CN104765070A - Method for determining optimum excitation well depth in homogeneous medium - Google Patents

Abstract

The invention relates to a petroleum geophysical exploration technology, and specifically relates to a method for determining the optimum excitation well depth in a homogeneous medium. The method comprises the steps of calculating the thrust F of underground explosion of an explosive according to the performance parameters of the used explosive, calculating the upward thrust F(upward thrust) of explosion of a one-shot combination well according to the principle that the upward thrust of explosion is 1/8 of the thrust of explosion, and calculating the optimum excitation well depth H based on the formula the explosion thrust F(upward thrust)=H*T*Rou*9.8N/kg+F3+F4. The invention provides a method which is used under the condition that there is no obvious ghost reflection interface, high-velocity layer or phreatic surface within 100m near the earth surface and the earth surface is dry loose loess, drift sand or other soil, and the result is very close to that of theoretical calculation.

Description

A kind of method determining best excitation well depth in uniform dielectric

Technical field:

The present invention relates to petroleum exploration technology, is a kind of method determining best excitation well depth in uniform dielectric.

Background technology

The design of the current excitation well depth of oil seismic exploration mainly contains two kinds of approach: (1) carries out well depth test, analyzes, determine excitation well depth to single big gun frequency, energy and the signal to noise ratio (S/N ratio) that well depth is tested; (2) investigated by micro logging, determine the speed of near surface, lithology and the water table degree of depth, under a certain velocity interface or constitute law, 3-5 rice, is defined as shooting depth.

First method is the method that petroleum exploration industry is continued to use for many years always, adopts identical well number with under the prerequisite of identical total dose, is excited by different well depths, contrasts the single big gun data frequency which kind of excitation well depth obtains high, and energy is strong, and signal to noise ratio (S/N ratio) is high.

Second method, must under certain physical interface of well depth design first it is contemplated that the energy excited will pass effectively down, and wherein ghosting interface is the interface of a very useful design excitation well depth.

Next considers velocity interface, worked by near-surface investigation, the near surface structure figure in work area can be drawn, therefore, it is possible to the high-velocity bed buried depth of each sp location is estimated, the excitation well depth of each shot point is instructed to design, the shortcoming done like this is, well depth is very large by the change in depth at high-velocity bed interface, causes excitation well depth to occur violent fluctuating.Also the consistance of excitation wavelet is made to be deteriorated.

Again consider the lithology excited exactly, according to geophysical general knowledge, well depth is darker, and the compaction of rock stratum is more obvious, and the density of country rock increases, and the property of water-bearing increases, and the coupling of the blasting charge and country rock is better; Such exciton wave frequency is higher, and excitation energy is strong.But the saying of this consideration excitation li thology is completely all qualitatively, can accurately not describe lithology is how to affect stimulation effect, especially in the exploratory area (huge thick soil layer/huge thick drift sand formation) that the lithological change of near surface is very faint, lithology does not more have the purposes instructing excitation well depth to design.

Be finally the impact that excitation well depth design will consider water table factor, in the geophysical survey study course of classics, below water table, 3-5m excites and can obtain extraordinary wavelet frequency, can obtain extraordinary excitation energy.Therefore adopting below water table to excite, is also use maximum seismic prospecting excitation well depth designing methods.

The deficiency of first method is that the gatherer process of single big gun of test exists manual operation error, test single shot record carry out frequency, energy and Analysis signal-to-noise ratio (SNR) time because the difference of analysis window size and position, there is greatest differences in analysis result.

The deficiency of second method, the interface of near surface alters a great deal, and for high-velocity bed, from the scope that 10m-100m is even larger, design result substantial deviation theoretical value, causes great difficulty to the excitation well depth design in work area.

But, there is no obvious ghosting interface near surface 100m, there is no high-speed interface, when water table, be dry loose loess or drift sand near surface, in said method, do not propose criterion and the method for determining excitation well depth.

Summary of the invention

The object of the invention is to provide a kind of does not have obvious ghosting interface on earth's surface, does not have high-speed interface, do not have the drying of water table loosen the soil body determination uniform dielectric in the method for best excitation well depth.

The present invention is realized by following steps:

1) according to using the performance parameter of explosive to calculate the thrust F of explosive at underground explosion; 1/8 calculating one big gun combination well blast thrust F upwards of explosive thrust to lifting force according to blast _{lifting force}:

F _{lifting force}=combination well number × F ÷ 8;

It is utilize following kinetic energy formula that the thrust F of described blast calculates:

Wherein, m is quality of explosive, and v is the explosion velocity of explosive, and F is the explosive thrust of a well, and S is half well spacing.

