# Turf Sprinkler Irrigation Systems

Chapter

## Sprinkler Patterns

Turf sprinklers generally have a decreasing application rate vs. distance from the sprinkler. Ideally, the overlapped patterns from two adjacent sprinklers results in a uniform application. In general, the best superposition of the two patterns is thought to take place with head-to-head coverage, which means that sprinkler spray from one sprinkler barely reaches the next sprinkler (Fig. 13.1). Because of the decreasing application rate vs. distance, sprinklers should always be placed in corners and along the sides of turf areas.

Application rate (mm/hr) is calculated as a function of flow rate, angle of coverage, and sprinkler spacing. Equations for full, half, and quarter circle sprinklers spaced at head-to-head spacing are shown in Fig. 13.1.

The sprinkler pattern in Fig. 13.1 is called a square pattern. Figure 13.2 shows a triangular pattern, which generally has higher uniformity than a square pattern for the same lateral spacing. Arranging sprinklers in odd shaped areas is sometimes difficult, and perfection may be impossible. It is unadvisable to mix sprinkler types in order to water all parts of odd shaped areas; however, a property may require more than one type of sprinkler. Small turf areas are generally watered by fixed spray heads, and large turf areas are generally watered by rotors or impact sprinklers. Triangular patterns fit in large fields, but it is difficult to arrange sprinklers in triangular patterns in small square turf areas because of the difficulty of offsetting sprinklers at the ends of the area.

The triangular pattern consists of equilateral triangles with 60° angles. The tangent of 60° is 1.73, thus sprinklers are spaced along laterals 2/1.73 = 1.156 times farther apart than lateral row spacing. For example, if sprinkler spacing is 12.2 m, then laterals are spaced 10.56 m apart.

## Sprinkler System Evaluation, Efficiency, and Uniformity

Application uniformity tends to decrease as sprinkler spacing increases beyond head to head coverage. A decrease in application uniformity can result in higher water costs over time if turf managers overirrigate in order to ensure that all parts of the turf area receive adequate water.

The economics of turf irrigation are different from that of agriculture because there is no yield. In general, turf is irrigated based on the principle that there is a minimum acceptable level of appearance. For example, an elite golf course would not allow any brown spots over the entire golf course. Thus, the golf course is over-irrigated on average in order to ensure that no part of the golf course is under-irrigated. One the other hand, a cash strapped school may irrigate just enough to keep all sections of the field barely alive. Regardless of the goal, the priority is to keep the mimimum water application depth above a threshold value.

Based on the criterion of a minimum acceptable water depth for turf appearance or survival, the Scheduling Coefficient (SC) is a useful method of economic analysis of turf irrigation.
$$\mathrm{S}\mathrm{C}=\mathrm{average}\ \mathrm{depth}\ \mathrm{applied}/\mathrm{minimum}\ \mathrm{depth}\ \mathrm{applied}.$$
(13.1)
The scheduling coefficient thus calculates the fraction of wasted water. The Chapter 13 Sprinkler Uniformity program calculates the scheduling coefficient and the coefficient of uniformity for specified triangular or square spacings. The The first step is to input the application rate vs. distance curve for sprinklers. Distances and depths are entered into the Principal worksheet in columns G:K (Fig. 13.3).
The next step is to select the sprinkler of interest in the Main worksheet (Fig. 13.4). The final step is to input the spacing in the Triangle or Square worksheet (Fig. 13.5).

### Example 13.1

Use the Sprinkler Uniformity program to calculate the CU and SC for the I-45 sprinkler for 10.5 m row spacing and 12.2 m head spacing for square and triangular pattern.

The square pattern is shown in the Square worksheet (Fig. 13.5). The left figure shows the application pattern for one sprinkler and the right figure shows the application pattern for square spacing. Although the CU is high, 91 %, the scheduling coefficient is also relatively high 1.32. This means that 32 % extra water is applied in order to maintain a minimum depth applied in the center of the four sprinklers. A high SC is common for square spacing because of the low application rate in the center.

