For an axisymmetric two-phase filtration model based on equation

$$\begin{array}{}{\displaystyle \mathit{\Delta}P=\frac{{Q}_{H}\mu \delta c}{\pi Dh\varphi k}}\end{array}$$(16)

where *Q*_{H} – oil production rate;

*μ* Poisson ratio;

*D*- well diameter;

*hϕ* – height of the filter;

*k* – filter permeability;

*δc*- cumulative thickness of filter cake and colmatation zone.

A computational algorithm for calculation of the nonstationary problems of mass transfer in the near-well zone has been developed. The model takes into account formation elastic capacity, capillary forces and spatial nonuniformity of the adjacent reservoir’s hydrophysical characteristics distribution.

The equation system (16), considering the conditions of the saturation sum (*s*_{v} + *s*_{n} = 1) and the capillary jump (*p*_{v} = *p*_{n} + *p*_{k}(*s*_{v})), can be transformed to an equivalent one consisting of a parabolic equation for the pressure (*p*_{v}) in the aqueous phase [2]

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}}{\mathrm{\partial}t}(rm)=\frac{\mathrm{\partial}}{\mathrm{\partial}r}(r({k}_{v}({s}_{v})+{k}_{n}(1-{s}_{v}))}\\ {\displaystyle \frac{\mathrm{\partial}{p}_{v}}{\mathrm{\partial}r}+r{k}_{n}(1-{s}_{v})\frac{\mathrm{\partial}{p}_{k}({s}_{v})}{\mathrm{\partial}r},}\\ {r}_{c}<r<L;\end{array}$$(17)

and the hyperbolic transfer equation for water saturation (*s*_{v})

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}}{\mathrm{\partial}t}(rm{s}_{v})=\frac{\mathrm{\partial}}{\mathrm{\partial}r}(r{k}_{v}({s}_{v})\frac{\mathrm{\partial}{p}_{v}}{\mathrm{\partial}r}),}\\ m=({m}_{0}(r)+\delta \phantom{\rule{thinmathspace}{0ex}}{p}_{v}),\\ {r}_{c}<r<L.\end{array}$$

In addition, boundary and initial conditions are given:

$$\begin{array}{}{\displaystyle {p}_{v}{|}_{r={r}_{c}}={p}_{v}^{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{p}_{v}{|}_{r=L}={p}_{v}^{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{s}_{v}{|}_{r={r}_{c}}={s}_{v}^{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{s}_{v}{|}_{r=L}={s}_{v}^{1};}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{p}_{v}{|}_{t=0}={p}_{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{s}_{v}{|}_{t=0}={s}_{0}.\end{array}$$

An iterative process has been used to solve this problem, which at the differential level can be written as follows

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}}{\mathrm{\partial}t}(rm({p}_{v}^{n}))=\frac{\mathrm{\partial}}{\mathrm{\partial}r}(r({k}_{v}({s}_{v}^{n-1})+{k}_{n}(1-{s}_{v}^{n-1}))\frac{\mathrm{\partial}{p}_{v}^{n}}{\mathrm{\partial}r})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+{F}^{n-1};\frac{\mathrm{\partial}}{\mathrm{\partial}t}(r{m}^{n}{s}_{v}^{n})=\frac{\mathrm{\partial}}{\mathrm{\partial}r}(r{k}_{v}({s}_{v}^{n})\frac{\mathrm{\partial}{p}_{v}^{n}}{\mathrm{\partial}r}),}\end{array}$$

$$\begin{array}{}{\displaystyle {m}^{n}=m({p}_{v}^{n})=({m}_{0}(r)+\delta \phantom{\rule{thinmathspace}{0ex}}{p}_{v}^{n}).}\end{array}$$

Here *n* - iteration step number,

$$\begin{array}{}{\displaystyle {F}^{n-1}=\frac{\mathrm{\partial}}{\mathrm{\partial}r}(r{k}_{n}(1-{s}_{v}^{n-1})\frac{\mathrm{\partial}{p}_{k}({s}_{v}^{n-1})}{\mathrm{\partial}r}).}\end{array}$$

There is pressure in the aqueous phase
$\begin{array}{}{p}_{v}^{n}\end{array}$ at the n-th iteration step of the first equation, and then the water saturation
$\begin{array}{}{s}_{v}^{n}\end{array}$ is determined from the second equation. If the conditions
$\begin{array}{}\underset{{r}_{c}<r<L}{max}|{s}_{v}^{n}-{s}_{v}^{n-1}|<{\epsilon}_{1},\underset{{r}_{c}<r<L}{max}|{p}_{v}^{n}-{p}_{v}^{n-1}|<{\epsilon}_{2}\end{array}$ are met, then iteration process stops.

