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