1. Characterization of Soil Moisture Status and the Movement of Water in Soils

2. Example: You collect a 200 cm 3 soil sample. Its moist weight is 150 g. After drying, the dry weight is 100 g. Gravimetric water content: Moist weight – Dry weight Dry weight = Water weight Dry weight 150 g - 100g 100g = 50 g = 0.5 or 50% 100g

3. Example: You collect a 200 cm 3 soil sample. Its moist weight is 150 g. After drying the dry weight is 100 g. Volumetric water content: 150 g - 100g 200 cm 3 = = 50 cm 3 water = 0.25 or 25% 200 cm 3 soil Volume Water Volume Soil Density of water 1 g/cm 3 50 g 200 cm 3

4. Energy-Based Characterizing Soil Moisture Status Relating water content and matric potential (suction)

5. Characterizing Soil Water Suction applied in discrete increments. Water Remaining In soil Suction applied (cm) 0 10,000 One soil saturated * Soil Core Moisture Release Curve

6. Texture, Density Water Remaining In soil Suction applied (cm) 0 10,000 saturated * A B Two Soils coarser finer

7. Pore Size Distribution Water Remaining In soil Suction applied (cm) 10,000 saturated *

8. Soil Water Content Suction applied (cm) 0 -15,000 saturation -330 F.C. Soil Water Energy PWP Saturation: Water content of soil when all pores are filled Suction equivalent: 0 cm water Field Capacity: Water content after drainage from saturation by gravity Suction equivalent: - 330 cm water Permanent:Water can no longer be accessed by plants Wilting pointSuction equivalent: - 15,000 cm water Plant available

9. Hydraulic Conductivity Strongly responsible for water distribution within the soil volume. Determines the rate of water movement in soil. Texture Density Structure Water content The ease with which water moves through soils

10. Texture Density Structure Water content Texture – small particles = small pores = poor conductivity Density – high density suggests low porosity and small pores Structure – inter-aggregate macropores improve conductivity Water content – water leaves large pores first. At lower water contents, smaller pores conduct water, reducing conductivity. Maximum conductivity is under saturated conditions.

11. Coarse uncompacted Fine compacted Hydraulic Conductivity

12. Clay Sand High conductivity low conductivity Ponded Water

13. h L A Flow Volume Volume time h * A L SoILSoIL WATERWATER Measuring Saturated Hydraulic Conductivity

14. Volume time h * A L Volume time = K h * A L A soil h L V water A K = V * L h * A * t

15. Approximate Ksat and Uses Ksat (cm/h) Comments 36 Beach sand/Golf Course Greens 18 Very sandy soils, cannot filter pollutants 1.8 Suitable for most agricultural, recreational, and urban uses 0.18 Clayey, Too slow for most uses <3.6 x 10 -5 Extremely slow; good if compacted material is needed Saturated hydraulic conductivity

16. Determining Saturated Flow Using Ksat and the Gradient Difference in total potential between points Distance between the points Gradient =

17. Determining Saturated Flow Darcy’s Equation Volume flow Area * time = Q A = Ksat * gradient

18. Reference level Ψ g = 0 Height (cm) 50 20 a b 10 Ψ Ta = -20 cm Ψ Tb =-100 cm Difference in total potential = 80 cm = 2 Distance between the points 40 cm = Gradient Difference in potential energy = -20 cm – (-100 cm) = 80 cm Gradient = Distance between points A and B = 40 cm

19. Darcy’s Equation Volume flow Area * time = Q = Ksat * gradient (Q) = 5 cm/hr * 2 = 10 cm/hr Difference in total potential = 80 cm = 2 Distance between the points 40 cm = Gradient = If Ksat = 5 cm/hr, calculate Q

20. Distance (cm) 0 Height (cm) 50 20 a b 10 Difference in total potential -100 - (-200) = 100 cm = 5 Distance between the points 20 cm 20 cm = 5 25 Ψma = -100 cm Ψga = 0 cm Ψmb = -200 cm Ψgb = 0 cm Ref. If Ksat = 5 cm/hr, then the flow (Q) = 5 cm/hr * 5 = 25 cm/hr