Cryogenic Valves for Liquefied Natural Gas Plants

By Karan Sotoodeh

Natural gas and liquefied natural gas (LNG) continue to grow as a part of the sustainable energy mix. While oil and gas companies look to lower emissions, one key refinery component that contributes up to 60% of emissions are valves, mainly due to poor design, sealing, and testing. Cryogenic Valves for Liquefied Natural Gas Plants delivers a much-needed reference that focuses on the design, testing, maintenance, material selection, and standards needed to stay environmentally compliant at natural gas refineries.

Covering technical definitions, case studies, and Q&A, the reference includes all ranges of natural gas compounds, including LPG, CNG, NGL, and PNG. Key design considerations are included that are specific for cryogenic services, including a case study on cryogenic butterfly valves. The material selection process can be more complex for cryogenic services, so the author goes into more detail about materials that adhere to cryogenic temperature resistance. Most importantly, testing of valves is covered in depth, including shell test, closure or seat test, and thermal shock tests, along with tactics on how to prevent dangerous cryogenic leaks, which are very harmful to the environment. The book is a vital resource for today’s natural gas engineers.

• Teaches LNG valve design, including sealing selection, wall thickness calculation of the valve body and bonnet, and proper material selection
• Provides tactics on how to prevent cryogenic leaks with compliant valve testing
• Applies natural gas calculations that will better support the LNG supply chain
• Enables readers to understand cryogenic valve standards, including EN, ISO, and MSS SP

• Language: English
• Publisher: Elsevier Science
• Release date: May 18, 2022
• ISBN: 9780323995856

Karan Sotoodeh

Karan Sotoodeh recently earned his PhD in Safety and Reliability in Mechanical Engineering from the University of Stavanger. Previously, Karan was the Senior / Lead Engineer in valves and actuators for Baker Hughes, one of the world’s largest oil field services company. He was responsible for engineering and delivering valves and actuators in subsea manifolds, working with valve suppliers, R&D activities, and maintaining the company’s valve database. He has also worked for AkerSolutions, NLI Engineering, and Nargan Engineers as a senior specialist in piping and valves, assisting with many projects around the world. He is the author of Prevention of Valve Fugitive Emissions in the Oil and Gas Industry and Subsea Valves and Actuators for the Oil and Gas Industry, both published by Elsevier. Karan earned a Master of Research in Mechanical Engineering and a Masters in Oil and Gas Engineering, both from Robert Gordon University of Aberdeen, and a Bachelors in Industrial Engineering from the Iran University of Science and Technology

Book preview

Chapter One: Natural gas technology

Abstract

This chapter is dedicated to natural gas technology. Different topics such as chemical composition and physical properties of natural gas are discussed in this chapter. Definitions such as lean gas, dry gas, saturated gas, sweet and sour gas are explained in this chapter. Various gas physical properties such as specific gravity, viscosity, critical pressure and temperature, Z or compressibility factor, formation volume factor and expansion factor, density are covered in this chapter. The chapter is rich in formulas and calculations related to natural gas such as specific gravity and viscosity, effect of temperature on viscosity, density, formation volume factor, pseudocritical pressure and temperature, etc. Besides, rules related to the relationship between pressure, temperature and volume for both ideal and real gases are explained. In addition, history of natural gas and how it forms in geological formation are explained in this chapter. Gas flaring that used to be common before and its negative effects on the environment like greenhouse effect and global warming are discussed further. Applications of natural gas in the industry, electricity generation and injection into the reservoir for advanced oil recovery is the last section of this chapter.