In numerical calculations by finite-difference method, implicit conservative difference schemes have been used, which have been solved by the sweep method [4]. On each time layer, the difference solution has been found using the described iterative algorithm. This iterative algorithm, when choosing sufficiently small values of time steps, converges and allows to solve the problem in a general formulation.

For an incompressible soil *δ* = 0, it follows from equation (17) that the total velocity

$$\begin{array}{}{\displaystyle V=(r{k}_{v}({s}_{v})\frac{\mathrm{\partial}{p}_{v}}{\mathrm{\partial}r}+r{k}_{n}(1-{s}_{v})\frac{\mathrm{\partial}{p}_{n}}{\mathrm{\partial}r})}\end{array}$$

is only a function of time *V* = *V*(*t*), in this case the computational algorithm can be simplified [2].

The approximate value of the total velocity *V*^{n} at the n-th iteration step is determined by the following formula

$$\begin{array}{}{\displaystyle {V}^{n}=\underset{{r}_{c}}{\overset{L}{\int}}{r}^{-1}({k}_{v}({s}_{v}^{n-1})}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+{k}_{n}(1-{s}_{v}^{n-1}){)}^{-1}dr\times [{p}_{v}^{1}-{p}_{v}^{0}+\mathit{\Phi}({s}_{v}^{1})-\mathit{\Phi}({s}_{v}^{0})].\end{array}$$

Here, the function,

$$\begin{array}{}{\displaystyle \mathit{\Phi}({s}_{v})=\underset{0}{\overset{{s}_{v}}{\int}}\frac{{k}_{n}(1-s)}{({k}_{v}(s)+{k}_{n}(1-s))}\frac{\mathrm{\partial}{p}_{k}(s)}{\mathrm{\partial}s}ds,}\end{array}$$

which in the case of coefficients dependences of the equations of the following form:

$$\begin{array}{}{\displaystyle {k}_{v}({s}_{v})={k}_{0}{s}_{v}^{2},}\\ {k}_{n}({s}_{n})={k}_{1}(1-{s}_{v}{)}^{2},{p}_{k}(s)={p}_{k}^{0}\sqrt{(1-{s}_{v})/{s}_{v}},\end{array}$$(18)

can be computed explicitly.

Further, using the representation

$$\begin{array}{}{\displaystyle r{k}_{v}({s}_{v})\frac{\mathrm{\partial}{p}_{v}}{\mathrm{\partial}r}=\frac{{k}_{v}({s}_{v})}{({k}_{v}({s}_{v})+{k}_{n}(1-{s}_{v}))}V}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle -\frac{r{k}_{v}({s}_{v}){k}_{n}({s}_{v})}{({k}_{v}({s}_{v})+{k}_{n}(1-{s}_{v}))}\frac{\mathrm{\partial}{p}_{k}}{\mathrm{\partial}r},}\end{array}$$

the nonlinear parabolic equation is solved numerically

$$\begin{array}{}{\displaystyle \frac{\mathrm{\partial}}{\mathrm{\partial}t}(r{m}^{n}{s}_{v}^{n})=\frac{\mathrm{\partial}}{\mathrm{\partial}r}(\frac{{k}_{v}({s}_{v}^{n})}{({k}_{v}({s}_{v}^{n})+{k}_{n}(1-{s}_{v}^{n}))}{V}^{n}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle -\frac{r{k}_{v}({s}_{v}^{n}){k}_{n}({s}_{v}^{n})}{({k}_{v}({s}_{v}^{n})+{k}_{n}(1-{s}_{v}^{n}))}\frac{\mathrm{\partial}{p}_{k}({s}_{v}^{n})}{\mathrm{\partial}r})}\end{array}$$

and the following iteration approximation for water saturation is determined.

