Capítulo 6: PRESENTACIÓN DE LOS ESTUDIOS DE CASO
6.1. Subsistema físico
6.1.2. El reconocimiento de la ciudad a partir de procesos alternativos
We now return to the task of deriving the single-phase flow equations for a compressible system. Throughout this derivation we refer to the control volume shown as block i in Figure 3 where certain terminology is also explained. In Figure 3, we have divided up the x-axis into increments of (constant) size, Δx, and constant cross-sectional area, A. Flow is considered to be in the positive x-direction (i increasing). The fluid has density, ρ, which may depend on pressure i.e. ρ(P). The porosity is denoted as φ and it too may depend on pressure, φ(P); in block i, the porosity is, φi. The volumetric flows across the boundaries of block i are given by: qi-1/2 and qi+1/2 as shown in Figure 3; the dimensions of these quantities are volume/time and typical units might be STB/day, m3/sec etc.
Let us now apply the mass conservation conditions as were applied in the introductory section of this chapter. Mass conservation states that:
The mass accumulation (increase or decrease)
in block i over a time step, ∆t
The mass that flows IN over
time ∆t
The mass that flows OUT over time ∆t
= - (1)
We can easily write mathematical expressions for each of the terms in equation 1 as follows:
The mass of fluid flowing IN to block i over time t
Volumetric rate of in flow to block i
density of
the fluid x t = q. t
∆ i-12
=
( )
x ∆ ρ .∆ (2a)
Llikewise:
The mass of fluid
flowing OUT of = (qρ)i+1/2.Δt (2b) block i over time Δt
5 5
The Flow Equations
Boundary
Boundary i-1/2
qi-1/2
qi+1/2 x
∆x
∆x
∆x i-1
i+1 Block i
i+1/2
Notation:
1 The boundaries of block i are denoted (i-1/2) between blocks (i-1)and i (i+1/2) between blocks i and i+1
2 q(i-1/2) denotes the volumetric flow rates across the (i-1/2) boundary q(i+1/2) denotes the volumetric flow rates across the (i+1/2) boundary
Note – units bbl/day, m3/s etc…
3 The porosity of block i = φi, the permeability is ki etc..
Therefore, the change in mass of fluid in block i over time Δt is given by:
Change in mass
over Δt in block i
= ( ) ( )
= [ ( ) − ( ) ]
= − [ ( ) − ( ) ]
+
− +
q t q t
q q t
q q t
i 12 i
i i+
i i
ρ ρ
ρ ρ
ρ ρ
-. - .
.
∆ ∆
∆
∆
12
12 1
2
12 1
2
(3)
Thus, equation 3 expresses the change in mass due to flow that occurs in block i using quantities – volumetric flow rates and densities – defined on the boundaries.
Note that we have changed the signs consistently in the final step of equation 3 for convenience below. Which quantity do you think we mean when we refer to ρi-1/2 : the density “on the boundary”, i-1/2? This seems a bit strange. However, you can think of ρi-1/2 being some sort of “average” density between ρi-1 and ρi, the densities in blocks (i-1) and i, respectively. In fact, we have discussed this matter of grid block to grid block average in Chapter 4.
We now turn to the alternative way of expressing the mass of fluid in a grid block i at the different times, t and t+Δt. This is shown in Figure 4 where we denote the mass of fluid in grid block i at times t and (t + Δt) by (ρφΔxA)t and (ρφΔxA)t+Δt , respectively. Therefore:
Figure 3
Control volume grid block for application of material balance for the single phase flow equation; notation is given below.
5 5
The Flow Equations
The change of mass of fluid
in block i over the time from = The mass at - The mass t to t+Δt (t + Δt) at t
= [(ρφ)t+Δt - (ρφ)t].ΔxA (4) where we have used the fact that Δx and A are constants in equation 4.
∆x Area = A
Rock SpacPore
e
The mass of fluid in block i
= (Pore volume of block i) x density
= (Volume of block x porosity) x density
= (∆x.A.φ.)ρ
= ρφ.∆x.A
We now apply the material balance condition (equation 1) by equating the expressions in equations 3 and 4 as follows:
ρφ ρφ ρ ρ
( ) − ( )
[
t + t∆ t] . ∆ xA = − [ ( ) q
i+ 12− ( ) q
i-12] . ∆ t
(5)Now divide through equation 5 by the (constant) Δx.AΔt to obtain:
ρφ ρφ
ρ ρ
( ) ( )
[ ] = − −
t + t
-
t i+ 12 i−t
q A
q A x
∆
∆ ∆
12 (6)
where we have taken the (constant) area, A, to the inner parenthesis on the RHS of equation 6.
As it stands, equation 6 is exactly true since it is simply a statement of mass balance.
