petroleum reservoir engineering ii lecture 10: material
TRANSCRIPT
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Tishk International UniversityEngineering FacultyPetroleum and Mining Engineering Department
Petroleum Reservoir Engineering II
Third Grade- Spring Semester 2020-2021
Lecture 10: Material Balance Equation (Gas Reservoirs)
Instructor: Sheida Mostafa Sheikheh
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Gas Reservoirs
β Reservoirs containing only free gas are termed gas reservoirs.
β Such a reservoir contains a mixture of hydrocarbons, which exists wholly in the
gaseous state.
β The mixture may be dry, wet, or condensate gas, depending on the composition of
the gas, along with the pressure and temperature at which the accumulation exists.
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Gas Reservoirs
β If the reservoir temperature T is greater than the critical temperature ππ of the
hydrocarbon fluid, the reservoir is considered as a gas reservoir.
β Based on the phase diagram and prevailing reservoir conditions, natural gases can
be classified into four categories:
1. Retrograde gas-condensate
2. Near-Critical gas-condensate
3. Wet gas
4. Dry gas
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Gas Reservoirs
β Gas reservoirs may have water influx from a contiguous water-bearing portion of the
formation or may be volumetric (i.e., have no water influx)
β Volumetric Gas Reservoir: A volumetric gas reservoir is completely enclosed
by low-permeability or completely impermeable barriers and does not receive
pressure support from external sources, such as an encroaching aquifer.
β Non-volumetric Gas Reservoir: Gas reservoirs with water influx from an
aquifer are non-volumetric reservoirs and they produce under the pressure
support provided by the encroaching water.
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Gas Reservoirs
β Most gas engineering calculations involve the use of gas formation volume factor π΅π
and gas expansion factor πΈπ.
β’ Gas Formation Volume Factor (π΅π): is defined as the actual volume occupied by n
moles of gas at a specified pressure and temperature, divided by the volume
occupied by the same amount of gas at standard conditions.
β’ Gas Expansion Factor (πΈπ): is simply the reciprocal of π΅π.
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Gas Reservoirs
β Through this lecture and the next lecture two approaches for estimating initial gas-
in-place G, gas reserves, and the gas recovery for volumetric and water-drive
mechanisms:
β’ Volumetric method
β’ Material balance approach
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Content:
β The Material Balance Method
β’ Form 1. In terms of p/z
β’ Form 2. In terms of π΅π
β Application of Material Balance Method
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Gas Reservoirs
β The Material Balance Method:
βͺ If enough production-pressure history is available for a gas reservoir, the initial gas-
in-place G, the initial reservoir pressure ππ, and the gas reserves can be calculated
without knowing A, h, ΙΈ, or ππ€.
βͺ This is accomplished by forming a mass or mole balance on the gas as:
ππ = ππ β ππ ββ β 1
Where ππ= moles of gas produced
ππ= moles of gas initially in the reservoir
ππ= moles of gas remaining in the reservoir
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Gas Reservoirs
β The Material Balance Method:
βͺ Representing the gas reservoir by an
idealized gas container, as shown
schematically in the figure, the gas moles
in equation (1) can be replaced by their
equivalents using the real gas law to give:
ππ ππΊπ
π ππ π=
πππ
π§ππ πβ
π π β ππ β ππ
π§π πββ β(2)
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Gas Reservoirs
β The Material Balance Method:
ππ ππΊπ
π ππ π=
πππ
π§ππ πβ
π π β ππ β ππ
π§π πββ β(2)
Where ππ= initial reservoir pressure πΊπ= cumulative gas production, scf
p= current reservoir pressure V= original gas volume, ππ‘3
π§π= gas deviation factor at ππ z= gas deviation factor at p
π€π= cumulative water influx, ππ‘3 π€π= cumulative water production, ππ‘3
T= temperature, Β°R
βͺ Equation (2) is essentially the general material balance equation (MBE).
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Gas Reservoirs
β The Material Balance Method:
βͺ Equation (2) can be expressed in numerous forms depending on the type of the application
and the driving mechanism.
