# Numerical Simulation Characteristics of Logging Response in Water Injection Well by Reproducing Kernel Method

Authors: Ming-Jing Du; Yu-Lan Wang; Temuer Chaolu
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

Reproducing kernel Hilbert space method (RKHSM) is an effective method. This paper, for the first time, uses the traditional RKHSM for solving the temperature field in two phase flows of multilayer water injection well. According to 2D oil-water temperature field mathematical model of two phase flows in cylindrical coordinates, selecting the properly initial and boundary conditions, by the process of Gram-Schmidt orthogonalization, the analytical solution was given by reproducing kernel functions in a series expansion form, and the approximate solution was expressed byn-term summation. The satisfied numerical results were carried out by Mathematica 7.0, showing that the larger the difference between injected water temperature and initial borehole temperature or water injection conditions, the more obvious the indication of water accepting zones. The numerical examples evidence the feasibility and effectiveness of the proposed method of the two phase flows in engineering.


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## Body

## 1. Introduction

As we known, during the middle and late stage of water flooding oil field it is difficult for temperature log to reflect actual water injection profile in normal water injection conditions; at the same time, with the deepening of oil field development and the extension of exploitation time, a phenomenon of the viscosity of oil in the well became stickier and the production is reduced, leading to a large amount of stored oil in underground which can not be mined. Gradually people found that water injection can save this problem. By injecting water, it can keep formation energy, enhance crude oil recovery ratio, and insure long-term high and stable production; at the same time, pressure in oil layer is improved, which can guarantee the high oil field production.In recent years, many researchers in order to improve temperature log to determine water injection profile, lots of mathematical methods were applied in temperature log such as the ADI method [1], the full implicit form of finite difference method [2], the finite difference method [3], the implicit finite difference method [4], the alternating direction implicit method (ADI), and speedup method [5].In this paper, RKHSM method is used to simulate the temperature model. It is well known that RKHSM is a simple and accurate method. In recent years, there has been a growing interest in using RKHSM to solve mathematical problems [6–14], but there is no one applying this method into characteristics of logging response in two phase flows problems. This paper tries to study the temperature field in water injection with RKHSM. By analyzing the simulated results we can conclude that temperature logging can distinguish water entry zones from non-water entry zones through hot-water injection for short periods of time, enlarging water injection velocity, selecting reasonable shut-in time. It is economical and reasonable way to use short, fast, hot-water injection in order to change the condition that temperature log can not delineate water injection profile.


## 2. Mathematical Modeling

### 2.1. The Mathematical Model

In water injection well, the temperature field model is [5](1)∂2λwT∂z2+∂2λwT∂r2+1r∂λwT∂r=∂cwρwT∂t+∂cwρwνzT∂z.T is the temperature, λw is the thermal conductivity of water, and Fw is water saturation. z is the depth of the oil wellbore, and r is the radial distance. ρw is the density of water, and cw are the specific heat of water. νz is the velocity of fluid inside the oil wellbore. The initial temperature of the formation and wellbore is at the geothermal condition which can be written as Tt=0=T0=a+bz; a is the formation temperature when z=0, b is the geothermal gradient, the boundary of adiabatic condition is (∂(λwT)/∂r)r=0=0, the upper boundary of adiabatic condition is (∂(λwT)/∂z)z=H=0, and the under boundary of adiabatic condition is (∂(λwT)/∂z)z=0=0.In order to solve (1), we introduce the reproducing kernel spaces.


