Phases of surface irrigation| Surge irrigation| Explanation with practical example
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- Опубліковано 6 лют 2025
- Know about surface irrigation and surge irrigation
An Insight into Agronomy book: amzn.to/3R41szn
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Very good effort bhai for the informative video.
Too good sir Ji
Cary on...
Sir make a video on water measurements methods.
Thank you so much for such informative lecture sir.
Excellent initiative sir🎉... Please make one video on climate resilient agronomy
Sir pl start lecture series on soil fertility
Good concepts and good explain sir ji 🙏🙏.. Practical knowledge good 👍
Informative sir 👌👏🙏
Very nice sir 💕
Very nice sir
Sir recession phase samjh me nehi aya
Plz sir tell, Surge irrigation may be apply on undulating lands?
Surge irrigation is just a modification of surface irrigation. Like other surface irrigation methods, it is not useful on undulating topography. Sprinkler or overhead irrigation is suitable for undulating topography.
@@AgricultureUnravelled
Sir, In wheat crop which type of irrigation method is suitable.
Because it's a confusion question for me...
Different different answers of this question from different different books📚📚
For solid row (without plant to plant spacing) closely spaced crop like wheat, among surface methods, border strip irrigation is suitable.
@@AgricultureUnravelled solid row means : without plant to plant distance.
And Without row to row distance?
A.What is the quantum of available water for a soil with field capacity 40%and wilting point 20% and actual moisture percent 35 ?
B.wheat crop require 40cm of irrigation water during 120 days irrigating period .How much land can be irrigated with a flow of 20l/sec for 12 hours a day?
Sir ... Kindly give solutions for these questions ..
1. As given, the permanent wilting point is 20% and field capacity is 40%. So, total available water when the soil is at field capacity is 40-20= 20%. But in question, the soil is not at field capacity, rather the actual moisture content is 35%. So, the actual available water present in soil is 35-20 = 15%
2. Given, delta = 40 cm
Irrigation period= 120 days
Bt irrigation is applied for 12 hours in a day i.e. 1/2 day
Means, base period (B) = 120/2 = 60 days
Duty= 864*B/Delta = 864*60/40= 1296 ha
Bt this duty will be achieved when the flow rate is 1 cumec i.e. 1000 l/sec
But here flow rate is 20 l/sec i.e 0.02 cumec
So, if 1 cumec flow rate cover 1296 ha, then 0.02 cumec will cover 1296*0.02 = 25.92 ha
So answer is 25.92 ha
Thank u soo much sir.....😊
Very nice Sir 👍👌