Radiant energy drives it and a lot of water is moved about annually.
This is the focus here --movement at a place-- and these are the paths. (any precipitation)
Principally, either infiltrate the soil surface or runoff from it. Of course, some of the water on the soil surface evaporates before it has time to infiltrate or runoff. Ignore this. Also, there is a proportion of any rain that is intercepted by canopy or surface residue that does not reach the soil surface and evaporates.
If the intensity of rainfall is less than the saturated hydraulic conductivity, all of the water that reaches the soil surface should infiltrate it. On the other hand if the intensity is greater, it would seem that the difference, i – K s would runoff. But not so. It depends.
See here. So why can there be behavior like in orange?
Recall Darcy’s Law for saturated flow, Q = K S A ([D + L] / L) This form is right for flow through a depth of porous material that is open To the atmosphere at the bottom and has water on the upper surface but It is a simplified form. The complete form would be, Q = K S A ([pressure @ top + elevation @ top] – [pressure @ bottom + elevation @ bottom]) / (elevation @ top – elevation @ bottom) The denominator is length of flow and the numerator is the difference in total water potential (pressure + gravitational) at the top and bottom, i.e., the hydraulic gradient the is causing water flow. The simplified form comes about because the pressure at the bottom is atmospheric (0), the elevation at the bottom is the reference point (0), the pressure at the top is the depth of water (D) and the gravitational potential at the top is L relative to the bottom. Now, let’s apply this way of thinking to a long column of soil that is unsaturated but that has become saturated just a the very surface due to rain at i > K s.
Take the reference point at 1 cm below the surface. There the soil is still unsaturated and the water is under tension, say a – 100 cm (recall that when water potential is expressed as energy per weight, the unit is length). At the surface there is neither a positive pressure potential nor tension, i.e., the pressure (or tension) is 0. Writing Darcy’s Law for this situation, Q = K s A ([0 + 1] – [-100 + 0]) / (1 – 0) = K s A (101 / 1) Divide this by A to give flux, q, which has the units of depth per time, q = K s 101. This is the maximum possible infiltration rate under these conditions. If the intensity of rain was > q = K s 101, there would be some runoff but if the intensity was only > K s, clearly there would not be runoff, just infiltration. However, as infiltrating water moves deeper into the soil, the water content at 1 cm increases and the tension at that point decreases, i.e., becomes less negative so q decreases. As infiltration proceeds, there will come a time when q decreases to i and thereafter q < i so that runoff occurs. When the soil at 1 cm is saturated, q = K s. So the orange curve makes sense. Red one, too.
Controls the soil water tension below the surface, so the answer to the below question is obviously, yes. This has to do with the hydraulic conduc- tivity right at the surface.
So the hydraulic conductivity decreases, typically, a lot. Consequently, more runoff, no?
This old study illustrates the importance of protecting the soil surface from the effect of raindrop impact destroying surface aggregates and creating a crust. Even with i < initial K s, a decrease in K s results in runoff.
Both are evaporative processes, just different paths. So, often the two a discussed jointly.
For the direction of arrows shown, ET increases. Make sense? By vapor pressure, what is meant is water vapor pressure in the air, i.e., humidity.
Curves look a lot like infiltration curves except they don’t approach a positive minimum, zero instead. The explanation is very similar to that for decrease in infiltration rate. See next slide.
Imagine a uniformly wet, say saturated, soil at time zero when evaporation begins. Write Darcy’s Law for the upper 1 cm like before in terms of matric (tension) and gravitational potential, q = K* ([-100 + 1] – [0 + 0]) / 1 = -K* 99 Here the flux, q, is not possible maximum, like with infiltration but real, and driven by radiant energy, wind speed, etc. (the external evaporativity). It won’t take long before very surface of the soil has lost water and the water there is under tension, say -100 cm. Also, the conductivity at the surface, K*, is less than the saturated conductivity. Notice the flux is negative, i.e., up, not down. Now, let this surface drying continue so that the tension in water at the surface increases. Also, let the upward movement of water from below the surface continue so that at 1 cm the soil dries and tension increases there. For the sake of argument, call the tension at the surface –2000 cm and the tension at 1 cm –1990 cm. At this point, the conductivity is a lot less, K**, so q = -K** 9 which is very most likely considerably less than the external evaporativity.
OK, runoff occurs when rainfall > infiltration. So, too, downward movement of water in the soil (percolation) occurs when infiltration > ET.
This is supposed to represent groundwater morphology, a saturated zone above an impervious layer and an unsaturated zone about this (called vadose zone). Arrows show percolation, recharging the saturated zone.
Due to capillarity, pores can remain filled with water up to a certain tension – the smaller the pores, the greater (more negative pressure) this tension. However, bore a well and water will fill it to depth of the free water surface, i.e., at atmospheric pressure, which is lower than the depth at which the soil is saturated.
This movement is due to a gradient in matric potential (tension) that exceeds the downward gradient in gravitational potential.
Curious phenomenon, radial flow to drains. The streamlines shown by the dye are perpendicular to the steepest gradient in water potential. So, would it take more closely spaced drain lines to lower a shallow water table in a sand or in a clay and why so?