01-Introduction to Well Testing

Introduction to Well Testing 1


Upon completion of this section, the student should be able to:Upon completion of this section, the student should be able to:

1. List 4 major objectives of well testing.

2. Define, give the units for, and specify typical sourcesfor each of the following variables: net pay thickness, porosity, saturation, viscosity, formation volume factor, total compressibility, wellbore radius.

3. Be able to compute the total compressibility for different reservoir systems.




A well test is conducted byA well test is conducted by

• Changing production rate at a well

• Measuring resulting pressure response at the same well or another well


• Exploration• Exploration

Is this zone economic?

How large is this reservoir?

• Reservoir engineering

What is the average reservoir pressure?

How do I describe this reservoir in order to

– estimate reserves?

– forecast future performance?

– optimize production?

• Production engineering

Is this well damaged?

How effective was this stimulation treatment?

Why is this well not performing as well as expected?


• Define reservoir limits• Define reservoir limits

– Distances to boundaries

– Drainage area

• Estimate average drainage area pressure

• Characterize reservoir

– Permeability

– Skin factor

– Dual porosity or layered behavior

• Diagnose productivity problems

– Permeability

– Skin factor

• Evaluate stimulation treatment effectiveness

– Skin factor

– Fracture conductivity

– Fracture half-length


Single well testsSingle well tests

Drawdown test –Produce a well at constant rate and measure the pressure response.

Buildup test – Shut in a well that has been producing and measure the pressure response.

Injection test – Inject fluid into a well at constant rate and measure the pressure response.

Injection-falloff test – Shut in an injection well and measure the pressure response.

Multi-well tests

Interference test – Produce one well at constant rate and measure the pressure response at one or more offset wells.

Pulse test – Alternately produce and shut in one well and measure the pressure response at one or more offset wells.



• Obtained by combining• Obtained by combining

– Continuity equation

– Equation of state for slightly compressible liquids

– Flow equation - Darcy’s law


The continuity equation is a restatement of the conservation of matter. The continuity equation is a restatement of the conservation of matter. That is, the rate of accumulation of fluid within a volume element is given by the rate at which the fluid flows into the volume minus the rate at which the fluid flows out of the volume.

Nomenclature

A = Cross-sectional area open to flow, ft2

= Rate of accumulation of mass within the volume, lbm/sec

v = Fluid velocity, ft/sec

ρ = Density of fluid, lbm/ft3

m&


This equation describes the change in density with pressure for This equation describes the change in density with pressure for a liquid with small and constant compressibility.

Nomenclature

c = Compressibility, psi-1

p = Pressure, psi

ρ = Density of fluid, lbm/ft3


NomenclatureNomenclature

A = Cross sectional area open to flow, cm2

k = Permeability, darcies

L = Length of flow path, cm

p = Pressure, atm

∆p = Pressure difference between upstream and downstream sides, atm

q = Flow rate, cm3/sec

ux = Flow velocity, cm/sec

x = Spatial coordinate, cm

µ = Viscosity, cp


• The diffusivity equation is obtained by combining• The diffusivity equation is obtained by combining

- The continuity equation

- The equation of state for a slightly compressible liquid

- Darcy’s law

• Other transient flow equations may be obtained by combining different equations of state and different flow equations

- Gas flow equation

- Multiphase flow equation



• The formation volume factor is the volume of fluid at • The formation volume factor is the volume of fluid at reservoir conditions necessary to produce a unit volume of fluid at surface conditions.

• Symbol – Bo, Bg, Bw

• Units – res bbl/STB, res bbl/ Mscf• Source – Lab measurements, correlations• Range and typical values

– Oil• 1 – 2 res bbl/STB, Black oil

• 2 – 4 res bbl/STB, Volatile oil

– Water• 1 – 1.1 res bbl/STB

– Gas• 0.5 res bbl/Mscf, at 9000 psi

• 5 res bbl/Mscf, at 680 psi

• 30 res bbl/Mscf, at 115 psi


• Viscosity is a measure of resistance to flow -- specifically, it • Viscosity is a measure of resistance to flow -- specifically, it is the ratio of the shear stress to the resulting rate of strain within a fluid.

