Petroleum Reservoir Engineering II Lecture 3: Fundamentals

Tishk International UniversityEngineering FacultyPetroleum and Mining Engineering Department

Petroleum Reservoir Engineering II

Third Grade- Spring Semester 2020-2021

Lecture 3: Fundamentals of Rock Properties (Surface and Interfacial Tension)

Instructor: Sheida Mostafa Sheikheh

Fundamentals of Rock Properties:

■ The material of which a petroleum reservoir rock may be composed can range from

very loose and unconsolidated sand to a very hard and dense sandstone, limestone,

or dolomite.

■ The grains may be bonded together with a number of materials, the most common

of which are silicate, calcite or clay.

■ Knowledge of the physical properties of the rock and the existing interaction

between hydrocarbon system and the formation is essential in understanding and

evaluating the performance of a given reservoir.


■ Rock properties are

determined by performing

laboratory analyses on cores

from the reservoir to be

evaluated.


■ There are basically two main categories of core analysis tests that are preformed on

core samples regarding physical properties of reservoir rocks.

Core Analysis Tests

Routine Core Analysis

Tests

Porosity Permeability Saturation

Special Tests

WettabilitySurface and Interfacial Tension

Capillary Pressure

Relative Permeability

Overburden Pressure

Surface and Interfacial Tension:

■ Interface:

❑ Interface: is the boundary between two or more phases exist together.

❑ The properties of the molecules forming the interface are different from those in the

bulk that these molecules are forming an interfacial phase.

❑ Several types of interface can exist depending on whether the two adjacent phases

are in solid, liquid or gaseous state.


■ Interface:


❑ In dealing with multiphase systems, it is necessary to consider the effect of the

forces at the interface when two immiscible fluids are in contact.

❑ When these two fluids are liquid and gas, the term surface tension is used to

describe the forces acting on the interface.

❑ When the interface is between two liquids, the acting forces are called interfacial

tension.

❑ The surface or interfacial tension has the units of force per unit of length, e.g.,

dynes/cm, and is usually denoted by the symbol σ.


❑ Surface tension is

defined as the force

per unit length

parallel to the

surface to

counterbalance the

net downward pull.


❑ Cohesive Forces: The force of cohesion is defined as the force of

attraction between molecules of the same substance.

❑ Adhesive Forces: The force of adhesion is defined as the force of

attraction between different substances, such as glass and water.



❑ In the case of interface, the

molecules at the interface will be

pulled by both faces into the bulk.

❑ Since Cohesive Forces (between

like molecules) are stronger than

Adhesive Force (between unlike

molecules) the net pull will be into

the bulk of same phase.


❑ Consider the two immiscible fluids, air (or gas)

and water (or oil) as shown in the figure.

❑ A liquid molecule, which is remote from the

interface, is surrounded by other liquid

molecules, thus having a resulting net

attractive force on the molecule of zero.

❑ A molecule at interface, however, has a force

acting on it from the air (gas) molecules lying

immediately above the interface and from

liquid molecules lying below the interface.

❑ Resulting forces are unbalanced and give rise

to surface tension.


❑ If a glass capillary tube is placed in a large open vessel containing water, the

combination of surface tension and wettability of tube to water will cause water to

rise in the tube above the water level in the container outside the tube.

❑ The water will rise in the tube until the total force acting to pull the liquid upward is

balanced by the weight of the column of liquid.

❑ Assuming the radius of the capillary tube is r, the total upward force 𝐅𝐮𝐩, which holds

the liquid up, is equal to the force per unit length of surface times the total length of

surface.


𝐹𝑢𝑝 = 2𝜋𝑟 𝜎𝑔𝑤 𝑐𝑜𝑠𝜃 −− −(1)

𝜎𝑔𝑤= surface tension between air (gas) and water

(oil), dynes/cm

𝜃= contact angle

r= radius, cm

𝐹𝑑𝑜𝑤𝑛 = 𝜋𝑟2ℎ 𝜌𝑤 − 𝜌𝑎𝑖𝑟 𝑔 = 𝜋𝑟2ℎ𝜌𝑤𝑔 −− −(2)

h= height to which the liquid is held, cm

g=acceleration due to gravity, cm/sec2

𝜌𝑤= density of water, gm/cm3

𝜌𝑎𝑖𝑟= density of gas, gm/cm3


Equating equation (1) and (2) and solving for the surface tension gives:

𝜎𝑔𝑤 =𝑟ℎ𝜌𝑤𝑔

2𝑐𝑜𝑠𝜃−− − 3

Because the density of oil is not negligible, equation (3) is used to calculate interfacial tension between oil and

water:

𝜎𝑜𝑤 =𝑟ℎ𝑔(𝜌𝑤 − 𝜌𝑜)

2𝑐𝑜𝑠𝜃−− − 4

Where 𝜌𝑜= density of oil, gm/cm3

𝜎𝑜𝑤= interfacial tension between the oil and the water, dynes/cm


■ Effect of Temperature on Surface Tension:

Do we desire to have

high interfacial tension or

low interfacial tension

between phases present in the reservoir?


■ Effect of Temperature on Surface Tension:

❑ Surface tension will be reduced when the temperature of the liquid increased.

