CN104989351A - Dryness, temperature and pressure coupling predicting method in oil-gas well gas injection process - Google Patents

Abstract

The invention belongs to the technical field of oil-gas reservoir development engineering management, and relates to a dryness, temperature and pressure coupling predicting method in an oil-gas well gas injection process, in particular to the coupling analysis of dryness and pressure of oil-gas two-phase instantaneous flow and shaft-stratum unsteady heat conduction, the mathematical modeling and the numerical simulation so as to predict dryness, temperature and pressure distribution situations of an ultra-deep well. The coupling predicting method is used for solving the problem that in a traditional scheme, dryness, temperature and pressure are not accurately predicted. The method includes the implementation steps of A, establishing a stratum heat conduction model; B, establishing a differential equation coupling model; C, solving the differential equation coupling model. The method is suitable for oil-gas reservoir development.

Description

Mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process

Technical field

The invention belongs to development of oil and gas reservoir administrative skill field, it relates to mass dryness fraction in Oil/gas Well gas injection process, temperature and coupling pressure Forecasting Methodology, be specifically related in HTHP Oil/gas Well gas injection process, the mass dryness fraction of oil gas two-phase transient state stream, pressure and pit shaft-stratum Geometry symmetry mode coupling analysis, mathematical modeling and numerical simulation, prediction ultradeep well mass dryness fraction, temperature, pressure distribution situation.

Background technology

In the gas injection operating mode of Oil-Gas Well Engineering, conventional method considers micro unit body conservation of mechanical energy and change of internal energy usually, but often ignore the interchange of heat with external environment condition, think that the change of internal steam work equals the change of outside, and also have ignored the friction loss of the steam of gas injection process, then calculate wellbore pressure and temperature, thus cause predicting extremely inaccurate to mass dryness fraction, temperature, pressure distribution in pit shaft.

Usually have ignored when building energy-balance equation the energy loss that frictional resistance causes and think that formation temperature is constant, just be subject to the impact of the degree of depth and geothermal gradient, the formation temperature that the heat trnasfer that also have ignored internal steam work change and the surrounding environment produced by frictional resistance causes can because of the change of conducting heat between pit shaft and stratum.The unsteady-state heat transfer on the friction loss in gas injection process and pit shaft-stratum is the key factor building differential equation coupling system model, gas-bearing formation infringement or improve the important parameter of degree, significant for mass dryness fraction in accurately predicting pit shaft, temperature, pressure distribution.

Oil/gas Well thermodynamic analysis: before Modling model, the hypothesis about steam flow and heat transfer is as follows:

(1) fluid physical property and formed by the impact of the degree of depth and temperature.

(2) all parameters (speed, pressure, temperature and aridity) of steam are constants.

(3) be stable from pipeline to the heat transfer of the second contact surface, from the second contact surface to the heat transfer on stratum be non-stable.

(4) vapor stream is regarded as one dimension two-phase homogeneous fluid.

(5) physical property (steam and some heat-barrier materials except) of other materials is not by the impact of time and temperature.

(6) there is a kind of linear relationship between the thermal conductivity of heat-barrier material and temperature.

(7) geothermal gradient is constant.

The hypothesis of formation temperature field is as follows:

(1) temperature in stratum around pit shaft distributes axisymmetricly.

(2) when injecting steam toward pit shaft, the formation temperature away from pit shaft axle center is identical.

(3) formation temperature field does not have inner-growth model and is transient state conduction.

Barometric gradient: because steam injection is constant concrete mass flow, that is: flow into micro unit quality and equal to flow out micro unit quality (see Fig. 1), it follows mass-conservation equation:

M＝ρ ₁υ ₁A＝ρ ₂υ ₂A＝ρ _mυ _mA (1)

It is followed in a period of time dt, the momentum that micro-body is subject to is Fdt=P ₁adt+ ρ _mag cos θ dzdt-P ₂adt-τ _fdt, in dt during this period of time, the change of momentum is ρ ₂υ ₂adt υ ₂-ρ ₁υ ₁adt υ ₁=Δ (m υ).It follows momentum theorem:

$\frac{\mathrm{dP}}{\mathrm{dz}}=\mathrm{\ρ}\mathrm{mg}\cos \mathrm{\θ}-{\mathrm{\ρ}}_m{\mathrm{\υ}}_m\frac{d{\mathrm{\υ}}_m}{\mathrm{dz}}-f_m\frac{{\mathrm{\ρ}}_m{\mathrm{\υ}}_m^2}{4r_{\mathrm{ti}}}---\left(2\right)$

Gas mass dryness fraction: in view of the energy loss caused by vapor stream frictional resistance, we draw following energy conservation equation:

$\frac{\mathrm{dQ}}{\mathrm{dz}}+\frac{\mathrm{dW}}{\mathrm{dz}}=-M\frac{d{H}_m}{\mathrm{dz}}-M\frac{d{\mathrm{\υ}}_m}{\mathrm{dz}}+\mathrm{Mg}\cos \mathrm{\θ}---\left(3\right)$

In this equation, dW represents that, because of the energy loss that frictional resistance causes between steam and tube wall, Hm represents the enthalpy of mixed flow, and it can be defined as:

H _m＝H _sX+H _w(1-X) (4)

In this equation, Hs represents the enthalpy of saturated vapour, and Hw represents saturation water, can be derived by equation (4):

$\frac{d{H}_m}{\mathrm{dz}}=(\frac{d{H}_s}{\mathrm{dz}}-\frac{d{H}_w}{\mathrm{dz}})X+(H_s-H_w)\frac{\mathrm{dX}}{\mathrm{dz}}+\frac{d{H}_w}{\mathrm{dz}}$

