M. Onur 09.09.2013

PCB 30313- Well Test Analysis September 2013/UTP Supplement No. 1

Subject: Derivation of continuity and diffusivity equations in radial coordinate system. Consider a horizontal cylindrical reservoir with height h and assume that we have flow only in the radial direction, r, as its cross-sectional view shown in Fig. 1. Using conservation of mass over the control volume shown in Fig. 1, derive the continuity equation, which will describe flow in the radial direction. Fluid velocity in the r-direction is vr [RB/(ft2-day)] and density of fluid is ρ. Use field units. Do not consider source/sink term in the continuity equation. Then derive the diffusivity equation for slightly compressible fluid of constant viscosity and homogeneous reservoir in terms of pressure replacing the velocity in continuity equation by Darcy’s equations and using the assumptions of slightly compressible fluid and homogeneous reservoir.

r-direction

r-direction

r-Δr/2 r + Δr/2

Controlvolume

Fig. 1. Cross-sectional view of cylindrical reservoir.

Solution 1: We will perform mass balance on the control volume shown in Figure 1 by considering flow only in the r (or radial) direction. Recall that mass balance can be expressed as

daccumulateoutin massmassmass =− (1.1) where mass in, mass out and accumulation terms are given by

( ) tAvmass rrrin Δρ= Δ− 2/615.5 (1.2)

( ) tAvmass rrrout Δρ= Δ+ 2/615.5 (1.3) and

( ) ( )[ ] btttaccumlated Vmass ρφ−ρφ= Δ+ (1.4) In Eqs. 1.2 and 1.3, A term denotes the cross-sectional area in ft2 perpendicular to flow in the r-direction. It is clear from Fig. 1 is that this cross-sectional area is a function of r as r increases the cross-sectional area open to flow increases. Note that (A)r-Δr/2 will indicate the

cross-sectional area at r-Δr/2. In addition, Vb is the bulk volume of the CV in ft3 shown in Fig. 1. Now, we will find the expressions for the cross-sectional areas at r-Δr/2 and r+Δr/2 and for the bulk volume Vb by taking into a consideration of the geometry of the control volume shown in Fig. 1. Because we have a cylinder, and flow in the r-direction will be perpendicular to the surface area of the cylinder, then the cross sectional areas:

hrrA rr )2/(22/ Δ−π=Δ− (1.5)

hrrA rr )2/(22/ Δ+π=Δ+ (1.6) and the bulk volume is given by

rhrhrrrrVb Δπ=⎥⎦⎤

⎢⎣⎡ Δ

−−Δ

+π= 2)2

()2

( 22 (1.7)

Using Eqs. 1.5, 1.6 and 1.7 in Eqs. 1.2, 1.3 and 1.4 gives

( )( ) trvhmass rrrin Δρπ= Δ− 2/2615.5 (1.8)

( )( ) trvhmass rrrout Δρπ= Δ+ 2/2615.5 (1.9) and

( ) ( )[ ]tttaccumlated rhrmass ρφ−ρφΔπ= Δ+2 (1.10) Using Eqs. 1.8, 1.9 and 1.10 in Eq. 1.1 gives

( )( ) ( )( ) ( ) ( )[ ]tttrrrrrr rhrtrvhtrvh ρφ−ρφΔπ=Δρπ−Δρπ Δ+Δ+Δ− 22615.52615.5 2/2/ (1.11) Dividing both sides of Eq. 1.11 by trhr ΔΔπ2 and rearranging the resulting equation gives

( ) ( )[ ] ( ) ( )[ ]tr

rvrvr

tttrrrrrr

Δρφ−ρφ

=Δ

ρ−ρ− Δ+Δ−Δ+ 2/2/615.5 (1.12)

Now, taking the limits as Δr and Δt go to zero in Eq. 1.12, we obtain

( ) ( )[ ] ( ) ( )[ ]tr

rvrvr

ttt

t

rrrrrr

r Δρφ−ρφ

=Δ

ρ−ρ− Δ+

→Δ

Δ−Δ+

→Δ 0

2/2/

0limlim615.5

(1.13)

Using the definition of derivative gives

( ) ( )tr

vrr

r

∂ρφ∂

=∂ρ∂− 615.5

(1.14)

or rearranging gives

( ) ( )tr

vrr

r

∂ρφ∂

−=∂ρ∂

615.511

(1.15)

which is the continuity equation in radial coordinates.

Darcy’s equation is given by

rpkvr ∂∂

μ×−= −310127.1 . (1.16)

Using Eq. 1.16 in Eq. 1.15 gives

( )txr

pkrrr ∂

ρφ∂=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

μρ

∂∂

−31033.611

(1.17)

As we have homogeneous permeability and constant viscosity, then Eq. 1.17 can be written as,

( )txr

prrr

k∂ρφ∂

=⎟⎠⎞

⎜⎝⎛

∂∂

ρ∂∂

μ −31033.611

(1.18)

Now, we expand the time derivative in the right-hand side of Eq. 1.18 by using the chain rule and considering that ρ and φ is a unique function of p at isothermal conditions; i.e., ρ = ρ (p) and φ = φ(p), where p = p(r,t). ( )

tp

dpd

dpd

tp

dpd

tp

dpd

ttt ∂∂

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∂∂

+∂∂

=∂

∂ φφ

ρρ

φρφρρφφρρφρφ 11 (1.19)

For a single-phase fluid, we define the isothermal fluid compressibility as:

Tdpdc ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ρρ1

(1.20)

and the isothermal effective formation (or pore) compressibility as:

Tdpdc ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

φφ1

(1.21)

Then, using Eqs. 1.20 and 1.21 in Eq. 1.19, we can express Eq. 1.19 as:

( ) ( )tpcc

t f ∂∂

+=∂

∂ φρρφ (1.22)

Further, we define the total compressibility for a single-phase fluid as:

ft ccc += (1.23) Then, Eq. 1.22 can be written as:

( )tpc

t t ∂∂

=∂

∂ φρρφ (1.24)

Similarly, we do the same thing for the spatial derivative (i.e., derivative w.r.t r) in the left-hand side of Eq. 1.18 as we did exactly in class to obtain

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∂∂ 2

rpcr

rpr

rrpr

rρρ , (1.25)

where c is the fluid isothermal compressibility, which is assumed to be small and constant so that the second term in Eq. 1.25 can be neglected to obtain

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

rpr

rrpr

rρρ , (1.26)

Using Eqs. 1.24 and 1.26 in Eq. 1.18 gives

3

1 16.33 10 t

k p pr cr r r x tρ φ ρ

μ −

⎡ ⎤∂ ∂ ∂⎛ ⎞ =⎜ ⎟⎢ ⎥∂ ∂ ∂⎝ ⎠⎣ ⎦ (1.22)

or canceling density from both sides and rearranging gives

3

16.33 10

tcp prr r r x k t

φ μ−

∂ ∂ ∂⎛ ⎞ =⎜ ⎟∂ ∂ ∂⎝ ⎠. (1.23)

which is radial diffusivity equation.