Keywords

1 Introduction

The transport and deposition of particles in porous media must be taken into consideration when designing tunnels and underground structures, preparing foundations and consolidating loose soils [1,2,3,4,5]. Fine particles transported by groundwater are usually heterogeneous and differ in shape and size. Knowledge of the composition of a suspension filtered in a porous sample allows us to calculate the distribution of sediment and its composition.

At deep bed filtration, suspended particles are transferred and retained throughout the entire porous medium [6,7,8]. A porous medium contains millions of pores of different sizes. The retention mechanisms are associated with hydrodynamic, electrical and gravitational forces. Sedimentation depends on the shape, mass, size and charge of suspension particles. Size-exclusion mechanism of particle retention means that transport and blocking of particles depends on particle to pore size ratio [9,10,11]. An analytical solution to the one-dimensional filtration problem with identical particles was obtained in [12].

When filtering a bidisperse suspension, the retention profiles of large and small particles are different [13]. A deep bed filtration scheme of a bidisperse suspension with particles of two sizes is presented in Fig. 1.

Fig. 1.
figure 1

Transport and retention of 2-particle suspension in porous medium

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The mathematical filtration model of a bidisperse suspension in a porous medium includes balance equations for the concentrations of suspended and deposited particles of each type. The growth of the sediment is proportional to the concentration of particles and depends on the formed sediment of particles of both types and their initial concentrations. The model under consideration was constructed and solved numerically in [14], the analytical solution was obtained in [15, 16]. 4x4 PDE-system with given initial concentrations and filtration coefficients determines the unknown concentrations of suspended and retained particles.

Laboratory equipment makes it possible to measure the total concentration of suspended particles of a suspension at the inlet and outlet of a porous medium. The inverse filtration problem consists of finding the initial partial concentrations and filtration coefficients from the measured total concentration. To obtain partial concentrations, the total concentration at the outlet of the porous medium at the moment of breakthrough is measured and compared with the analytical solution to the direct filtration problem. By collecting the suspension at the outlet and passing it through a new sample of the porous medium, we obtain a new value of the total concentration at the moment of breakthrough. Based on several experiments, a closed system of algebraic equations for finding unknown concentrations of particles of different types and their filtration coefficients has been compiled. This problem is close to the discrete moment problem, studied in probability theory and mathematical statistics.

Inverse filtration problems, as a rule, do not have an analytical solution and are solved numerically [17]. The instability of such problems leads to a significant change in the solution with small errors in the initial data. The presence of an exact solution allows one to determine the parameters of the direct problem with high accuracy.

The inverse filtration problem is reduced to a nonlinear algebraic system of equations. Conditions for the existence of a solution are obtained, an exact solution to the inverse problem is given explicitly.

2 Mathematical Model of Direct and Inverse Filtration Problem

Consider 1-D deep bed filtration model in dimensionless form. Filtration of a bidisperse suspension in a homogeneous porous medium is determined by the system

$$ \frac{{\partial c_{i} }}{\partial t} + \frac{{\partial c_{i} }}{\partial x} + \frac{{\partial s_{i} }}{\partial t} = 0, $$
(1)
$$ \frac{{\partial s_{i} }}{\partial t} = (1 - b)\lambda_{i} c_{i} ,\,\,\,\,b = c_{1}^{0} s_{1} + c_{2}^{0} s_{2} ,\,\,\,\,\,\,i = 1,2 $$
(2)

with conditions

$$ x = 0:\;c_{i} = c_{i}^{0} ,\;c_{i}^{0} > 0,\quad t = 0:\;c_{i} = 0,\;s_{i} = 0,\quad i = 1,2. $$
(3)

Here \(\lambda_{i} ,\,\,c_{i}^{0}\) are positive constants and \(\lambda_{1} > \lambda_{2} > 0,\quad c_{1}^{0} + c_{2}^{0} = 1\).

Before the concentration front at \(0 < t < x\) the solution is zero: \(c = 0,\;\;s = 0,\;\;b = 0;\) after the front at \(t > x\) the solution is positive. At the concentration front \(s_{i} = 0,\;\;i = 1,2\).