Keywords:

Natural gas; Specific gravity; Viscosity; Z or compressibility factor; Density; Formation volume factor and expansion factor; Real gas; Ideal gas; Gas flaring; Greenhouse effect and global warming

1.1: Chemical composition

Natural gas, also called fossil gas or just gas for short, is a non-renewable source of energy. Natural gas is the combination of hydrocarbon and non-hydrocarbon gasses found in the porous formation called the reservoir under the earth’s surface with oil, and the principal compound of natural gas is methane. Methane is a colorless and odorless gas with the chemical formula of CH4 (one carbon atom and four hydrogen atoms, as illustrated in Fig. 1.1). It is very flammable, burns rapidly in the air, and forms mainly carbon dioxide and water vapor during combustion. Although methane is a stable gas, a mixture of a small percentage of methane, between 5% and 14%, and air is extremely explosive. Methane can be released from the coal in coal mines and cause an explosion as a result of mixture with the air. A hydrocarbon is an organic molecule compound that is made of carbon and hydrogen. Hydrocarbons are divided into three main categories: alkane, alkene, and alkyne. Alkane has single bonds in its atomic structure. Alkene has double bonds in its atomic structure, and alkyne has triple bonds. The ten main hydrocarbons are methane (CH4), ethane (C2H6),  propane (C3H8),  butane (C4H10), pentane (C5H12), hexane (C6H14), heptane (C7H16), octane (C8H18), nonane (C9H20), and decane (C10H22). All of these hydrocarbons are alkane, with a single bond between carbon and hydrogen atoms, and the general formula of alkane compounds is CnH2n + 2.

Fig. 1.1

Fig. 1.1 Methane atom structure. (Courtesy: Shutterstock.)

Natural gas is a colorless, highly flammable mixture of gases. It is a type of petroleum and hydrocarbon that is produced as a byproduct of oil in deep underground reservoirs. Fig. 1.2 illustrates two oil wells, one in the land and the other in the sea, that are drilled into reservoirs where gas is trapped above the oil and water. The typical composition of natural gas includes both hydrocarbon and non-hydrocarbon compounds. The main hydrocarbon gas in natural gas is methane, which can comprise 70–90% of the natural gas’s volume. The non-hydrocarbon gases in natural gas are nitrogen, hydrogen, oxygen, carbon dioxide, and hydrogen sulfide, which are minor in percentage of volume and inorganic. The inorganic compounds in natural gas are not desirable since they are not combustible and cause corrosion and other types of problems in the production and processing of natural gas. Natural gas is used for the generation of electricity, heating, cooking, and as a fuel for certain vehicles. Table 1.1 displays the chemical composition of natural gas. It is important to know that methane, ethane, propane, and butane exist as gases at normal temperatures and pressures, while other hydrocarbon compounds are liquid at normal temperatures and pressures.

Fig. 1.2

Fig. 1.2 Drilling a land-based well and a subsea well into reservoirs where natural gas is trapped above oil and water. (Courtesy: Shutterstock.)

Fig. 1.3

Fig. 1.3 The different atomic arrangements of n-butane and isobutane.

Table 1.1

Note 1: n-butane is the same as butane, which is an alkane with a formula of (C4H10). Isobutane is an isomer of butane. Isomers in chemistry refer to molecules with identical formulas, meaning that isomers have the same number of atoms but in different arrangements. Fig. 1.3 illustrates the different atomic arrangements of n-butane and isobutane.

Note 2: n-pentane and ISO pentane are also isomers.

1.2: Gas definitions

It is important to know that raw natural gas produced from a well must be processed to be a marketable product. In short, processing a natural gas involves the removal of water vapor, toxic and corrosive compounds such as hydrogen sulfide (H2S), and carbon dioxide (CO2), as well as the separation of the condensable hydrocarbon compounds. Natural gas can be classified into several broad categories, which are explained below:

Wet gas: Wet gas contains heavy, condensable hydrocarbons such as propane, butane, and pentane. More precisely, if more than 5% of a gas is composed of the above-mentioned heavy hydrocarbon molecules and less than 95% is methane, it is classified as a wet gas. Wet gas should not be confused with saturated gas, which is explained in the next paragraph.

Saturated gas: Saturated gas refers to gas before the dehydration process in which the gas is saturated with water. Gas dehydration is the process of extracting the moisture and water from the natural gas and gasses mixture.