The triangular CU (Fig. 13.6) is only slightly higher (92 %) than that for the square spacing (91 %); however, the SC (1.16) is much lower than the SC for square spacing. Thus, only 16 % of water is wasted for the triangular spacing, which is half of the wasted water for the square spacing.

The Economic summary worksheet finds the optimal spacing for the selected and spacing. Financial costs are entered in the worksheet. Scheduling coefficients and costs are calculated for a range of spacings in order to find the optimal design.

### Example 13.2

Determine the optimal economic spacing and pattern for the I-41 sprinkler if trenching + pipe costs are $2.00 per m and each sprinkler + fittings costs$25.00 per unit. The cost of irrigation water is $0.10 per m3. The required depth of water/year is 1.8 m (including evapotranspiration). Assume that square spacing uses equal head and row spacings and that triangular spacing uses equilateral triangles. An equation can be established for the cost of pipe and sprinklers per ha. The length of pipe is area/row spacing. $$\begin{array}{l}\mathrm{Pipe}\ \mathrm{length}/\mathrm{ha}=10,000/{\mathrm{s}}_{\mathrm{d}}\hfill \\ {}\mathrm{Number}\ \mathrm{of}\ \mathrm{sprinklers}/\mathrm{ha}=10,000/\left({\mathrm{s}}_{\mathrm{l}}\ {\mathrm{s}}_{\mathrm{d}}\right)\hfill \end{array}$$ where • sl = sprinkler spacing along lateral, m • sd = row spacing, distance between laterals, m. The cost of materials and installation is calculated as follows: $$\/\mathrm{ha}=10,000\left(\2/{\mathrm{s}}_{\mathrm{d}}+\25/\left({\mathrm{s}}_{\mathrm{l}}\ {\mathrm{s}}_{\mathrm{d}}\right)\right)$$ The water cost is calculated as follows: $$\/\mathrm{ha}=\left(1.8\ \mathrm{m}\right)\left(\mathrm{S}\mathrm{C}\right)\left(10,000\ {\mathrm{m}}^2/\mathrm{ha}\right)\left(\0.10/{\mathrm{m}}^3\right)=1,800\ \mathrm{S}\mathrm{C}$$ The desired spacings are entered in Column D and the economic parameters in Column B in the Economic summary worksheet (Fig. 13.7). Click the Run Economics Button to run the VBA calculation program. Scheduling coefficients for a range of spacings are output in column F. The scheduling coefficient for triangular spacing increases between 19 m and 20 m. The square spacing sprinkler coefficient rises and falls and is less stable than the triangular spacing coefficient. The square spacing scheduling coefficient is more sensitive to minor variations in the one-dimensional application distribution. Water cost follows the same trend as the scheduling coefficient (Fig. 13.7). The minimum present value for the triangle geometry,$22,562/ha, is at 15 × 17.3 m spacing. The minimum present value for the square geometry, \$22,608/ha, is at 14 m spacing. Thus, both have comparable costs. The sprinkler used in this analysis has a wetted diameter slightly greater than 15 m. Thus, the recommended spacing is approximately equal to head to head spacing.

Once systems are installed, periodic uniformity tests should be conducted in order to evaluate system degradation. Catch cans should be placed between sprinklers and next to sprinklers. If a grid is used, then approximately 20 cans should be placed between four sprinklers in a square spacing or three sprinklers in a triangular spacing. The sprinkler system should run until at least 25 ml are collected by each can.