• Symbols

µo, µg, µw

• Units – cp

• Source – Lab measurements, correlations

• Range and typical values

- 0.25 – 10,000 cp, Black oil

- 0.5 – 1.0 cp, Water

- 0.012 – 0.035 cp, Gas


• Compressibility is the fractional change in volume due to a • Compressibility is the fractional change in volume due to a unit change in pressure.

• Symbol – co, cg, cw

• Units – psi-1, microsips (1 microsip = 1x10-6 psi-1)• Source – Lab measurements, correlations

Typical Values• Oil

– 15x10-6 psi-1, undersaturated oil

– 180x10-6 psi-1, saturated oil

• Water– 4x10-6 psi-1

• Gas– 1/p, Ideal gas

– 60x10-6 psi-1, at 9000 psi

– 1.5x10-3 psi-1, at 680 psi

– 9x10-3 psi-1, at 115 psi


• Porosity is the ratio of volume of pore space to bulk volume • Porosity is the ratio of volume of pore space to bulk volume of rock.

• Symbol - φ

• Units

– Equations - fraction

– Reports - % (or fraction)

• Source

– Logs, cores

• Range or Typical Value

– 30%, unconsolidated well-sorted sandstone

– 20%, clean, well-sorted consolidated sandstone

– 8%, low permeability reservoir rock

– 0.5%, natural fracture porosity


• Permeability is the measure of capacity of rock to transmit • Permeability is the measure of capacity of rock to transmit fluid.

• Symbol

– k

• Units

– Darcy or millidarcy (md or mD)

• Source

– Well tests, core analysis

• Range

– 0.001 md - 10,000 md


• Pore volume compressibility is the fractional change in • Pore volume compressibility is the fractional change in porosity due to unit change in pressure.

• Symbol – cf

• Units – psi-1, microsips

• Source – Lab measurement, correlation, guess


– 4x10-6 psi-1, well-consolidated sandstone

– 30x10-6 psi-1, unconsolidated sandstone

– 4 to 50 x 10-6 psi-1 consolidated limestones


• The net pay thickness is the total thickness of all productive • The net pay thickness is the total thickness of all productive layers in communication with the well.

– NOTE: Also includes any rock that has sufficient vertical permeability to allow fluid to move to a layer from which it may be produced.

• Symbol – h

• Units – ft

• Source – logs


– May be as small as 5 ft or even less

– May be as large as 1,000 ft or more


• Saturation is the fraction of pore volume occupied by a • Saturation is the fraction of pore volume occupied by a particular fluid.

• Symbol – So, Sw, Sg

• Units – fraction or %

• Source – logs


– 15 to 25% – connate water saturation in well-sorted, coarse sandstones

– 40 to 60% – connate water saturation in poorly sorted, fine-grained, shaly, low-permeability reservoir rock


• Wellbore radius is the size of wellbore.• Wellbore radius is the size of wellbore.

• Symbol

– rw

• Units

– feet

• Source

– Bit diameter/2

– Caliper log


– 2 to 8 in.


• The total compressibility is the sum of pore compressibility • The total compressibility is the sum of pore compressibility and saturation weighted fluid compressibilities.

• Symbol – ct

• Units

– psi-1, microsips

• Source

– Calculated


– See exercises


Exercise 1

List 4 Objectives of Well Testing

List 4 objectives of well testing. List as many as possible without referring to the notes.

1.

2.

3.

4.4.


Exercise 2

Define Variables Used In Well Testing

Define, give the units for, and name a common source for each of the following variables used in well testing. Complete as much of this exercise as possible before referring to the notes.

1. Porosity

2. Water saturation

3. Total compressibility

4. Oil compressibility

5. Formation volume factor

6. Viscosity

7. Wellbore radius

8. Net pay thickness

9. Permeability


Exercise 3

Calculate Compressibility for Undersaturated Oil Reservoir

Calculate total compressibility for the following situation. Assume solution gas/oil ratios do not include stock tank vent gas.

