Introduction to Transient Testing

Azeb D Habte

May 2016

UTP

Lesson Content

•Basic concepts of well testing

•Diffusivity equation and its boundary conditions

•Exponential integral (line source) solution and its logarithmic approximation

•Radius of investigation

Lesson Outcomes

• List objectives of pressure transient testing.

•Describe the basic types well tests.

•Understand fundamental theory and physics of fluid flow through porous media and its application to pressure transient analysis.

BASIC CONCEPTS OF WELL TESTING

What is Well Testing?

•During a well test, a transient response is created by a temporary and controlled change in production/injection rate.

• Then the response (pressure, temperature and/or flow rate at the bottom hole) of the well to changing production (or injection) is monitored/measured.

• The response is controlled by the characteristic of the well and reservoir properties and thus it is possible to infer well/reservoir parameters from the response.

How is PTT conducted?

Pressure Transient Testing (PTT)

• It includes generating and measuring pressure variations with time in wells and analyse the data to estimate rock, fluid, and well properties:

• Wellbore volume

• Wellbore damage and stimulation

• Reservoir average pressure

• Permeability

• Porosity

• Reserves

• Reservoir fluid type

• Reservoir and fluid discontinuities,…

PTT data analyses or interpretation methods

• Interpretation of observed pressure and rate data based on mathematical well/reservoir model involves inverse problem. i.e.,

• Matching of the observed data with the model data allows estimation of ϕ, k, s, C …

Input (production/injectio

n rate)

Real system (reservoir)

Output-Observed data (P vs Time)

Input (production/injectio

n rate)

Model (ϕ, k, s, C,

etc)Output-Model data

(P vs Time)

Objectives of PTTTypical objectives of testing and analyzing are to determine:

Initial pressure Current average reservoir pressure Absolute open flow potential (AOF) Formation flow capacity kh (ability of reservoir to transmit fluid) Reservoir storativity or porosity Presence of formation damage or stimulation Size of drainage area Reservoir boundaries Existence, nature and extent of boundaries (faults, WOC, …) Existence and extent of fracturing (natural or artificial) Necessity for formation treatment Effectiveness of formation treatment, …

The primary objective is to estimate productivity of a well and formation properties.

It reflects formation properties such as permeability and porosity under in-situ dynamic condition.

Types of PTT

• Productivity test

• Drawdown test

• Buildup test

• Injection test

• Falloff test

• Variable rate test

• Multi-well test

• Interference test

• Pulse test

Drawdown test• Is conducted by producing a well at a known rate or rates while

measuring changes in bottomhole pressure (BHP) as a function of time.

• It is designed to determine permeability and skin

• If the pressure transient is affected by outer reservoir boundary, drawdown test can be used to establish the outer limits of the reservoir and to estimate the hydrocarbon volume (reservoir-limit test).

• If properly designed and analyzed, it can be an alternative to productivity test. (in case of minimum loss of production time is required)

Drawdown Cont’d

Buildup Test• In buildup test, a well which is already flowing (ideally at constant

rate) is shut in, and the bottomhole pressure measured as the pressure builds up.

Variable rate test

Multiple well test

Diffusivity Equation and Its Boundary Conditions

The Diffusivity Equation

• Diffusivity equation is a basic differential equation that describes the diffusion (transmission) of pressure in a porous media.

• For a 1D radial system, it has a form1

𝑟

𝜕

𝜕𝑟𝑟𝜕𝑝

𝜕𝑟=

ϕ𝜇𝑐𝑡

2.637×10−4𝑘

𝜕𝑝

𝜕𝑡(1)

where, p=pressure, psi

r=radial distance, ft

t=time, days

𝜇=viscosity, cp

𝑘=permeability, mD

𝑐𝑡=total compressibility, 𝑝𝑠𝑖−1

η=hydraulic diffusivity constant

1/η

• Equation 1 describes the flow of:

slightly compressible fluid

Laminar (Darcy) flow

Small and constant fluid compressibility

Isothermal conditions

Negligible gravity effects

Homogeneous porous media

Diffusivity equation is derived from three fundamental physical principles:

1. Conservation of mass (continuity equation)

1

𝑟

𝜕 𝑟 𝑣𝜌

𝜕𝑟=𝜕 ϕ𝜌

𝜕𝑡(2)

2. Equation of motion (Darcy’s equation)

𝑣 = −𝑘

𝜇

𝑑𝑝

𝑑𝑟(3)

3. Equation of state (EOS)

For slightly compressible fluid:

𝜌 = 𝜌𝑏𝑒𝑥𝑝 𝑐 𝑝 − 𝑝𝑏 (4)

Boundary and Initial Conditions

The diffusivity equation (Eq. 1) is a second order PDE and it requires initial (@ t=0) and boundary conditions to obtain solutions for the different flow regimes.