The Sprinkler Uniformity program does not consider wind. If an area is windy, then spacing is generally decreased (Table 13.1). It is best to run lateral pipes in a direction perpendicular to the general wind direction and to place sprinklers closer together along laterals. This helps avoid strips of dry ground, which are common in windy areas if sprinklers are widely spaced in the direction perpendicular to the wind direction. Triangular spacing works well in windy areas with pipes perpendicular to the wind because of better downwind overlap.
Table 13.1

Recommended sprinkler spacings for different wind speeds

Rectangular spacing

Wind speed

Sprinkler spacing – percent of diameter

Lateral spacing – percent of diameter

(kph)

(mph)

0–5

0–3

50

60

6–11

4–7

45

60

12–20

8–12

40

60

Square spacing

0–5

0–3

55

55

6–11

4–7

50

50

12–20

8–12

45

45

Equilateral triangle spacing

0–5

0–3

60

6–11

4–7

55

12–20

8–12

50

## System Design

There are four steps in the design of a turf sprinkler irrigation system. The first step is to perform a complete assessment of the site. Time of watering may be constrained by public use. Pressure should be measured or calculated at the expected flow rate. It also may fluctuate due to water usage in the city. An outline of the property (Fig. 13.8) with design constraints and the location of the water source is drawn. Once the map is completed, the next step is to draw the positions of all the sprinklers, which might be more art than science on an irregularly shaped property. A sprinkler should be selected that meets the spatial dimensions of the project and also is appropriate for the use; for example, a commercial sprinkler (vandal resistant) should be used on a commercial property. Then, the pipe system is laid out, grouping common types of sprinklers if possible, and keeping the total flow rate in any one zone below the maximum allowable flow rate. Pipe sizes are selected based on friction loss. The last step is to compile a report that lists the parts and pipe lengths and can be used by contractors or homeowners for cost estimation.
• Site information

• Water source, ET, map, pressure, obstacles, soils, time of watering

• Sprinkler selection and placement

• Spacing – flow rate

• Pipe system design

• Zones, laterals, mains, valves, irrigation schedule

• Preparation for bidding and installation

• Picking parts and preparing for contractor bids.

### Example 13.3

Design a sprinkler system for the area shown in Fig. 13.8.

Site information for public site.
• Maximum soil infiltration rate = 0.5 in/hr.

• TAW = 2 inches.

• Pressure at water source = 80 PSI.

• Minimum sprinkler pressure = 50 PSI.

• Max ETc (middle of summer) = 0.26 in/day.

• Eight hours available for water application (between 10 p.m. and 6 a.m.).

• Maximum system flow rate = 35 gal/min.

• Average night atmospheric conditions: RH = 75 %, T = 18 °C, U = 8 km/h (2.22 m/sec).

• Based on irregular geometry, assume that the scheduling coefficient is 1.5.

Sprinkler selection

Because wind speed is in the 6–11 km/hr range, choose the 50 % of diameter coverage (Table 13.1) for square spacing. The side lengths are in the range of 100 ft so select a sprinkler with approximately 50 foot wetted radius. Table 13.2 lists the flow rate and radius of an appropriate rotor sprinkler with number 4, 6, and 8 nozzles. The wetted radius is approximately 50 ft, which matches the requirements of the project. The application rates are in the range of the maximum application rate (application rate <0.5 in/hr) or less. This sprinkler is rated for commercial applications (vandal resistant with a steel cap), which is an important characteristic in public facilities: brilliant people like to do thing like shoot BB guns at the tops of sprinklers in order to turn the sprinklers into water fountains.
Table 13.2

Sprinkler parameters

Nozzle

gal/min

Half-circle application rate (in/hr)

8

51

7.4

0.55

6

49

5.5

0.44

4

41

3.7

0.21

Sprinkler placement

The next step is to locate the sprinklers on the property (Fig. 13.9). Because the geometry of the site is irregular, not all sprinklers are spaced 50 ft apart, but they are placed as close to 50 ft intervals as possible with sprinklers in all corners.

Zones

With 35 gal/min, a maximum of five number 6 nozzles (5.5 gal/min) can be placed in each zone. A maximum of four sprinklers with number 8 nozzles (7.4 gal/min) can be placed in each zone.

The center full-circle sprinklers can be placed in two zones with four sprinklers in each zone (Fig. 13.10). Number 6 nozzles are selected for these zones; however, a number 8 nozzle is selected for the lower left sprinkler in zone 3 because it covers a larger area than the other sprinklers in zone 3.

