Theory and practice of micro water-oil displacement efficiency based on mathematical models

Abstract

Theoretical study about quantitative influence of the micro parameters (interfacial tension, wetting angle, distance, unit length, bottom radius, boundary distance, etc.) on the macro parameters (oil displacement efficiency, pressure gradient, etc.) in water-oil displacement process is very important. This paper focuses on the micro oil displacement efficiency calculation method in oil reservoir, building a new bridge between these parameters. Combining fluid motion equation, continuity equation to pressure gradient equation, the occurrence radius equation of remaining oil in different capillary is established; Taking mercury intrusion porosimetry experiment and constructing the mathematical model, new normal probability distribution of pore radius is achieved, and then a new probability function for injected water entering pore throat is established based on the occurrence radius equation and pore radius distribution equation; By using the Darcy’s Law and definite integral theory, new micro oil displacement efficiency equation is obtained. The Nuclear Magnetic Resonance experiments show: new method gets high calculating accuracy, and can be used to guide oil reservoir development.

1 Introduction

Experts and scholars focus on water-oil displacement efficiency for years. Richardson [1] shows the oil displacement efficiency in weak hydrophilic rock samples is higher than that in strong hydrophilic rock samples. And then, Morrow [2] find there is a negative correlation between the water-oil displacement efficiency and rock sample hydrophilic characteristics. Li [3] and He [4] study the Permeability influence on water-oil displacement efficiency in two different stages. Wang [5] gives the Positive correlation of feature value function and water-oil displacement efficiency. Cai [6] and Shen [7] and Chen [8] show the negative correlation of inhomogeneity of pore structure and water-oil displacement efficiency. Yu [9] studied the Oil and water viscosity ratio influence on water-oil displacement efficiency. Lu [10] studies the indication phenomenon is a key factor influence the oil displacement efficiency. Yan [11] gives that the water-oil displacement efficiency increased by Injection multiples. Jiao’s research [12] shows that the EOR is logarithmic to the injected pore volume. Fu [13] shows the replacement pressure is positive correlative to the oil displacement efficiency in the early development stage. Fan Li [14], Rogério Soares Silva [15], Zheng [16] and Dong [17] also work on how to improve the EOR. However, there is seldom theoretical research on micro factors influences (such as interfacial tension) on macro factors (oil displacement efficiency), and the explanation of oil field development phenomena is usually acted by our experiences.

In previous time, my team studied oil displacement efficiency from different aspects: based on the water flooding location, the water drive characteristic curve, the statistic methods ANN and $GM(1,n)$ and the time-varying system [18,19,20,21,22,23,24]. After years’ study, we find there is a necessary quantitative connection between water-oil displacement efficiency and these micro factors. This manuscript aims to establish new relationships between macro parameter water-oil displacement efficiency, pressure gradient with the micro parameters interfacial tension, wetting angle, distance, unit length, bottom radius, boundary distance, etc. Then, we will get the interaction mechanism between the macro parameters and the micro parameters.

2 Remaining oil occurrence radius in capillary

For the incompressible single-phase liquid, the seepage process in a homogeneous reservoir is isothermal steady.

Fluid motion equation:

$$ \overline{v} = \frac{K}{\mu} \nabla p. $$

(1)

Continuity equation:

$$ \nabla \cdot \overline{v} = 0. $$

(2)

Combining Eq. (1) to Eq. (2), Eq. (3) is established:

$$ \nabla^{2}p = 0. $$

(3)

For some particular reservoir with bottom water or lumpy oil layer, etc, the liquid flow in the bottom well is spherical directed flow.

Using space polar coordinate form in Eq. (3), Eq. (4) is established:

$$ R = \sqrt{x^{2} + y^{2} + z^{2}}. $$

(4)

Combing Eq. (4) to Eq. (3) gives:

$$ \begin{aligned}& \frac{1}{R}\frac{d}{dR}\biggl(R\frac{dp}{dR} \biggr) = 0, \\ &\textstyle\begin{cases} R = r_{e},\qquad p = p_{e}, \\ R = r_{w},\qquad p = p_{w}. \end{cases}\displaystyle \end{aligned} $$

(5)

By solving Eq. (5) gives Eq. (6) and Eq. (7):

$$\begin{aligned} &p = p_{e} - \frac{p_{e} - p_{w}}{r_{w}^{ - 1} - r_{e}^{ - 1}} \bigl( R^{ - 1} - r_{e}^{ - 1} \bigr), \end{aligned}$$