Because $\frac{d{H}_s}{\mathrm{dz}}=C_{\mathrm{Ps}}\frac{\mathrm{dT}}{\mathrm{dz}}-C_{\mathrm{Js}}{C}_{\mathrm{Ps}}\frac{\mathrm{dP}}{\mathrm{dz}}$ And $\frac{d{H}_w}{\mathrm{dz}}=C_{\mathrm{Pw}}\frac{\mathrm{dT}}{\mathrm{dz}}-C_{\mathrm{Jw}}{C}_{\mathrm{Pw}}\frac{\mathrm{dP}}{\mathrm{dz}},$ So can derive:

$\begin{array}{c}\frac{d{H}_m}{\mathrm{dz}}=[(C_{\mathrm{Ps}}-C_{\mathrm{Pw}})\frac{\mathrm{dT}}{\mathrm{dz}}-(C_{\mathrm{Js}}{C}_{\mathrm{Ps}}-C_{\mathrm{Jw}}{C}_{\mathrm{Pw}})\frac{\mathrm{dP}}{\mathrm{dz}}]X\\ +(H_s-H_w)\frac{\mathrm{dX}}{\mathrm{dz}}+C_{\mathrm{Pw}}\frac{\mathrm{dT}}{\mathrm{dz}}-C_{\mathrm{Jw}}{C}_{\mathrm{Pw}}\frac{\mathrm{dP}}{\mathrm{dz}}\end{array}$

In addition, the rate equation of mixed flow is:

${\mathrm{\υ}}_m={\mathrm{\υ}}_s+{\mathrm{\υ}}_m=\frac{\mathrm{MX}}{{\mathrm{\ρ}}_{s}A}+\frac{M(1-X)}{{\mathrm{\ρ}}_{w}A}---\left(5\right)$

So then have,

$\frac{d{\mathrm{\υ}}_m}{\mathrm{dz}}=R\frac{\mathrm{dX}}{\mathrm{dz}}-S\frac{\mathrm{dP}}{\mathrm{dz}}---\left(6\right)$

In this equation, $R=\frac{M}{A}(\frac{1}{\mathrm{\ρs}}-\frac{q}{\mathrm{\ρw}}),S=\frac{M}{A}(\frac{X}{{\mathrm{\ρ}}_S^2}\frac{\mathrm{d\ρs}}{\mathrm{dP}}+\frac{1-X}{{\mathrm{\ρ}}_W^2}\frac{\mathrm{d\ρw}}{\mathrm{dP}}).$ Because the direction of steam flow and frictional force is contrary, so the energy loss dW caused because of frictional resistance between steam and tube wall is negative value.In unit interval, the frictional force of dz is:

$\mathrm{dW}=\frac{{\mathrm{\τ}}_f\mathrm{dz}}{\mathrm{dt}}\frac{{\mathrm{\τ}}_f\mathrm{dz}}{\mathrm{dz}/{\mathrm{\υ}}_m}{\mathrm{\τ}}_f{\mathrm{\υ}}_m---\left(7\right)$

Equation (1) gives the definition of quality, thus we obtain the aridity equation of following gas model:

$A_1\frac{\mathrm{dX}}{\mathrm{dz}}+A_{2}X+A_3=0---\left(8\right)$

In this equation,

$A_1=H_s-H_w+R{\mathrm{\υ}}_m,A_2=(C_{P_S}-C_{P_W})\frac{\mathrm{dT}}{\mathrm{dz}}-(C_{J_S}{C}_{P_S}-C_{J_w}{C}_{P_W})\frac{\mathrm{dP}}{\mathrm{dz}},$

$A_3=C_{P_W}\frac{\mathrm{dT}}{\mathrm{dz}}-(C_{J_w}{C}_{P_W}+{\mathrm{\υ}}_{m}S)\frac{\mathrm{dP}}{\mathrm{dz}}-g\cos \mathrm{\θ}+\frac{1}{M}(\frac{\mathrm{dQ}}{\mathrm{dz}}+\frac{{\mathrm{\τ}}_f{\mathrm{\υ}}_m}{\mathrm{dz}})$

Wellbore heat: note, assuming that from pipeline to the heat transfer of the second contact surface be stable (see Fig. 2).It is followed

$\frac{\mathrm{dZ}}{\mathrm{dz}}=\mathrm{\π}{D}_{\mathrm{to}}{U}_{\mathrm{to}}(T-T_{\mathrm{ref}})---\left(9\right)$

Second contact surface to the radial direction heat transfer of surrounding formation is:

$\frac{\mathrm{dQ}}{\mathrm{dz}}=\frac{\mathrm{\π}{K}_e}{f\left(t_D\right)}(T_{\mathrm{ref}}-T_e)---\left(10\right)$

In conjunction with equation (9) and (10), thus draw the heat transfer model amount between vapor stream and surrounding formation, as follows:

$\frac{\mathrm{dQ}}{\mathrm{dz}}=\frac{\mathrm{\π}{D}_{\mathrm{to}}{U}_{\mathrm{to}}{K}_e}{0.5D_{\mathrm{to}}{U}_{\mathrm{to}}f\left(t_D\right)+K_e}(T-T_e)---\left(11\right)$

If α=π is D _tou _tok _e/ (0.5D _tou _tof (t _d)+K _e), then

Summary of the invention

Technical problem to be solved by this invention is: propose mass dryness fraction, temperature and coupling pressure Forecasting Methodology in a kind of Oil/gas Well gas injection process, solves in traditional scheme the inaccurate problem of mass dryness fraction, temperature and pressure prediction.

The technical solution adopted for the present invention to solve the technical problems is: mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process, comprising:

A. stratum heat transfer model is built;

B. differential equation coupling model is built;

C. differential equation coupling model is solved.

Further, based on from the second contact surface to the hypothesis of stratum unsteady heat transfer in steps A, build stratum heat transfer model, specifically comprise:

$\frac{\∂T_e}{\∂t}=\frac{{\mathrm{\λ}}_e}{C_{P_e}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{\∂r^2}+\frac{1}{r}\frac{\∂T_e}{\∂r})---\left(12\right)$

Primary condition:

T _e=T _α+ γ ^zif, t=0.