The explosion velocity of described explosive is 5000-6000m/s, and average 5500m/s.

2) best excitation well depth H is calculated with following formula:

Explosive thrust F _{lifting force}=H × T × ρ × 9.8N/kg+F3+F4;

In formula: F _{lifting force}the explosive charge calculated in-step 1) thrust upwards;

H-excitation well depth;

The floor projection area that T-combination well surrounds;

ρ-soil body density;

F ₃-the soil body moves upward the friction force produced with surrounding country rock;

F ₄the horizontal stress that-the soil body is subject to when equilibrium state.

The calculating of described soil body density p is the container using known weight and volume, gathers the primary soil sample of more than 2, utilizes density formula m ₂=ρ v ₂calculate the average density of soil sample, in formula: m ₂the weight of soil sample, v ₂it is the volume of soil sample.

The described soil body moves upward the friction force F produced with surrounding country rock ₃adopt following formulae discovery:

F ₃=all combination wells surround the intergranular free frictional resistance of area × soil body of the soil body and surrounding country rock;

The intergranular free frictional resistance measurement of the described soil body is: the same volume soil body slowly topples over the cone of formation to surface level, angle used when the side of cone and the angle theta of surface level are the decomposition being object gravity G on inclined-plane, the intergranular free frictional resistance fc:fc=G × sin θ of the soil body.

Described same volume is bottom surface circular diameter 6.5cm, the right cylinder of height 7cm.

The horizontal stress F that the described soil body is subject to when equilibrium state ₄adopt following formulae discovery:

$F_4=\frac{(H\×T\×\mathrm{\ρ}\×\frac{9.8N}{\mathrm{kg}})\×\mathrm{\μ}}{1-\mathrm{\μ}}$

Wherein: μ is the Poisson ratio of soil medium.

The invention provides near surface 100m and do not have obvious ghosting interface, do not have high-speed interface, when water table, is the method for best excitation well depth in the soil body determination uniform dielectric such as dry loose loess or drift sand near surface.By one group of excitation well depth test, do energy spectrometer to the single big gun data gathered, the well depth obtained and the result of theory calculate are closely.

Accompanying drawing explanation

Fig. 1 is the volume of the prismatoid that 11 shooting on group wells are formed in underground.

Fig. 2 utilizes loess cone to calculate the intergranular friction force of loess.

Fig. 3 is 1m ²the computation model of the intergranular free friction force of loess on inclined-plane.

Fig. 4 prismatoid moves upward the downward friction force that four sides are subject to.

Fig. 5 is the cylindrical volume that 2 shooting on group wells are formed in underground.

Fig. 6 utilizes drift sand cone to calculate the intergranular friction force of drift sand.

Fig. 7 is 1m ²the computation model of the free friction force between the sand grain of upper reaches, inclined-plane.

Fig. 8 right cylinder moves upward the downward friction force that side is subject to.

Fig. 9 oess hills 11 mouthfuls × (8-21) m × 3kg instigated recordings road average energy spirogram.

Embodiment

The present invention is described in detail below in conjunction with accompanying drawing and experiment embodiment.

The present invention illustrates for oess hills 11 shooting on group factor:

1) with oess hills excitation condition 11 mouthfuls × H × 3kg, well spacing 3m, array pitch 3m are that example calculates, and use the performance parameter of explosive to calculate the thrust F(kN of explosive at underground explosion);

The calculating of a well explosive thrust utilizes kinetic energy formula:

$\frac{1}{2}m{v}^2=\mathrm{FS}$

Wherein, m is quality of explosive, and v is the explosion velocity of explosive, and F is the thrust that a well (3kg explosive) blast produces, and S is half well spacing (explosive is at the burst radius of loess);

The explosion velocity of explosive is 5000-6000m/s, and average 5500m/s.

Calculate the thrust that a well (3kg explosive) blast produces: F=30250kN;

Well (3kg explosive) blast thrust upwards (be 1/8 of whole space thrust, the energy that the sphere that namely centre of sphere angle is 90 degree distributes, remaining energy passes down the energy with side) and be: F/8=3781.25kN;

One big gun 11 combines (33kg explosive) blast thrust upwards: F _{lifting force}=11 × F/8=41593.75kN.

2) best excitation well depth H is calculated with following formula:

Explosive thrust F lifting force=H × T × ρ × 9.8N/kg+F3+F4; (innovative technology)

In formula: the explosive charge thrust upwards calculated in F lifting force-step 1);

H-excitation well depth;

T-11 well combination surround at floor projection area;

ρ-loess density;

F ₃-prismatoid side moves upward the friction force produced with surrounding country rock;

F ₄the horizontal stress that-prismatoid is subject to when equilibrium state.