Lean gas: In general, lean gas represents a kind of natural gas with the absence of condensable hydrocarbons. More precisely, lean gas is the opposite of wet gas, containing more than 95% methane and less than 5% condensable or heavy hydrocarbon molecules.

Dry gas: Water moisture exists in every type of gas, but the quantity of it varies. Dry gas is a type of gas with negligible water content, less than 7 pounds per 1 million cubic feet of gas. One million cubic feet of gas is abbreviated as MMCF, where the two M’s mean one thousand thousand or one million. A related term is absolutely dry natural gas, which is defined as natural gas with zero water vapor.

Sour gas: One of the undesirable byproducts of oil and gas is hydrogen sulfide (H2S). elemental sulfur is another undesirable compound that can make the gas sour. Hydrogen sulfide is colorless, flammable, toxic, and corrosive. Therefore, hydrogen sulfide should be removed from hydrocarbon. Natural gas is considered sour if the amount of hydrogen sulfide is more than 5.7 mg/m³ of natural gas, which is equivalent to approximately 4 ppm by volume under standard temperature and pressure.

Sweet gas: Sweet gas refers to a type of natural gas with less than 4 ppm of hydrogen sulfide. However, sweet gas can contain carbon dioxide, which creates a sweet and uniform corrosion type.

1.3: Physical properties

It is essential for engineers to understand these gas properties in order to be able to design and analyze gas production and processing systems and plants. The physical properties of gas are largely dependent on the chemical composition of the gas. The properties of natural gas may be determined either directly from laboratory tests or through calculations based on the known chemical composition of the gas. The process of calculating the physical properties of natural gas based on the physical properties and composition of the individual components is sometimes called mixing rules. This section provides some information about the properties of natural gas.

The properties of natural gas include the specific gravity, viscosity, density, compressibility of the Z-factor, pseudocritical pressure, and temperature; they must be known in order to design and analyze natural gas production and processing systems. As explained before, natural gas is a combination of light hydrocarbons with a small number of inorganic elements or compounds, such as nitrogen, carbon dioxide, etc. It is always ideal and important to determine the chemical composition of natural gas through measurement. Gas composition, as will be shown in different examples in this chapter, is typically reported as the mole fractions of the different components. If the gas composition is known, it is usually possible to calculate the gas properties through certain rules, formulas, and correlations. The next section focuses on the calculation and estimation of the different natural gas properties mentioned above.

1.3.1: Specific gravity of gas

Specific gravity of gas: The specific gravity of a gas, parameter SG, is a dimensionless unit that is defined as the ratio of the molecular weight of the natural gas to the molecular weight of air. Specific gravity is also called relative density. Eq. (1.1) is used to calculate the specific gravity of natural gas based on the above-mentioned definition.

Equation 1.1: Specific gravity calculation

si3_e    (1.1)

Where:

γ = SGg = Specific gravity of the gas (dimensionless).

Mg = Molar mass of gas si4_e .

Ma= Molar mass of air, which is equal to 28.97 si4_e .

The mole is the unit of measurement for the amount of a substance in the international system of units. A mole is used as a unit in particle counting. A mole contains approximately 6 × 10²³ particles, which can be atoms, molecules, ions, or electrons. The molar mass of a substance refers to the mass or weight of 1 mol of that substance. The molecular weight of air given that the air is dry and contains approximately 79% nitrogen and 21% oxygen is equal to 28.97 si4_e (molar mass). But the question is how is the molar mass of air calculated? Oxygen’s molar mass is approximately 15.9994 si4_e , and nitrogen has a molar mass of 14.0067 si4_e . But it is important to know that both nitrogen and oxygen are diatomic, meaning they contain two atoms, like O2 and N2. Therefore, the mass molar of oxygen gas in air is approximately 32 si4_e , and the molar mass of nitrogen gas in air is approximately 28 si4_e . The mixture or average of molar mass is the sum of the mole fractions of each gas multiplied by its fraction according to Eq. (1.2).

