3.1: Introduction
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Need for Water Measurement
In Chapter 2, we discussed soil water storage and related this storage to an “equivalent depth” of water on the soil surface. You can envision this “equivalent depth” as the water measured in a rain gauge. In Chapter 4 we will present how this depth relates to plant water needs. In this chapter we discuss how we can determine the depth of water applied with an irrigation system.
Have you ever wondered what it would be like to drive an automobile without a speedometer and an odometer? You might feel somewhat lost. You would not know how fast you were going, nor how far you've traveled. Irrigating without water measurement is much the same way. Without knowing the water flow rate, you do not know how fast you are applying water. And, without measured volumes, you cannot determine the depth of application. Good water management begins with accurate water measurement. Unfortunately, because of the regulatory implications, some water users have an unfavorable attitude towards water measurement. Good water managers use water measurement to evaluate how efficiently they are using the water that they apply
Energy management is another reason to measure water. To evaluate the energy efficiency of pumping systems, you need to know both the energy input and the output from the pumping system. The output includes the water flow rate.
Depth Volume Relationships
Irrigators commonly measure and discuss rainfall depth. Since irrigation is artificial rainfall, it is also useful to express irrigation water application as a depth. Equations 3.1 and 3.2 relate the depth applied and applied volume to the land area irrigated:
\(d=\dfrac{V}{A}\) (3.1)
\(V=A\times{d}\) (3.2)
where: d = depth of water applied
V = volume of water applied, and
A = area irrigated.
The concepts of Equations 3.1 and 3.2 are shown in Figure 3.1.
Figure 3.1 Relationship between volume of water applied, land area, and depth applied.
Since the volume of water applied is the product of system flow rate and the time of application, Equation 3.2 is often expressed as:
Q × t = A × d (3.3)
where: Q = system flow rate and
t = time of water application.
This equation assumes that the flow rate is constant over the application time. Use of Equation 3.3 is basic to efficient irrigation management. Although straightforward, Equation 3.3 requires that the user apply the appropriate conversion factors to make the units consistent. When using Equation 3.1 and 3.2, it is most convenient to convert volume units to acre-inch (ac-in) when working in agriculture. Table 3.1 lists common unit conversion factors for volume and flow rate. When using U.S. units, approximately 450 gallons per minute (gpm) equals 1 ac-in/hr. An ac-in is the volume of water that covers 1 acre 1 inch deep. Before using Equation 3.3 for an agricultural application, the system flow rate (Q) should be converted to ac-in/hr. One ac-in/hr is also equal to 1 cubic foot per second (cfs), another common flow unit used in agricultural irrigation.
| U.S. Customary System of Units (USCS) | Conversion between USCS and SI Systems | International System of Units (SI) |
|---|---|---|
| Volume Units | ||
|
1 gal = 8.33 lb 1 ft3 = 7.48 gal 1 ac-in = 3,630 ft3 1 ac-in = 27,154 gal 1 ac-ft = 43,560 ft3 1 ac-ft = 325,851 gal |
1 ft3 = 0.02832 m3 1 L = 0.264 gal 1 gal = 3.79 L 1 m3 = 264.2 gal 1 ac-in = 1.028 ha-cm |
1 L = 1,000 cm3 1 cm3 = 0.001 L 1 m3 = 1,000 L 1 L = 0.001 m3 1 ha-cm = 100 m3 |
| Flow Units | ||
|
1 cfs = 449 gpm (450 for practical purposes) 1 cfs = 1 ac-in/hr 452 gpm (450 for practical purposes) = 1 ac-in/hr 1 gpm = 0.00223 cfs 1 gpm = 0.00221 ac-in/hr |
1 cfs = 0.02832 cms 1 cms = 35.31 cfs 1 gpm = 0.06309 L/s 1 L/s = 15.85 gpm 1 gal/h = 63.1 mL/s |
1 L/s = 1,000 mL/s 1 cms = 16.7 L/min |
| Length Units | ||
|
1 mile = 5280 ft 1 rod = 16.5 ft |
1 ft = 0.3048 meters 1 meter = 3.281 ft |
1 cm = 0.01 meter 1 meter = 100 cm 1 km = 1,000 meter |
| Area Units | ||
| 1 ac = 43,560 ft2 |
1 ac = 0.4047 ha 1 ha = 2.471 ac |
1 ha = 10,000 m2 |
|
cfs = cubic feet per second gpm = gallons per minute cms = cubic meters per second L/s = liters per second |
||
Equation 3.3 can be rearranged to calculate both depth per unit of time (d/t) and average application intensity.
\(\dfrac{d}{t}=\dfrac{Q}{A}\) (3.4)
For sprinkler heads and other water emitters, Equation 3.4 often takes the form:
\(A_r=\dfrac{96.3 q}{A}\) (3.5)
where: Ar = application rate, application intensity, or precipitation rate (in/hr),
q = discharge rate (gpm),
A = in effect, the area irrigated by the device (ft2), and
96.3 = constant for unit conversion.
The area irrigated by an individual sprinkler head equals the spacing between heads on the lateral (ft) multiplied by the spacing between laterals (ft).
The flow rate (Q) is a volume per unit of time and can be metered with flow measuring devices. For a known flow rate, Equation 3.3 can then be used to determine depth applied. Using Equation 3.3 to determine depth requires that the flow rate remain constant over the entire application time. Records must be kept of both time of application and flow rate. If the flow measuring device includes a volume totalizer, record keeping is much simpler. Volume totalizers register the total volume that has passed through the device much like an odometer measures total miles traveled in an automobile. Equation 3.1 would be used to calculate depth as shown in the next example.
An irrigation system delivers 900 gpm. If 30 acres is irrigated every 24 hr, determine the total depth applied in inches.
Given: t = 24 hr
A = 30 ac
Find: Depth applied in inches
Solution
Using Equation 3.3: \(Q\times t=A\times d \text{ or }d=\dfrac{Qt}{A}\)
First, convert the flow rate from gpm to acre-inch/hour:
\(900 \text{ gpm} \times \dfrac{1\text{ ac-in/hr}}{450 \text{ gpm}}=2 \text{ ac-in/hr}\)
\(d=\dfrac{(2 \text{ ac-in/hr})(24 \text{ hr})}{30 \text{ ac}}=1.6 \text{in}\)
A flow meter has a volume totalizer. If 90 acres were irrigated and the totalizer registered 12,590,900 gallons after an irrigation and 8,925,100 gallons before the irrigation, what was the depth of application in inches?
Given: Volume after = 12,590,900 gal
Volume before = 8,925,100 gal
Find: Depth in inches
Solution
\(d=\dfrac{V}{A}\)
Volume applied \(= 12,590,900 \text{ gal}-8,925,100 \text{ gal}=3,665,800 \text{ gal}\)
\(V=3,665,800 \text{ gal}\dfrac{(1 \text{ ac-in})}{(27,154 \text{ gal})}= 135 \text{ ac-in}\)
\(d=\dfrac{135 \text{ ac-in}}{90 \text{ac}}=1.5 \text{ in}\)