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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
Gas Reservoirs
■ Reservoirs containing only free gas are termed gas reservoirs.
■ Such a reservoir contains a mixture of hydrocarbons, which exists wholly in the
gaseous state.
■ 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.
Gas Reservoirs
■ If the reservoir temperature T is greater than the critical temperature 𝑇𝑐 of the
hydrocarbon fluid, the reservoir is considered as a gas reservoir.
■ 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
Gas Reservoirs
■ 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)
■ 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.
■ 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.
Gas Reservoirs
■ Most gas engineering calculations involve the use of gas formation volume factor 𝐵𝑔
and gas expansion factor 𝐸𝑔.
• Gas Formation Volume Factor (𝐵𝑔): 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.
• Gas Expansion Factor (𝐸𝑔): is simply the reciprocal of 𝐵𝑔.
Gas Reservoirs
■ 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:
• Volumetric method
• Material balance approach
Content:
■ The Material Balance Method
• Form 1. In terms of p/z
• Form 2. In terms of 𝐵𝑔
■ Application of Material Balance Method
Gas Reservoirs
■ The Material Balance Method:
▪ If enough production-pressure history is available for a gas reservoir, the initial gas-
in-place G, the initial reservoir pressure 𝑝𝑖, and the gas reserves can be calculated
without knowing A, h, ɸ, or 𝑆𝑤.
▪ This is accomplished by forming a mass or mole balance on the gas as:
𝑛𝑝 = 𝑛𝑖 − 𝑛𝑓 −− − 1
Where 𝑛𝑝= moles of gas produced
𝑛𝑖= moles of gas initially in the reservoir
𝑛𝑓= moles of gas remaining in the reservoir
Gas Reservoirs
■ The Material Balance Method:
▪ 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:
𝑝𝑠𝑐𝐺𝑝
𝑅𝑇𝑠𝑐=
𝑝𝑖𝑉
𝑧𝑖𝑅𝑇−
𝑝 𝑉 − 𝑊𝑒 − 𝑊𝑝
𝑧𝑅𝑇−− −(2)
Gas Reservoirs
■ The Material Balance Method:
𝑝𝑠𝑐𝐺𝑝
𝑅𝑇𝑠𝑐=
𝑝𝑖𝑉
𝑧𝑖𝑅𝑇−
𝑝 𝑉 − 𝑊𝑒 − 𝑊𝑝
𝑧𝑅𝑇−− −(2)
Where 𝑝𝑖= initial reservoir pressure 𝐺𝑝= cumulative gas production, scf
p= current reservoir pressure V= original gas volume, 𝑓𝑡3
𝑧𝑖= gas deviation factor at 𝑝𝑖 z= gas deviation factor at p
𝑤𝑒= cumulative water influx, 𝑓𝑡3 𝑤𝑒= cumulative water production, 𝑓𝑡3
T= temperature, °R
▪ Equation (2) is essentially the general material balance equation (MBE).
Gas Reservoirs
■ The Material Balance Method:
▪ Equation (2) can be expressed in numerous forms depending on the type of the application
and the driving mechanism.
▪ In general, dry gas reservoirs can be classified into two categories:
• Volumetric gas reservoirs
• Water-drive gas reservoirs
Volumetric Gas Reservoirs
■ For a volumetric reservoir and assuming no water production, equation (2) is reduced to:
𝑝𝑠𝑐𝐺𝑝
𝑇𝑠𝑐=
𝑝𝑖
𝑧𝑖𝑇𝑉 −
𝑝
𝑧𝑇𝑉 −− − 3
■ Equation (3) is commonly expressed in the following two forms:
Form 1. In terms of p/z
Form 2. In terms of 𝐵𝑔
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
■ Rearranging equation (3) and solving for p/z gives:
𝑝
𝑧=
𝑝𝑖
𝑧𝑖−
𝑝𝑠𝑐𝑇
𝑇𝑠𝑐𝑉𝐺𝑝 −− −(4)
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
■ Equation (4) is an equation of a straight
line when (p/z) is plotted versus the
cumulative gas production 𝐺𝑝, as shown
in the figure.
■ This straight-line relationship is perhaps
one of the most widely used relationships
in gas-reserve determination.