The program implementation of the model has been performed on the basis of Visual C ++ 6.0, the program provides a dialog mode of entering the initial information and provides the possibility of graphical interpretation of the calculation results. The program allows solving problems in an axisymmetric and one-dimensional settings, with the conditions of the first and second kind on the right border (r = L). Hydrophysical characteristics of the reservoir can vary depending on the distance from the well, when setting them in the form of piecewise constant functions.

Local changes in the properties of the formation near the well as a result of acid treatment are given according to the formulas:

$$\begin{array}{}{\displaystyle \frac{1}{r}\frac{\mathrm{\partial}}{\mathrm{\partial}r}(rv)=\frac{1}{r}\frac{\mathrm{\partial}}{\mathrm{\partial}r}(rV)=0}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\sigma}_{r}=Pe+Q\epsilon -2N\frac{U}{r};}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}P=-\frac{\sigma}{f};\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sigma =Qe+R\epsilon}\end{array}$$

When solving the problem the results of the calculations are displayed in tables or graphs. After analyzing and modifying the data, the calculations can be continued using the counted results, interpreted as a new initial data.

On the basis of the developed program, calculations were performed that estimate the flow rate of oil entering the extraction well depending on the type of model (with and without capillary forces) and with acid treatment of the wellbore zone.

Coefficients dependences on water saturation were given in the form (18). The calculations were carried out with the following data of an axisymmetric problem.

Constant parameters of the problem: borehole radius *r*_{c} = 0.1; void structure *m* = 0.15; filtration coefficient *k*_{0} = 0.2*m*/*day*; field length *L* = 30*m*; initial water saturation *s*_{v} = 0.4; the ratio of viscosities of water and oil *μ*_{0} = 0.2; lowering the pressure in the well compared to the reservoir
$\begin{array}{}\mathit{\Delta}p={p}_{v}^{1}-{p}_{v}^{0}=20({p}_{v}^{0}=0).\end{array}$

The following parameters were varied: the parameter in the capillary jump formula
($\begin{array}{}{p}_{k}^{0}\end{array}$)
and the compressibility coefficient of the reservoir (*δ*).

The results of calculations of the untreated well flow are given in , where the numerator corresponds to the flow rate of the oil inflow, the denominator is the water flow rate.

Table 1 The results of calculations of the flowrate of untreated wells

Numerical modeling of two-phase flow while acid treatment of the well was carried out at the hydrophysical parameters of the formation *δ* = 0.0001, $\begin{array}{}{p}_{k}^{0}\end{array}$ = 0, $\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.8, and the treated well bore zone *a*_{0} = 0.05, *λ* = 1, *R*_{*} = 2 ().

The increase in the flowrate the treated well is observed both for the model taking into account the capillary forces ($\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.8), and without taking them into account ($\begin{array}{}{p}_{k}^{0}\end{array}$ = 0). The last two columns of the show the increase in the costs of acid treatment (the ratio of relevant costs with acid treatment and without it), respectively, the numerator $\begin{array}{}{q}_{n}^{0}\end{array}$ -for oil, the denominator $\begin{array}{}{q}_{v}^{0}\end{array}$ - for water.

Table 2 Increase in the costs of acid treatment

Figures 2, 3 show the graphs of the distribution of oil saturation and pressure for option parameters *δ* = 0.0001, $\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.8 without acid treatment.

Figure 2 Distribution of oil saturation

Figure 3 Pressure distribution

It should be particularly noted the effect of compressibility of the reservoir on the two-phase flow, which qualitatively changes its character.

In this case, compared to rigid filtration mode the dynamics of the pressure change and its distribution in the reservoir significantly influence the appearance of the solution.

