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Introduction to Well Testing 1
Introduction to Well Testing 2
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.
Introduction to Well Testing 3
Introduction to Well Testing 4
Introduction to Well Testing 5
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
Introduction to Well Testing 6
• 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?
Introduction to Well Testing 7
• 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
Introduction to Well Testing 8
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.
Introduction to Well Testing 9
Introduction to Well Testing 10
• Obtained by combining• Obtained by combining
– Continuity equation
– Equation of state for slightly compressible liquids
– Flow equation - Darcy’s law
Introduction to Well Testing 11
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&
Introduction to Well Testing 12
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
Introduction to Well Testing 13
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
Introduction to Well Testing 14
• 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
Introduction to Well Testing 15
Introduction to Well Testing 16
• 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
Introduction to Well Testing 17
• 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
Introduction to Well Testing 18
• 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
Introduction to Well Testing 19
• 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
Introduction to Well Testing 20
• 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
Introduction to Well Testing 21
• 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
• Range or Typical Value
– 4x10-6 psi-1, well-consolidated sandstone
– 30x10-6 psi-1, unconsolidated sandstone
– 4 to 50 x 10-6 psi-1 consolidated limestones
Introduction to Well Testing 22
• 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
• Range or Typical Value
– May be as small as 5 ft or even less
– May be as large as 1,000 ft or more
Introduction to Well Testing 23
• 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
• Range or Typical Value
– 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
Introduction to Well Testing 24
• Wellbore radius is the size of wellbore.• Wellbore radius is the size of wellbore.
• Symbol
– rw
• Units
– feet
• Source
– Bit diameter/2
– Caliper log
• Range or Typical Value
– 2 to 8 in.
Introduction to Well Testing 25
• 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
• Range or Typical Value
– See exercises
Introduction to Well Testing 26
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.
Introduction to Well Testing 27
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
Introduction to Well Testing 28
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
Introduction to Well Testing 29
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
From fluid properties correlations,
pb = 2803 psi
co = 1.158 x 10-5
cw = 2.277 x 10-6
ct = cf + So co + Sw cw + Sg cg
= 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
Introduction to Well Testing 30
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
From fluid properties correlations,
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
Introduction to Well Testing 31
Exercise 4
Calculate Compressibility for Saturated Oil Reservoir
Solution
ct = cf + So co + Sw cw + Sg cg
cf = 3.6 x 10-6 psi-1
Sw = 0.17
Sg = 0.05
So = 1 - 0.17 - 0.05 = 0.78
From fluid properties correlations,
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
ct = cf + So co + Sw cw + Sg cg
= 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
Introduction to Well Testing 32
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]
From fluid properties correlations,
cg = 8.144 x 10-3 psi-1
cw = 4x10-6 psi-1
Introduction to Well Testing 33
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
From fluid properties correlations,
cg = 8.144 x 10-3 psi-1
cw = 4 x 10-6 psi-1cw = 4 x 10 psi
ct = cf + So co + Sg cg + Sw cw
= 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
Introduction to Well Testing 34
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
From fluid properties correlations,
cg = 1.447 x 10-4 psi-1
cw = 3.512 x10-6 psi-1
Introduction to Well Testing 35
Exercise 6
Calculate Compressibility for High-Pressure, Low-Permeability Gas Reservoir
Solution
ct = cf + So co + Sg cg + Sw cw
cf = 2.0 x 10-5 psi-1
So = 0
Sw = 0.35
Sg = 1 - So - Sw = 1 - 0 - 0.35 = 0.65
From fluid properties correlations,
cg = 1.447 x 10-4 psi-1
cw = 3.512 x 10-6 psi-1
ct = cf + So co + Sg cg + Sw cw
= 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