(6)

$$\begin{aligned} &v = \frac{K}{\mu} \frac{p_{e} - p_{w}}{r_{w}^{ - 1} - r_{e}^{ - 1}}\frac{1}{R^{2}}. \end{aligned}$$

(7)

For any stratigraphic micro unit, $\delta_{R}$ is the unit length, when the driving force in the place R is big enough to displace the crude oil in the pore throat, the Eq. (8) of pressure gradient is established:

$$ \Delta p = \vert p_{R + \delta_{R}} - p_{R} \vert = \frac{p_{e} - p_{w}}{r_{w}^{ - 1} - r_{e}^{ - 1}}\frac{\delta_{R}}{R ( R + \delta_{R} )}. $$

(8)

The pressure gradient in any pore throat in place R and $R + \delta_{R}$ equals to the overcoming resistance force, Eq. (9):

$$ \Delta p = \frac{2\sigma}{r} \vert \cos \theta \vert . $$

(9)

Combing Eq. (8) to Eq. (9) gives the maximum occurrence radius equation of remaining oil in different capillary, see Eq. (10):

$$ r_{D} = \frac{r_{w}^{ - 1} - r_{e}^{ - 1}}{p_{e} - p_{w}}\frac{2\sigma R ( R + \delta_{R} )}{\delta_{R}} \vert \cos \theta \vert . $$

(10)

Hence, the maximum occurrence radius changed by pressure gradient $p_{e} - p_{w}$, wetting angle θ, the unit distance, the unit length $\delta_{R}^{ - 1}$, the bottom radius $r_{w}$, the boundary distance $r_{e}$ and the interfacial tension σ.

For polymer flooding, the polymerized substance can change the wetting angle and interfacial tension, and then the occurrence radius changed. By increasing displacement PV, the $r_{D}$ will also be changed, we can also quantitatively compare the difference of water drive and polymer flooding efficiency with parameter $r_{D}$.

For heterogeneous oil reservoirs, the fluid motion equation and continuity equation will be changed, we can also use the above-mentioned steps to obtain $r_{D}$.

3 New oil displacement efficiency calculation method

The MIP experiment data show that the probability distribution of pore radius is normal. When using numerical fitting method, gives probability distribution equation of pore radius, Eq. (11):

$$ \varphi (r) = C ( r_{\max} - r ) ( r - r_{\min} )e^{ - \frac{(r - \overline{R})^{2}}{2\sigma_{R}^{2}}}. $$

(11)

For any capillary in the given unit length $\delta_{R}$, the radius is $r_{i}$, the pore volume is:

$$ V ( r_{i} ) = \pi r_{i}^{2}\delta_{R}. $$

(12)

In the displacement process, the water first enters the large capillary pores, and then gradually enters the small capillary pores corresponding with the increase of pressure; for the imbibition process, the phenomenon is just reversed. Assume that the water can complete displace the oil in the capillary, and then we get the oil displacement efficiency $E_{D1}(r_{D})$:

$$ E_{D1}(r_{D}) = \frac{\sum_{i = 1}^{m} \varphi (r_{i})V ( r_{i} )}{\sum_{j = 1}^{n} \varphi (r_{j})V ( r_{j} )} = \frac{\sum_{i = 1}^{m} \varphi (r_{i})\pi r_{i}^{2}\delta_{R}}{\sum_{j = 1}^{n} \varphi (r_{j})\pi r_{j}^{2}\delta_{R}}. $$

(13)

Where i counts from 1 to the number of capillary in the interval $[r_{D},r_{\max}]$, and j counts from 1 to the number of capillary in the interval $[r_{\min},r_{\max}]$.

Introducing definite integral theory to Eq. (13), gives

$$ E_{D1}(r_{D}) = \frac{\lim_{r_{i} \to 0}\sum_{i = 1}^{m} \varphi (r_{i})\pi r_{i}\delta_{R}(r_{i})\Delta r_{i}}{\lim_{r_{j} \to 0}\sum_{j = 1}^{n} \varphi (r_{j})\pi r_{j}\delta_{R}(r_{j})\Delta r_{j}} = \frac{\int_{r_{D}}^{r_{\max}} \varphi (r)r\delta_{R}(r)\,dr}{ \int_{r_{\min}}^{r_{\max}} \varphi (r)r\delta_{R}(r)\,dr}. $$

(14)

When fixing the unit length in any capillary, it is

$$ \delta_{R}(r) = \delta_{R}. $$

(15)