### 2.2. Reproducing Kernel Hilbert Spaces Method (RKHSM)

There are several reproducing kernel spaces (RKHS) to introduce:W21[0,1], W23[0,1], W2.13[0,1], W23[0,1]⊗W2.13[0,1], and W21[0,1]⊗W21[0,1].(1) SpaceW21[0,1]=u∣u is one-variable absolutely continuous function, u′∈L2[0,1].An inner product is(2)ux,vxW21=u0v0+∫01u′x·v′xdx,ux,vx∈W210,1.The reproducing kernel is(3)
R
x
1
y
=
1
+
y
,
x
>
y
1
+
x
,
y
>
x
.(2) SpaceW23[0,1]=u∣u, u′, u′′ are one-variable absolutely continuous functions, u(0)=u′(0)=0, u′′′∈L2[0,1].An inner product is(4)ux,vxW23=∑i=12ui0vi0+∫01u′′′x·v′′′xdx,ux,vx∈W230,1.The reproducing kernel is(5)
R
x
3
y
=
y
2
-
5
x
y
2
+
y
3
+
10
x
2
3
+
y
120
,
x
>
y
,
x
2
x
3
-
5
x
2
y
+
30
y
2
+
10
x
y
2
120
,
y
>
x
.(3) SpaceW2.13[0,1]=u∣u, u′, u′′ are one-variable absolutely continuous functions, u′(0)=u′(1)=0, u′′′∈L20,1.An inner product is(6)ux,vxW2.13=∑i=12ui0vi0+∫01u′′′x·v′′′xdx,ux,vx∈W2.130,1.The reproducing kernel is(7)
R
x
3
y
=
-
1
-
4
x
+
18
x
2
1
-
4
y
+
18
y
2
768
+
120
-
5
x
y
4
+
y
5
+
10
x
2
y
2
3
+
y
120
,
x
>
y
,
-
1
-
4
x
+
18
x
2
1
-
4
y
+
18
y
2
768
+
120
+
x
5
-
5
x
4
y
+
30
x
2
y
2
+
10
x
3
y
2
120
,
y
>
x
.(4) The reproducing kernel space is(8)W=W230,1⊗W2.130,1=ux,y∣∂n+m∂xn∂ymux,yare two-variable complete continuous functions,n=0,1,2,m=0,1,2,∂p+q∂xp∂yqux,y∈L2D,p=0,1,2,q=0,1,2,ux,0=u′x,0=u0,y=u′1,y=0=ux,y∣∑i=1∞∑j=1∞cijv~1ix·v2jy:∀∑i=1∞∑j=1∞cij2<∞,where D=[0,1]×[0,1] and v~k(i)(k=1,2) is a complete orthogonal function system of W23[0,1] and W2.13[0,1].For anyu(x,y)=∑i=1∞‍∑j=1∞‍cijv~1(i)(x)·v2(j)(y)∈W, v(x,y)=∑i=1∞‍∑j=1∞‍dklv~1(k)(x)·v2(l)(y)∈W. An inner product u,v=∑i=1∞‍∑j=1∞‍∑i=1∞‍∑j=1∞‍cijd¯klv~1(i),v~1(k)1v2(j),v2(l)2=∑i=1∞‍∑j=1∞‍cijd¯ij, so the reproducing kernel of W is R(ξ,η){3}(x,y)=Rξ{3}(x)×Rη{3}(y); by the same way, we continue to get W1=W21[0,1]⊗W21[0,1], and the reproducing kernel of W1 is R(ξ,η){1}(x,y)=Rξ{1}(x)×Rη{1}(y).


### 2.3. Analytical Solution and Approximate Solution

On defining the linear operatorL, L-1 is existent: W(D)→W1(D) as [7](9)LTr,z=∂2λlT∂z2+∂2λlT∂r2+1r∂λlT∂r,Fr,z=∂clρlT∂t-∂clρlνzT∂z,and model problem (1) changes to the following problem:(10)LTr,z,t=Fr,z,t.W
(
D
) is a RKHS with the reproducing kernel R(ξ,η){3}(r,z), W1(D) is a RKHS with the reproducing kernelR(ξ,η){1}(r,z). According to [11], it can be noted that ϕi(r,z)=R(ri,zi){1}(r,z), and ψi(r,z)=L∗ϕi(r,z), where L∗ is the adjoint operator of L. The orthonormal system {ψ¯i(r,z)}i=1∞ of W(D) can be derived from the process of Gram-Schmidt orthogonalization of {ψi(r,z)}i=1∞ as ψ¯i(r,z)=∑k=1i‍βikψk(r,z). If {(ri,zi)}i=1∞ is dense in W(D) and βik is the coefficient of orthogonalization, then T(r,z,t)=∑i=1∞‍∑k=1i‍βikF(rk,zk,t)ψ¯i(r,z) is an analytical solution of (10). The approximate solution is Tn(r,z,t)=∑i=1n‍∑k=1i‍βikF(rk,zk,t)ψ¯i(r,z), i=1,…,n. If (10) is nonlinear, the approximate solution to (10) can be expressed by Tn(r,z,t)=∑i=1n‍∑k=1i‍βikF(rk,zk,Tk-1(rk,zk),t)ψ¯i(r,z), i=1,…,n.The convergence analysis, error analysis, and stability have been done by many researchers [6–14], and this paper no longer discusses them.