Corner sprinklers (90+ degree arc) can be grouped in one zone or they can be incorporated into the side sprinkler groups. If corner sprinklers are grouped together, then they can use the same nozzle diameter (flow rate) as other sprinklers on the project because they are controlled by one valve and have a shorter watering time. However, extra pipe is needed to link all the corner sprinklers in one zone. In this case, the decision is made to group the corner sprinklers with the side sprinklers (180 degree arc), and to use number 4 nozzles in the corners (3.7 gal/min) and number 8 nozzles (7.4 gal/min) on the sides. The corner sprinklers with larger than 90 degree arcs are given number 6 nozzles (Fig. 13.9).

Once the sprinklers are selected and the zones are defined, then the pipe locations can be laid out. The solid lines represent sprinkler lateral lines. Even though valves are shown within the turf area, they are installed as close to the edge of the turf area as possible, or preferably outside the turf area.

Pipe system design

It is generally recommended that landscape irrigation pipes should be placed 30 cm below the ground surface. Sprinklers should be connected to the pipes with swing joints. People and equipment apply downward pressure on sprinklers; the force can break the lateral pipe if the sprinklers are installed directly over the pipe. Swing joints allow sprinklers to be pushed down without breaking the pipe. The second advantage of swing joints is that they allow adjustment of sprinkler elevation and location during installation. Complete swing joints can be purchased, or they can be made with Marlex (crystalline polypropylene) 90s.

The rule of thumb for agricultural systems is that lateral pressure loss for sprinkler systems should not exceed 20 %, resulting in 10 % flow variation. However, a more conservative approach may be appropriate for landscape irrigation systems where water costs are higher.

A backflow prevention device is placed after the water source in order to prevent the possible siphoning of chemicals from the irrigation system back into the potable water supply. In the case of this project, which has a level turf area and for which chemigation will not be used, a pressure vacuum breaker (PVB) is sufficient. However, more expensive backflow prevention devices such as reduced pressure backflow preventers or double check valves are needed on some projects, depending on topography and chemigation. City regulations should be consulted for specification of the correct backflow prevention device for each project. The 1½″ PVB selected for this project has a 5 PSI pressure loss at 30–35 gal/min.

A pressure regulator is placed after the backflow prevention device. City water pressure is not constant. If city water pressure fluctuates, then, without a pressure regulator, water application rate to the project is unknown. Although not required by law, a pressure regulator can save money by reducing water waste. On this project, a 2″ pressure regulator is selected with 6 PSI pressure loss at 30 gal/min.

Because pressure regulators have a significant pressure loss, it might not be possible to use pressure regulators on projects with marginal pressure; if sprinkler pressure drops below a threshold value (typically 40 PSI for conventional sprinklers), then a “doughnut” shaped water application pattern results with a dry spot near the sprinkler.

Solenoid valves are generally sized one size smaller than mainline pipe. Thus, 1½″ solenoid valves are selected for the 2″ mainline pipe. A 1½″ solenoid valve has a pressure loss of 3.6 PSI at 30 gal/min flow rate. An extra solenoid valve, called a master valve, is placed before the other solenoid valves on the mainline. The master valve is turned on when any of the other valves are activated. A master valve prevents water waste if one of the zone valves does not shut down properly or if the mainline is damaged and has a leak. In this example, estimate that the two solenoid valves (master and zone valve) have a pressure loss of 7.2 PSI prior to each zone.