This is especially clearly seen in the absence of the capillary jump. The Buckley-Leverett model at *δ* = 0, homogeneous flooding of the reservoir at the initial time moment (*s*_{t=0} = *s*(*r*, 0) = *s*_{0} = *const*) and the boundary condition *s*_{r=L} = *s*_{0} has a solution in the form of a constant *sv*(*r*, *t*) = *s*_{0}. However, the solution has a completely different form for the compressible soil: water saturation and oil saturation are non-monotonic functions that depend on time.

Figure 4 shows the calculation results for option $\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.0, *δ* = 0.005.

Figure 4 Oil saturation distribution

The effect of compressibility of the reservoir is also influenced by considering capillary jump, especially for large values of the compressibility coefficient of the reservoir (*δ* = 0.005), which may be due to the presence of the gas phase in the reservoir. In this case, when the pressure surge wave passes, a local maximum of the oil saturation distribution is formed near the borehole (the graphs are shown in Figure 5 ($\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.8) and, in addition, at the initial
moments of time near it, the pressure gradients are higher (Figure 6 ($\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.8)). These factors allow to explain higher values of oil inflow into the well at the initial moment of time () compared to the results obtained for slightly compressible soils *δ* = 0.0001.

Figure 5 Oil saturation distribution

Figure 6 Pressure distribution ($\begin{array}{}{p}_{k}^{0}\end{array}$ = 0.8)

Table 3 Oil inflow into the well at the initial time

From the series of calculations performed, it is possible to draw a general conclusion.

Acid treatment of the near-well zone of the production well, under other equal conditions, can be an effective method of increasing the flow of oil. Naturally, the effectiveness of this method depends on processing technology and physicochemical parameters of reservoir formations. For example, the results presented in , it is clear that a significant role in increasing the inflow of oil (≈1.5 times) plays reducing the negative impact of capillary forces.

The implementation of hydraulic fracturing Main form

The possibility of using the parallel sweep method for well systems is considered in the article. The proposed system of equations includes, firstly, a system of difference equations defined on each segment as a result of approximating differential equations, in the sense of Stefan and Verigin, and, secondly, equations defined at the vertices, which can be considered as boundary conditions.

The mathematical models under consideration were investigated in domains with complex geometry. For the comparative analysis, the following solutions to the problem of determining the radius of displacement of formation fluid were used when technological measures aimed at reducing the depth of penetration of the drilling fluid filtrate during the opening of the productive formation. We use the basic filtration equations describing the laws of fluid motion in the reservoir for the solution

We write down the equation of continuity in the case of the radially symmetric motion of the filter (liquid), assuming that the filtrate penetration process occurs without diffusion mixing; an expression for the displacement radius is obtained:

$$\begin{array}{}{\displaystyle 2\pi hm\frac{d{R}_{\varphi}}{dt}=Q(t)}\end{array}$$(19)

If the fluid flow is a known function, then regardless of the type of formation and the nature of the filtration, we have:

$$\begin{array}{}{\displaystyle {R}_{\varphi}={R}_{c}\sqrt{1+\frac{W(T)}{\pi hm{R}_{c}^{2}}},}\end{array}$$(20)

where *h* - bed formation thickness,

*R*_{ϕ} - radius of the displacement front of formation fluid, *m* - void structure,

*Q*(*t*) - flow rate at *r* = *R*_{c} (*r* - current formation radius, *Rc* - borehole radius),

$\begin{array}{}W(T)={\int}_{0}^{T}Q(t)dt\end{array}$ -liquid volume entering to the reservoir in a time.

Usually (for example, during repression on a reservoir), the flow *Q*(*t*) is an unknown function, depending on the pressure drop *Δp*(*t*). In this case, it is necessary to determine the flow rate *Q*(*t*) from the solution of the corresponding non-stationary filtration problem:

$$\begin{array}{}{\displaystyle Q(t)=\frac{4\pi \epsilon \mathit{\Delta}P}{ln\frac{2.25xt}{{R}_{0}^{2}}+2S}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{\Delta}{P}_{0}=const}\end{array}$$

where $\begin{array}{}{\displaystyle \epsilon =\frac{kh}{\mu},\chi =\frac{k}{2m\mu},}\end{array}$

*k* - permeability

*μ* - fluid viscosity

*S* - degree of growth of the surface resistance at *S* > 0 or its decrease at *S* < 0