Equation 1.2: Average molar mass calculation of gas mixtures such as air

si11_e

   (1.2)

Where:

Mmg = Molar mass of the mixture of gasses such as air si4_e ;

Xi = Mole fraction or percentage of each gas in the mixture or air (dimensionless);

Mi = The molar mass of each gas si4_e ;

Using the Eq. (1.2), it is possible to calculate the molar mass of the air as follows:

si14_e

The specific gravity of air is approximately equal to one. If the specific gravity of a gas is more than that of air, which is approximately one, then that gas is heavier than air. On the other hand, if the specific gravity of a gas is less than one, then that gas is lighter than air. Natural gas is lighter than air, and its specific gravity value is usually in the range of 0.55 to 0.87. A natural gas including rich gases or a higher wet gas content and condensates typically has higher specific gravity compared to lean gas with a smaller number of condensates and higher rates of methane. Carbon dioxide (CO2) which causes sweet and uniform corrosion in non-corrosion resistant alloys (CRAs) such as carbon steel, has a specific gravity of 1.5189. This means that carbon dioxide is heavier than air and the presence of carbon dioxide can increase the specific gravity of natural gas.

Example 1.1

Gas compositions, molar fractions, and molar masses are provided in Table 1.2. Calculate the relative density or specific gravity of each gas.

Table 1.2

Answer

The first step is to calculate the molar weight of the natural gas (Mg) using the data provided in the table and using Eq. (1.2) as follows:

si15_e

The molar weight of the air is equal to 28.97 si16_e , so it is now possible to calculate the specific gravity using Eq. (1.1) as follows:

si17_e

A light gas contains mainly methane and some ethane. The specific gravity of pure methane as the lightest gas is 0.55. A rich or heavy gas has a specific gravity of approximately 0.75 and in some cases more than 0.9. Based on the given information, it can be concluded that the given gas in this example is light to medium weight.

There are two important laws that should be discussed here regarding the pressure, volume and mole fraction of ideal gases: the first is Dalton’s law, and the second is Amagat’s law. Dalton’s law states that the partial pressure of a gas in a mixture of gases is defined as the pressure that the gas would exert if it alone were present at the same temperature and volume as the mixture. Dalton’s law states that the sum of the partial pressures of the gases in a mixture is equal to the total pressure of the mixture. The second law is only true for ideal gases. Definitions of ideal and real gases are provided later in this chapter. Eq. (1.3) indicates the relationship between the partial pressure of a gas mixture and the total pressure for ideal gases according to Dalton’s law. Eq. (1.4) provides a calculation of the partial pressure of each gas in a mixture.

Equation 1.3: Relationship between the partial pressure of a gas mixture and the total pressure for ideal gases according to Dalton’s law

si18_e

   (1.3)

Where:

PTotal: Total pressure of the gas mixture

PGas1: Partial pressure of Gas #1 in the gas mixture

PGas2: Partial pressure of Gas #2 in the gas mixture

PGas3: Partial pressure of Gas #3 in the gas mixture

PGasn: Partial pressure of Gas #n in the gas mixture

X1: mole fraction of Gas #1

X2: mole fraction of Gas #2

X3: mole fraction of Gas #3

Xn: mole fraction of Gas #n

Equation 1.4: Partial pressure of gas calculation

si19_e    (1.4)

Where:

PGas: Partial pressure of the gas

XGas: Mole fraction of the gas

PTotal: Total pressure of the gas

Example 1.2

Air contains 20.95% oxygen, 78.08% nitrogen, and 0.97% argon. The pressure of air is equal to 14.7 psi. Calculate the partial pressure of the other gases inside the air.