The major design constraint for turf sprinkler systems is that the worst-case sprinkler must have adequate pressure; thus, if the worst-case zone has adequate pressure, then the rest of the sprinklers will have more than adequate pressure. On this project, Zone 1 or Zone 5 are the zones that are most likely to have the lowest pressure since they have the largest flow rate and the longest pipe networks (Fig. 13.10). Zone 5 has better hydraulic characteristics within the zone because the valve connection is in the middle of the zone; however, Zone 1 has a better position on the mainline because it is closer to the water source. In order to select the worst-case zone, evaluate the hydraulics of both zones. Sprinkler flows are added from the end to the beginning in order to find the flow rate in each pipe section. Pipe flows for zone 1 are shown in Fig. 13.11. For example, the second to last pipe supplies two sprinklers so its flow rate is 7.4 + 5.5 = 12.9 gal/min.
A sprinkler flow vs. pressure equation can be developed based on catalog flow rates and the assumption that flow vs. pressure exponent is 0.5. The coefficient k for the number 4, 6, and 8 nozzles is calculated as follows where the flow rates Q are the listed flows at 50 PSI pressure:
$$\begin{array}{l}\mathrm{k}8=\mathrm{Q}/\mathrm{H}\mathrm{x}=7.4/50{}^{\circ}.5=1.05\hfill \\ {}\mathrm{k}6=\mathrm{Q}/\mathrm{H}\mathrm{x}=5.5/50{}^{\circ}.5=0.778\hfill \\ {}\mathrm{k}4=\mathrm{Q}/\mathrm{H}\mathrm{x}=3.7/50{}^{\circ}.5=0.523\hfill \end{array}$$
Pipe friction loss is calculated with the Hazen-Williams equation. On this project, use Class 200 pipe for 1″ and 1.25″, and Class 125 pipe for 1.5″ diameter and greater. Start with 1.25″ pipe in the last section in zone 1. The inside diameter of the 1.25″ Cl 200 pipe is 1.50″.
$${H}_L= kL\left(\frac{{\left(\frac{Q}{C}\right)}^{1.852}}{D^{4.87}}\right)=10.46(60)\left(\frac{{\left(\frac{7.42}{140}\right)}^{1.852}}{1.50^{4.87}}\right)=0.375\kern0.5em ft$$
where
• HL = head loss in pipe, ft

• k = constant, 10.46

• Q = flow rate, gal/min

• C = pipe roughness coefficient, 140 for PVC pipe

• D = inside diameter of pipe, inches

• L = pipe length, ft.

The Sprinkler Uniformity program has pipe calculation worksheets. In this case, use the Lateral_Simple_US_k worksheet (Fig. 13.12). The pipe diameters in column F are linked to the selections in column E. The sprinkler flow rate coefficients (cells I6:J8) are linked to the Nozzle ID numbers in column G. With the selected pipe diameters, the total pressure loss is 1.1 PSI in the zone. Thus, the pipe pressure just downstream from the solenoid valve is 51.1 PSI.
The next step is to evaluate zone 5 (Fig. 13.13 and Table 13.3). Only the upper half of zone 5 is analyzed. The pressure drop is 0.7 PSI. The pressure loss in 200 ft of 2″ mainline pipe at a flow rate of 31.4 gal/min is 2.6 ft or 1.1 PSI. Thus, the total pipe pressure loss when zone 5 is running is 1.1 + 0.7 PSI = 1.8 PSI. This is greater than the 1.1 PSI drop in zone 1 so zone 5 is the worst-case zone.
Table 13.3

Flow rates for upper half of zone 5

Pressure (PSI)

Nozzle number

Sprinkler flow (gal/min)

Pipe flow (gal/min)

Pipe ID Nominal (in)

Length (ft)

Pressure loss (ft)

50

8

7.400

7.400

1.5

60

0.376

50.16

8

7.412

14.812

1.78

50

0.492

50.38

4

3.712

18.524

1.78

50

0.745

50.70

The total pressure required for the project is calculated by summing the required sprinkler pressure plus all of the friction losses to the worst-case sprinkler, which is the upper sprinkler in zone 5. Four PSI is added as a safety factor to account for losses due to system degradation over time.
 Pressure vacuum breaker 5 PSI Pressure regulator 6 PSI 2 Solenoid valves 7.2 PSI Mainline loss friction loss 1.1 PSI Zone 5 friction loss 0.7 PSI Safety factor 4 PSI Sprinkler pressure 50 PSI Total 74 PSI

The pressure available to the project is 80 PSI. Thus, the design is acceptable.