Substitute (2) into Eq. (4):

$$ \frac{{\partial c_{i} }}{\partial t} + \frac{{\partial c_{i} }}{\partial x} + \lambda_{i} c_{i} = 0 $$
(4)

Solution to Eq. (4) with the first condition (3) sets partial suspended concentrations at the front \(c_{i}^{f} (x) = c_{i}^{0} e^{{ - \lambda_{i} x}} ,\;\;i = 1,2\).

The breakthrough concentration at the porous medium outlet x = 1

$$ c_{i}^{1} = c_{i}^{0} e^{{ - \lambda_{i} }} ,\;\;i = 1,2 $$
(5)

To find the initial partial concentrations \(c_{1}^{1} ,\,\,c_{2}^{1}\) and filtration coefficients \(\lambda_{1} ,\,\,\lambda_{2}\), Formula (5) is used several times. If you collect a suspension at the outlet of a porous medium at the breakthrough moment, measure its total concentration and filter it again through a porous sample, then at the outlet the breakthrough partial concentrations are \(c_{i}^{2} = c_{i}^{0} e^{{ - 2\lambda_{i} }} ,\;\;i = 1,2\).

After the third filtration the breakthrough partial suspended concentrations at the outlet \(c_{i}^{3} = c_{i}^{0} e^{{ - 3\lambda_{i} }} ,\;\;i = 1,2\).

Using 3 measurements of the total suspended concentration at the outlet, we obtain a non-linear system of algebraic equations

$$ \begin{gathered} c_{1}^{0} + c_{2}^{0} = 1, \hfill \\ c_{1}^{0} e^{{ - \lambda_{1} }} + c_{2}^{0} e^{{ - \lambda_{2} }} = m_{1} , \hfill \\ c_{1}^{0} e^{{ - 2\lambda_{1} }} + c_{2}^{0} e^{{ - 2\lambda_{2} }} = m_{2} , \hfill \\ c_{1}^{0} e^{{ - 3\lambda_{1} }} + c_{2}^{0} e^{{ - 3\lambda_{2} }} = m_{3} . \hfill \\ \end{gathered} $$
(6)

System (6) is a discrete finite moment problem complicated by additional conditions \(\lambda_{1} > \lambda_{2} > 0,\,\,\,\,c_{1}^{0} > 0,\,\,\,\,c_{2}^{0} > 0\).

3 Analytical Solution to the Inverse Problem

Denote

$$ k_{1} = e^{{ - \lambda_{1} }} ,\quad k_{2} = e^{{ - \lambda_{2} }} ,\quad 0 < k_{1} < k_{2} < 1. $$
(7)

The system (6) takes the form

$$ c_{1}^{0} + c_{2}^{0} = 1, $$
(8)
$$ c_{1}^{0} k_{1} + c_{2}^{0} k_{2} = m_{1} , $$
(9)
$$ c_{1}^{0} k_{1}^{2} + c_{2}^{0} k_{2}^{2} = m_{2} , $$
(10)
$$ c_{1}^{0} k_{1}^{3} + c_{2}^{0} k_{2}^{3} = m_{3} . $$
(11)

Using (8), (9), express unknowns \(c_{1}^{1} ,\,\,c_{2}^{1}\) through \(k_{1} ,\,\,k_{2}\):

$$ c_{1}^{0} = \frac{{m_{1} - k_{2} }}{{k_{1} - k_{2} }},\quad c_{2}^{0} = \frac{{m_{1} - k_{1} }}{{k_{2} - k_{1} }}. $$
(12)

Substitute (12) into Eqs. (10) and (11) and transform the equations

$$ \begin{gathered} m_{1} (k_{1} + k_{2} ) - k_{1} k_{2} = m_{2} , \hfill \\ m_{1} (k_{2}^{2} + k_{1} k_{2} + k_{1}^{2} ) - k_{1} k_{2} (k_{2} + k_{1} ) = m_{3} . \hfill \\ \end{gathered} $$
(13)

Denote

$$ u = k_{1} + k_{2} ,\quad v = k_{1} k_{2} . $$
(14)

System (13) takes the form

$$ \begin{gathered} m_{1} u - v = m_{2} , \hfill \\ m_{1} (u^{2} - v) - uv = m_{3} . \hfill \\ \end{gathered} $$
(15)

Solution to system (15)

$$ u = \frac{{m_{3} - m_{1} m_{2} }}{{m_{2} - m_{1}^{2} }},\quad v = \frac{{m_{1} m_{3} - m_{2}^{2} }}{{m_{2} - m_{1}^{2} }}. $$
(16)