### 2.4. Numerical Experiment

In order to show RKHSM is effective, a nonlinear equation is used as a test to prove the high accuracy of the method. Select the above conditions and boundary conditions and letλl=1 and D=(0,1)×(0,1); the equation has an exact solution of the form T(r,z,t)=e-t·sin⁡10r·cos⁡10z. By Mathematica 7.0, using our method, we choose the number of nodes as 23 and obtain the approximate solution T23; the comparison of the results with the exact solution is shown in Table 1, and they are found to be in good agreement with each other. Figure 1 is the exact solution T(r,z,1), and Figure 2 is the error (T-T23).Table 1
Numerical results att=1.
(r, z)
Approximate solution
Exact solution
Absolute error
(0.1, 0.1)
0.165583
0.167256
0.001673
(0.2, 0.2)
−0.373791
−0.373791
0.001392
(0.3, 0.3)
−0.050882
−0.051396
0.000514
(0.4, 0.4)
0.180162
0.181982
0.001820
(0.5, 0.5)
−0.099066
−0.100067
0.001001
(0.6, 0.6)
−0.097710
−0.098697
0.000987
(0.7, 0.7)
0.180390
0.182212
0.001822
(0.8, 0.8)
−0.052427
−0.052957
0.000530
(0.9, 0.9)
−0.136755
−0.138136
0.001381
(1.0, 1.0)
0.166248
0.167927
0.001679Figure 1
Exact solutionT(r,z,1).Figure 2
Errors(T-T23).


## 2.1. The Mathematical Model

In water injection well, the temperature field model is [5](1)∂2λwT∂z2+∂2λwT∂r2+1r∂λwT∂r=∂cwρwT∂t+∂cwρwνzT∂z.T is the temperature, λw is the thermal conductivity of water, and Fw is water saturation. z is the depth of the oil wellbore, and r is the radial distance. ρw is the density of water, and cw are the specific heat of water. νz is the velocity of fluid inside the oil wellbore. The initial temperature of the formation and wellbore is at the geothermal condition which can be written as Tt=0=T0=a+bz; a is the formation temperature when z=0, b is the geothermal gradient, the boundary of adiabatic condition is (∂(λwT)/∂r)r=0=0, the upper boundary of adiabatic condition is (∂(λwT)/∂z)z=H=0, and the under boundary of adiabatic condition is (∂(λwT)/∂z)z=0=0.In order to solve (1), we introduce the reproducing kernel spaces.


