Introduction to soil water dynamics

1. Introduction to soil water dynamics

2. Analogies • Potential difference in electrical circuits • Flow in pipes

3. Soil water potential • Many forces at work which contribute to soil water potential / energy density. • Sum called total soil water potential. • Differences in temperature and distribution of electrical forces will contribute to the total potential. • For “simple” analyses of water movement in the soil, forces which influence water movement are grouped into what is termed “hydraulic potential”.

4. As there is no membrane in soil, osmotic potential is not included. However, it would need to be added to hydraulic potential if we were considering water movement at the soil-root interface. • Temperature and electro-magnetic forces are also excluded.

5. We will briefly consider the components of the hydraulic potential. Note : water moves in response to the gradient in space of the specific potential. This is the hydraulic potential per quantity of water. Soil water moves from higher (or less negative) to lower specific potentials (or more negative).

6. The quantity of water referred to may be mass, volume or weight. The mathematics are simplified if we use the specific weight potential - the energy per unit weight - hereafter called “potential”. Energy per unit weight has the units of distance. (Energy is force x distance and weight is a force). Potential is sometimes called “suction”. This is to be avoided as there will inevitably be confusion over the sign.

7. Gravitational potentialfg Water tends to “fall” downwards. Reference level often arranged to be soil surface so potential is negative below surface. Sometimes water table is chosen Consider potential energy of water particle with volume V, mass M... Eg = Mgz = rVgz Energy per unit mass,fg = gz Energy per unit volume= rw gz Potential Energy per unit weight = z

8. Think of the capillary movement upwards from a suddenly occurring water table. Upward movement ceases when gravitational downward forces equal matric forces which are pulling the water upwards. As it is the net potential gradient that causes movement, the datum for the gravitational component is irrelevant.

9. Pressure potential Pressure potential results from the total fluid pressure on a point. It may be considered as combining the submergence potential due to depth of point in a submerged soil below the free water surface and the pneumatic potential resulting from changes in gas pressure. Normally, this is assumed to be zero but it may not be.

10. Submergence potential Pressure (force per unit area) below free water is given by: P = rgz (piezometric head) [relative to atmospheric] Submergence potential per unit weight is simply the depth below the free water surface.

11. Pneumatic potential Pneumatic potential is the pressure resulting from gas (or air) pressure. In field soils - usually considered constant and so no potential gradient. It is usually ignored. However, in the pressure plate equipment, the additional pressure (above atmospheric) required to "blow out' the matric water is in effect a pneumatic potential. Also when air is trapped in pockets in the soil, pressure will build up and tend to prevent water from moving into the high pressure spots.

12. Matric potential Matric potential results from capillary & adsorption forces associated with soil matrix. = energy required to remove unit quantity of water from soil against matric forces holding onto water. In pressure plate apparatus, a pressure difference is required to remove the water. When pneumatic potential applied equals matric potential, no more water can be removed.

13. Osmotic potential As already pointed out, apart from the soil-root interface, there are no membranes in the soil and it is not considered as part of the hydraulic potential which is concerned with flow of water through the soil. However, you will need to understand it in order to think about movements of water into plants from the soil.

14. Osmosis - rate of water L to R lower than R to L until pressure builds up when rates in each direction are the same Osmotic pressure P is hydrostatic pressure required to ensure net flow is zero

15. fo = - nc cRT For non-ionising solutions n = 1; For ionising solutions, it is the number of ions per molecule c =concentration (moles/kg) c is the osmotic coefficient R is the universal gas constant T is the temperature (°K) If mixtures are present, the osmotic potential is the sum of contributions from components. Interaction between types of solute is small.

16. To a close approximation, fos 36 Cs wherefososmotic pressure in saturated soil,is in J/kg Cs (dS/m) is the electrical conductivity of the saturation extract (dS = mmho) Ignoring changes in concentration and precipitation of sparingly soluble salts, osmotic pressure dependence on moisture content is given by fo = fosqs/q

17. A note on Ohm’s Law V/I = R so I = V/R The resistance, R is related to the length of the wire and the resistivity (specific resistance) and the cross-section area by: R = sL A Thus I/A = 1/s x V/L and so i = c V/L where i is the current per unit cross-section and c is the conductivity - the reciprocal of the resistivity