By Vieta's theorem and Formula (14) the unknowns \(k_{1} ,\,\,\,k_{2}\) are determined by quadratic equation \(k^{2} - uk + v = 0\) with two roots

$$ k_{1,2} = \frac{{u \pm \sqrt {u^{2} - 4v} }}{2} = \frac{{(m_{3} - m_{1} m_{2} ) \pm \sqrt {(m_{3} - m_{1} m_{2} )^{2} - 4(m_{1} m_{3} - m_{2}^{2} )(m_{2} - m_{1}^{2} )} }}{{2(m_{2} - m_{1}^{2} )}}. $$
(17)

Let us formulate the conditions on the measured total concentrations \(m_{i}\) under which the solution to system (8)–(11) satisfy the inequalities

$$ c_{1}^{0} > 0,\;c_{2}^{0} > 0\;{\text{and}}\;0 < k_{1} < k_{2} < 1. $$
(18)

Necessary conditions follow from positive-semi definiteness of Hankel matrices:

$$ 0 < m_{3} < m_{2} < m_{1} < 1,\quad m_{2} > m_{1}^{2} ,\quad m_{3} > m_{1} m_{2} ,\quad m_{1} m_{3} > m_{2}^{2} . $$
(19)

These conditions can be derived directly from the inequalities (18):

$$ \begin{gathered} 1 = c_{1}^{0} + c_{2}^{0} > c_{1}^{0} k_{1} + c_{2}^{0} k_{2} = m_{1} > c_{1}^{0} k_{1}^{2} + c_{2}^{0} k_{2}^{2} = m_{2} > c_{1}^{0} k_{1}^{3} + c_{2}^{0} k_{2}^{3} = m_{3} > 0, \hfill \\ m_{2} - m_{1}^{2} = c_{1}^{0} c_{2}^{0} (k_{2} - k_{1} )^{2} > 0, \hfill \\ m_{3} - m_{1} m_{2} = c_{1}^{0} c_{2}^{0} (k_{1} + k_{2} )(k_{2} - k_{1} )^{2} > 0, \hfill \\ m_{1} m_{3} - m_{2}^{2} = c_{1}^{0} c_{2}^{0} k_{1} k_{2} (k_{2} - k_{1} )^{2} > 0. \hfill \\ \end{gathered} $$
(20)

From the necessary conditions follows an inequality

$$ k_{1} < m_{1} < k_{2} . $$
(21)

Formulae (21) and (12) ensure the inequalities \(c_{1}^{0} > 0,\,\,\,\,c_{2}^{0} > 0\).

Sufficient conditions specify restrictions on measurements \(m_{i}\) under which the inequalities \(0 < k_{1} < k_{2} < 1\) are satisfied. According to the properties of the parabola \(y = k^{2} - uk + v\), its roots satisfy the relation \(0< k_{1} < k_{2} < 1\), if

$$ y(0) > 0,\quad y(1) > 0,\quad 0 < \frac{{k_{1} + k_{2} }}{2} = \frac{u}{2} < 1,\quad y\left( \frac{u}{2} \right) < 0 \Leftrightarrow D > 0. $$
(22)

Figure 2 demonstrates the sufficiency of the inequalities (22) for finding two roots in the interval (0;1).

Fig. 2.
figure 2

Parabola graph.

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In the variables u, v, inequalities (22) take the form

$$ 0 < u < 2,\quad v > 0,\quad v > u - 1,\quad u^{2} - 4v > 0. $$
(23)

The inequalities \(u > 0,\,\,\,\,v > 0\) follow from (20). Other sufficient conditions are also necessary:

$$ \begin{gathered} u - 2 = k_{1} + k_{2} - 2 < 0, \hfill \\ v - u + 1 = (k_{1} - 1)(k_{2} - 1) > 0, \hfill \\ u^{2} - 4v = (k_{2} - k_{1} )^{2} > 0. \hfill \\ \end{gathered} $$
(24)

Sufficient conditions (24) do not follow from the necessary conditions (19). Let \(m_{1} = 0.5,\,\,\,m_{2} = 0.275,\,\,\,m_{3} = 0.25\).