## 2.2. Reproducing Kernel Hilbert Spaces Method (RKHSM)

There are several reproducing kernel spaces (RKHS) to introduce:W21[0,1], W23[0,1], W2.13[0,1], W23[0,1]⊗W2.13[0,1], and W21[0,1]⊗W21[0,1].(1) SpaceW21[0,1]=u∣u is one-variable absolutely continuous function, u′∈L2[0,1].An inner product is(2)ux,vxW21=u0v0+∫01u′x·v′xdx,ux,vx∈W210,1.The reproducing kernel is(3)
R
x
1
y
=
1
+
y
,
x
>
y
1
+
x
,
y
>
x
.(2) SpaceW23[0,1]=u∣u, u′, u′′ are one-variable absolutely continuous functions, u(0)=u′(0)=0, u′′′∈L2[0,1].An inner product is(4)ux,vxW23=∑i=12ui0vi0+∫01u′′′x·v′′′xdx,ux,vx∈W230,1.The reproducing kernel is(5)
R
x
3
y
=
y
2
-
5
x
y
2
+
y
3
+
10
x
2
3
+
y
120
,
x
>
y
,
x
2
x
3
-
5
x
2
y
+
30
y
2
+
10
x
y
2
120
,
y
>
x
.(3) SpaceW2.13[0,1]=u∣u, u′, u′′ are one-variable absolutely continuous functions, u′(0)=u′(1)=0, u′′′∈L20,1.An inner product is(6)ux,vxW2.13=∑i=12ui0vi0+∫01u′′′x·v′′′xdx,ux,vx∈W2.130,1.The reproducing kernel is(7)
R
x
3
y
=
-
1
-
4
x
+
18
x
2
1
-
4
y
+
18
y
2
768
+
120
-
5
x
y
4
+
y
5
+
10
x
2
y
2
3
+
y
120
,
x
>
y
,
-
1
-
4
x
+
18
x
2
1
-
4
y
+
18
y
2
768
+
120
+
x
5
-
5
x
4
y
+
30
x
2
y
2
+
10
x
3
y
2
120
,
y
>
x
.(4) The reproducing kernel space is(8)W=W230,1⊗W2.130,1=ux,y∣∂n+m∂xn∂ymux,yare two-variable complete continuous functions,n=0,1,2,m=0,1,2,∂p+q∂xp∂yqux,y∈L2D,p=0,1,2,q=0,1,2,ux,0=u′x,0=u0,y=u′1,y=0=ux,y∣∑i=1∞∑j=1∞cijv~1ix·v2jy:∀∑i=1∞∑j=1∞cij2<∞,where D=[0,1]×[0,1] and v~k(i)(k=1,2) is a complete orthogonal function system of W23[0,1] and W2.13[0,1].For anyu(x,y)=∑i=1∞‍∑j=1∞‍cijv~1(i)(x)·v2(j)(y)∈W, v(x,y)=∑i=1∞‍∑j=1∞‍dklv~1(k)(x)·v2(l)(y)∈W. An inner product u,v=∑i=1∞‍∑j=1∞‍∑i=1∞‍∑j=1∞‍cijd¯klv~1(i),v~1(k)1v2(j),v2(l)2=∑i=1∞‍∑j=1∞‍cijd¯ij, so the reproducing kernel of W is R(ξ,η){3}(x,y)=Rξ{3}(x)×Rη{3}(y); by the same way, we continue to get W1=W21[0,1]⊗W21[0,1], and the reproducing kernel of W1 is R(ξ,η){1}(x,y)=Rξ{1}(x)×Rη{1}(y).


## 2.3. Analytical Solution and Approximate Solution

On defining the linear operatorL, L-1 is existent: W(D)→W1(D) as [7](9)LTr,z=∂2λlT∂z2+∂2λlT∂r2+1r∂λlT∂r,Fr,z=∂clρlT∂t-∂clρlνzT∂z,and model problem (1) changes to the following problem:(10)LTr,z,t=Fr,z,t.W
(
D
) is a RKHS with the reproducing kernel R(ξ,η){3}(r,z), W1(D) is a RKHS with the reproducing kernelR(ξ,η){1}(r,z). According to [11], it can be noted that ϕi(r,z)=R(ri,zi){1}(r,z), and ψi(r,z)=L∗ϕi(r,z), where L∗ is the adjoint operator of L. The orthonormal system {ψ¯i(r,z)}i=1∞ of W(D) can be derived from the process of Gram-Schmidt orthogonalization of {ψi(r,z)}i=1∞ as ψ¯i(r,z)=∑k=1i‍βikψk(r,z). If {(ri,zi)}i=1∞ is dense in W(D) and βik is the coefficient of orthogonalization, then T(r,z,t)=∑i=1∞‍∑k=1i‍βikF(rk,zk,t)ψ¯i(r,z) is an analytical solution of (10). The approximate solution is Tn(r,z,t)=∑i=1n‍∑k=1i‍βikF(rk,zk,t)ψ¯i(r,z), i=1,…,n. If (10) is nonlinear, the approximate solution to (10) can be expressed by Tn(r,z,t)=∑i=1n‍∑k=1i‍βikF(rk,zk,Tk-1(rk,zk),t)ψ¯i(r,z), i=1,…,n.The convergence analysis, error analysis, and stability have been done by many researchers [6–14], and this paper no longer discusses them.


