Calculate The Drainage Volume Integral for The Infinite Conductivity Fracture

The drainage volume integral for an infinite conductivity fracture describes the volume of fluid drained from a reservoir through a fracture with infinite conductivity. This calculation is essential in reservoir engineering for evaluating fluid recovery efficiency and fracture performance.

Introduction

In reservoir engineering, fractures with infinite conductivity are idealized models where the fracture width is much larger than the reservoir thickness, allowing fluid to flow freely without resistance. The drainage volume integral quantifies the total volume of fluid that can be drained from the reservoir through such a fracture.

This calculation is based on the assumption of radial flow and constant pressure at the fracture face. The result provides insights into reservoir performance and helps engineers optimize recovery strategies.

Formula

The drainage volume integral for an infinite conductivity fracture is calculated using the following formula:

V = 2πkh(Δp/μB) * ∫[r_e to r_w] (ln(r_e/r) / ln(r_e/r_w)) dr

Where:

V = Drainage volume (bbl)
k = Permeability (md)
h = Net pay thickness (ft)
Δp = Pressure difference (psi)
μ = Viscosity (cp)
B = Formation volume factor (RB/STB)
r_e = External radius (ft)
r_w = Wellbore radius (ft)

The integral evaluates to a logarithmic relationship between the external and wellbore radii, providing the volume of fluid drained through the fracture.

Calculation Process

To calculate the drainage volume integral:

Determine the reservoir properties: permeability, net pay thickness, and pressure difference.
Identify the fluid properties: viscosity and formation volume factor.
Define the geometric parameters: external radius and wellbore radius.
Apply the formula to compute the drainage volume.
Analyze the result to assess reservoir performance and recovery efficiency.

Note: This calculation assumes idealized conditions of infinite conductivity and radial flow. Real-world scenarios may involve additional factors such as skin effect and non-radial flow patterns.

Worked Example

Consider a reservoir with the following properties:

Parameter	Value	Unit
Permeability (k)	100	md
Net pay thickness (h)	50	ft
Pressure difference (Δp)	500	psi
Viscosity (μ)	1	cp
Formation volume factor (B)	1.2	RB/STB
External radius (r_e)	1000	ft
Wellbore radius (r_w)	0.5	ft

Using the formula:

V = 2π * 100 * 50 * (500 / (1 * 1.2)) * ∫[1000 to 0.5] (ln(1000/r) / ln(1000/0.5)) dr

The integral evaluates to approximately 0.5, so:

V ≈ 2π * 100 * 50 * 416.67 * 0.5 ≈ 132,733 bbl

This example demonstrates the significant volume of fluid that can be drained through an infinite conductivity fracture.

Interpreting Results

The drainage volume integral provides several key insights:

Recovery Efficiency: Higher drainage volumes indicate better reservoir performance and potential for increased fluid recovery.
Fracture Design: The result helps evaluate the effectiveness of fracture stimulation treatments.
Reservoir Management: Understanding drainage volumes aids in optimizing well spacing and production strategies.

Engineers should consider the assumptions and limitations of the model when interpreting results and applying them to real-world scenarios.

FAQ

What is an infinite conductivity fracture?: An infinite conductivity fracture is an idealized model where the fracture width is much larger than the reservoir thickness, allowing fluid to flow freely without resistance.
How does the drainage volume integral differ from other reservoir calculations?: The drainage volume integral specifically quantifies the volume of fluid drained through a fracture, whereas other calculations may focus on pressure distribution or flow rates.
What factors can affect the accuracy of this calculation?: Real-world factors such as skin effect, non-radial flow patterns, and fracture conductivity variations may affect the accuracy of the calculation.
How can I use this calculation in reservoir engineering?: This calculation helps evaluate reservoir performance, optimize fracture design, and improve production strategies in reservoir engineering applications.
Is this calculation suitable for all types of reservoirs?: The calculation assumes idealized conditions and may not be suitable for all reservoir types. Engineers should consider the specific characteristics of their reservoir when applying this method.