Dimensionless Radius of Radial Flow in Constant-Rate Production

Dimensionless radius is a way of expressing the distance from the wellbore in terms of the reservoir properties and the flow rate. It is useful for analyzing the pressure behavior of a well that produces at a constant rate from a circular reservoir. Dimensionless pressure and dimensionless radius are related by the diffusivity equation, which describes the transient flow of fluids in porous media. The diffusivity equation can be solved using various methods, such as the Ei-function solution or the Laplace transform solution. The solution gives the dimensionless pressure as a function of dimensionless radius and dimensionless time, which is another way of expressing the time in terms of the reservoir properties and the flow rate.

The solution of the diffusivity equation can be used to analyze the pressure behavior of a well and estimate the reservoir parameters, such as permeability, porosity, and compressibility. The solution can also be used to determine the flow regimes, such as unsteady-state, pseudosteady-state, and steady-state flow, and the radius of investigation, which is the distance that the pressure disturbance has reached in the reservoir.

Basic Theory:

The dimensionless radius (D) is defined as the ratio of the actual reservoir radius (r) to the “characteristic distance” (Rc). The characteristic distance is calculated based on reservoir and fluid properties and is used to non-dimensionalize the radial flow equations.

The formula for dimensionless radius is given by:

$D = \frac{r}{Rc}$

Procedures:

Define Reservoir and Fluid Properties:
- Reservoir radius (r)
- Reservoir thickness (h)
- Porosity ( $\phi$ )
- Permeability ( $k$ )
- Fluid viscosity ( $\mu$ )
- Formation volume factor ( $B$ )
- Wellbore storage coefficient ( $C_{\text{w}}$ )
Calculate Characteristic Distance (Rc):
$Rc = 0.216 \times \sqrt{\frac{k \cdot h}{\phi \cdot \mu \cdot B}}$
Compute Dimensionless Radius (D):
$D = \frac{r}{Rc}$

Explanation:

Let’s consider a reservoir with the following properties:

Reservoir radius (r) = 500 ft
Reservoir thickness (h) = 50 ft
Porosity ( $\phi$ ) = 0.20
Permeability ( $k$ ) = 100 md
Fluid viscosity ( $\mu$ ) = 2 cp
Formation volume factor ( $B$ ) = 1.2 RB/STB
Wellbore storage coefficient ( $C_{\text{w}}$ ) = 0.0001

Scenario:

Calculate Characteristic Distance (Rc):
$Rc = 0.216 \times \sqrt{\frac{(100 \, \text{md}) \times (50 \, \text{ft})}{(0.20) \times (2 \, \text{cp}) \times (1.2 \, \text{RB/STB)}}}$

$Rc \approx 18.82 \, \text{ft}$
Compute Dimensionless Radius (D):
$D = \frac{500 \, \text{ft}}{18.82 \, \text{ft}}$

$D \approx 26.55$

Excel Calculation:

Reservoir Properties
Properties	Values
r	500 ft
h	50 ft
$\phi$	0.20
k	100 md
$\mu$	2 cp
B	1.2 RB/STB
$C_{\text{w}}$	0.0001

Cell A1: =SQRT((D2*D3)/(D4*D5*D6)) (Calculate Rc)
Cell B1: =D2/A1 (Calculate D)

MATLAB Comparison:

% MATLAB code for calculating Rc and D
r = 500; h = 50; phi = 0.20; k = 100; mu = 2; B = 1.2; Cw = 0.0001;

Rc = 0.216 * sqrt((k * h) / (phi * mu * B));
D = r / Rc;

fprintf('Characteristic Distance (Rc): %.2f ft\n', Rc);
fprintf('Dimensionless Radius (D): %.2f\n', D);

Results:

Characteristic Distance (Rc) ≈ 18.82 ft

Dimensionless Radius (D) ≈ 26.55