Irrigation Water Quality

1. Irrigation Water Quality Suspended solids dissolved solids

2. Water salinity • Total salinity • Chloride salinity • Units: • mg/L • ppm • mmhos/cm • mmole/L • Measurements: Evaporation, EC meters • EC: dS/m, mmhos/cm • 1dS/m = 640 mg/l

3. EC-meters

4. Salinity effects

5. Crops classification according to salinity

6. Rising Up Of Salts

7. Salt Accumulation In Furrow Irrigation

8. Salt Accumulation In Furrow Irrigation

9. Drip Irrigation

10. Salinity Accumulated On The Border Between The Wet To Dry Zones In Drip Irrigation

11. Saline Spots Created By Subsoil Drip Irrigation

12. Distribution Of Salts In SoilAs Influenced By Different Irrigation System

13. Soil Type And Water Movement. The application Of Water Is By Drippers

14. Leaching • Leaching consist of applying irrigation water in excess of the soil moisture depletion level to remove salts from the root zone. The excess water flows down below the root zone, carrying salts with it. • The leaching requirement is the leaching fraction (the amount of excess water) needed to keep the root zone salinity level within that tolerated by the crop. This requirement is determined by the crop`s tolerance to salinity and by the salinity of the irrigation water. • This excess, water expressed as a percent of the applied irrigation water, is the leaching fraction.

15. Water Balance At The Root Zone

16. Leaching With Different Types Of Water

17. Roots Distribution And Water Use Affected By Depth

18. Salt and Water Balance • Water balance • Depth of irrigation water = ET + Depth of drainage water • Diw = ET + Ddw • Salt balance • Diw * Eciw =Ddw * Ecdw • LF= Eciw/Ecdw = Ddw/Diw

19. Distribution of salts in the soil profile • Assumptions: • ET : 40% from upper 25% • 30% from second 25% • 20% from third 25% • 10% from fourth 25%

20. Example • ET = 800 mm • Eciw= 1.2 dS/m • LF=0.15 • Find salinity distribution along the profile?

21. Solution for example problem • Diw * Eciw=Ddw*Eciw • LF=Ddw/diw=Eciw/Ecdw • Diw=ET+Ddw • Diw = ET/(1-LF)= 800/(1-.15)=941 mm • Water use: • 0.4*800 = 320 mm first quarter • 0.3*800= 240 mm second quarter • 0.2*800= 160 mm third quarter • 0.1 *800 = 80 mm fourth quarter

22. Solution of example problem • Ecsw0 = Eciw = 1.2 dS/m • LF1= Ddw1/Diw = (941-320)/941 = 0.66 • ECsw1 = Eciw/LF1 = 1.2/.66 =1.82 dS/m • LF2= Ddw2/Diw = (941-320-240)/941=0.405 • ECsw2=1.2/.405=2.96 dS/m • LF3=(941-320-240-160)/941=0.235 • ECsw3=1.2/0.235=5.11 dS/m • LF4=(941-320-240-160-80)/941=0.15 • ECsw4=Ecdw=1.2/0.15=8 dS/m • Average Ecsw=(1.2+1.82+2.96+5.11+8)/5=3.82 dS/m • Average ECe= 3.82/2=1.9 dS/m

23. Design for Leaching Requirement Lr: Leaching requirement ECi: EC of irrigation water ECe : required EC for soil extract

24. Example for leaching requirment calculation • ET = 800 mm • ECi = 1.2 dS/m • ECe = 1.9 dS/m • Find required irrigation depth?

25. Example • Lr = ECi/(5 ECe-ECi) = 1.2/(5*1.9-1.2) = 0.145 • Diw = ET/(1-Lr) = 800/(1-.145) = 935 mm • Ddw = Diw-ET = 935-800 = 135 mm

26. Exchangeable sodium

27. Permeability

28. Water toxicity