Interplay between soil drying and root exudation in rhizosheath development

Abstract

Background and Aims

Wetting-drying cycles are important environmental processes known to enhance aggregation. However, very little attention has been given to drying as a process that transports mucilage to inter-particle contacts where it is deposited and serves as binding glue. The objective of this study was to formulate and test conceptual and mathematical models that describe the role of drying in soil aggregation through transportation and deposition of binding agents.

Methods

We used an ESEM to visualize aggregate formation of pair of glass beads. To test our model, we subjected three different sizes of sand to multiple wetting-drying cycles of PGA solution as a mimic of root exudates to form artificial aggregates. Water stable aggregate was determined using wet sieving apparatus.

Results

A model to predict aggregate stability in presence of organic matter was developed, where aggregate stability depends on soil texture as well as the strength, density and mass fraction of organic matter, which was confirmed experimentally. The ESEM images emphasize the role of wetting-drying cycles on soil aggregate formation.

Conclusions

Our experimental results confirmed the mathematical model predictions as well as the ESEM images on the role of drying in soil aggregation as an agent for transport and deposition of binding agents.

Appendices

Appendix A Geometrical considerations

Recall the dimensionless of Young–Laplace equation (Eq. 5)

$$ \psi = \frac{1}{r_1}-\frac{1}{r_2} $$

(21)

From the right angle triangle in Fig. 9 using Pythagorean equation we have

$$ ({r_1}+{r_2})^2+1=({r_2}+1)^2 $$

(22)

Fig. 9

Dimensionless geometric definitions of capillary water

By solving Eqs. 21 and 22 simultaneously using the Solve function of Mathematica (Wolfram Inc., Champaign, IL), analytical expression for r ₁ and r ₂ can be written as

$$ r_1= \frac{4}{3+\sqrt{9-8 \psi }} $$

(23)

$$ r_2=-\frac{r_1}{r_1 \psi -1} = \frac{4}{3+ \sqrt{9-8 \psi }-4 \psi } $$

(24)

From Fig. 9 we can calculate the half wetting angle (φ) as

$$ \varphi = ArcCos(\frac{1}{1 + r_2}) $$

(25)

Appendix B Pendular liquid volume

The liquid-vapor interface is described by equation of circle in the x-z plane as

$$ f(z)= \sqrt{r_2^2-z^2}+r_1+r_2 $$

(26)

the total volume (V _t ) retained behind the interface is a solid of revolution around the z axis, given by

$$ V_t = \pi \int_0^c (f(z))^2 dz $$

(27)

which can be calculated by substituting Eq. 26 in Eq. 27

$$\begin{array}{rll} V_t&=&\pi \left(-\frac{c^3}{3}+c \left(\text{r1} \left(\sqrt{\text{r2}^2-c^2}+2 \text{r2}\right)\right.\right. \\ && \quad\left.\left. +\,\text{r2} \left(\sqrt{\text{r2}^2-c^2}+2 \text{r2}\right)+\text{r1}^2\right)\right. \\ &&\quad\left. +\,\text{r2}^2 (\text{r1}+\text{r2}) \sin ^{-1}\left(\frac{c}{\text{r2}}\right)\right) \end{array}$$

(28)

The volume of solid (V _s ) behind the meniscus represented by spherical cap of radius R and cap c, given by

$$ V_s = \frac{1}{3}\pi c^2 (3-c) $$

(29)

where c is obtained from simple geometric relation from Fig. 9 as

$$ c = \frac{r_2}{1 + r_2} $$

(30)

The volume of liquid (V _l ) can be found by subtracting Eq. 29 from Eq. 28, given by

$$ V_l=\frac{\pi r_2 \left(3 (r_2+1)^2 \left(2 r_1+\sqrt{\frac{r_2^3 (r_2+2)}{(r_2+1)^2}}+2 r_2+r_2 (r_2+1) \sec ^{-1}(r_2+1)\right)-2 r_2 (2 r_2+3)\right)}{3 (r_2+1)^3} $$

(31)

The ratio of the liquid volume to the solid volume is obtained by substituting Eqs. 23 and 24 in Eq. 31, then divided by the solid volume (4/3 π)

$$ \begin{array}{lll} &&\frac{V_l}{V_s} \\&& =\frac{3 \left(\frac{4}{-4 \psi +\sqrt{9-8 \psi }+3}+1\right)^2 \left(\frac{4 \sqrt{2} \sqrt{-4 \psi +\sqrt{9-8 \psi }+5}}{8 \psi ^2-4 \left(\sqrt{9-8 \psi }+6\right) \psi +5 \left(\sqrt{9-8 \psi }+3\right)}+\frac{8}{-4 \psi +\sqrt{9-8 \psi }+3}+\frac{8}{\sqrt{9-8 \psi }+3}+\frac{4 \left(-4 \psi + \sqrt{9-8 \psi }+7\right) \sec ^{-1}\left(\frac{4}{-4 \psi +\sqrt{9-8 \psi }+3}+1\right)}{\left(-4 \psi +\sqrt{9-8 \psi }+3\right)^2}\right)-\frac{8 \left(-12 \psi +3 \sqrt{9-8 \psi }+17\right)}{\left(-4 \psi +\sqrt{9-8 \psi }+3\right)^2}}{\left(\frac{4}{-4 \psi +\sqrt{9-8 \psi }+3}+1\right)^3 \left(-4 \psi +\sqrt{9-8 \psi }+3\right)} \end{array} $$

(32)

The ratio of the liquid volume to the solid volume (Eq. 32) was fitted over a wide range of potential using the sum of squared error (sse) to the following equation:

$$ \frac{\emph{v}_L}{\emph{v}_S} =\left(1 + (c \psi)^d \right)^{-1} $$

(33)

where c = 7/5 and d = 5/4

Similarly, the width of the capillary meniscus (r ₁) can be expressed as a function of the capillary pressure by solving Eq. 31 for r ₁ then substitute r ₂ from Eq. 24

$$ r_1 = \frac{16 \left(-128 \psi ^6+64 \left(3 \sqrt{9-8 \psi }+28\right) \psi ^5-272 \left(5 \sqrt{9-8 \psi }+29\right) \psi ^4+32 \left(113 \sqrt{9-8 \psi }+507\right) \psi ^3-8 \left(574 \sqrt{9-8 \psi }+2167\right) \psi ^2+60 \left(47 \sqrt{9-8 \psi }+156\right) \psi -675 \left(\sqrt{9-8 \psi }+3\right)\right)}{\left(\sqrt{9-8 \psi }+3\right) \left(-4 \psi +\sqrt{9-8 \psi }+7\right) \left(-8 \psi ^2+4 \left(\sqrt{9-8 \psi }+6\right) \psi -5 \left(\sqrt{9-8 \psi }+3\right)\right) \left(-16 \psi ^3+12 \left(\sqrt{9-8 \psi }+5\right) \psi ^2-4 \left(5 \sqrt{9-8 \psi }+18\right) \psi +9 \left(\sqrt{9-8 \psi }+3\right)\right)} $$

(34)

The width of the capillary meniscus (r ₁) was fitted over a wide range of potential using the sum of squared error (sse) to the following equation:

$$ r_1 = \frac{2}{3}\left(1 + (a \psi)^b \right)^{-1/2} $$

(35)

where a = 3/8 and b = 9/10.