There are two boundary conditions and one initial condition.

o The two boundary conditions are:

1. Inner boundary condition:The formation produces at a constant rate into the wellbore .

2. Outer boundary condition:i. Infinite-acting reservoir: the reservoir is so large that the outer boundary is not felt

from the well.ii. No-flow boundary: the reservoir may be bounded by a closed (no-flow) boundary.iii. Constant pressure boundary: the reservoir is bounded by a reservoir with strong

pressure support (aquifer system).

o Initial condition

The reservoir is at a uniform pressure when production begins, i.e., time = 0

Mathematically,

• Inner boundary condition:

𝑟𝜕𝑝

𝜕𝑟𝑟=𝑟𝑤

=141.2𝑞𝐵𝜇

𝑘ℎ

• Outer boundary condition:

i. Infinite acting reservoir: 𝑝 𝑟 → ∞, 𝑡 = 𝑝𝑖

ii. No-flow boundary: 𝜕𝑝

𝜕𝑟 𝑟𝑒= 0

iii. Constant-pressure boundary: 𝑝 𝑟 = 𝑟𝑒 , 𝑡 = 𝑝𝑖 OR ∆𝑝 𝑟 = 𝑟𝑒 , 𝑡 = 0

• Initial condition:

𝑝 𝑟, 𝑡 = 0 = 𝑝𝑖

Exponential Integral (line source) Solution and Its Logarithmic Approximation

Ei solutionEi-function solution (line-source solution) is first proposed by Matthews and Russell in 1967. It is based on the following assumptions:

Infinite acting reservoir, i.e., the reservoir is infinite in size.

The well is producing at a constant flow rate.

The reservoir is at a uniform pressure, 𝑝𝑖, when production begins.

The solution has the following form:

𝑝 𝑟, 𝑡 = 𝑝𝑖 +70.6𝑄𝑜𝜇𝑜𝐵𝑜

𝑘ℎ𝐸𝑖

−948ϕ𝜇𝑜𝑐𝑡𝑟2

𝑘𝑡(6)

where, p(r,t) = pressure at radius r from the well after t hours

t=time, hrs

k=permeability, md

Qo=flow rate, STB/D

𝐸𝑖 −𝑥 = − 𝑥

∞ 𝑒−𝑦

𝑦𝑑𝑦

• For x<0.01, the 𝐸𝑖 function has the following logarithmic approximation:

𝐸𝑖 −𝑥 = ln(1.781𝑥) (7)

where, 𝑥 =948𝜇ϕ𝑐𝑡𝑟

2

𝑘𝑡

Substituting Eq. 7 into Eq. 6 gives:

• For the bottomhole flowing pressure, i.e., @r=rw, at any time, Eq. 8 can be rewritten as:

23.3log

6.1622

wto

oooiwf

rc

kt

kh

BQpp

Logarithmic approximation

(8)

(9)

For 0.01<x<10.9, the Table 1.1

can be used.

For x>10.9, 𝐸𝑖 −𝑥 = 0

Example 1

Determine the pressure distribution as a function of radius for time 0.1, 1, 10, and 100 hours after a well begins to produce from a formation originally at 2000 psi. The well and formation characteristics are given below:

Q=177 STB/D h=150 ft

𝜇=1 cp 𝜙=0.15

B=1.2 RB/STB 𝑐𝑡 = 70.3 × 10−6 𝑝𝑠𝑖−1

k=10md 𝑟𝑒 = 3000𝑓𝑡

𝑟𝑤 = 0.1 𝑓𝑡

SolutionStep 1: Calculate x for different radial distances between rw and re at a given time.

