Irrigation Time Calculation for Border Strip
This solution calculates the time required to irrigate a border strip based on the provided parameters.
Input Parameters Analysis
- Border Strip Dimensions: Width $W = 8 \text{ m}$, Length $L = 250 \text{ m}$
- Stream Flow Rate: $Q = 50 \text{ lps}$
- Infiltration Capacity: $i = 25 \text{ mm h}^{-1}$
- Average Depth of Water Sheet: $d_{avg} = 70 \text{ mm}$
Unit Conversions
To ensure consistent calculations, convert units to meters and hours:
- Area $A = L \times W = 250 \text{ m} \times 8 \text{ m} = 2000 \text{ m}^2$
- Flow Rate $Q = 50 \text{ lps} = 50 \times \frac{1 \text{ m}^3}{1000 \text{ L}} \times \frac{3600 \text{ s}}{1 \text{ h}} = 180 \text{ m}^3/\text{h}$
- Average Depth $d_{avg} = 70 \text{ mm} = 0.07 \text{ m}$
- Infiltration Rate $i = 25 \text{ mm h}^{-1} = 0.025 \text{ m h}^{-1}$
Irrigation Time Formula
The time required for border irrigation ($T$, in hours) can be estimated using the following formula, which considers the volume needed for surface storage and infiltration, assuming the average time water is ponded on the field is approximately half the total irrigation time ($T/2$):
$ T = \frac{A \times d_{avg}}{Q - (A \times i / 2)} $
Calculation Steps
- Calculate the numerator (Volume for surface storage): $V_{storage} = A \times d_{avg} = 2000 \text{ m}^2 \times 0.07 \text{ m} = 140 \text{ m}^3$
- Calculate the term related to infiltration loss rate: $Loss \, Rate = A \times i / 2 = 2000 \text{ m}^2 \times 0.025 \text{ m h}^{-1} / 2 = 25 \text{ m}^3/\text{h}$
- Calculate the denominator (Net flow rate after infiltration): $Net \, Q = Q - Loss \, Rate = 180 \text{ m}^3/\text{h} - 25 \text{ m}^3/\text{h} = 155 \text{ m}^3/\text{h}$
- Calculate the total irrigation time in hours: $T = \frac{V_{storage}}{Net \, Q} = \frac{140 \text{ m}^3}{155 \text{ m}^3/\text{h}} \approx 0.9032 \text{ h}$
- Convert the time to minutes: $T_{minutes} = T \times 60 \text{ min/h} \approx 0.9032 \times 60 \text{ min} \approx 54.19 \text{ min}$
Conclusion
The calculated irrigation time is approximately $54.19$ minutes, which falls within the expected range of 52 to 57 minutes.
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.
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 ______________.
