 \text{ gpm}} \times \frac{24 \text{ hr}}{\text{day}} \times \frac{1}{130 \text{ ac}} \times 0.85 = 0.17 \text{ in/day}\)

The 500 gpm capacity (1.7 in/d in Figure 5.10) falls short of meeting the ET in late June and soil water would be depleted. The 500 gpm capacity continues to fall short of the 10-d deficit from early July through late August, resulting in a cumulative depletion of 4 in. Suppose that the AD before stress occurs is 3 in for the crop and soil in Figure 5.10. With the 500 gpm capacity system the soil water would be depleted below the allowable level in late July and the crop would suffer yield reduction. Obviously, 500 gpm is inadequate for maximum yield at this site. The C_n for a 700 gpm system is also shown in Figure 5.10. Here the system can supply the 10-d deficit for all but 20 d in late July. The cumulative soil water deficit for the 700 gpm system would be about 1.25 in with proper management. That depletion is well above the AD and should not reduce crop yield. This example shows that the maximum cumulative soil water depletion would be approximately 4, 1.25, and 0 in for gross capacities of 500, 700, and 950 gpm, respectively. Clearly the opportunity to utilize available soil water substantially reduces the required system capacity. Simulation programs using daily time steps to predict the soil water content have been used to determine the C_n when soil water is intentionally depleted. Some models such as by Heermann et al. (1974) and Bergsrud et al. (1982) use the soil water balance equation to predict daily soil water content. von Bernuth et al. (1984) and Howell et al. (1989) used crop simulation models to predict the C_n to maintain soil water above the specified AD or the C_n needed to maintain yields above a specified percentage of the maximum crop yield. University extension services have also created guides for determining C_n and C_g (e.g., Kranz et al., 2008). The capacities determined using soil water and/or crop yield simulation are usually very dependent on the available water capacity (AWC) of the soil. An example from the results of Heermann et al. (1974) is shown in Figure 5.11 and is illustrated in the subsequent example problem for a sandy loam soil.

Figure 5.10. Diagram of the 10-d ET, rain and the corresponding water deficit, plus the soil water depletion pattern over a growing season as affected by gross system capacity. Based on a 130-ac field and 85% ELQ.

Figure 5.11. Design net capacity required for corn grown in Eastern Colorado to maintain soil water depletion above a specified depletion for 3 design probabilities (adapted from Heermann et al., 1974).

Example 5.7

Given: A sandy loam soil that holds 1.5 in of available water per ft of soil depth.

Corn root zone depth of 4 ft. Allowable fraction depleted = 0.50.

Find: The net system capacity needed at a 95% probability level.

Solution

The allowable depletion is computed as:

1.5 in/ft x 4 ft x 0.5 = 3.0 in

From Figure 5.11, the C_n is approximately 0.22 in/d.

To use this procedure, the AD of the soil profile must be determined; the AD is the product of the allowable fraction depleted and the total AWC in the crop root zone. The gross system capacity does not include on-farm conveyance losses. If the delivery system for the farm contains major losses, then the capacity at the delivery point on the farm should be increased. The conveyance efficiency (E_c) is used to compute the farm capacity (Q_f):

\(Q_f = \frac{Q_g}{\left(\frac{E_c}{100\%}\right)}\) (5.21)

where: Q_f = farm system capacity (gpm)

Q_g = gross system capacity (gpm), and

E_c = conveyance efficiency (%).

The example below illustrates the use of the procedure to compute Q_f for two fields supplied by a network of canals (Figure 5.12).

Figure 5.12. Example of a farm layout with seepage losses between the source of the water and delivery to the field.

Example 5.8

Given: A farm has an irrigation system (Figure 5.12) with a net capacity of 0.3 in/d. Each field is 80 ac, and both are furrow-irrigated with siphon tubes. The E_LQ is 65% for both fields. The system is shut down about 10% of the time

Find: Determine the discharge needed from the well.

Solution:

1. Net capacity for the farm is expressed in in/d, so convert to flow rate per unit area (gpm/ac):

\(C_n = 0.30 \text{ in/d} \times \frac{452 \text{ gpm}}{1 \text{ ac-in/hr}} \times \frac{1 \text{ d}}{24 \text{ hr}} = 5.7 \text{ gpm/ac}\)

2. The gross capacity for each field is:

\(C_g = \frac{5.7 \text{ gpm/ac}}{0.65 (1 - 0.1)} = 9.7 \text{ gpm/ac}\)

3. System capacity is then:

\(Q_g = C_g \times \text{area}\)

\(Q_g = 9.7 \text{ gpm/ac} \times 80 \text{ ac} = 780 \text{ gpm}\)

4. However, the losses in the conveyance system must also be supplied by the pump.

Discharge needed for Field 1 is: \(Q_{f1} = 780 \text{ gpm}/0.8 = 975 \text{ gpm}\)

Discharge for Field 2 would be: \(Q_{f2} = 780 \text{ gpm}/0.9 = 867 \text{ gpm}\)

The well must supply the flow to each field plus the loss in the main supply canal:

\(Q_f = (975 + 867)/0.9 = 2,047 \text{ gpm}, \text{ or about } 2,050 \text{ gpm}\)