Step 2: Determine Ei(-x)

if x<0.01, Ei(-x)=ln(1.781x)

if 0.01<x<10.9, read Ei(-x) from Table 1.1

if x>10.9, Ei(-x)=0

Step 3: Calculate pressure using

𝑝 𝑟, 𝑡 = 𝑝𝑖 +70.6𝑄𝑜𝜇𝑜𝐵𝑜

𝑘ℎ𝐸𝑖−948ϕ𝜇𝑜𝑐𝑡𝑟

2

𝑘𝑡

𝑝 𝑟, 𝑡 = 𝑝𝑖 +70.6𝑄𝑜𝜇𝑜𝐵𝑜

𝑘ℎ𝐸𝑖(−𝑥)

r, ft x Ei(-x) p(r,0.1), psi

0.1 9.99666E-05 -8.6335 1913.691

0.5 0.002499165 -5.41462 1945.87

5 0.2499165 -1.044 1989.563

10 0.999666 -0.219 1997.811

50 24.99165 0 2000

100 99.9666 0 2000

500 2499.165 0 2000

1000 9996.66 0 2000

1500 22492.485 0 2000

2000 39986.64 0 2000

2500 62479.125 0 2000

t=0.1 hr

r, ft x Ei(-x) p(r,1), psi

0.1 9.99666E-06 -10.936 1890.672

0.5 0.000249917 -7.717 1922.851

5 0.02499165 -3.137 1968.639

10 0.0999666 -4.038 1959.632

50 2.499165 -2.49E-02 1999.751

100 9.99666 4.15E-06 2000

500 249.9165 0 2000

1000 999.666 0 2000

1500 2249.2485 0 2000

2000 3998.664 0 2000

2500 6247.9125 0 2000

t=1 hr

r, ft x Ei(-x) p(r,10), psi

0.1 9.99666E-07 -13.238 1867.654

0.5 2.49917E-05 -10.019 1899.833

5 0.002499165 -5.414 1945.87

10 0.00999666 -4.028 1959.729

50 0.2499165 -1.044 1989.563

100 0.999666 -2.19E-01 1997.811

500 24.99165 0 2000

1000 99.9666 0 2000

1500 224.92485 0 2000

2000 399.8664 0 2000

2500 624.79125 0 2000

t=10 hrsr, ft x Ei(-x) p(r,100), psi

0.1 9.99666E-08 -15.5413 1844.635

0.5 2.49917E-06 -12.3224 1876.814

5 0.000249917 -7.71721 1922.851

10 0.000999666 -6.33091 1936.71

50 0.02499165 -3.137 1968.64

100 0.0999666 -4.04E+00 1959.632

500 2.499165 -2.49E-02 1999.751

1000 9.99666 -4.15E-06 2000

1500 22.492485 0 2000

2000 39.98664 0 2000

2500 62.479125 0 2000

t=100 hrs 1820

1840

1860

1880

1900

1920

1940

1960

1980

2000

2020

0.1 1 10 100 1000 10000

pre

ssu

re,

psi

r, ft

t=0.1 hr

t=1 hr

t=10 hrs

t=100 hrs

Radius of Investigation

Radius of Investigation

• Radius of Investigation (𝑟𝑖) is the distance that a pressure transient has moved into a formation following rate change in a well.

• It is related to formation rock and fluid properties and time.

𝑟𝑖 =𝑘𝑡

948𝜙𝜇𝑐𝑡(10)

Example 2Determine the radius of investigation for t=0.1, 1, 10,100 hrs in Example 1.

Solution:

t=0.1hr

𝑟𝑖 =𝑘𝑡

948𝜙𝜇𝑐𝑡=

10∗0.1

948∗0.15∗1∗70.3𝑒−6= 10.002 𝑓𝑡

t=1hr

𝑟𝑖 =𝑘𝑡

948𝜙𝜇𝑐𝑡=

10∗1

948∗0.15∗1∗70.3𝑒−6= 31.628 𝑓𝑡

t=10hr

𝑟𝑖=𝑘𝑡

948𝜙𝜇𝑐𝑡=

10∗10

948∗0.15∗1∗70.3𝑒−6= 100.017 𝑓𝑡

t=100hr

𝑟𝑖 =𝑘𝑡

948𝜙𝜇𝑐𝑡=

10∗100

948∗0.15∗1∗70.3𝑒−6= 316.281 𝑓𝑡