Answer

The percentage of each element in air is equal to the mole fraction. Thus, the mole fractions of oxygen, nitrogen, and argon are 20.95%, 78.08%, and 0.97% respectively. The partial pressure of each gas in the air (a mixture of gases) is obtained by multiplying the total pressure, which is equal to 14.7 psi, by the mole fraction of each gas. Thus, the partial pressures of each of these three elements are calculated as follows:

si20_esi21_esi22_e

According to Dalton’s law, the sum of the partial pressure values for all 3 gases calculated above is equal to the pressure of air, meaning that:

si23_e

Amagat’s law states that the volume of a gas in a mixture of gases is defined as the volume that the gas would occupy if it alone were present at the same pressure and temperature as the mixture of gases. For ideal gases, the sum of the partial volume of the gases that make up a gas is equal to the total volume of the main gas. This rule can be stated in another way: under the same conditions of temperature and pressure, equal volumes of ideal gases have the same number of molecules. In fact, one-pound mole of an ideal gas contains 2.7333 × 10²⁶ molecules, occupies 378.6 ft³ at 60 °F, and has a pressure equal to 14.73 psi.

1.3.2: Gas viscosity

Gas viscosity (parameter μ) is defined as a measurement of a gas’s resistance to flow. Viscosity is derived from the Latin word viscum meaning anything sticky. A fluid with low viscosity is called thin, while a fluid with high viscosity is called thick. Viscosity describes the internal friction of moving fluid, and a fluid with a higher amount of viscosity resists motion because the molecules of a high viscosity fluid create a lot of internal friction. Low viscosity fluids, on the other hand, flow better. Fig. 1.4 compares the viscosity of water and honey with regard to their gravitational flow rates. Think about fluid flow, specifically the layers of moving molecules. These layers of molecules are rubbing against each other, and greater friction means slower movement and higher viscosity. Honey is more viscous than water, so it resists more against the gravitational forces and flows harder and more slowly than water. There are two types of viscosities explained in this section: dynamic viscosity and kinetic viscosity.

Fig. 1.4

Fig. 1.4 Viscosity comparison between water and honey. (Courtesy: Shutterstock.)

Viscosity is sometimes defined using Isaac Newton’s equation for fluids, or more precisely, the second law of motion. Newton’s second law of motion states that when a force acts on an object, it will cause the object to accelerate. There is a relationship between the force required for acceleration and the mass of the object, meaning that a greater force is required to move and accelerate a more massive object. Newton’s second law is formulated by Eq. (1.5). The greater the force required to move the fluid, the higher the viscosity.

Equation 1.5: Second law of newton formulation (Relationship between the force and mass)

si24_e    (1.5)

Where:

F: Force (Newton or si25_e )

m: Mass (Kilogram or kg)

a: Acceleration si26_e )

Acceleration (parameter a) is defined as the rate of velocity change with time. Speeding up increases the acceleration, and slowing down reduces the acceleration. Acceleration is calculated according to Eq. (1.6) as follows:

Equation 1.6: Acceleration calculation

si27_e    (1.6)

Where:

a: Acceleration si26_e )

V: Final velocity ( si29_e )

Vo: Initial velocity ( si29_e )

t: Duration or time (s)

Velocity (parameter V) is a quantity that measures displacement or change in position over a change in time or over a specific period of time. This definition is one of the fundamental concepts of mechanics. Eq. (1.7) is used to calculate velocity based on two other important parameters: distance and time.

Equation 1.7: Velocity calculation

si31_e    (1.7)

Where:

V: Velocity ( si29_e )

d: Distance (m)

t: time (s)

There is no correlation between force and viscosity based on the equations provided above. A better approach for the mathematical formulation of dynamic viscosity is the use of the two terms "Shear Stress and Shear rate." Dynamic viscosity is the tangential force required to move one horizontal layer or plane of a fluid relative to another layer. As will be explained later in this section, dynamic viscosity is calculated as a ratio of shear stress to shear rate. The other name for dynamic viscosity is absolute viscosity.

Underlying the theory of fluid movement is the internal friction between molecules. Shear stress is defined as a force (parameter F) pushing one layer of fluid flow across another layer within the cross-section area of A. Shear stress is a kind of force that causes deformation of a material by slippage along the planes that run parallel to the imposed stress (Fig. 1.5).

