NGEN10; Ecohydrology Physical Geography and Ecosystem Analysis, Lund University Exercise soil water and soil physics

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1 NGEN10; Ecohydrology Physical Geography and Ecosystem Analysis, Lund University Exercise soil water and soil physics March/April 2012 (updated 2016) Thomas Holst (based on H. Lankreijer) Soil water potential and soil water flow Aim The aim of this exercise is to analyse some of the physical characteristics of soils and estimation of water flows in the vadose or unsaturated zone. This exercise addresses the concept of soil moisture potential energy and soil water movement as a dynamic system. We will limit us to the movement of water in a soil profile in one dimension. The first step is to learn more about the system: to determine the variables, parameters and functions, which are needed. Answer all questions in this exercise and describe clearly how you found the answers. Getting acquainted with the soil characteristics Plants take up water from the soil. Water is the basis for growth and forms the transport medium for nutrients. Actually less than 1% of the water taken up by the plants remains in the plant. Water in the soil is also important for other processes like decomposition, which is limited when the soil is too dry or completely saturated with water. If there is too much water in the soil, oxygen supply to the roots is limited, and plants can eventually die or at least show a decreased growth rate. Water in the soil is bound to soil particles by forces, and not all present soil water is available to the plant. The forces are adsorption (London-van der Waals forces), capillary and osmotic binding. They differ for each soil type depending on the distribution of the particle sizes (texture), organic matter content and they vary with time due to changes in soil structure and salt content. Soil texture: the distribution of the soil particles in different size-classes (sand, silt, clay) Soil structure: the arrangement and organisation of soil particles Aggregates or peds: quasi stable small clods of soil particles Soil characteristics Before describing the concept of potential, some characteristic quantities of soils are repeated: Total soil volume V t (m³) is the total of all volumes of the soil constituents and equal to the thickness z times area A (in m and m² resp.): V t =V m +V h +V a +V w V m = total volume of mineral particles V h = total volume of organic particles A A V a = total volume of air (filled pores) z V w = total volume of water (filled pores) V p = total volume of pores=v a +V w Porosity f p = volume fraction of voids= empty space fraction =pore fraction=v p /V t Volume fraction of air f a =V a /V t z

2 Volume fraction solid phase (humus and mineral particles) f m+h =(V m +V h )/V t, under ice-free conditions. Water content as volume fraction θ=v w /V t Water content as fraction of weight of solid phase, or gravimetric water content: w=m w /m m+h, where m is the mass of constituents. Of course all the fractions can also be expressed in percentages %=fraction x 100% From the above ratios it follows that f p =f a + θ. And f a + θ+f m+h =1 Another unit for water content is volume per area: d w =V w /A expressed in m. A is often expressed as 1 m², and the value d w is expressed in mm. This is a very common way to express the amount of water on a surface: the precipitation is usually expressed in mm. 1 mm over 1 m² is equal to 1 litre, and it is (almost) equal to 1 kg/m² as the density of water is close to 1kg/litre (or 1000kg/m³). In the soil this unit is an equivalent unit, because a water layer of x mm does not exist in the soil; the water is distributed in the pores. If the soil is saturated with water also the air phase will be filled with water and V w =V p and θ sat =V p /V t. Then it is possible to express the characteristics of the soil as density=mass per volume: Wet bulk density: density of the total moist soil ρ t (kg/m³) Dry bulk density: density of the dry soil ρ d (kg/m³) Organic density: density of the organic part of the soil ρ h (kg/m³) The pores fraction can be estimated from the density of the completely dry soil (usually dried in an oven at 105 C for 24h) and the density of the solid phase: f p =1-ρ d /ρ m+h Exercise 1 Question 1A: the soil consists of 3 phases. Under which condition does the soil only consist of two phases? Is it possible that soil only consists of one phase? Question 1B: A soil volume of 0.33 high x 1 x 1 m has at the moment a water content of 50mm. At saturation the soil water content is 158.4mm. What are the porosity and the soil water content at the moment expressed in %?

3 Exercise 2 We take a soil sample with a volume of V t =100cm³. The sample is field-humid, i.e. it contains a certain amount of water and its weight is 177.7g. It is dried for 24 hours at 105 C to remove all water. Then it has a weight of 139.9g. the mass humus fraction is f h = The density of the organic fraction is ρ h =1470 kg/m³ and the density of the mineral fraction is ρ m =2650 kg/m³. Question 2: what are the volume fractions of organic matter, mineral matter, solid matter, water and air of this soil sample? The concept of potential energy Water is bound to soil particles by different forces. To extract the water from the soil energy is needed, which amount increases with decreasing water content. To characterize the energy level, the concept of energy potential is introduced. The energy potential does not only describe how much water is available at different locations in the soil, but also if movement of water in the soil is possible. Everybody knows that to fall from 1 meter height is less dangerous than to fall from 8 m height. The state of energy at 8m height is much higher than at 1m, and this is valid for humans, roof tiles or rain drops. The potential energy U p of any object relative to the Earth surface is (keep an eye on the units): U p =m g z m= mass of the object (kg) g= gravitational field 10 N/kg = 10 kg m s -2 kg -1 = 10 m s -2 z= the vertical distance from the reference level (m) The potential energy of a roof tile of 1 kg at 8 m height above the Earth surface is U p =1 kg * 10N kg -1 * 8 m= 80 N m = 80 J At 1 m height this is reduced to U p =1 kg * 10N kg -1 * 1 m= 10 N m = 10 J So the potential energy depends on the height in the gravity field, and energy is needed to increase the height. The energy needed to increase the height is stored as potential energy in the object. From this example we can derive a quantity named energy level, state of energy or potential, when the energy is divided by the mass of the object: Ψ= U p /m This is the common way to express the energy state of water in the soil in J/g. However, there are two other ways of expressing the energy state, which can be handy in other situations. The first option would be to divide the energy by volume V. this results in expressing the energy by pressure or tension: p=u p /V