Fringe conditions:

$\frac{\∂T_e}{\∂r}=0,\mathrm{if\; r}\→\∞$

$\mathrm{dQ}=-2\mathrm{\π}{r}_{\mathrm{cem}}{\mathrm{\λ}}_{\mathrm{cem}}\frac{\∂T_e}{\∂r}\mathrm{dz}{|}_{r=r_{\mathrm{cem}}}$

Nondimensional number with substitute into equation (12),

$\frac{\∂T_e}{\∂t_D}=\frac{{\mathrm{\λ}}_e}{C_{P_e}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{\∂r_D^2}+\frac{1}{r_D}\frac{\∂T_e}{\∂r_D})---\left(13\right)$

Fringe conditions is converted into:

$\begin{array}{c}\frac{{\∂}^2{T}_e}{\∂r_D}{|}_{r_D=1}=-\frac{\mathrm{dQ}}{\mathrm{dz}}{\left(2\mathrm{\π}{\mathrm{\λ}}_e\right)}^{-1}\\ \frac{{\∂}^2{T}_e}{\∂r_D}{|}_{r_D\→\∞}=0\end{array}---\left(14\right)$

Relation between saturated saturated-steam temperature and pressure:

$\frac{\mathrm{dT}}{\mathrm{dz}}=44.15P^{-0.79}\frac{\mathrm{dP}}{\mathrm{dz}}---\left(15\right).$

Further, in step B, described differential equation coupling model considers friction loss and pit shaft-stratum unsteady-state heat transfer, and the differential equation coupling model of structure is:

$\left\{\begin{array}{c}\frac{\mathrm{dP}}{\mathrm{dz}}={\mathrm{\ρ}}_{m}g\cos \mathrm{\θ}-p_m{\mathrm{\υ}}_m(R\frac{\mathrm{dX}}{\mathrm{dz}}-S\frac{\mathrm{dP}}{\mathrm{dz}})-f_m\frac{p_m{\mathrm{\υ}}_m^2}{4r_{\mathrm{ti}}}\\ A_1\frac{\mathrm{dX}}{\mathrm{dz}}+A_{2}X+A_3=0\\ \frac{\mathrm{dT}}{\mathrm{dz}}=44.15P^{-0.79}\frac{\mathrm{dP}}{\mathrm{dz}}\\ \frac{\mathrm{dQ}}{\mathrm{dz}}=\mathrm{\α}(T-T_e)\\ \frac{\∂T_e}{{\∂t}_D}=\frac{{\mathrm{\λ}}_e}{C_{P_e}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{\∂r_D^2}+\frac{1}{r_D}\frac{\∂T_e}{\∂r_D})\end{array}\right.---\left(16\right).$

Further, in step C, before solving model, the parameter in model is handled as follows:

1) vapor (steam) velocity V _m:

${\mathrm{\υ}}_m={\mathrm{\υ}}_s+{\mathrm{\υ}}_w=\frac{\mathrm{MX}}{{\mathrm{\ρ}}_{s}A}+\frac{M(1-X)}{{\mathrm{\ρ}}_{w}A}---\left(17\right);$

2) vapour density Pm: because water vapour air-flow is biphase gas and liquid flow, therefore applies the averag density that Beggs-Brill method calculates mixture;

3) frictional force T _f:

${\mathrm{\τ}}_f=\frac{1}{4}\mathrm{\π}{f}_m{r}_{\mathrm{ti}}{\mathrm{\ρ}}_m{\mathrm{\υ}}_m^2\mathrm{dz}---\left(18\right);$

4) the friction factor f of gas-liquid mixture _m, f _mthe function about Reynolds number and absolute roughness e,

$f_m=\left\{\begin{array}{cc}\frac{\mathrm{Re}}{64}& \mathrm{if\; Re}\≤2000\\ {[1.14-2\ln (\frac{\mathrm{\ϵ}}{2r_{\mathrm{ti}}}+21.25{\mathrm{Re}}^{-0.9})]}^{-2}& \mathrm{if\; Re}>2000\end{array}\right.---\left(19\right);$

5) thermal transmittance U _to:

$\frac{1}{U_{\mathrm{to}}}=r_{\mathrm{ti}}\·\frac{1}{{\mathrm{\λ}}_{\mathrm{ins}}}\ln \left(\frac{r_{\mathrm{ci}}}{r_{\mathrm{to}}}\right)+\frac{1}{h_c+h_r}+r_{\mathrm{ti}}\·\frac{1}{{\mathrm{\λ}}_{\mathrm{cem}}}\ln \left(\frac{r_{\mathrm{cem}}}{r_{\mathrm{co}}}\right)---\left(20\right)$

Wherein, λ _insand λ _cembe respectively the thermal conductivity of thermal insulation material and cement sheath.Hc and hr is respectively convective heat-transfer coefficient and radiation heat transfer coefficient;

6) nondimensional time function:

$f\left(t_D\right)=\left\{\begin{array}{cc}1.1281\sqrt{t_D}(1-0.3\sqrt{t_D})& \mathrm{if}{t}_D\≤1.5\\ (0.5\ln \left(t_D\right)+0.4063)(1+\frac{0.6}{t_D})& \mathrm{if}{t}_D>1.5\end{array}\right.---\left(21\right).$

Further, in step C, adopt interactive quadravalence Runge-Kutta finite difference method to solve differential equation coupling model, comprising:

Step 1. provides θ respectively ₀, T ₀, P ₀, X ₀, Q ₀and T _e0initial value;

Step 2. calculates all coefficients in coupling model;

Step 3. makes the differential method be converted into function f _i(i=1,2,3,4), can obtain following coupling function system:

$\left\{\begin{array}{c}{f}_1={\mathrm{\ρ}}_{m}g\cos \mathrm{\θ}-{\mathrm{\ρ}}_m{\mathrm{\υ}}_m(R{f}_2-S{f}_1)-f_m\frac{{\mathrm{\ρ}}_m{\mathrm{\υ}}_m^2}{4r_{\mathrm{ti}}}\\ A_1{f}_2+A_{2}X+A_3=0\\ f_3=\mathrm{\α}(T-T_e)\\ f_4=44.15P^{-0.79}{f}_1\end{array}\right.---\left(22\right)$

The equation group that step 4. solves above obtains P _k, X _k, Q _k, T _k, and T _ek, can coefficient be drawn:

a _i＝f _i(P _k，X _k，Q _k，T _k)，j＝1，2，3，4；

Step 5. makes te is obtained by separating following equation,

$\left\{\begin{array}{c}\frac{\∂T_e}{\∂t_D}-\frac{{\mathrm{\λ}}_e}{C_{P_e{\mathrm{\ρ}}_e}}\frac{{\∂}^2{T}_e}{\∂r_D^2}-\frac{{\mathrm{\λ}}_e}{r_D{C}_{P_e{\mathrm{\ρ}}_e}}\frac{\∂T_e}{\∂r_D}=0\\ T_e{|}_{t_D=0}=T_0+\mathrm{\γz}\\ \frac{\∂T_e}{\∂r_{\mathrm{De}}}{|}_{r_D=1}=-\mathrm{\α}(T-T_e){\left(2\mathrm{\π}{\mathrm{\λ}}_e\right)}^{-1}\\ \frac{\∂T_e}{\∂r_{\mathrm{De}}}{|}_{r_D\→\∞}=0\end{array}\right.---\left(23\right)$

As time j, radial i when degree of depth z, T ^j _irepresent temperature, i=1; 2 ... M, j=1,2 ... N; Wherein M and N represents time and last radial node respectively; Finite difference method is adopted to carry out discrete obtaining to equation (23):

In this equation, τ represents the time interval, and ζ represents spaced radial; It can convert canonical form to, as follows:

Then calculus of finite differences discrete boundary condition is used; If rD=1, then:

$\frac{\∂T_e}{\∂r_{\mathrm{De}}}{|}_{r_D=1}-\frac{\mathrm{\α}}{2\mathrm{\π}{\mathrm{\λ}}_e}{T}_e{|}_{r_D=1}=-\frac{\mathrm{\αT}}{2\mathrm{\π}{\mathrm{\λ}}_e}---\left(26\right)$

It meets

If rD=N, then:

$T_N^{i+1}-T_{N-1}^{i+1}=0---\left(28\right)$

In conjunction with equation (25), (27) and (28), the numerical solution of formation temperature Te can be calculated;

Step 6., as rD=1, substitutes into equation system (23) Te, can obtain:

$b_j=f_j(P_k+\frac{h{a}_1}{2},X_k+\frac{h{a}_2}{2},Q_k+\frac{h{a}_3}{2},T_k+\frac{h{a}_4}{2})$

Step 7. calculates $c_j=f_j(P_k+\frac{h{b}_1}{2},X_k+\frac{h{b}_2}{2},Q_k+\frac{h{b}_3}{2},T_k+\frac{h{b}_3}{2})$ With

d _j＝f _j(P _k+hc ₁，X _k+hc ₂，Q _k+hc ₃，T _k+hc ₄)(j＝1，2，3，4)

Step 8. calculates gas-liquid mixture in (K+1) this point, and mass dryness fraction, the pressure and temperature of gas can obtain:

$P_{k+1}=P_k+\frac{h(a_1+2b_1+2c_1+d_1)}{6},X_{k+1}=X_k+\frac{h(a_2+2b_2+2c_2+d_2)}{6},$

$Q_{k+1}=Q_k+\frac{h(a_3+2b_3+2c_3+d_3)}{6},T_{k+1}=T_k+\frac{h(a_4+2b_4+2c_4+d_4)}{6}.$

Step 9. is T=T _k+1substitute into the fringe conditions of equation (23), can T be obtained by finite difference method _e, _k+1;

Step 10. repeats step 2-9, until calculate P _nx _nq _n, T _nand T _en.

The invention has the beneficial effects as follows: when setting up coupling model, considering formation temperature that the internal steam work change that produced by frictional resistance and the heat trnasfer of surrounding environment cause can because of the change of conducting heat between pit shaft and stratum, thus accurately predicting ultradeep well mass dryness fraction, temperature, pressure distribution situation.

Accompanying drawing explanation

Fig. 1 is the force analysis figure of micro unit;

Fig. 2 is shaft structure schematic diagram;

Fig. 3 is Forecasting Methodology flow chart of the present invention;

Fig. 4 is the change curve of pressure with injection length of steam;

Fig. 5 is the change curve of mass dryness fraction with injection length of steam;

Fig. 6 is the change curve of the temperature in well with injection length;

Fig. 7 is the change curve of near-bottom temperature with injection length.

Detailed description of the invention

The present invention is intended to propose mass dryness fraction, temperature and coupling pressure Forecasting Methodology in a kind of Oil/gas Well gas injection process, solves in traditional scheme the inaccurate problem of mass dryness fraction, temperature and pressure prediction.

As shown in Figure 3, this Forecasting Methodology comprises:

A. stratum heat transfer model is built;

B. differential equation coupling model is built;

C. differential equation coupling model is solved.

Be described in detail for each step from concrete enforcement below:

Conduct heat in stratum: according to from the second contact surface to the hypothesis of stratum unsteady heat transfer, and we draw with sub-surface heat transfer model:

$\frac{\∂T_e}{\∂t}=\frac{{\mathrm{\λ}}_e}{C_{P_e}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{\∂r^2}+\frac{1}{r}\frac{\∂T_e}{\∂r})---\left(12\right)$

Primary condition:

T _e=T _α+ γ ^zif, t=0.