If *Δp* is a variable of time values, we can use the approximate formula

$$\begin{array}{}{\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{Q}_{n}=\frac{4\pi \epsilon \mathit{\Delta}{P}_{n}-\sum _{j=1}^{n-1}{Q}_{j}{\phi}_{nj}}{{\phi}_{nn}+2S}}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\phi}_{nn}=ln\frac{2.25\chi ({t}_{n}-{t}_{n-1)}}{{R}_{0}^{2}}}\\ \mathit{\Delta}{P}_{u}={P}_{c(t)}-{P}_{i}l,\phantom{\rule{1em}{0ex}}(n\ge 2;\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j=1,2,\dots n-1),\end{array}$$

where the time intervals *Δt*_{j} = *t*_{j} − *t*_{j−1} must satisfy the condition $\begin{array}{}\mathit{\Delta}t>>{R}_{c}^{2}/4\chi .\end{array}$

The formula $\begin{array}{}s=(\frac{{k}_{2}}{{k}_{1}}-1)\end{array}$ ln $\begin{array}{}\frac{{R}_{1}}{{R}_{c}}\end{array}$ in the case *ΔP* = *ΔP*_{0} = *const* takes the form:

$$\begin{array}{}{\displaystyle {R}_{\varphi}={R}_{c}\sqrt{1+\frac{a\tau}{{\tau}_{\chi}}{L}_{i}({\tau}_{\chi})}}\end{array}$$

where $\begin{array}{}a=\frac{1.8k\mathit{\Delta}{P}_{0}}{m\chi \mu},\tau =\frac{2.25\chi t}{{R}_{0}^{2}},{\tau}_{\ast}=\tau {e}^{2s},\end{array}$

*L*_{i}(*τ*_{ϕ}) = $\begin{array}{}{\int}_{0}^{{\tau}_{\varphi}}\frac{du}{lnu}\end{array}$
integral logarithm, the tabulated function

At *τ*_{ϕ} >> 1, using the asymptote

$$\begin{array}{}{\displaystyle {L}_{i}({\tau}_{\ast})\approx \frac{{\tau}_{\ast}}{ln{\tau}_{\ast}}}\end{array}$$

we obtain a formula for finding the front radius of the computation in time:

$$\begin{array}{}{\displaystyle {R}_{\varphi}={R}_{c}\sqrt{1+\frac{a\tau}{ln\tau +2S}}}\end{array}$$

If the flow rate is calculated using formula (20), we proceed as follows.

By value, the parameters *R*_{c}/4_{χ} choose a time step $\begin{array}{}\mathit{\Delta}\phantom{\rule{thinmathspace}{0ex}}>\phantom{\rule{negativethinmathspace}{0ex}}>\phantom{\rule{thinmathspace}{0ex}}\frac{{R}_{c}^{2}}{4\chi},\end{array}$ satisfying condition, compile the table of values *ΔP*_{u} = *P*(*t*_{u}) − *P*_{m} for corresponding moments of time *t*_{u} = *nΔt*_{0} and for a given value *δ* we compute sequentially *Q*_{u} = *Q*(*t*_{n}).

Then the time interval *T*is divided into *n* equal intervals, where it is determined from the condition $\begin{array}{}n\phantom{\rule{thinmathspace}{0ex}}>\phantom{\rule{negativethinmathspace}{0ex}}>\phantom{\rule{thinmathspace}{0ex}}[\frac{4T\mathit{\Delta}E}{{R}_{c}}]\end{array}$
([] - is the integer part of the number E). Then the function *W*(*t*) is approximately computed by the formula:

$$\begin{array}{}{\displaystyle W(t)\approx \frac{T}{n}\sum _{j=1}^{n}{Q}_{j}}\end{array}$$

If the viscosity of the solution filtrate is not equal to the viscosity of the fluid, then in determining the flow rate *Q*(*t*) it is possible to use the quasi-steady-state method by which the filtrate flow is calculated:

$$\begin{array}{}{\displaystyle Q=\frac{2\pi \mathit{\Delta}P}{{\epsilon}_{1}^{-1}!ln(\frac{{R}_{1}}{{R}_{c}})+{\epsilon}^{-1}ln\frac{{R}_{k}}{{R}_{1}}},}\end{array}$$

where $\begin{array}{}{\epsilon}_{1}=\frac{kh}{\mu},\epsilon =\frac{kh}{\mu}\end{array}$ - Hydraulic conductivity in the zones of reservoir fluid and reservoir filters, respectively

*R*_{k} - radius

*R*_{1} - invasion range of the filtrate into the formation.