(1) T described in is that 11 wells combine the floor projection trapezoidal area surrounded, well spacing 3m, array pitch 3m, and a row is 6 wells, and a row is 5 wells.First the surface projection area T of calculation combination well, 5 wells be upper base, be the length of 5 well spacings; 6 wells while for going to the bottom, be the length of 6 well spacings; Trapezoidal height is the length of 2 well spacings; Then surface projection area T is trapezoidal (see figure 1), and T can accurately calculate:

T=(1/2)×（15m+18m）×6m=99m ²。

(2) ρ described in is the density of loess.Gather the primary soil sample (loess) in combination well area, calculate density of soil sample;

Described calculating density of soil sample is the container using known weight and volume, gathers 6 primary soil samples, utilizes density formula (m ₂=ρ v ₂) calculate the average density of soil sample.

In formula: m ₂the weight of soil sample, v ₂be the volume of soil sample, calculating density p is=1.613668g/cm ³;

(3) F described in ₃, be that the soil body (prismatoid) moves upward the friction force produced with surrounding country rock;

F ₃the intergranular free frictional resistance of lateral area × loess of=prismatoid;

1. the intergranular free frictional resistance of loess is measured;

Measuring method is: measure the soil body (drift sand and loess) when same volume, the cone of formation is slowly toppled over to surface level, drift sand and loess centrum have different basal diameter and vertebral height, the different physical phenomenon of this centrum shape, be drift sand and loess ground particle between the performance of free frictional resistance difference in size.

Same volume described in this patent is bottom surface circular diameter 6.5cm, the right cylinder of height 7cm.

Test findings:

Drift sand cone: base diameter 15.2 centimetres, height 4.8cm;

Loess cone: base diameter 13.7 centimetres, height 6cm;

The side of its intermediate cam centrum and the angle of surface level are exactly θ; This angle θ is exactly the angle used when doing the decomposition of object gravity G on inclined-plane.The height of loess cone is 6cm, and bottom surface radius is that 6.85cm(is shown in Fig. 2).

$\sin \mathrm{\θ}=6/\sqrt{6^2+{6.85}^2}=0.66$

This relation represents: at θ=arcsin(0.66), the intergranular friction power of loess reaches balance, and namely θ angle increases again, and loess particle will glide.

2. so, 1m ²on inclined-plane, the intergranular free friction force of loess is (see figure 3):

f _1m2=(1/2)×0.66m×0.75m×1m×1614kg/m ³×9.8N/kg×0.66=2583.74N

3. calculate the soil body (prismatoid) to move upward the friction force F3 produced with surrounding country rock;

F3=f1+f2+f3+f4

F1, f2, f3, f4---be the end with trapezoidal, H is 4 sides of the prismatoid of thickness, produce when moving upward with the friction force (see figure 4) of country rock.

f1=H×18×f _1m2=46.507H(kN)

f2=H×15×f _1m2=38.756H(kN)

f3=H×6.18×f _1m2=15.968H(kN)

f4=H×6.18×f _1m2=15.968H(kN)

F3=117.2H（kN）

4. the horizontal stress F4 that described prismatoid is subject to when equilibrium state adopts following formulae discovery:

The horizontal stress that F4=prismatoid is subject to;

$F_4=\frac{(H\×T\×\mathrm{\ρ}\×\frac{9.8N}{\mathrm{kg}})\×\mathrm{\μ}}{1-\mathrm{\μ}}$

H × T × ρ × 9.8N/kg---be the perpendicular stress that loess is subject to;

H-excitation well depth;

The floor projection area that T-11 combination well surrounds;

ρ-loess density;

μ---be the Poisson ratio of Loess Medium.

Value μ is between 0.40 ~ 0.50 for loess Poisson ratio, and we get intermediate value μ=0.45.

Then: F4=H × 99m ³× 1614kg/m ³× 9.8N/kg × μ/(1-μ)

=（0.45/0.55）×1565.903H

=1281.19H（kN）

5. the best excitation well depth in loess is calculated:

Explosive thrust F lifting force=H × T × ρ × 9.8N/kg+F3+F4;

41593.75kN=H×99m ³×1614kg/m ³×9.8N/kg+117.2H+1281.19H

H=14.031592m

Then " H+ burst radius 1.5m=15.531592m " is exactly that well spacing 3m, array pitch 3m, individual well dose 3kg excites, and when the average explosion velocity of explosive is 5500m/s, loess density is ρ=1.613668g/cm in 11 combinations ³medium in best theoretical well depth.