Fringe conditions:

$\frac{\∂T_e}{\∂r}=0,\mathrm{if\; r}\→\∞$

$\mathrm{dQ}=-2\mathrm{\π}{r}_{\mathrm{cem}}{\mathrm{\λ}}_{\mathrm{cem}}\frac{\∂T_e}{\∂r}\mathrm{dz}{|}_{r=r_{\mathrm{cem}}}$

Nondimensional number with substitute into equation (12),

$\frac{\∂T_e}{\∂t_D}=\frac{{\mathrm{\λ}}_e}{C_{P_e}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{\∂r_D^2}+\frac{1}{r_D}\frac{\∂T_e}{\∂r_D})---\left(13\right)$

Fringe conditions is converted into:

$\begin{array}{c}\frac{{\∂}^2{T}_e}{\∂r_D}{|}_{r_D=1}=-\frac{\mathrm{dQ}}{\mathrm{dz}}{\left(2\mathrm{\π}{\mathrm{\λ}}_e\right)}^{-1}\\ \frac{{\∂}^2{T}_e}{\∂r_D}{|}_{r_D\→\∞}=0\end{array}---\left(14\right)$

In addition, Ni etc. (2005) propose the relation between saturated-steam temperature and pressure:

$\frac{\mathrm{dT}}{\mathrm{dz}}=44.15P^{-0.79}\frac{\mathrm{dP}}{\mathrm{dz}}---\left(15\right)$

Derivative scalar coupling builds: the present invention is that in prediction HTHP Oil/gas Well gas injection process, mass dryness fraction, temperature, pressure distribution propose a kind of differential equation coupling system model considering friction loss and pit shaft-stratum unsteady-state heat transfer:

$\left\{\begin{array}{c}\frac{\mathrm{dP}}{\mathrm{dz}}={\mathrm{\ρ}}_{m}g\cos \mathrm{\θ}-p_m{\mathrm{\υ}}_m(R\frac{\mathrm{dX}}{\mathrm{dz}}-S\frac{\mathrm{dP}}{\mathrm{dz}})-f_m\frac{p_m{\mathrm{\υ}}_m^2}{4r_{\mathrm{ti}}}\\ A_1\frac{\mathrm{dX}}{\mathrm{dz}}+A_{2}X+A_3=0\\ \frac{\mathrm{dT}}{\mathrm{dz}}=44.15P^{-0.79}\frac{\mathrm{dP}}{\mathrm{dz}}\\ \frac{\mathrm{dQ}}{\mathrm{dz}}=\mathrm{\α}(T-T_e)\\ \frac{\∂T_e}{{\∂t}_D}=\frac{{\mathrm{\λ}}_e}{C_{P_e}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{\∂r_D^2}+\frac{1}{r_D}\frac{\∂T_e}{\∂r_D})\end{array}\right.---\left(16\right)$

Interactive quadravalence Runge-Kutta finite difference method: before proposition solving model, partial parameters is handled as follows:

(1) vapor (steam) velocity V _m.

${\mathrm{\υ}}_m={\mathrm{\υ}}_s+{\mathrm{\υ}}_w=\frac{\mathrm{MX}}{{\mathrm{\ρ}}_{s}A}+\frac{M(1-X)}{{\mathrm{\ρ}}_{w}A}---\left(17\right)$

(2) vapour density Pm. is because water vapour air-flow is biphase gas and liquid flow, and we apply the averag density that Beggs-Brill (Beggs and Brill, 1973) method calculates mixture.

(3) frictional force T _f.It can be calculated by following equations:

${\mathrm{\τ}}_f=\frac{1}{4}\mathrm{\π}{f}_m{r}_{\mathrm{ti}}{\mathrm{\ρ}}_m{\mathrm{\υ}}_m^2\mathrm{dz}---\left(18\right)$

(4) the friction factor f of gas-liquid mixture _m.F _mthe function about Reynolds number and absolute roughness e,

$f_m=\left\{\begin{array}{cc}\frac{\mathrm{Re}}{64}& \mathrm{if\; Re}\≤2000\\ {[1.14-2\ln (\frac{\mathrm{\ϵ}}{2r_{\mathrm{ti}}}+21.25{\mathrm{Re}}^{-0.9})]}^{-2}& \mathrm{if\; Re}>2000\end{array}\right.---\left(19\right)$

(5) thermal transmittance U _to(from the diverse location of pit shaft axle to the second contact surface).

$\frac{1}{U_{\mathrm{to}}}=r_{\mathrm{ti}}\·\frac{1}{{\mathrm{\λ}}_{\mathrm{ins}}}\ln \left(\frac{r_{\mathrm{ci}}}{r_{\mathrm{to}}}\right)+\frac{1}{h_c+h_r}+r_{\mathrm{ti}}\·\frac{1}{{\mathrm{\λ}}_{\mathrm{cem}}}\ln \left(\frac{r_{\mathrm{cem}}}{r_{\mathrm{co}}}\right)---\left(20\right)$

The right half part of equation (20) is the thermal resistance of insulation tube, is respectively target chamber and cement sheath.λ _insand λ _cembe respectively the thermal conductivity of thermal insulation material and cement sheath.Hc and hr is respectively convective heat-transfer coefficient and radiation heat transfer coefficient.

(6) nondimensional time function.It can be calculated by equation (21).

$f\left(t_D\right)=\left\{\begin{array}{cc}1.1281\sqrt{t_D}(1-0.3\sqrt{t_D})& \mathrm{if}{t}_D\≤1.5\\ (0.5\ln \left(t_D\right)+0.4063)(1+\frac{0.6}{t_D})& \mathrm{if}{t}_D>1.5\end{array}\right.---\left(21\right)$

Because differential equation coupling system model not only comprises ODE, and comprise partial differential equation, so we propose interactive quadravalence Runge-Kutta finite difference method solve problem.Detailed algorithm can be summarized as with under type:

Step 1. provides θ respectively ₀, T ₀, P ₀, X ₀, Q ₀and T _e0initial value.

