Irrigation Water Conveyance and Application Methods
Subtopic: Design and evaluation of irrigation methods
Question Statement:
A furrow irrigation system has a furrow length of $L = 200.0\text{ m}$ and furrow spacing of $W = 0.75\text{ m}$. An inflow stream of $Q = 1.80\text{ L/s}$ ($0.00180\text{ m}^3/\text{s}$) is admitted into each furrow for a duration of $t = 2.50\text{ hours}$ ($9000.0\text{ s}$). Runoff measured at the downstream tail-end of the furrow is $V_{\text{ro}} = 2.70\text{ m}^3$, and deep percolation below the crop root zone is $V_{\text{dp}} = 3.50\text{ m}^3$. The water application efficiency ($\eta_a$) of the furrow in percent ($\text{\%}$) is ________ (round off to one decimal place).
Answer Key:
61.7
Step-by-Step Solution & Explanation
**Method 1: Furrow Water Balance Formulation**
1. Total volume of water applied at furrow inlet ($V_{\text{in}}$):
$V_{\text{in}} = Q \times t = 0.00180\text{ m}^3/\text{s} \times 9000.0\text{ s} = 16.20\text{ m}^3$
2. Water stored in crop root zone ($V_s$):
$V_s = V_{\text{in}} - V_{\text{ro}} - V_{\text{dp}} = 16.20 - 2.70 - 3.50 = 10.00\text{ m}^3$
3. Water application efficiency ($\eta_a$):
$\eta_a = \frac{V_s}{V_{\text{in}}} \times 100\% = \frac{10.00}{16.20} \times 100\% = 61.728\% \approx 61.7\%$
**Method 2: Loss Fraction Complement**
$\text{Loss Fraction} = \frac{2.70 + 3.50}{16.20} = \frac{6.20}{16.20} = 0.38272$
$\eta_a = (1 - 0.38272) \times 100\% = 61.73\% \approx 61.7\%$