Another example of the present invention is that uniform saturation water drift sand illustrates near surface:

Drift sand is also approaches uniformity medium common in a kind of seismic prospecting, for saturation water drift sand:

1) excitation condition 2 mouthfuls × H × 6kg that saturation water drift sand is main at present, well spacing 5.4m are that example calculates, and use the performance parameter of explosive to calculate the thrust F(kN of explosive at underground explosion);

The calculating of a well explosive thrust utilizes kinetic energy formula:

$\frac{1}{2}{\mathrm{mv}}^2=\mathrm{FS}$

Wherein, m is quality of explosive, and v is the explosion velocity of explosive, and F is the thrust that a well (6kg explosive) blast produces, and S is half well spacing (the burst radius 2.7m of explosive in saturation water drift sand);

The explosion velocity of explosive is 5000-6000m/s, and average 5500m/s.

Calculate the thrust that a well (6kg explosive) blast produces: F=33611.111kN;

Well (6kg explosive) blast thrust upwards (be 1/8 of whole space thrust, the energy that the sphere that namely centre of sphere angle is 90 degree distributes, remaining energy passes down the energy with side) and be: F/8=4201.389kN;

One big gun 2 combines (12kg explosive) blast thrust upwards: F lifting force=2 × F/8=8402.778kN.

2) best excitation well depth H is calculated with following formula:

Explosive thrust F lifting force=H × T × ρ × 9.8N/kg+F3+F4; (innovative technology)

In formula: the explosive charge thrust upwards calculated in F lifting force-step 1);

H-excitation well depth;

T-2 well combines the circle that surrounds at floor projection area;

ρ-saturation water drift sand density;

F ₃-cylindrical sides moves upward the friction force produced with surrounding country rock;

F ₄the horizontal stress that-right cylinder is subject to when equilibrium state.

(1) T described in is for 2 combination wells (well spacing 5.4m), one row totally 2 wells, each well is that (because well spacing is comparatively large, well number is few for circle in the projected area on earth's surface, rectangle is thought so can not be similar to), T is the (see figure 5) that can accurately calculate:

T=2×π×2.7m×2.7m=45.78m ²。

(2) ρ described in is the density of saturation water drift sand.Gather the primary rock sample (saturation water drift sand) in combination well area, calculate drift sand density;

Described calculating drift sand density is the container using known weight and volume, gathers the primary drift sand sample of more than 6, utilizes density formula (m ₂=ρ v ₂) calculate the average density of saturation water drift sand.

In formula: m ₂the weight of saturation water drift sand, v ₂be the volume of drift sand, calculating saturation water drift sand density p is=1.890056g/cm ³;

(3) F described in ₃that right cylinder moves upward the friction force produced with surrounding country rock;

F ₃=cylindrical sides amasss the × intergranular free frictional resistance of drift sand;

1. the intergranular free frictional resistance of drift sand is measured;

Measuring method is: measure the soil body (drift sand and loess) when same volume, the cone of formation is slowly toppled over to surface level, drift sand and loess centrum have different basal diameter and vertebral height, the different physical phenomenon of this centrum shape, be drift sand and loess ground particle between the performance of free frictional resistance difference in size.

Same volume described in this patent is bottom surface circular diameter 6.5cm, the right cylinder of height 7cm.

Test findings:

Loess cone: base diameter 13.7 centimetres, height 6cm;

Drift sand cone: base diameter 15.2 centimetres, height 4.8cm;

The side of its intermediate cam centrum and the angle of surface level are exactly θ; This angle θ is exactly the angle used when doing the decomposition of object gravity G on inclined-plane.The height of drift sand cone is 4.8cm, and bottom surface radius is that 7.6cm(is shown in Fig. 6).

$\sin \mathrm{\θ}=\frac{4.8}{\sqrt{{4.8}^2+{7.6}^2}}=0.53$

This relation represents: at θ=arcsin(0.53), the intergranular friction power of drift sand reaches balance, and namely θ angle increases again, and drift sand particle will glide.

2. so, 1m ²free friction force between the sand grain of upper reaches, inclined-plane is (see figure 7):

f1m2=(1/2)×0.53m×0.85m×1m×1890kg/m3×9.8N/kg×0.53=2211.2N

3. calculate drift sand (right cylinder) to move upward the friction force F3 produced with surrounding country rock;

F3=f side

F side---take circle the end of as, H is the cylindrical side of thickness, produce when moving upward with the friction force (see figure 8) of country rock.