## 2.4. Numerical Experiment

In order to show RKHSM is effective, a nonlinear equation is used as a test to prove the high accuracy of the method. Select the above conditions and boundary conditions and letλl=1 and D=(0,1)×(0,1); the equation has an exact solution of the form T(r,z,t)=e-t·sin⁡10r·cos⁡10z. By Mathematica 7.0, using our method, we choose the number of nodes as 23 and obtain the approximate solution T23; the comparison of the results with the exact solution is shown in Table 1, and they are found to be in good agreement with each other. Figure 1 is the exact solution T(r,z,1), and Figure 2 is the error (T-T23).Table 1
Numerical results att=1.
(r, z)
Approximate solution
Exact solution
Absolute error
(0.1, 0.1)
0.165583
0.167256
0.001673
(0.2, 0.2)
−0.373791
−0.373791
0.001392
(0.3, 0.3)
−0.050882
−0.051396
0.000514
(0.4, 0.4)
0.180162
0.181982
0.001820
(0.5, 0.5)
−0.099066
−0.100067
0.001001
(0.6, 0.6)
−0.097710
−0.098697
0.000987
(0.7, 0.7)
0.180390
0.182212
0.001822
(0.8, 0.8)
−0.052427
−0.052957
0.000530
(0.9, 0.9)
−0.136755
−0.138136
0.001381
(1.0, 1.0)
0.166248
0.167927
0.001679Figure 1
Exact solutionT(r,z,1).Figure 2
Errors(T-T23).


## 3. Numerical Simulation Results and Discussion

Using the common body structure of vertical wells [5], assuming that the fluid is Newton fluid, the radius of oil well is 0.1m, and the geothermal gradient is 0.025°C/m; ignoring the influence of casing and cement, the initial formation temperature is 15°C. The interlayer between the two injections is 20m. The thermal conductivity of water and oil is 4182 J/kg·K and 2234.2 J/kg·K, the density of water and oil is 1000kg/m3 and 800kg/m3, and the specific heat of water and oil is 0.618 W/m·K and 0.156 W/m·K. The boundary of adiabatic condition is Tz=0=15°C, (∂(λwT)/∂r)r=0=0, the upper boundary of adiabatic condition is (∂(λwT)/∂z)z=1=0, and the under boundary of adiabatic condition is (∂(λwT)/∂z)z=0=0.


### 3.1. Effect of Different Injection Rate

The injection temperature is80°C; temperatures of numerical simulations are given in Figure 1 at t=1h when the injection rate is 0.0059m/s, 0.095m/s, and 0.0368m/s.From Figure3, it can be seen that the temperature of injection layer is more changed than others when νz=0.0059m/s. The temperature in injection layer is obviously changed when the injection rate is small, because the fluid and borehole wall exchange the heat transfer sufficiently in the wellbore.Figure 3
Effect of different injection rates0.0059m/s, 0.095m/s, and 0.0368m/s.


### 3.2. Effect of Different Injection Temperature

When injection rateνz=0.0059m/s, the temperature of numerical simulation when injection temperature is 20°C, 60°C, and 80°C is given in Figure 4 at t=1h.Figure 4
Effect of different injection temperatures20°C, 60°C, and 80°C.Figure4 shows the temperature profile is less obvious when the injection temperature is close to the original formation temperature (42.5°C), and the temperature profile is more obviously changed when the injection temperature is further deviated from the original formation temperature (20°C, 80°C).