4 This quantity is used for gas or liquids and expressed in J/m³=N m/m³=n m -2 =Pa The second option is to express the energy by dividing it by its weight, which is the force of gravity (G) on the object (G=mass x gravity): H=U p /G Because G is expressed in Newton (N), H is expressed in J/N and equal to m. in hydrology/soil physics it is defined to express water potential as the potential energy expressed in relation to weight. H is thus the energy expressed in height or in cm water column, where 1m= Pa. Compare this to another common unit for potential energy: MPa=10 6 Pa= 10 4 cm water column=100 m water or more precisely 9810 cm water column. (Weight of 1 m³ is 1000kg*9.81 m s -2 =9810 N). in the example of the roof tile it is only the gravity and height, which causes the energy difference between the two locations. This energy due to gravity can also be expressed as height: H g = Ψ/g = g z/g = z Matrix forces However, besides gravity also all matrix forces (adhesion, osmotic etc.) have effect on the potential energy of water in the soil. So if we consider water in the unsaturated soil layer both the gravity as well as the matrix forces determine the final potential energy. Usually the soil water potential is expressed with the phreatic ground water level as reference level. The ground water level is defined as the level where the pressure is equal to atmospheric pressure. When the soil water pressure is higher than atmospheric pressure it is positive, and lower than atmospheric pressure it is negative. Above the groundwater level, water is bound by capillary forces and the potential is negative (= negative pore pressure, tension, vacuum). The pf curve With decreasing water content in the soil pores, more pressure is needed to extract water. In this respect it is interesting to know how much water is available to the plant, because there are different soils with different texture. The sucking-tension plants can create is 1.6 MPa, or equivalent to a 160 m water column! If the water is bound to the soil with higher tension like this, the plant cannot extract the water and will start to wilt. This is called the wilting point of the soil. The pressure is independent of the soil type, but the absolute amount of water in the soil at this pressure is different for each soil type. For each soil type you can draw a figure where the pressure of the soil is set against the amount of water in the soil (θ, ratio of water volume to total volume). The y-axis will however become very large (from 0 up to cm) so therefore we make this scale logarithmic: pf= 10 log (-h/cm) with h expressed in cm The wilting point of a soil is at 160 m = cm = pf 4.2. When the soil is saturated with water, all pores are filled with water and pf=- and set to 0. A pf curve is shown in the

5 figure below. Exercise 3 Question 3: if the pf curve in the figure is found for a soil volume of 1x0.5x0.3m, and the actual pf is at 3, define graphically the volumes of the soil phases in m³. The potential energy by gravity (H g ) and the potential energy due to the matrix forces (h m ): H h =H g +h m The estimation of the potential energy is dependent on where you put the datum (reference height): usually at the soil surface level or at groundwater level. For the difference in potential between two points relative to the datum it gives the same result: If the reference height is set to the soil surface and the pf value at (A) 30 cm below surface is 2 (field capacity) then the total potential energy at that location in the same profile is H g +h m = = -130 cm. If the same soil profile the pf at (B) 50cm below the surface is equal to 40 cm then the total potential energy is equal to = -90 cm. At 30cm depth the potential is smaller than at 40cm depth and thus the water flow is directed upwards as it flows from high to low potential. The difference between the potential is (-90) = -40 cm. pf Pressure height Condition - 0 cm Saturated 0-1 cm Very wet 1-10 cm Wet cm Humid cm Dry (germination limited) cm Very dry (wilting point) cm Air dry cm Dust If the reference height is put to the ground water level at (e.g.) 1 m then at point A the potential is = -30 cm and at B it is = +10 cm. The difference is = - 40cm. Exercise 4 If you place a narrow tube (capillary) in water then the water will rise into the tube due to capillary forces. The water rises to that level where the weight of the water column is equal to the capillary forces between the tube and the water molecules. If the water column height is given by h and d is the diameter of the tube (both in cm) then the following relation can be given: h=0.3/d=0.15/r The more narrow the tube, the larger is h. This capillary rise happens also in the soil. You should assume that the pores in the soil are small capillary tubes with all different diameters. This means that if you put on an increasing pressure or suction on the soil, the pores will be emptied from the water where pores with a larger diameter will be emptied before the smaller ones.

