Part 4 Diffusivity Equation Line Source Radial Flow Radius Invest

Diffusivity Equation - WTA 1

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• Obtained by combiningy g– Continuity equation– Equation of state for slightly compressible liquids– Flow equation - Darcy’s law

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The continuity equation is a restatement of the conservation of matter. y qThat 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.

NomenclatureA = Cross-sectional area open to flow, ft2

m = Rate of accumulation of mass within the volume, lbm/secv = Fluid velocity, ft/secρ = Density of fluid, lbm/ft3

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This equation describes the change in density with pressure for a q g y pliquid with small and constant compressibility.

Nomenclaturec = Compressibility, psi-1

P ip = Pressure, psiρ = Density of fluid, lbm/ft3

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NomenclatureA = Cross sectional area open to flow, cm2

k = Permeability, darciesL = Length of flow path, cmp = Pressure, atmΔp = Pressure difference between upstream and

downstream sides, atmq = Flow rate, cm3/secux = Flow velocity, cm/secx = Spatial coordinate, cmx Spatial coordinate, cmμ = Viscosity, cp

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• The diffusivity equation is obtained by combiningy q y g- The continuity equation- The equation of state for a slightly compressible liquid- Darcy’s law

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

- Gas flow equation- Multiphase flow equation

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L is here lengthg

T is time

In oil field units, η is in units of ft2/hr

h is the formation thickness (in field units in ft)

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P is here pressure in psi.p p

In oil field units, the units of ct is 1/psi.

cr is solid rock compressibility in 1/psi.

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Diffusivity Equation - PTT Interpretation and Analysis 12

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Here, pi is the initial pressure, which is assumed to be uniform in , p p ,the reservoir

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The Ei-function solution to the diffusivity equation assumes line y qsource well (finite size of wellbore can be neglected).This solution is valid only for r > rw .It predicts the pressure response in the reservoir as a function of both time t and distance from the center of the wellbore r.

The above pressure equation is in oil field units; k is in md, ct is in 1/psi, h is in ft etc., time t is in hours.

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The argument of the Ei-function, x, is given by:g , , g y

Short times or large distances ⇒ large x

ktrc948x

2tμφ

=

Long times or short distances ⇒ small x

For short times, x > 10, pressure response predicted by the Ei-function is negligible.For long times, x < 0.01, pressure response may be calculatedFor long times, x 0.01, pressure response may be calculated using the logarithmic approximation to the Ei-function.For intermediate times, 0.01 < x < 10, the full Ei-function must be used to calculate the pressure response.

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At any given point in the reservoir, at sufficiently early times, the y g p , y y ,pressure response is essentially negligible.This approximation applies whenever

10rc948 2t >

μφ .10kt

>

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At any given point in the reservoir, at sufficiently late times, the y g p , y ,pressure response is approximately logarithmic in time.This approximation applies whenever

rc948 2φμ .01.0kt

rc948 t <φμ

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Consider the pressure profile in an infinite-acting reservoir during p p g gdrawdown. At t = 0 the pressure is uniform throughout the reservoir.At t = 0.01 hours only a small region within 10 ft of the wellbore has shown the effects of the transient.Sometime later at t =1 hour the pressure transient has movedSometime later, at t =1 hour, the pressure transient has moved into a larger region, perhaps 100 ft from the wellbore.Still later, at t =100 hours, the pressure transient has moved even further from the wellbore.As production continues, the pressure transient continues to move through the reservoir until it has reached all of themove through the reservoir until it has reached all of the boundaries of the reservoir.

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• Assumptionsp– Radial flow– Infinite-acting reservoir– Homogeneous reservoir

• Effect of reservoir propertiesEffect of reservoir properties – Increasing porosity, viscosity, or total compressibility

increases the time required to reach a given radius of investigation.

– Increasing permeability decreases the time required to reach a given radius of investigationreach a given radius of investigation.

– Changing the rate has no effect on the radius of investigation.

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Consider what happens when we shut in the well from the ppprevious slide for a buildup.At the instant of shutin, Δt=0, the pressure begins to build up in the well. However, this rise in pressure does not affect the entire reservoir at once.At Δt = 0.01 hours, the pressure buildup has affected only that , p p ypart of the reservoir within about 10 ft of the wellbore. A pressure gradient still exists in the bulk of the reservoir. This means that fluid continues to flow in most of the reservoir, even during buildup.At Δt =1 hour, the pressure has built up in a larger area, within

b t 100 ft f th llbabout 100 ft of the wellbore.As the shutin period continues, the region within which the pressure has built up grows until the entire reservoir is at uniform pressure.

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Exercise 6R di l Fl d R di f I ti tiRadial Flow and Radius of InvestigationCalculate Radius of Investigation for an

Undersaturated Oil Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 2 in the section “Part-2 Basics_WTA,” with the following additional information.Undersaturated oil reservoir (above the bubblepoint)

φ = 0.17

μ = 1.06 cp

ct = 1.36x10-5 psi-1

ko = 250 md

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Exercise 7R di l Fl d R di f I ti tiRadial Flow and Radius of InvestigationCalculate Radius of Investigation for a

Saturated Oil Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 3 in the section “Part_2_Basics_WTA” with the following additional information.

Saturated oil reservoir (below the original bubblepoint)( g p )

φ = 0.17

μ = 1.185 cp

ct = 1.42 x 10-4 psi-1

kro = 0.8

k = 250 md (absolute permeability)

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Low-Pressure, High-Permeability Gas Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 4 in the section “Part-2_Basics_WTA” with the following additional information.Low-pressure, high-permeability gas reservoir

φ = 0.12

μ = 0.01151 cp

ct = 6.52 x 10-3 psi-1

k = 100 md

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High-Pressure, Low-Permeability Gas Reservoir

Calculate the time required to reach a radius of investigation of 745 feet for the following situation. Use the data and results from Exercise 5 in the section “Part-2 Basics_WTA,” with the following additional information.

High-pressure, low-permeability gas reservoir

φ = 0.04

μ = 0.02514 cp

ct = 1.151 x 10-4 psi-1

k = 0.08

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Part 4 Diffusivity Equation Line Source Radial Flow Radius Invest

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