Watering schedule

The calculation of the watering schedule begins with the evaporation calculation. The Evaporation worksheet (Fig. 13.14) uses the Frost and Schwalen algorithm (Eqs. and ) to estimate the percent evaporation during irrigation. Calculate the evaporation based on the average nozzle diameter (nozzle 6, 6/64″). The nighttime relative humidity, wind speed, and temperature are used for the calculation because the system operates at night. Thus, 11 % of gross application evaporates before it hits the soil.
With SC = 1.5, the daily depth of water required is
$$\mathrm{Depth}\ \mathrm{required}={\mathrm{ET}}_{\mathrm{c}}/\left(1-{\mathrm{L}}_{\mathrm{e}}\right)\mathrm{S}\mathrm{C}=0.26/\left(1-0.11\right)1.50=0.44\ \mathrm{in}/\mathrm{day}$$
Zones 3 and 4 have full circle sprinklers with number 6 nozzles. The full-circle application rate with number 6 nozzles is 0.22 in/h. Thus, zones 3 and 4 must operate for 2 hours/day. In general, zones 1, 2, and 5 have half-circle sprinklers with number 8 nozzles, with an application rate of 0.55 in/h. Thus, the operation time for these zones is
$$\mathrm{H}\mathrm{r}\mathrm{s}/\mathrm{day}=\left(0.44\ \mathrm{in}/\mathrm{day}\right)/\left(0.55\ \mathrm{in}/\mathrm{hr}\right)=0.8\ \mathrm{hrs}/\mathrm{day}.$$
The total required watering time is (2 zones) (2 hrs/d) + (3 zones) (0.8 hrs/d) = 6.4 hours per day.

Thus, the area can be watered in the specified length of time (8 hours/day).

### Example 13.4

An athletic field has a maximum summer ET rate of 0.24 in/day. The field dimensions are 240 ft by 312 ft. The water pressure is 90 PSI and maximum flow rate is 80 gal/min. The maximum infiltration rate is low, 0.22 in/hours. The turf should be watered at night between 9 p.m. and 5 a.m. The reclaimed water source is at the southwest corner of the field

Large sprinklers are selected in order to minimize trenching and minimize application rate. The sprinkler has an 80 ft radius at 40 gpm. A 5 × 4 sprinkler grid (312 ft/4 = 78 ft and 240/3 = 80 ft) divides the field into approximately square spacing with 80 ft between sprinklers (Fig. 13.15). The sprinklers have a solenoid valve in the sprinkler so only one pipe is needed in each trench (Fig. 13.15). It is not possible to arrange the field with triangular spacing because of the small number of sprinklers and the square field dimensions.

Even with sprinklers operating alone (no adjacent sprinklers running), runoff will probably occur because of the low infiltration rate of the soil. In this case, a technique is used that is similar to the high velocity center pivots. Sprinklers are cycled on and off, which allows ponded surface water on the surface to soak into the soil before more water is applied. This process is called “cycle and soak.”

If it is assumed that the scheduling coefficient is 1.3 and evaporation is 10 %, then the required depth of water per day is
$$\left(0.24\ \mathrm{in}/\mathrm{day}\right)(1.3)/0.9=0.35\ \mathrm{in}/\mathrm{day}.$$
The application rate for the half circle sprinklers is 0.59 in/hr. Thus, the application time is
$$Calculated\kern0.5em watering\kern0.5em time=\frac{ET\kern0.5em requirement\kern0.5em \left( in/ day\right)}{Application\kern0.5em rate\kern0.5em \left( in/ hr\right)}=\frac{0.35\kern0.5em \left( in/ day\right)}{0.59\kern0.5em \left( in/ hr\right)}=0.6\kern0.5em hr.$$
The full circle sprinklers require twice as much time, 1.2 hours, and the quarter circle sprinklers require half as much time, 0.3 hr.
There are three full circle zones, five half circle zones, and two quarter circle zones. The total watering time is
$$3(1.2)+5(0.6)+2(0.3)=7.2\ \mathrm{hours}$$
Thus, the area can be watered within the time constraint (8 hours).