βͺ In general, dry gas reservoirs can be classified into two categories:
β’ Volumetric gas reservoirs
β’ Water-drive gas reservoirs
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Volumetric Gas Reservoirs
β For a volumetric reservoir and assuming no water production, equation (2) is reduced to:
ππ ππΊπ
ππ π=
ππ
π§πππ β
π
π§ππ ββ β 3
β Equation (3) is commonly expressed in the following two forms:
Form 1. In terms of p/z
Form 2. In terms of π΅π
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
β Rearranging equation (3) and solving for p/z gives:
π
π§=
ππ
π§πβ
ππ ππ
ππ πππΊπ ββ β(4)
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
β Equation (4) is an equation of a straight
line when (p/z) is plotted versus the
cumulative gas production πΊπ, as shown
in the figure.
β This straight-line relationship is perhaps
one of the most widely used relationships
in gas-reserve determination.
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
β The straight-line relationship provides the
engineer with the reservoir
characteristics:
β’ Slope of the straight line is equal to:
π ππππ =ππ ππ
ππ ππββ β(5)
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
β The straight-line relationship provides the
engineer with the reservoir
characteristics:
β’ The original gas volume V can be
calculated from the slope and used to
determine the areal extent of the
reservoir from:
π = 43,560 π΄βπ 1 β ππ€π
Where A is the reservoir area in acres.
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
β The straight-line relationship provides the
engineer with the reservoir
characteristics:
β’ Intercept at πΊπ = 0 gives Ξ€ππ π§π
β’ Intercept at p/z=0 gives the gas initially in
place G in scf
β’ Cumulative gas production or gas
recovery at any pressure
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
Example: A volumetric gas reservoir has the following production history.
The following data are also available:
ΙΈ= 13% ππ€π = 0.52 A=1060 acres h= 54 ft T=164 Β°F
Calculate the gas initially in place volumetrically and from the MBE.
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
Solution:
Step 1. Calculate π΅ππ from equation (1):
π΅ππ = 0.02827(0.869)(164 + 460)
1798= 0.00853 ππ‘3/π ππ
Step 2. Calculate the gas initially in place volumetrically by applying equation (3):
πΊ = 43,5601060 54 0.13 1 β 0.52
0.00853= 18.2 ππππ ππ
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Volumetric Gas Reservoirs
Form 1. In terms of p/z:
Solution:
Step 3. Plot p/z versus πΊπ as shown in
following figure and determine G.
G= 14.2 MMMscf
This checks the volumetric calculations.
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Volumetric Gas Reservoirs
Form 2. In terms of π©π:
β From the definition of the gas formation volume factor, it can be expressed as:
π΅ππ =π
πΊ
β Combining the above expression with π΅π equation:
π΅π =ππ π
ππ π
π§π
π= 0.02827
π§π
π
Gives:
ππ π
ππ π
π§ππ
ππ=
π
πΊββ β(6)
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Volumetric Gas Reservoirs
Form 2. In terms of π©π:
ππ π
ππ π
π§ππ
ππ=
π
πΊββ β(6)
Where V= volume of gas originally in place, ππ‘3
G= volume of gas originally in place, scf
ππ= original reservoir pressure
π§π= gas compressibility factor at ππ
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Volumetric Gas Reservoirs
Form 2. In terms of π©π:
β Combining equation (6) with equation (3):
ππ π
ππ π
π§ππ
ππ=
π
πΊ
ππ ππΊπ
ππ π=
ππ
π§πππ β
π
π§ππ
Gives:
πΊ =πΊπ π΅π
π΅π β π΅ππββ β(7)
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Volumetric Gas Reservoirs
Form 2. In terms of π©π:
Example: after producing 360 MMscf of gas from a volumetric gas reservoir, the pressure has
declined from 3,200 psi to 3,000 psi, given:
π΅ππ = 0.005278 Ξ€ππ‘3 π ππ
π΅π = 0.005390 Ξ€ππ‘3 π ππ
a. Calculate the gas initially in place.
b. Recalculate the gas initially in place assuming that the pressure measurements were
incorrect, and the true average pressure is 2,900 psi. The gas formation volume factor at
this pressure is 0.00558 Ξ€ππ‘3 π ππ.
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Volumetric Gas Reservoirs
Form 2. In terms of π©π:
Solution:
a. Using equation (7), calculate G:
πΊ =πΊπ π΅π
π΅π β π΅ππ=
360 β 106(0.00539)
0.00539 β 0.005278= 17.325 ππππ ππ
b. Recalculate G by using the correct value of π΅π:
πΊ =360 β 106(0.00668)
0.00558 β 0.005278= 6.652 ππππ ππ