❑ This is due to the thermal expansion of liquids.

❑ When temperature increases, the molecular thermal activity increases causing a decrease in

cohesive interaction.

❑ This causes decrease in surface tension.

❑ This continues till the temperature of the liquid reaches the Critical temperature of the liquid.

❑ At this point, surface tension becomes zero.


■ Effect of Pressure on Surface Tension:

❑ Another factor affecting the surface tension is pressure.

❑ In most applications, the pressure does not play a role but needs to be considered when

processes that happen under high pressures are studied.

❑ These include for example enhanced oil recovery (EOR).

❑ In general, dissolution of the gas in liquid increases with the increasing pressure, decreasing the

surface tension.


■ Surface Active Agents (Surfactant):

❑ Agents which used to lower surface tension of liquid & reduces interfacial tension between two

liquids.

❑ They contain “hydrophilic” & “hydrophobic” groups.

❑ Surfactants which have both polar & non-polar groups called “Amphiphiles”.



❑ Surfactants are amphiphilic molecules that include both hydrophobic (non-polar, water insoluble)

and hydrophilic (polar, water soluble) segments.

❑ When surfactants are dissolved in water, they orientated at the surface so that hydrophilic parts

are in water and hydrophobic parts are in air.

❑ The surface tension is reduced as some of the water molecules are replaced by the surfactant

molecules and interaction forces between surfactant and water is less than between two water

molecules.



❑ The effectiveness of surfactant molecule is determined by the amount of surfactant needed and

the minimum surface tension value that can be reached.

❑ In addition to efficiency of the surfactant, the speed of the surface-active agent is also of high

importance in industrial processes.

❑ That is the rate at which the molecules are able to arrange themselves at the interface.



Why Surfactants are Important in

Petroleum Engineering?

Content (Practical):

■ Measurement of Surface Tension:

✓ Capillary Rise Method

• Aim

• Apparatus/ Materials Required

• Theory

• Procedure

• Observation

• Calculation

• Result

Measurement of Surface Tension:

■ Capillary Rise Method:

Aim: To find the surface tension of water by capillary rise method.

Apparatus/ Materials Required:

✓ Three capillary tubes of different radii

✓ A tipped pointer clamped in a metallic plate with a handle

✓ Travelling microscope

✓ Adjustable height stand

✓ A flat bottom open dish

✓ Thermometer

✓ Clean water in a beaker

✓ Clamp and a stand




Theory: The surface tension of water is given by the formula:

σ =r(h +

r3)ρg

2cosθ



Procedure:

(a) Arranging the apparatus

1. Place the adjustable height stand on the table and make its base horizontal by levelling the screws.

2. Take dirt and grease free water in an open dish with a fat bottom and put it on top of the stand.

3. Take three capillary tubes of different radii.

4. Clean the tubes and dry them and then clamp them to a metallic plate to increase the radius. Clamp a pointer after third capillary tube.

5. Clamp the horizontal handle of the metallic plate in a vertical stand so that the capillary tube and the pointer become vertical.

6. Adjust the height of the metallic plate that the capillary tubes dip in the water in open dish.

7. Adjust the position of the pointer such that the tip touches the water surface.



Procedure:

(b) Measurement of capillary rise

1. Calculate the least count of the travelling microscope for vertical and horizontal scale.

2. Raise the microscope to a suitable height pointed towards the capillary tube with a horizontal axis.

3. Focus the microscope to the first capillary tube.

4. Make the horizontal cross wire touch the central part of the concave meniscus seen convex through the microscope

5. Note the reading of the microscope on the vertical scale.

6. Move the microscope horizontally and bring it in front of the second capillary tube.

7. Lower the microscope and repeat steps 11 and 12

8. Likewise, repeat steps 11 and 12 for the third capillary

9. Lower the stand for the pointer tip to be visible.

10. Move the microscope horizontally and bring it in front of the pointer.

11. Lower the microscope and make the horizontal cross wire touch the tip of the pointer.



Procedure:

(c) Measurement of the internal diameter of the capillary tube

1. Place the first capillary tube horizontally on the adjustable stand.

2. Focus the microscope on the end dipped in water. A white circle with a green strip will be visible.

3. Make the horizontal cross-wire touch the inner circle at point A.



Procedure:



Procedure:



Observation:

5.1 0.0155.1

4.95 0.021

5.15 0.013

3

3

3

0.008

0.008

0.008



Observation: Table for Determination of radius

Tube No. Reading of the left-hand side Reading of the right-hand side

MSR (a)

in cm

VSR (a) in

cm

Total (cm) MSR (b)

in cm

VSR (b) in

cm

Total (cm) D(a-b) r=D/2

1 6.05 0.012 5.9 0.018

2 6 0.015 5.85 0.023

3 5.9 0.02 5.75 0.03



Calculation:

Put the value h and r for each capillary tube separately and find the values of

σ using the following formula:

σ =r(h +

r3)ρg

2cosθ

Find the mean value of the obtained 𝜎 values as follows:

𝜎 = 𝜎1 + 𝜎2 + 𝜎3/3

The unit of 𝜎 is dynes.cm−1

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Petroleum Reservoir Engineering II Lecture 3: Fundamentals