Pipes, up to the last pipe section, should be sized to carry the full 80 gal/min flow rate. Water source pressure is 90 PSI so 10 PSI can be allocated to pipe friction loss and valve losses.

The pipe friction loss is calculated for the sprinkler that is the maximum distance (worst-case) from the water source (upper right in Fig. 13.15). The distance from the water source to the second to the last sprinkler on the line is 480 ft. The distance to the last sprinkler is then 78 ft.

Flow

Distance

Diameter

Friction loss

80 gal/min

480 ft

3 inch (3.23 ID)

2.55 PSI

40 gal/min

78 ft

2 inch

0.64 PSI

Total

3.19 PSI

A backflow prevention device is not needed in this case for the reclaimed water system. If a master solenoid valve and zone solenoid valve are installed, then there is enough pressure to operate the system with a 4 PSI loss in a pressure regulating valve.
 1 Solenoid valve 3 PSI Pipe friction loss 3.2 PSI Safety factor 4 PSI Sprinkler pressure 80 PSI Total 90 PSI

Small residential turf areas are irrigated with spray heads. In general, the flow meter is 5/8″ (16 mm), and the maximum flow rate is 12 gal/min (45 L/min). For these systems, use one pipe size ( 1″ (25 mm)), because the price of pipe and fittings is less than the cost of going to the store to purchase specific diameter parts.

### Example 13.5

Irrigate a 30 ft × 30 ft turf area with spray heads.

The turf area shown in Fig. 13.16 is set up with 1 gal/min spray heads in the corners, 2 gal/min spray heads on the sides, and a 4 gal/min spray head in the middle. All of the pipes are 1″ (25 mm). The area must be broken into two zones because the upper zone has the maximum flow.
 Two quarter-circle 1 + 1 gal/min Three half-circle 2 + 2 + 2 gal/min One full-circle 4 gal/min Total flow rate 12 gal/min
The application rate is calculated based on the full-circle sprinkler flow rate (Fig. 13.16)
$$in/ hr=\frac{(96.3)\left( gal/ \min \right)}{(ft)(ft)}=\frac{96.3(4)}{(15)(15)}=1.7\kern0.5em in/ hr$$

## Subsurface Drip Irrigation

Subsurface drip irrigation is an alternative to sprinkler irrigation that may be appropriate for odd-shaped turf areas, climates with high evaporation, and when sprinkler spray is not allowable. Drip irrigation tubing with internal emitters is laid approximately 0.15 m below the ground surface. Spacing is dependent on soil texture. The application rate is calculated as a function of emitter flow rate and emitter and line spacing. Equations are derived in Chap. .

### Example 13.6

Design a subsurface drip system to irrigate a turf area with summer reference ET = 12 mm/day. Emitter spacing and tubing spacing are 0.3 m. Irrigation efficiency is 76 %. The AWC = 15 %. The crop coefficient, Kc, is 0.7. Soil is sandy loam. MAD is 0.6. Turf rooting depth is 0.25 m. Each emitter has an 8 L/h flow rate. The total irrigated area is 140 m2.

The daily water volume applied per emitter is
$${\mathrm{V}}_{\mathrm{emitter}}=\left(\mathrm{A}\right)\left({\mathrm{ET}}_0\right)\left({\mathrm{K}}_{\mathrm{c}}\right)/\mathrm{Eff} = \left(0.3\ \mathrm{m}\right)\left(0.3\ \mathrm{m}\right)\left(12\ \mathrm{m}\mathrm{m}/\mathrm{day}\right)(0.7)/0.76=1\mathrm{L}/\mathrm{day}$$
The soil water holding capacity for each emitter is

SWHC = (A)(Z)(AWC)(MAD) = (0.3 m) (0.3 m) (0.25 m) (0.15) (0.6) = 0.002 m3 = 2 L

Thus, there should be a maximum of 2 days between irrigations. The emitter flow rate is 8 L/h so the daily watering time should be 0.25 hours or 15 minutes every other day in order to apply an average of 1 L/d/emitter.