18. Darcy's law We have already suggested that water flows through soil as a result of a gradient in hydraulic potential (like current flows in a wire as a result of a gradient in electrical potential) This is expressed mathematically as Darcy’s Law wherefis the (specific weight) hydraulic potential q is usually the flux of water per unit area x is distance

19. Pressure potential at top surface is h2 - z2 Gravitational potential is z2 Hydraulic potenital is sum = h2 At bottom, gravitational potential is z1, pressure potential is h1 - z1. Total hydraulic potential is h1 Q / A = K(h2-h1)/L

20. For horizontal flow, we need only consider the pressure potentials h2 and h1

21. At D, pressure potential is hp and gravitational potential is hg Total is h2. At E, gravitational potential is higher, pressure potential is less, total is the same Q/A = K (h2-h1)/L

22. Saturated and unsaturated hydraulic conductivity Water flows only through those pores which contain water at a particular matric potential. As the soil dries out, the size and number of pores the water can flow through becomes smaller Hydraulic conductivity saturation Moisture content

23. Diffusivity For many problems concerned with unsaturated soil, it simplifies the mathematics if we imagine the water flowing not as a result of a gradient in potential, but as a result of differences in water content. In the horizontal direction: wherefmis the matric suction. So:

24. We define the diffusivity as: This is the hydraulic conductivity at the particular moisture content multiplied by the reciprocal of the slope of the moisture characteristic curve at the same moisture content slope of moisture characteristic curve at q1 q1 Moisture content 0 -1500 kPa Matric potential

25. So we can write: That is, the flow equals the diffusivity multiplied by the moisture gradient

26. Richards' equation In three dimensions, Darcy’s law for unsaturated conditions can be written: This means the total flow is the sum of the flows in each direction Since as noted before, expressing pressure as a height: f = fm + z where fmis the matric potential and z is the gravitational potential, we can rewrite Darcy’s law as: We need now be concerned only with the matric potential

27. Continuity equation & Laplaces equation Consider a leaky bucket The rate of change of liquid in the bucket is the difference between the inflow rate and the outflow rate. In calculus terms:

28. In the same way, we can consider the rate at which water flows into and out of a small element of soil and the rate of change of the moisture content within it. If q is the moisture content per unit volume in a soil, then the amount of water in a given volume V isqV and the rate of change of moisture in the volume is :- Consider now a very small volume of soil which is 1 unit high, 1 unit wide and dx units thick where dx is a very very thin slice in which there is a volume flux of water q flowing through unit area in unit time.

29. The volume of the slice is 1 x 1 xdx = dx The rate of change of moisture content in the slice = dq/dt x dx (rate of change of moisture fraction multiplied by the volume)

30. Like the leaky bucket, the rate of change of moisture content in the slice must be equal to the difference between rate at which water enters and rate at which it leaves. Suppose q is a continuous function of x, as shown.

31. Since we have chosen a unit cross-section, flow rate into slice is q(x) x 1 and flow rate out of slice is q(x + dx) Difference in the rate of inflow into the slice from the rate of outflow q(x) - q(x+dx). From the diagram, q(x+dx) = q(x) - dq which is: where is the slope of the curve at the point q,x. Thus the difference between the inflow rate into and the outflow rate out from the slice is simply

32. Thus for volume of thickness dx :- Thus to find rate on unit volume basis, we need to divide each side by volume of slice and so For three dimensions it can be shown similarly that: This is called the continuity equation and is essential to the solution of all unsaturated soil water flow problems

33. etc. Since We can also write :- If K is isotropic (the same in all directions) and the soil is uniform (K does not vary with distance) ...

34. Below a water table the water content does not change so if K is isotropic and the soil is uniform This is Laplace’s equation. The shorthand for this is Many drainage problems are solved using this equation. It may look simple, but in practice is rather difficult to solve. Many solutions are approximations using numerical analysis methods.

35. Effect of texture and structure on hydraulic conductivity Saturated hydraulic conductivity may be estimated from: where Ks is the saturated conductivity, C is a constant given as 4 x 10-3 kg s m-3, ms and mc are the silt and clay mass fractions respectively Bulk density can be allowed for by multiplying C by (1.3/rb)1.3b [density in kg/m3] b is the same b used in the equation relating f to moisture content and pore size : -- b = 2+fes + 0 2 sg