Combing Eq. (11), Eq. (15) to Eq. (14) gives

$$ E_{D1}(r_{D}) = \frac{\int_{r_{D}}^{r_{\max}} r ( r_{\max} - r ) ( r - r_{\min} )e^{ - \frac{(r - \overline{R})^{2}}{2\sigma_{R}^{2}}}\,dr}{\int_{r_{\min}}^{r_{\max}} r ( r_{\max} - r ) ( r - r_{\min} )e^{ - \frac{(r - \overline{R})^{2}}{2\sigma_{R}^{2}}}\,dr}. $$

(16)

However, not all of the injected water enters the capillary pores, the residual part, the discontinuous part and the possible formation pathway should be taken into consideration. In order to get the probability function for injected water entering pore throat, the fluid conductivity function $\eta (r)$ should be established. According to the Poiseuille Law, the flow equation in single capillary can be established:

$$ q = \frac{\pi r^{4}\Delta p}{8\mu L},\quad r_{w} \le L \le r_{e}. $$

(17)

The fluid conductivity function in single capillary is achieved:

$$ \eta (r,L) = \frac{q}{\Delta p} = \frac{\pi r^{4}}{8\mu L}. $$

(18)

Hence, the probability function $J_{k}(r_{D})$ is obtained

$$ J(r_{D}) = \frac{\sum_{r_{D} \le r \le r_{e}} \eta (r)}{\sum_{r_{\min} \le r \le r_{e}} \eta (r)} = \frac{\int_{r_{D}}^{r_{\max}} r^{3}\,dr}{ \int_{r_{\min}}^{r_{\max}} r^{3}\,dr} = \frac{ ( r_{\max} )^{4} - ( r_{D} )^{4}}{ ( r_{\max} )^{4} - ( r_{\min} )^{4}}. $$

(19)

Hence, the oil displacement efficiency $E_{D}$ can be calculated

$$\begin{aligned} E_{D}(r_{D}) &= J(r_{D})E_{D1}(r_{D}) \\ &= \frac{ ( r_{\max} )^{4} - ( r_{D} )^{4}}{ ( r_{\max} )^{4} - ( r_{\min} )^{4}}\frac{\int_{r_{D}}^{r_{\max}} r ( r_{\max} - r ) ( r - r_{\min} )e^{ - \frac{(r - \overline{R})^{2}}{2\sigma_{R}^{2}}}\,dr}{\int_{r_{\min}}^{r_{\max}} r ( r_{\max} - r ) ( r - r_{\min} )e^{ - \frac{(r - \overline{R})^{2}}{2\sigma_{R}^{2}}}\,dr}. \end{aligned}$$

(20)

Here, $E_{D}(r_{D})$ is an error function, $E_{D}(r_{D}) = \operatorname{erf}(r_{D})$, math software-Matlab can be used to calculate the numerical solution.

Hence, $E_{D}$ is a function of $\Delta p,r_{w},r_{e},\sigma,\delta_{R},\theta$, it is

$$ E_{D} = E_{D}(r_{D}) = E_{D}(\Delta p,r_{w},r_{e},\sigma,\delta_{R},\theta ). $$

(21)

By introducing related reservoir engineering measurement to change related factors can lead to the change of $E_{D}$.

4 Results and discussion

(1) PIM experiment and pore radius distribution. By analyzing lots of PIM experiment data, we find the distribution of pore radius follows the normal distribution, such as the data from GD7-42-J195TG block of SL oilfield. The data figure between mercury saturation, accumulated permeability contribution values and pore radius is given, see Fig. 1.

Figure 1

PIM experiment data figure

By using the numerical fitting method, the normal distribution fitting equation is achieved, see Fig. 2:

$$ \varphi (r) = 0.0478 ( r_{\max} - r ) ( r - r_{\min} )e^{ - 0.0396r^{1.12}}. $$

(22)

Figure 2

Normal distribution fitting curve

(2) Calculation of oil displacement efficiency. The related parameter values in GD7-42-J195TG block see Table 1.

Table 1 Parameter values

By using Eq. (10) gives:

$$ r_{D} = \frac{0.2586R}{ ( P_{e} - P_{w} )_{i}}. $$

(23)

The Occurrence radius distribution curve in different places and pressure gradient can be achieved, see Fig. 3 and Fig. 4.

Figure 3

Occurrence radius in different well place

Figure 4

Occurrence radius in different pressure gradient

Assume that the displacement pressure gradient is 0.2 Mpa, and then parameter values are obtained: see Table 2, Fig. 5 and Fig. 6.