In the formula, $\begin{array}{}\frac{{\mathrm{\partial}}^{2}P}{\mathrm{\partial}{\xi}_{1}^{2}}+\frac{{\mathrm{\partial}}^{2}P}{\mathrm{\partial}{\xi}_{2}^{2}}=0,\end{array}$ assuming *R*_{1} = *R*_{ϕ}, *R*_{k} = *R*_{t}, we write it in the form:

$$\begin{array}{}{\displaystyle Q(t)=\frac{2\pi \epsilon \mathit{\Delta}P}{\overline{\mu}ln\frac{{R}_{\varphi}}{{R}_{c}}+ln(\frac{{R}_{t}}{{R}_{\varphi}})},}\end{array}$$(21)

where $\begin{array}{}\overline{\mu}=\frac{{\mu}_{1}}{\mu},{R}_{t}=\sqrt{2.25\chi t},{R}_{\varphi}(t)<{R}_{t}.\end{array}$

Substituting (21) into the right-hand side of the expression (19) we obtain the differential equation with respect to

$$\begin{array}{c}{\displaystyle \overline{{R}_{\varphi}}=\frac{{R}_{\varphi}}{{R}_{c}}}\\ \\ {\displaystyle \frac{d\overline{{R}_{\varphi}}}{d\tau}=\frac{a}{2\overline{{R}_{\varphi}}}[ln\tau +2\overline{\mu}S-2(1-\overline{\mu})ln\overline{{R}_{\varphi}}{]}^{-1}}\end{array}$$

Calculations show that with increasing parameters *μ* and, *S*, invasion range of the filtrate decreases significantly, for example, when *S* ≥ 10^{3} it is practically independent of the viscosity of the formation fluid.

Practical interest represent the following limit cases. If at the opening of the permeable formation as the drilling fluid to use a liquid containing no colloidal particles (water, oil) and its viscosity is equal to the viscosity of the fluid, then, putting the equation, we obtain the following calculation formula:

$$\begin{array}{}{\displaystyle {R}_{\varphi}=\sqrt{\frac{k\mathit{\Delta}PT}{\pi m\mu}}}\end{array}$$(22)

Therefore, in this case, the front radius of the computation depends, in addition to the process parameters F and T only on the properties of the formation, for example, at *K* = 2 · 10^{−13}, *m*^{2}, *m* = 0.15, *μ* = 10^{−3} *PA*, *ΔP* = 5 MPa and *T* = 8.64 · 10^{−4} s will receive *R*_{ϕ} = 13.5 m.

The equation (22) can also be used to estimate the penetration radius of a grouting mortar when fixing the absorbing horizon or acid solutions during treatment of the bottom hole zone if the colmatation zone has been destroyed. A simple calculation from the formula (22) shows that if the clay crust is removed or destroyed during the cementing of the well, the penetration radius of the filtrate of the component solution can be quite significant.

The results of numerical simulation are shown in Figure 7-12. To illustrate the results, a set of programs with real technological indicators of the above deposits was created. A series of calculations were performed to adapt mathematical models.

Figure 7 Test examples of calculating the fracture volume

Figure 8 Evaluation of the cooled thermal front of the formation

Figure 9 Changing the cooled thermal front of the formation

Figure 10 Baring of the well bore zone

Figure 11 Reservoir response

Figure 12 Baring of the well bore zone

At the same time, data on the history of field development were used. Similar packages can be used for uranium deposits *i.e*. their technological parameters with the parameters of the Kenbai deposit are in many respects identical.

For pressure calculation during the well bore zone bearing of the formation, the window for parameter entry is shown in Figure 10, and the solved test example is shown in Figure 11.

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