Graded furrows of 80 m length and 0.75 m spacing are used for irrigating a field with an initial furrow stream of 100 L $min^{-1}$. The initial furrow stream flow reaches the lower end of the field in 40 min. Thereafter, the furrow stream flow is reduced to 30 L $min^{-1}$ and the cutback stream flow is continued for 1 hour. The average depth of irrigation over the field in cm will be ______________.
Irrigation Depth Calculation
This solution calculates the average irrigation depth for graded furrows based on flow rates and irrigation durations.
Irrigation Parameters
- Furrow length: $L = 80 \, m$
- Furrow spacing: $S = 0.75 \, m$
- Initial furrow stream: $q_{init} = 100 \, L \, min^{-1}$
- Time for initial flow advance: $t_{advance} = 40 \, min$
- Cutback furrow stream: $q_{cutback} = 30 \, L \, min^{-1}$
- Cutback duration: $t_{cutback} = 1 \, hour = 60 \, min$
Total Volume Applied Calculation
The total volume of water applied to a single furrow is the sum of the volumes applied during the initial flow phase and the cutback phase.
- Volume during initial flow:
- $V_1 = q_{init} \times t_{advance}$
- $V_1 = 100 \, L \, min^{-1} \times 40 \, min = 4000 \, L$
- Volume during cutback flow:
- $V_2 = q_{cutback} \times t_{cutback}$
- $V_2 = 30 \, L \, min^{-1} \times 60 \, min = 1800 \, L$
- Total volume per furrow:
- $V_{total, furrow} = V_1 + V_2$
- $V_{total, furrow} = 4000 \, L + 1800 \, L = 5800 \, L$
Volume Unit Conversion
Convert the total volume from liters to cubic meters for consistency in units.
- $V_{total, furrow} = 5800 \, L \times \frac{1 \, m^3}{1000 \, L} = 5.8 \, m^3$
Irrigated Area Calculation
Determine the area of the field irrigated by a single furrow.
- Area per furrow ($A_{furrow}$) = Furrow length $\times$ Furrow spacing
- $A_{furrow} = L \times S$
- $A_{furrow} = 80 \, m \times 0.75 \, m = 60 \, m^2$
Average Depth Calculation
The average depth of irrigation is the total volume applied per furrow divided by the area irrigated by that furrow.
- Average depth ($d$) = $\frac{V_{total, furrow}}{A_{furrow}}$
- $d = \frac{5.8 \, m^3}{60 \, m^2} = 0.09666... \, m$
- Convert the depth from meters to centimeters:
- $d_{cm} = 0.09666... \, m \times 100 \, cm/m = 9.666... \, cm$
The calculated average depth of irrigation is approximately 9.67 cm.
Important Questions from Surface Irrigation Methods
- Uniformity of irrigation is checked in a border strip by measuring water penetration depth at 16 equally spaced stations along a border strip. The depths of penetration of water recorded in mm are as follows:
$0.65, 0.83, 0.79, 0.87,$
$0.67, 0.85, 0.86, 0.68,$
$0.88, 0.77, 0.73, 0.59,$
$0.89, 0.63, 0.72, 0.70$
The distribution uniformity low-quarter (DULQ) (in fraction) is ________
(Rounded off to two decimal places) The difference between a given surface and the actual time of advance of water in a surface irrigation event is known as ____________.
- Furrows of 120 m length with 0.5% slope are made at 90 cm spacing. The maximum non-erosive stream flow rate is applied in a furrow that takes 1.0 hour to reach the lower end. Then this flow rate is reduced to half of its size and, subsequently, continued for another 1.0 hour. The average depth of applied water is ________ cm.
- A border strip of $8 \times 250 \text{ m}$ is being irrigated by a border stream of $50 \text{ lps}$. The infiltration capacity of the soil is $25 \text{ mm h}^{-1}$ (assumed to be constant throughout the period of irrigation). The average depth of the advancing sheet of water over the land is $70 \text{ mm}$. The time required to irrigate the border strip, in minutes, will be