Undersaturated oil reservoir (above the bubblepoint)

Sw = 17%, TDS = 18 wt %, oil gravity = 27°API,

Rso = 530 scf/STB, gas gravity = 0.85, Tf = 185°F,

p = 3500 psi, cf = 3.6×10-6 psi-1

Tsep = 75°F, p sep = 115 psia

From fluid properties correlations,

pb = 2803 psi

co = 1.158 x 10-5 psi-1co = 1.158 x 10 psi

cw = 2.277 x 10-5 psi-1


Exercise 3

Calculate Compressibility for Undersaturated Oil Reservoir

Solution

ct = cf + So co + Sw cw + Sg cg

cf = 3.6 x 10-6 psi-1

Sw = 0.17

Sg = 0

So = 1 - Sw - Sg = 1 - 0.17 - 0.0 = 0.83


pb = 2803 psi

co = 1.158 x 10-5

cw = 2.277 x 10-6


= 3.6 x 10-6 + (0.83) (1.158 x 10-5)

+ (0.17) (2.277 x 10-6) + (0) (?)

= 1.36 x 10-5 psi-1


Exercise 4

Calculate Compressibility for Saturated Oil Reservoir

Calculate total compressibility for the following situation. Assume solution gas/oil ratios do not include stock tank vent gas.

Saturated oil reservoir (below the original bubblepoint)

Sw = 17%, Sg = 5%, TDS = 18 wt %, oil gravity = 27°API,

Rso = 530 scf/STB, gas gravity = 0.85, Tf = 185°F,

p = 2000 psi, cf = 3.6×10-6 psi-1

Tsep = 75°F, p sep = 115 psia


pb = 2803 psi

co = 1.429 x 10-4 psi-1

cg = 5.251 x 1-4 psi-1

cw = 4.995 x 10-6 psi-1


Exercise 4

Calculate Compressibility for Saturated Oil Reservoir

Solution


cf = 3.6 x 10-6 psi-1

Sw = 0.17

Sg = 0.05

So = 1 - 0.17 - 0.05 = 0.78


pb = 2803 psi

co = 1.429 x 10-4 psi-1

cg = 5.251 x 10-4 psi-1

cw = 4.995 x 10-6 psi-1cw = 4.995 x 10-6 psi-1


= 3.6 x 10-6 + (0.78) (1.429 x 10-4) + (0.17) (4.995 x 10-6)

+ (0.05) (5.251 x 10-4 )

= 1.42 x 10-4 psi-1


Exercise 5

Calculate Compressibility for Low-Pressure, High-Permeability Gas Reservoir

Calculate total compressibility for the following situation. Assume a dry gas.

Low-pressure, high-permeability gas reservoir

Sw = 20%, gas gravity = 0.74, Tf = 125°F, p = 125 psi,

cf = 3.6×10-6 psi-1, cw = 4 x 10-6 psi [Tf is outside range of correlations]


cg = 8.144 x 10-3 psi-1

cw = 4x10-6 psi-1


Exercise 5

Calculate Compressibility for Low-Pressure, High-Permeability Gas Reservoir

Solution

ct = cf + So co + Sg cg + Sw cw

cf = 3.6 x 10-6 psi-1

Sw = 0.2

Sg = 0

Sg = 1 - Sw - So = 1 - 0.2 - 0 = 0.8


cg = 8.144 x 10-3 psi-1

cw = 4 x 10-6 psi-1cw = 4 x 10 psi


= 3.6 x 10-6 + (0) ( ? ) + (0.8) (8.144 x 10-3)

+ (0.2) (4 x 10-6)

= 6.52 x 10-3 psi-1


Exercise 6

Calculate Compressibility for High-Pressure, Low-Permeability Gas Reservoir

Calculate total compressibility for the following situation. Assume a dry gas.

High pressure, low permeability gas reservoir

Sw = 35%, TDS = 22 wt %, gas gravity = 0.67, Tf = 270°F,

p = 5,000 psi, cf = 20×10-6 psi-1


cg = 1.447 x 10-4 psi-1

cw = 3.512 x10-6 psi-1


Exercise 6

Calculate Compressibility for High-Pressure, Low-Permeability Gas Reservoir

Solution


cf = 2.0 x 10-5 psi-1

So = 0

Sw = 0.35

Sg = 1 - So - Sw = 1 - 0 - 0.35 = 0.65


cg = 1.447 x 10-4 psi-1

cw = 3.512 x 10-6 psi-1


= 2.0 x 10-5 + (0) ( ? ) + (0.65) (1.447 x 10-4)

+(0.35) (3.512 x 10-6)

= 1.15 x 10-4 psi-1

01-Introduction to Well Testing

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