Surface irrigation is the oldest and most common method of applying water to crops. It involves the distribution of water across the soil surface by gravity.
The design of surface irrigation systems relies on the Volume Balance Equation and Infiltration Models.
The rate at which water enters the soil is critical for determining how long water must remain on the surface. \[I = k \cdot t^a\] Where: - \(I\) = Cumulative infiltration (mm or m) - \(t\) = Intake opportunity time (min) - \(k, a\) = Soil-specific empirical constants
Used to calculate the velocity of water flow in furrows or borders. \[V = \frac{1}{n} R^{2/3} S^{1/2}\] Where: - \(V\) = Velocity (m/s) - \(n\) = Manning’s roughness coefficient - \(R\) = Hydraulic radius (m) - \(S\) = Longitudinal slope (m/m)
For a given time \(t\), the volume of water applied equals the volume on the surface plus the volume infiltrated. \[Q \cdot t = A_s \cdot L + z_{avg} \cdot L\] Where: - \(Q\) = Inflow rate per unit width (\(m^3/s/m\)) - \(A_s\) = Average cross-sectional area of surface storage (\(m^2\)) - \(L\) = Advance distance (m) - \(z_{avg}\) = Average depth of infiltrated water (\(m\))
When designing a furrow system, we must determine the Maximum Unit Stream (\(Q_{max}\)) to avoid erosion and the Minimum Unit Stream (\(Q_{min}\)) to ensure water reaches the end of the field.
\[Q_{max} = \frac{C}{S}\] *Typically, \(C = 0.6\) to \(0.63\) for many soils (where \(Q\) is in L/s and \(S\) is in %).
A farmer wants to design a furrow irrigation system for a maize field. - Soil Type: Loam (\(k = 0.0051\) m/min^a, \(a = 0.5\)) - Field Length (\(L\)): 200 m - Furrow Spacing (\(W\)): 0.75 m - Required Net Depth (\(d_{net}\)): 75 mm (0.075 m) - Slope (\(S\)): 0.1% - Inflow Rate (\(Q\)): 1.5 L/s per furrow
We will calculate the Intake Opportunity Time (\(t_{req}\)) required to deliver 75mm of water and the Total Application Time.
# Constants
k <- 0.0051 # Kostiakov constant
a <- 0.5 # Kostiakov exponent
d_req <- 0.075 # Required depth in meters
L <- 200 # Length in meters
W <- 0.75 # Spacing in meters
Q_ls <- 1.5 # Inflow in Liters per second
# 1. Calculate required intake opportunity time (t_req)
# From d = k * t^a -> t = (d/k)^(1/a)
t_req <- (d_req / k)^(1/a)
# 2. Calculate Total Volume required per furrow (m3)
vol_req <- d_req * L * W
# 3. Calculate Application Time (T_app) in minutes
# T_app = Vol / Q
Q_m3_min <- (Q_ls / 1000) * 60 # Convert L/s to m3/min
t_app <- vol_req / Q_m3_min
# Results
cat("Required Opportunity Time (min):", round(t_req, 2), "\n")
## Required Opportunity Time (min): 216.26
cat("Total Application Time (min):", round(t_app, 2), "\n")
## Total Application Time (min): 125
\[E_a = \frac{\text{Water Stored in Root Zone}}{\text{Water Delivered to Field}} \times 100\]
Occurs when the water remains at the head of the field longer than necessary, causing water to move below the root zone.
Water that reaches the end of the furrow and exits the field. This can be recaptured using a Tailwater Recovery System.
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