6 Question 4: Below in table I the pf curve of a peat soil is given (nr 1). What is the volume fraction of pores with equivalent diameter between 0.1 and 1 mm? Exercise 5: We have determined the average pf curves of the main three layers of a typical podzol. The first layer Ah is a humus containing, loam poor, fine sandy soil (nr 2 in table I) and 5cm thick. The second layer in the soil is the B layer, a humus poor, loam poor fine sand (nr 3 in table I) with a 30cm thickness, and layer C is very humus poor, fine sand also loam poor (nr 4 in table I). The last layer continues to the ground water level. We found that most roots are in the first 5cm, but the root zone extents to almost 70cm. The ground water table is at 100cm depth. Make figures of the pf curves and use those to solve the following questions with help of the figures. Question 5A: It rained a couple of hours ago, and the soil profile is in balance with the ground water level and drainage has just stopped. Estimate the total soil water content of the root zone in mm. The next weeks the weather is fine, sunny with no rain, around 15 C. Three installed tensiometer show that after 20 days (480 hours) the water potential has lowered with 140 cm at 5 cm depth, 100 cm at 20 cm depth and 50cm water potential at 70 cm depth. Ground water is still at 1 m depth. Question 5B: Calculate the average rate of evaporation and transpiration from the soil in mm/hour and in W/m². Volume fraction of water (%) with following pf f p (%) Table I: pf curves of some soils with different texture Flow of water The flow of water in the unsaturated soil in general follows the law of Darcy, and flows from high potential to low potential: Q=-k dψ/dz Here, z is not the thickness of a soil layer but the distance between the mid-point of each layer or distance between the points of given potential difference. dψ is the difference between the potentials of two layers

7 The k value is the hydraulic conductivity, usually expressed in m/s. When the soil is saturated with water the k value is called the saturated hydraulic conductivity and it is constant regardless of the pressure. In the unsaturated soil the k value changes with the soil water pressure in the pores and thus with the soil water content. The more pores are filled with water, the higher the conductivity. Usually, the relationship between k and soil water pressure (h) are presented in tables for different soils. There are equations available, but often limited to ranges of h. A function is given by Mualum (1976), but in this exercise the conductivity is assumed constant. The actual flow rate of water in the soil is higher than predicted by Darcy s law. This is due to the labyrinthine route the water goes though a porous medium. This is expressed by the tortuosity factor. Tortuosity (tortuous=labyrinthine): average ratio of actual roundabout path to the apparent straight flow path. A dimensionless parameter larger than 1 and may even exceed 2. The direction of flow It is very important to look carefully at the direction of flow. The flow will be from high to low potential, but it is needed to define what is the positive direction and what the negative direction in the soil, and in order to know if the flow is subtracted from the water content in a layer or added to the water content of a layer. The general convention is that the flow is positive when the flow is upward and negative when the flow is directed downwards. This is the reason why k is having a minus sign in Darcy s law. Exercise 6 Soil heat flux The following table below is measured data from a station located in a forest gap, showing air temperature and a soil temperature profile for a summer day. The soil temperatures were measured between 1 cm and 40 cm depth. Data is in hourly resolution. Question 6: a) provide a graph showing the temporal development of the temperatures over the course of the day. What can you say about temperature amplitudes, minima and maxima in different depths? What does this mean? b) Calculate the soil heat flux according to the equation provided in the energy balance lecture: dt G dz In this case, λ is the thermal conductivity of the soil (layer). Depending on soil type and moisture, λ varies with values of W/(m K) (dry sandy soil to saturated sandy soil), W/(m K) (dry to saturated clay) or W/(m K) (dry/saturated peat soil) (see Monteith & Unsworth 2013). A good estimate probably is λ =1.5 W/(m K) to be applied in this case. For the temperature gradient, use different layers: one with a focus on the surface (e.g. a gradient spanning between 1cm and 3cm depth), a second that uses a larger (deeper) gradient (e.g. 3-10cm). How does the diurnal pattern look like, is the result the same?

8 c) Calculate the soil heat flux and its course over the day for the data given using tautochrones (a graph showing temperature profiles for each hour of the day; compare Fig 15.5 below taken from Monteith & Unsworth 2008): 0 0 qz c p z q T dz t Assume that T_40cm remains constant, so the calculate the heat flow in slices down to 40cm for each hour. Assume that C T = c p =const.=2.0 MJ/m³ K, which is a value corresponding to mineral soil. How does the diurnal change of the calculated soil heat flux look like? What is your interpretation? hour T_air ( C) T_1cm ( C) T_3cm ( C) T_5cm ( C) T_10cm ( C) T_20cm ( C) T_40cm ( C)

9 Reference: Mualum, Y A new model for predicting the hydraulic conductivity of unsaturated porous media. Water Resource Research 12: Useful book: Fundamentals of Soil Physics. Daniel Hillel, Academic Press, New York.

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