### 3.3. Effect of Different Injection Time

When injection rateνz=0.0059m/s, the injection temperature is 80°C, and the temperature of numerical simulation when injection time is 0.1d, 10d, and 100d is given in Figure 5.Figure 5
Effect of different injection times0.1d, 10d, and 100d.Figure5 presents that the temperature of injection layer is less changed compared with others when t=0.1d; this is because the fluid and borehole wall do not exchange the heat transfer sufficiently. As the temperature of fluid and borehole wall becomes consistent when injection time is long, temperature profile can not be distinguished. Therefore, reasonable injection time is good for temperature field of the recovery.


### 3.4. Effect of Different Oil Layer Thickness

When injection rateνz=0.0059m/s, the injection temperature is 80°C, and injection time is t=1h; the temperature of numerical simulation when oil layer thickness is 1m, 6m, and 10m is given in Figure 6.Figure 6
Effect of different oil layer thicknesses1m, 6m, and 10m.In Figure6, when oil layer thickness is 1m, the temperature of injection layer is less changed compared with others. When the oil layer is thick, the temperature is high, because the fluids are fully blended in the entrance of the injection layer.


## 3.1. Effect of Different Injection Rate

The injection temperature is80°C; temperatures of numerical simulations are given in Figure 1 at t=1h when the injection rate is 0.0059m/s, 0.095m/s, and 0.0368m/s.From Figure3, it can be seen that the temperature of injection layer is more changed than others when νz=0.0059m/s. The temperature in injection layer is obviously changed when the injection rate is small, because the fluid and borehole wall exchange the heat transfer sufficiently in the wellbore.Figure 3
Effect of different injection rates0.0059m/s, 0.095m/s, and 0.0368m/s.


## 3.2. Effect of Different Injection Temperature

When injection rateνz=0.0059m/s, the temperature of numerical simulation when injection temperature is 20°C, 60°C, and 80°C is given in Figure 4 at t=1h.Figure 4
Effect of different injection temperatures20°C, 60°C, and 80°C.Figure4 shows the temperature profile is less obvious when the injection temperature is close to the original formation temperature (42.5°C), and the temperature profile is more obviously changed when the injection temperature is further deviated from the original formation temperature (20°C, 80°C).


## 3.3. Effect of Different Injection Time

When injection rateνz=0.0059m/s, the injection temperature is 80°C, and the temperature of numerical simulation when injection time is 0.1d, 10d, and 100d is given in Figure 5.Figure 5
Effect of different injection times0.1d, 10d, and 100d.Figure5 presents that the temperature of injection layer is less changed compared with others when t=0.1d; this is because the fluid and borehole wall do not exchange the heat transfer sufficiently. As the temperature of fluid and borehole wall becomes consistent when injection time is long, temperature profile can not be distinguished. Therefore, reasonable injection time is good for temperature field of the recovery.


## 3.4. Effect of Different Oil Layer Thickness

When injection rateνz=0.0059m/s, the injection temperature is 80°C, and injection time is t=1h; the temperature of numerical simulation when oil layer thickness is 1m, 6m, and 10m is given in Figure 6.Figure 6
Effect of different oil layer thicknesses1m, 6m, and 10m.In Figure6, when oil layer thickness is 1m, the temperature of injection layer is less changed compared with others. When the oil layer is thick, the temperature is high, because the fluids are fully blended in the entrance of the injection layer.


## 4. Conclusions

In this paper, RKHSM is used to solve a class of linear or nonlinear partial differential equations which can describe the temperature field in water injection well; it can be seen that the method is feasible. According to the numerical simulation results in Section3, the following conclusions are given.During the water injection wellbore production, different injection-production conditions make different temperature profile. By analyzing the simulated results we can conclude that temperature logging can distinguish water entry zones from non-water entry zones through hot-water injection for short periods of time, enlarging water injection velocity, selecting reasonable shut-in time, so we can use the short, fast, hot-water injection in order to change the condition that temperature log can not delineate water injection profile.



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*Source: 272856-2015-11-22.xml*