Step 2. calculates all coefficients in coupled system.

Step 3. makes the differential method be converted into function f _i(i=1,2,3,4).Then we obtain following coupling function system:

$\left\{\begin{array}{c}{f}_1={\mathrm{\ρ}}_{m}g\cos \mathrm{\θ}-{\mathrm{\ρ}}_m{\mathrm{\υ}}_m(R{f}_2-S{f}_1)-f_m\frac{{\mathrm{\ρ}}_m{\mathrm{\υ}}_m^2}{4r_{\mathrm{ti}}}\\ A_1{f}_2+A_{2}X+A_3=0\\ f_3=\mathrm{\α}(T-T_e)\\ f_4=44.15P^{-0.79}{f}_1\end{array}\right.---\left(22\right)$

The equation group that step 4. solves above obtains P _k, X _k, Q _k, T _k, and T _ek, then we draw coefficient

a _i＝f _i(P _k，X _k，Q _k，T _k)，j＝1，2，3，4。

Step 5. makes then by separating following equation, we obtain Te,

$\left\{\begin{array}{c}\frac{\∂T_e}{\∂t_D}-\frac{{\mathrm{\λ}}_e}{C_{P_e{\mathrm{\ρ}}_e}}\frac{{\∂}^2{T}_e}{\∂r_D^2}-\frac{{\mathrm{\λ}}_e}{r_D{C}_{P_e{\mathrm{\ρ}}_e}}\frac{\∂T_e}{\∂r_D}=0\\ T_e{|}_{t_D=0}=T_0+\mathrm{\γz}\\ \frac{\∂T_e}{\∂r_{\mathrm{De}}}{|}_{r_D=1}=-\mathrm{\α}(T-T_e){\left(2\mathrm{\π}{\mathrm{\λ}}_e\right)}^{-1}\\ \frac{\∂T_e}{\∂r_{\mathrm{De}}}{|}_{r_D\→\∞}=0\end{array}\right.---\left(23\right)$

As time j, radial i when degree of depth z, T ^j _irepresent temperature, i=1; 2 ... M, j=1,2 ... N.Wherein M and N represents time and last radial node respectively.We apply finite difference method and carry out discrete obtaining to equation (23):

In this equation, τ represents the time interval, and ζ represents spaced radial.It can convert canonical form to, as follows:

Then calculus of finite differences discrete boundary condition is used.If rD=1, then:

$\frac{\∂T_e}{\∂r_{\mathrm{De}}}{|}_{r_D=1}-\frac{\mathrm{\α}}{2\mathrm{\π}{\mathrm{\λ}}_e}{T}_e{|}_{r_D=1}=-\frac{\mathrm{\αT}}{2\mathrm{\π}{\mathrm{\λ}}_e}---\left(26\right)$

It meets

If rD=N, then:

$T_N^{i+1}-T_{N-1}^{i+1}=0---\left(28\right)$

In conjunction with equation (25), (27) and (28), we can calculate the numerical solution of formation temperature Te.

Step 6. is as rD=1, and Te is substituted into equation system (23), and we obtain:

$b_j=f_j(P_k+\frac{h{a}_1}{2},X_k+\frac{h{a}_2}{2},Q_k+\frac{h{a}_3}{2},T_k+\frac{h{a}_4}{2})$

The similar method of step 7. calculates $c_j=f_j(P_k+\frac{h{b}_1}{2},X_k+\frac{h{b}_2}{2},Q_k+\frac{h{b}_3}{2},T_k+\frac{h{b}_3}{2})$ With

d _j＝f _j(P _k+hc ₁，X _k+hc ₂，Q _k+hc ₃，T _k+hc ₄)(j＝1，2，3，4)

Step 8. calculates gas-liquid mixture in (K+1) this point, and mass dryness fraction, the pressure and temperature of gas can obtain:

$P_{k+1}=P_k+\frac{h(a_1+2b_1+2c_1+d_1)}{6},X_{k+1}=X_k+\frac{h(a_2+2b_2+2c_2+d_2)}{6},$

$Q_{k+1}=Q_k+\frac{h(a_3+2b_3+2c_3+d_3)}{6},T_{k+1}=T_k+\frac{h(a_4+2b_4+2c_4+d_4)}{6}.$

Step 9. is T=T _k+1substitute into the fringe conditions of equation (23), by finite difference method, we can obtain T _{e, k+1}.

Step 10. repeats step 2-9, until calculate P _nx _nq _n, T _nand T _en.

Embodiment: for Y gas well at HTHP, utilizes the above model set up, calculates parameters such as the mass dryness fraction in gas injection process, pressure, temperature.As described in above-mentioned model analysis and solution procedure, well section is split into some unit from bottom, the time also will do similar division, then calculates according to above-mentioned calculation procedure.

Relevant parameter and survey data: internal flow density is 1000kg/m3; External fluid density is 1000kg/m3; The degree of depth is 7100 meters; Surface temperature is 20c; Ground level heat conductance parameter is 2.06; Surface temperature gradient is 0.0218 ₁c/m; The length of every section be 1m. in addition, pipe parameter, inclined shaft parameter, inclination, orientation and vertical depth parameter, respectively at table 1, are listed in table 2 and table 3.

Table 1

Table 2

Table 3

Result of calculation trend analysis: based on Runge-Kutta method and finite difference method, we obtain a series of result about this well.Pressure and temperature and the formation temperature of steam quality, pit shaft are as shown in table 4.