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
■ The straight-line relationship provides the
engineer with the reservoir
characteristics:
• Slope of the straight line is equal to:
𝑠𝑙𝑜𝑝𝑒 =𝑝𝑠𝑐𝑇
𝑇𝑠𝑐𝑉−− −(5)
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
■ The straight-line relationship provides the
engineer with the reservoir
characteristics:
• The original gas volume V can be
calculated from the slope and used to
determine the areal extent of the
reservoir from:
𝑉 = 43,560 𝐴ℎ𝜑 1 − 𝑆𝑤𝑖
Where A is the reservoir area in acres.
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
■ The straight-line relationship provides the
engineer with the reservoir
characteristics:
• Intercept at 𝐺𝑝 = 0 gives Τ𝑝𝑖 𝑧𝑖
• Intercept at p/z=0 gives the gas initially in
place G in scf
• Cumulative gas production or gas
recovery at any pressure
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:
ɸ= 13% 𝑆𝑤𝑖 = 0.52 A=1060 acres h= 54 ft T=164 °F
Calculate the gas initially in place volumetrically and from the MBE.
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
Solution:
Step 1. Calculate 𝐵𝑔𝑖 from equation (1):
𝐵𝑔𝑖 = 0.02827(0.869)(164 + 460)
1798= 0.00853 𝑓𝑡3/𝑠𝑐𝑓
Step 2. Calculate the gas initially in place volumetrically by applying equation (3):
𝐺 = 43,5601060 54 0.13 1 − 0.52
0.00853= 18.2 𝑀𝑀𝑀𝑠𝑐𝑓
Volumetric Gas Reservoirs
Form 1. In terms of p/z:
Solution:
Step 3. Plot p/z versus 𝐺𝑝 as shown in
following figure and determine G.
G= 14.2 MMMscf
This checks the volumetric calculations.
Volumetric Gas Reservoirs
Form 2. In terms of 𝑩𝒈:
■ From the definition of the gas formation volume factor, it can be expressed as:
𝐵𝑔𝑖 =𝑉
𝐺
■ Combining the above expression with 𝐵𝑔 equation:
𝐵𝑔 =𝑝𝑠𝑐
𝑇𝑠𝑐
𝑧𝑇
𝑝= 0.02827
𝑧𝑇
𝑝
Gives:
𝑝𝑠𝑐
𝑇𝑠𝑐
𝑧𝑖𝑇
𝑝𝑖=
𝑉
𝐺−− −(6)
Volumetric Gas Reservoirs
Form 2. In terms of 𝑩𝒈:
𝑝𝑠𝑐
𝑇𝑠𝑐
𝑧𝑖𝑇
𝑝𝑖=
𝑉
𝐺−− −(6)
Where V= volume of gas originally in place, 𝑓𝑡3
G= volume of gas originally in place, scf
𝑝𝑖= original reservoir pressure
𝑧𝑖= gas compressibility factor at 𝑝𝑖
Volumetric Gas Reservoirs
Form 2. In terms of 𝑩𝒈:
■ Combining equation (6) with equation (3):
𝑝𝑠𝑐
𝑇𝑠𝑐
𝑧𝑖𝑇
𝑝𝑖=
𝑉
𝐺
𝑝𝑠𝑐𝐺𝑝
𝑇𝑠𝑐=
𝑝𝑖
𝑧𝑖𝑇𝑉 −
𝑝
𝑧𝑇𝑉
Gives:
𝐺 =𝐺𝑝 𝐵𝑔
𝐵𝑔 − 𝐵𝑔𝑖−− −(7)
Volumetric Gas Reservoirs
Form 2. In terms of 𝑩𝒈:
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:
𝐵𝑔𝑖 = 0.005278 Τ𝑓𝑡3 𝑠𝑐𝑓
𝐵𝑔 = 0.005390 Τ𝑓𝑡3 𝑠𝑐𝑓
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 Τ𝑓𝑡3 𝑠𝑐𝑓.
Volumetric Gas Reservoirs
Form 2. In terms of 𝑩𝒈:
Solution:
a. Using equation (7), calculate G:
𝐺 =𝐺𝑝 𝐵𝑔
𝐵𝑔 − 𝐵𝑔𝑖=
360 ∗ 106(0.00539)
0.00539 − 0.005278= 17.325 𝑀𝑀𝑀𝑠𝑐𝑓
b. Recalculate G by using the correct value of 𝐵𝑔:
𝐺 =360 ∗ 106(0.00668)
0.00558 − 0.005278= 6.652 𝑀𝑀𝑀𝑠𝑐𝑓
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