F3=f side=H × 4 × π × 2.7 × f1m2=75.024H (kN)

F3=75.02H（kN）

4. the horizontal stress (known technology-rock-soil mechanics) that is subject to when equilibrium state of right cylinder

The horizontal stress that F4=right cylinder drift sand is subject to;

$F_4=\frac{(H\×T\×\mathrm{\ρ}\×\frac{9.8N}{\mathrm{kg}})\×\mathrm{\μ}}{1-\mathrm{\μ}}$

H × T × ρ × 9.8N/kg---be the perpendicular stress that saturation water drift sand is subject to;

H-excitation well depth;

The floor projection circular area that T-2 combination well surrounds;

ρ-saturation water drift sand density;

μ---be the Poisson ratio of drift sand medium.

Drift sand Poisson ratio value μ is (less than the Poisson ratio 0.30-0.35 of sand and the Poisson ratio 0.35-0.40 of silt) between 0.20 ~ 0.30, gets intermediate value μ=0.25.

Then: F4=H × 45.78m3 × 1890kg/m ³× 9.8N/kg × μ/(1-μ)

=（0.25/0.75）×847.937H

=282.646H（kN）：

Explosive thrust F lifting force=H × T × ρ × 9.8N/kg+F3+F4;

8402.778kN=H×45.78m ²×1890kg/m ³×9.8N/kg+75.02H+282.646H

H=6.969777m

Then " H+ burst radius 2.7m=9.669777m " is exactly in 2 combinations, well spacing 5.4m, and individual well dose 6kg excites, and when the average explosion velocity of explosive is 5500m/s, saturation water drift sand density is theoretical well depth best in the medium of ρ=1.8900g/cm3.

In order to the reliability that argumentation theory calculates, in loess, arranged the test of one group of excitation well depth, adopted 11 mouthfuls of well shootings on group, well depth from 8-21m, individual well dose 3kg.Do energy spectrometer to the single big gun data gathered, the conclusion obtained is between well depth 14-16m, and the energy passed down is obviously strengthened, with the result closely (see figure 9) of theory calculate.

Claims (8)

1. determine a method for best excitation well depth in uniform dielectric, feature adopts following steps to realize:

1) according to using the performance parameter of explosive to calculate the thrust F of explosive at underground explosion; 1/8 calculating one big gun combination well blast thrust F upwards of explosive thrust to lifting force according to blast _{lifting force}:

F _{lifting force}=combination well number × F ÷ 8;

2) best excitation well depth H is calculated with following formula:

Explosive thrust F _{lifting force}=H × T × ρ × 9.8N/kg+F3+F4;

In formula: F _{lifting force}the explosive charge calculated in-step 1) thrust upwards;

H-excitation well depth;

The floor projection area that T-combination well surrounds;

ρ-soil body density;

F ₃-the soil body moves upward the friction force produced with surrounding country rock;

F ₄the horizontal stress that-the soil body is subject to when equilibrium state.

2. method according to claim 1, feature is that the thrust F of blast described in step 1) calculates is utilize following kinetic energy formula:

Wherein, m is quality of explosive, and v is the explosion velocity of explosive, and F is the explosive thrust of a well, and S is half well spacing.

3. method according to claim 2, feature is the explosion velocity of described explosive is 5000-6000m/s, and average 5500m/s.

4. method according to claim 1, feature is step 2) described in the calculating of soil body density p be the container using known weight and volume, gather the primary soil sample of more than 2, utilize density formula m ₂=ρ v ₂calculate the average density of soil sample, in formula: m ₂the weight of soil sample, v ₂it is the volume of soil sample.

5. method according to claim 1, feature is step 2) described in the soil body to move upward the friction force F produced with surrounding country rock ₃adopt following formulae discovery:

F ₃=all combination wells surround the intergranular free frictional resistance of area × soil body of the soil body and surrounding country rock.

6. method according to claim 1, feature is step 2) described in the intergranular free frictional resistance measurement of the soil body be: the same volume soil body slowly topples over the cone of formation to surface level, angle used when the side of cone and the angle theta of surface level are the decomposition being object gravity G on inclined-plane, the intergranular free frictional resistance fc:fc=G × sin θ of the soil body.

7. method according to claim 6, feature is described same volume is bottom surface circular diameter 6.5cm, the right cylinder of height 7cm.

8. method according to claim 1, feature is step 2) described in the horizontal stress F that is subject to when equilibrium state of the soil body ₄adopt following formulae discovery:

$F_4=\frac{(H\×T\×\mathrm{\ρ}\×\frac{9.8N}{\mathrm{kg}})\×\mathrm{\μ}}{1-\mathrm{\μ}}$

Wherein: μ is the Poisson ratio of soil medium.