Table 4

Sensitivity analysis: main result as shown in FIG. 4,5,6, 7.The pressure of steam, temperature and mass dryness fraction increase along with the increase of injection length.In fact, it becomes close to actual conditions.Along with the increase of injecting quantity of steam, the water in pit shaft reduces gradually, so the aridity of steam increases.Meanwhile, along with the increase of heat transfer, the temperature on stratum also can increase, but the speed increased becomes slow.In order to study gas-liquid mixture injection rate and surface temperature gradient to the impact of gas aridity, pressure and temperature, we use different gas-liquid mixture injection rates and surface temperature gradient to reform former calculating.

Claims (5)

1. mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process, is characterized in that, comprising:

A. stratum heat transfer model is built;

B. differential equation coupling model is built;

C. differential equation coupling model is solved.

2. mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process as claimed in claim 1, is characterized in that,

Based on from the second contact surface to the hypothesis of stratum unsteady heat transfer in steps A, build stratum heat transfer model, specifically comprise:

$\frac{\∂T_e}{\∂t}=\frac{{\mathrm{\λ}}_e}{C_{\mathrm{Pe}}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{{\∂r}^2}+\frac{1}{r}\frac{\∂T_e}{\∂r})---\left(12\right)$

Primary condition:

T _e=T _α+ γ z, if t=0.

Fringe conditions:

$\frac{{\∂T}_e}{\∂r}=0,\mathrm{if\; r}\→\∞$

$\mathrm{dQ}=-2{\mathrm{\πr}}_{\mathrm{cem}}{\mathrm{\λ}}_{\mathrm{cem}}\frac{{\∂T}_e}{\∂r}\mathrm{dz}{|}_{r=r_{\mathrm{cem}}}$

Nondimensional number $r_D=\frac{r}{r_{\mathrm{cem}}}$ With $t_D=\frac{{\mathrm{\λ}}_{e}t}{r_{\mathrm{cem}}^2}$ Substitute into equation (12),

$\frac{\∂T_e}{{\∂t}_D}=\frac{{\mathrm{\λ}}_e}{C_{\mathrm{Pe}}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{{\∂r}_D^2}+\frac{1}{r_D}\frac{{\∂T}_e}{{\∂r}_D})---\left(13\right)$

Fringe conditions is converted into:

$\begin{array}{c}\frac{{\∂}^2{T}_e}{{\∂r}_D}{|}_{r_D=1}=-\frac{\mathrm{dQ}}{\mathrm{dz}}{\left(2\mathrm{\π}{\mathrm{\λ}}_e\right)}^{-1}\\ \frac{{\∂}^2{T}_e}{{\∂r}_D}{|}_{r_D\→\∞}=0\end{array}---\left(14\right)$

Relation between saturated saturated-steam temperature and pressure:

$\frac{\mathrm{dT}}{\mathrm{dz}}=44.15P^{-0.79}\frac{\mathrm{dP}}{\mathrm{dz}}---\left(15\right).$

3. mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process as claimed in claim 2, is characterized in that,

In step B, described differential equation coupling model considers friction loss and pit shaft-stratum unsteady-state heat transfer, and the differential equation coupling model of structure is:

$\left\{\begin{array}{c}\frac{\mathrm{dP}}{\mathrm{dz}}=\mathrm{\ρ}\mathrm{mg}\cos \mathrm{\θ}-p_m{\mathrm{\υ}}_m(R\frac{\mathrm{dX}}{\mathrm{dz}}-S\frac{\mathrm{dP}}{\mathrm{dz}})-f_m\frac{p_m{v}_m^2}{4r_{\mathrm{ti}}}\\ A_1\frac{\mathrm{dX}}{\mathrm{dz}}+A_{2}X+A_3=0\\ \frac{\mathrm{dT}}{\mathrm{dz}}=44.15P^{-0.79}\frac{\mathrm{dP}}{\mathrm{dz}}\\ \frac{\mathrm{dQ}}{\mathrm{dz}}=\mathrm{\α}(T-T_e)\\ \frac{{\∂T}_e}{{\∂t}_D}=\frac{{\mathrm{\λ}}_e}{C_{\mathrm{Pe}}{\mathrm{\ρ}}_e}(\frac{{\∂}^2{T}_e}{{\∂r}_D^2}+\frac{1}{r_D}\frac{{\∂T}_e}{{\∂r}_D})\end{array}\right..---\left(16\right)$

4. mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process as claimed in claim 3, is characterized in that,

In step C, before solving model, the parameter in model is handled as follows:

1) vapor (steam) velocity V _m:

$v_m=v_s+v_w=\frac{\mathrm{MX}}{{\mathrm{\ρ}}_{s}A}+\frac{M(1-X)}{{\mathrm{\ρ}}_{w}A}---\left(17\right);$

2) vapour density Pm: because water vapour air-flow is biphase gas and liquid flow, therefore applies the averag density that Beggs-Brill method calculates mixture;

3) frictional force T _f:

${\mathrm{\τ}}_f=\frac{1}{4}\mathrm{\π}{f}_m{r}_{\mathrm{ti}}{\mathrm{\ρ}}_m{v}_m^2\mathrm{dz}---\left(18\right);$

4) the friction factor f of gas-liquid mixture _m, f _mthe function about Reynolds number and absolute roughness e,

$f_m=\left\{\begin{array}{cc}\frac{\mathrm{Re}}{64}& \mathrm{if\; Re}\≤2000\\ [1.14-2\ln {(\frac{\mathrm{\ϵ}}{{2r}_{\mathrm{ti}}}+21.25{\mathrm{Re}}^{-0.9})]}^{-2}& \mathrm{if\; Re}>2000\end{array}\right.---\left(19\right);$

5) thermal transmittance U _to:

$\frac{1}{U_{\mathrm{to}}}=r_{\mathrm{ti}}\·\frac{1}{{\mathrm{\λ}}_{\mathrm{ins}}}\ln \left(\frac{r_{\mathrm{ci}}}{r_{\mathrm{to}}}\right)+\frac{1}{h_c+h_r}+r_{\mathrm{ti}}\·\frac{1}{{\mathrm{\λ}}_{\mathrm{cem}}}\ln \left(\frac{r_{\mathrm{cem}}}{r_{\mathrm{co}}}\right)---\left(20\right)$