All the necessary conditions are fulfilled:

$$ \begin{gathered} 0 < 0.25 < 0.275 < 0.5 < 1,\quad 0.275 > 0.5^{2} = 0.25, \hfill \\ 0.25 > 0.5 \cdot 0.275 = 0.1375,\quad 0.5 \cdot 0.25 = 0.125 > 0.275^{2} = 0.075625, \hfill \\ \end{gathered} $$

but sufficient condition \(u < 2\) is not valid:

$$ u = \frac{{m_{3} - m_{1} m_{2} }}{{m_{2} - m_{1}^{2} }} = 4.5 > 2. $$
(25)

In this case \(0 < k_{1} < 1 < k_{2}\) and conditions (18) are not true.

Thus, the inequalities (19) and (23) do not follow from each other, these conditions are necessary and sufficient to solve the inverse problem.

If all inequalities (19) and (23) are true, then \(c_{1}^{0} ,\,\,\,c_{2}^{0}\) are expressed by (12) and \(\lambda_{1} ,\,\,\lambda_{2}\) are obtained from (7).

4 Numerical Results

For example, let the measurements of the total breakthrough suspended concentration at the porous medium outlet are \(m_{1} = 0.5,\,\,\,m_{2} = 0.33,\,\,\,m_{3} = 0.25\).

All the necessary conditions (19) are valid:

$$ \begin{gathered} 0 < 0.25 < 0.33 < 0.5 < 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.33 > 0.5^{2} = 0.25, \hfill \\ 0.25 > 0.5 \cdot 0.33 = 0.165,\,\,\,\,\,\,\,\,\,\,\,\,0.5 \cdot 0.25 = 0.125 > 0.33^{2} = 0.1089. \hfill \\ \end{gathered} $$
(26)

The variables \(u = 1.0625,\,\,\,\,v = 0.20125\) and sufficient conditions (23) are fulfilled also:

$$ \begin{gathered} 0 < 1.0625 < 2,\quad 0.20125 > 0,\quad 0.20125 > 1.0625 - 1, \hfill \\ 1.0625^{2} - 4 \cdot 0.20125 = 0.324 > 0. \hfill \\ \end{gathered} $$
(27)

The roots \(k_{1} ,\,\,k_{2}\) are obtained from (17):

$$ k_{1} = 0.479,\quad k_{2} = 0.584. $$
(28)

The required parameters are determined by Formulae (7) and (12):

$$ \lambda_{1} = 0.736,\quad \lambda_{2} = 0.538,\quad c_{1}^{0} = 0.8,\quad c_{2}^{0} = 0.2. $$
(29)

Figure 3 shows the partial and total suspended concentrations at the front.

Fig. 3.
figure 3

Suspended concentrations \(c^{f} = c_{1}^{f} + c_{2}^{f} ,\,\,\,c_{1}^{f} ,\,\,\,c_{2}^{f}\) at the front.

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5 Discussion

Inverse problems, as a rule, are unstable, that is, small measurement errors lead to large deviations in the solution. Such problems are solved numerically; they require a large number of calculations. The use of exact solutions makes the inverse problem correct and significantly simplifies the calculations [18, 19].

Filtration of a bidisperse suspension leads to uneven distribution of partial sediment: large particles are mainly deposited closer to the inlet, and small ones - at the outlet of the porous medium. Solving the inverse filtration problem allows to calculate the distribution of large and small suspension particles. This makes it theoretically possible to non-chemically separate different types of particles.

The refined filtration model of a bidisperse suspension contains additional parameters that determine the size of the places occupied by particles of different types on the porous medium frame. These parameters are not included in system (6) and can be determined separately using the asymptotics of the solution at the exit of the porous medium.

The generalization of the inverse filtration problem to the case of a polydisperse suspension is important for theory and practice [20, 21]. For n types of particles, only the necessary conditions for solving the inverse problem are known [22,23,24]. Finding sufficient conditions for solving the inverse problem of filtration of a polydisperse suspension and its solution is a separate complex problem.

6 Conclusion

The study of the inverse filtration problem of a 2-particle suspension in a porous medium lead to the following conclusions:

  • A model for solving the inverse problem was constructed based on measurements of the concentration at the outlet of a porous medium.

  • Necessary and sufficient conditions for the existence of a solution to the inverse problem in the form of a system of inequalities are derived.

  • An exact solution to the inverse problem has been obtained in explicit form.

  • The exact solutions regularize the inverse problem and make it resistant to measurement and calculation errors.