Figure 5

Percentage of water entries into pore capillary changed by occurrence radius

Figure 6

Oil displacement efficiency changed by the occurrence radius in GD7-42-J195TG block

Table 2 Parameter values in a constant displacement pressure gradient

Figure 5 shows that the Percentage of water entries into rock pore decreased by occurrence radius, when the occurrence radius is small enough, the water can 100% enter the pore capillary.

Figure 6 shows: the oil displacement efficiency decreased by occurrence radius. When the occurrence radius is smaller than 25 μm, the oil displacement efficiency is over 80%, and when $R=5$ m, $r_{D} = 6.465$ μm, the oil displacement efficiency is near to 0.937788 in the GD7-42-J195TG block. Hence, the oil displacement efficiency decreased by occurrence radius.

According to the definition of oil displacement efficiency, gives

$$ E_{D}(r_{D}) = \frac{1 - s_{\mathrm{or}} - s_{\mathrm{wc}}}{1 - s_{\mathrm{wc}}}. $$

(24)

When the displacing time and pressure gradient is big enough, the remaining oil saturation equals to the residual oil saturation, it is $s_{\mathrm{or}} = s_{o}$, the oil displacement efficiency maximized.

Hence, the oil saturation in any place can be calculated

$$ s_{\mathrm{or}} = ( 1 - s_{\mathrm{wc}} ) \bigl( 1 - E_{D}(r_{D}) \bigr). $$

(25)

Here, statistics method is used to calculate the average oil saturation $\overline{s_{o}}$

$$ \overline{s_{o}} = \frac{\sum_{r_{D}} \phi ( r_{D} ) ( 1 - s_{\mathrm{wc}} ) ( 1 - E_{D}(r_{D}) )}{m}. $$

(26)

Using Eq. (26) and data of Table 2, water saturation is achieved, $s_{o} ( \Delta p = 0.2~\mathrm{Mpa} ) = 0.22744$.

By using the same method, the oil saturation in different displacement rate (0.05 mL/min; 0.5 mL/min; 2 mL/min) in different rock sample-A7, B9, C112 can be achieved, see Table 3.

Table 3 Theoretical calculating results of oil saturation

(3) NMR experiment: verification and discussion. By carrying Out Nuclear Magnetic Resonance experiments and using rock samples A7, B9, C12, T2 distribution curve Fig. 7, Fig. 8 and Fig. 9 and MRI Fig. 10, Fig. 11 and Fig. 12, oil saturation Table 4 are obtained.

Figure 7

T2 distribution curve of sample A7 in different flow speed

Figure 8

T2 distribution curve of sample B9 in different flow speed

Figure 9

T2 distribution curve of sample C112 in different flow speed

Figure 10

MRI of sample A7 in different flow speed

Figure 11

MRI of sample B9 in different flow speed

Figure 12

MRI of sample C112 in different flow speed

Table 4 Remaining oil saturation with NMR experiment

In Fig. 10, Fig. 11 and Fig. 12 the oil is red and the water is blue, separately. Figure 10, Fig. 11 and Fig. 12 show: the remaining oil saturation can be achieved and the residual mechanism can be concluded.

By analyzing the data in Table 3 and Table 4, Fig. 13 is obtained.

Figure 13

Comparison of remaining oil saturation from the prospective of theory and experiment

Figure 13 show that the theoretical calculating results of oil saturation is close to the experimental results. The oil saturation variance of sample-A7 is 2.116367, sample-B9 is 3.3992, and sample-C112 is 2.978333. Hence, new method can be used to guide the oilfield development.

5 Conclusions

(1) Occurrence radius of remaining oil is decided by pressure gradient, wetting angle, well distance, the unit length, the bottom radius, the boundary distance and the interfacial tension, influencing the oil displacement efficiency, it collects micro factors and macro factor in a special way.

(2) MIP experiments data show that the distribution of pore radius is normal, and the probability expression can be worked out by numerical fitting method, and it can be used to calculate pore radius distributions of new blocks in the same oil reservoir.

(3) Not all of the injected water enters the capillary pores, the residual part, the discontinuous part and the possible formation pathway should be taken into consideration.

(4) The micro oil displacement efficiency decreased by occurrence radius.

(5) NMR experiments can be used to test the accuracy of the new method, and the experiments data of rock samples A7, B9, C12 show: the NMR experiments values (oil saturations) are in accordance with the values calculated by new method.