Wherein, λ _insand λ _cembe respectively the thermal conductivity of thermal insulation material and cement sheath; Hc and hr is respectively convective heat-transfer coefficient and radiation heat transfer coefficient;

6) nondimensional time function:

$f\left(t_D\right)=\left\{\begin{array}{ccc}1.1281\sqrt{t_D}(1-0.3\sqrt{t_D})& \mathrm{if}& t_D\≤1.5\\ (0.5\ln \left(t_D\right)+0.4063)(1+\frac{0.6}{t_D})& \mathrm{if}& t_D>1.5\end{array}\right.---\left(21\right).$

5. mass dryness fraction, temperature and coupling pressure Forecasting Methodology in Oil/gas Well gas injection process as claimed in claim 4, is characterized in that,

Adopt interactive quadravalence Runge-Kutta finite difference method to solve differential equation coupling model, comprising:

Step 1. provides θ respectively ₀, T ₀, P ₀, X ₀, Q ₀and T _e0initial value;

Step 2. calculates all coefficients in coupling model;

Step 3. makes the differential method be converted into function f _i(i=1,2,3,4), can obtain following coupling function system:

$\left\{\begin{array}{c}{f}_1={\mathrm{\ρ}}_{m}g\cos \mathrm{\θ}-{\mathrm{\ρ}}_m{v}_m(R{f}_2-S{f}_1)-f_m\frac{{\mathrm{\ρ}}_m{v}_m^2}{4r_{\mathrm{ti}}}\\ A_1{f}_2+A_{2}X{+A}_3=0\\ f_3=\mathrm{\α}(T-T_e)\\ f_4=44.15P^{-0.79}{f}_1\end{array}\right.---\left(22\right)$

The equation group that step 4. solves above obtains P _k, X _k, Q _k, T _k, and T _ek, can coefficient be drawn:

a _j＝f _i(P _k，X _k，Q _k，T _k)，j＝1，2，3，4；

Step 5. makes te is obtained by separating following equation,

$\left\{\begin{array}{c}\frac{\∂T_e}{{\∂t}_D}-\frac{{\mathrm{\λ}}_e}{C_{P_e\mathrm{\ρe}}}\frac{{\∂}^2{T}_e}{{\∂r}_D^2}-\frac{{\mathrm{\λ}}_e}{r_D{C}_{P_e\mathrm{\ρe}}}\frac{{\∂T}_e}{{\∂r}_D}=0\\ T_e{|}_{t_D=0}=T_0+\mathrm{\γz}\\ \frac{{\∂T}_e}{{\∂r}_{\mathrm{De}}}{|}_{r_D=1}=-\mathrm{\α}(T-T_e){\left(2\mathrm{\π}{\mathrm{\λ}}_e\right)}^{-1}\\ \frac{{\∂T}_e}{{\∂r}_{D_e}}{|}_{r_D\→\∞}=0\end{array}\right.---\left(23\right)$

As time j, radial i when degree of depth z, T ^j _irepresent temperature, i=1; 2 ... M, j=1,2 ... N; Wherein M and N represents time and last radial node respectively; Finite difference method is adopted to carry out discrete obtaining to equation (23):

In this equation, τ represents the time interval, and ζ represents spaced radial; It can convert canonical form to, as follows:

Then calculus of finite differences discrete boundary condition is used; If rD=1, then:

$\frac{{\∂T}_e}{{\∂r}_{\mathrm{De}}}{|}_{r_D=1}-\frac{\mathrm{\α}}{2\mathrm{\π}{\mathrm{\λ}}_e}{T}_e{|}_{r_D=1}=-\frac{\mathrm{\αT}}{2\mathrm{\π}{\mathrm{\λ}}_e}---\left(26\right)$

It meets

If rD=N, then:

$T_N^{i+1}-T_{N-1}^{i+1}=0---\left(28\right)$

In conjunction with equation (25), (27) and (28), the numerical solution of formation temperature Te can be calculated;

Step 6., as rD=1, substitutes into equation system (23) Te, can obtain:

$b_j=f_j(P_k+\frac{{\mathrm{ha}}_1}{2},X_k+\frac{{\mathrm{ha}}_2}{2},Q_k+\frac{{\mathrm{ha}}_3}{2},T_k+\frac{{\mathrm{ha}}_4}{2})$

Step 7. calculates $c_j=f_j(P_k+\frac{{\mathrm{hb}}_1}{2},X_k+\frac{{\mathrm{hb}}_2}{2},Q_k+\frac{{\mathrm{hb}}_3}{3},T_k+\frac{{\mathrm{hb}}_4}{2})$ With

$d_j=f_j(P_k+{\mathrm{hc}}_1,X_k+{\mathrm{hc}}_2,Q_k+{\mathrm{hc}}_3,T_k+{\mathrm{hc}}_4)(j=\mathrm{1,2,3,4})$ Step 8. calculates gas-liquid mixture in (K+1) this point, and mass dryness fraction, the pressure and temperature of gas can obtain:

$P_{k+1}=P_k+\frac{h(a_1+{2b}_1+{2c}_1+d_1)}{6},X_{k+1}=X_k+\frac{h(a_2+{2b}_2+{2c}_2+d_2)}{6},$

$Q_{k+1}=Q_k+\frac{h(a_3+{2b}_3+{2c}_3+d_3)}{6},T_{k+1}=T_k+\frac{h(a_4+{2b}_4+{2c}_4+d_4)}{6},$

Step 9. is T=T _k+1substitute into the fringe conditions of equation (23), can T be obtained by finite difference method _{e, k+1};

Step 10. repeats step 2-9, until calculate P _nx _nq _n, T _nand T _en.