There are no assumptions in this equation. Indeed, we can simplify equation 6 a little more by noting that q/A is the Darcy velocity, u. This becomes:
ρφ ρφ ρ ρ
( ) ( )
[
t + t-
t] = − [ ( )
i+ 12( )
i-12]
t
u. - u.
x
∆
∆ ∆
(7) Figure 4
Expressions for the mass of single-phase fluid in a grid block
5 5
The Flow Equations
which is still an exact statement of fact. We can take the limits of the difference equation 7 at Δt → 0 and Δx → 0 to obtain the equivalent differential equation.
Clearly:
Lim
t
.
∆ t
∆∆
-t
t + t t
→
( ) ( )
[ ] = ∂ ( )
∂ 0
ρφ ρφ ρφ
(8)and
Lim.
∆ x ∆
u. - u.
x
u.
x
i+ 12 i
→ − [ ( ) ( ) ] = − ∂ ( )
∂
−
0
12
ρ ρ ρ
(9)and therefore:
∂ ( )
∂ = − ∂ ( )
∂
ρφ ρ
t u.
x
(10)Equation 10 is the differential form of the conservation equation and again, it is
“exact” in the sense discussed above. Before going on to use Darcy’s law (for u), look at the structure of equation 10; in particular look at the symmetry between ρ and φ. These two quantities appear in an identical manner in the LHS of this equation.
Hence, if the mass in a grid block stayed the same with time, i.e. ∂(ρφ)/∂t = 0, then this could be because the fluid density went down (the fluid expands) and the rock porosity went up (the rock contracts or compresses).
We now assume Darcy’s law for u, that is:
u - k . P
= ∂ x
∂
µ
(11)(as given in the Glossary etc.). We may substitute this form of the Darcy law directly into equation 10 (taking care with the signs) to obtain:
∂ ( )
∂ = ∂
∂
∂
∂
ρφ ρ
µ t
x
k P
x
(12)This equation is now inexact, in that its validity depends on whether Darcy’s law is or is not a good assumption. However, it was necessary to use a flow law such as Darcy’s law since pressure (i.e. P(x,t)) does not explicitly appear in equation 10. Thus, Darcy’s law is our “link” between fluid velocities and pressure gradients. Equation 12 still does not have pressure (P(x,t)) explicitly shown on the LHS; it simply has a term ∂(φρ)/∂t. However, we know that φ and ρ depend locally only on pressure i.e.
they can be written ρ(P) and φ(P). We therefore manipulate the LHS of equation 12 as follows using the chain rule of differentiation:
∂ ( )
∂ = ∂ ( )
∂
∂
∂
ρφ ρφ
t P
P
. t
(13)5 5
The Flow Equations
Thus, substituting equation 13 back into equation 12 we obtain a “true” single phase pressure equation as follows:
∂ ( )
∂
∂
∂
= ∂
∂
∂
∂
ρφ ρ
µ P
P
t x
k P
. . x
(14)where we can consider the term ∂(φρ)/∂P as a generalised fluid and rock compressibility, term, C(P). Equation 14 would then become:
C P P
t x
k P
( ) ∂ ∂ = ∂ ∂ ∂ ∂ x
. ρ .
µ
(15)Some points should be noted about equation 15, as follows:
(i) It is a non-linear partial differential equation (PDE). By “non-linear” we mean that the coefficients in the equation – the “input”, if you like- depend on the quantity we are trying to find, the unknown pressure, P(x,t). In other words, quantities such as C(P), ρ(P), μ(P) etc. “depend on the answer”. We will not go into mathematical detail but such problems are notoriously difficult to solve analytically for general cases.
(ii) Following on from point (i) above, such equations can only usually be solved by one or two approaches:
• By solving them numerically where we can handle the non-linearities using certain types of iterative methods (see Chapter 6).
• We may simplify the equation to the extent that it becomes soluble.
Clearly, by its nature, a numerical solution will be approximate although it is an approximation to the full equation (equation 15, in this case). With a simplified equation, we may have an exact solution, but the simplifications may have
“thrown away” some of the important physics. In fact, in the following section, we will take the second approach.
(iii) Going back to equation 10, we could include gravity effects by taking the Darcy equation with gravity as follows:
u
= − ∂
∂
− ∂
∂
k P
x g z
µ ρ x
(16)(instead of equation 11). If the 1D system is at constant incline, z(x), then (∂z/∂x) = cos θ where θ is the angle of incline, therefore, we obtain equation 17 below:
∂
k P
5 5
The Flow Equations
This could simply lead to the generalised single phase equation as follows:
∂ ( )
∂
∂
∂
= ∂
∂
∂
∂
−
ρφ ρ ρ
µ
ρ ρ θ
P t x
k
x g cos
.
(18)but this is fairly straightforward.