Fig. 1.5

Fig. 1.5 Viscosity of a fluid with different layers of fluid flow. (Courtesy: Shutterstock.)

Shear stress is shown with parameter τ. Shear stress is calculated according to Eq. (1.8), as follows:

Equation 1.8: Shear stress calculation

si33_e    (1.8)

Where:

τ: Shear stress ( si34_e ) or si35_e

F: Force (N) or (dyne)

A: Area (m²) or (cm²)

The shear rate can also be defined as the rate at which fluid layers are moving past each other. The shear rate depends on both geometry and the speed of the flow. The shear rate is calculated according to Eq. (1.9), as follows:

Equation 1.9: Shear rate calculation

si36_e    (1.9)

Where:

γ: Shear rate si37_e

Δ V: Difference in velocity between two layers si29_e or si39_e

Δ h: The distance between two fluid layers (m or cm)

Dynamic or absolute viscosity is calculated by dividing the shear stress by the shear rate, according to Eq. (1.10):

Equation 1.10: Dynamic viscosity calculation

si40_e    (1.10)

Where:

si41_e : Dynamic viscosity ( si42_e equal to Pascal second) or ( si43_e equal to poise)

The given units are in both the international system (SI) and centimeter-gram-second system (CGS). CGS is a variant of the metric system that uses centimeters as the unit of length, grams as the unit of mass, and seconds as the unit of time. The CGS system has largely been replaced by SI, which is also known as the metric system or the MKS system. MKS is an abbreviation for meter, kilogram, second. In many scientific fields, SI is the only system used, but there are still some fields in which the CGS system is used. In order to convert CGS system units to MKS, the unit conversion factors are normally powers of 10, except for the unit of time, which is seconds in both systems. For example, 1 m in the MKS system is equal to 100 cm in the CGS system. The SI unit of force is a Newton, which is equivalent to si25_e . The unit of force in the CGS system is a dyne and is equivalent to si45_e . The main question is how to convert Newtons to dynes. In fact, 1 Newton is equal to 10⁵ dynes, which is calculated as follows:

si46_e

Correspondingly, 1 dyne is equal to 10− 5 Newton.

The most common unit of gas viscosity is a poise or centipoise, which is named after the French physician, Jean Louise Marie Poiseuille, who lived from 1799 to 1869. A poise unit is typically expressed using the CGS system. The SI unit of viscosity is a Pascal second, abbreviated as Pa.s. A Pascal is a unit used to quantify internal pressure, and 1 Pascal is equal to ( si47_e ).

1 Pascal = si47_e (1 Newton is equal to si25_e ), so 1 Pascal is equal to si50_e . Since the viscosity unit is a Pascal second, the SI unit of viscosity is si51_e .

Kinetic viscosity (parameter Vg) is calculated according to Eq. (1.11) by dividing the dynamic viscosity by the density of the gas. Kinetic viscosity is not typically calculated for natural gases so it will not be discussed further.

Equation 1.11: Kinetic viscosity calculation

si52_e    (1.11)

Fluids are divided into two categories with regards to viscosity: Newtonian fluids, in which shear stress is linearly connected to shear rate so that the viscosity of a fluid has a constant value, and non-Newtonian fluids, for which viscosity is subject to change and is a dynamic function of certain variables such as time and shear rate. When shaking a non-Newtonian fluid, the viscosity could decrease or increase during the shaking process. It is easier to work with and model Newtonian fluids as their behaviors with regard to viscosity are more predictable. It is worth noting that both air and water are considered Newtonian fluids. Examples of non-Newtonian fluids are blood, paints, and some polymers. In addition, most gases are considered Newtonian fluids.

Viscosity in gases comes from gas molecules transferring momentum between each other. The viscosity of natural gas is much lower than that of oil and water. This lower viscosity means that the gas is more mobile and moves faster than oil and water. This is especially true in petroleum production, where three phases of oil, water, and gas are produced from reservoirs. An important point about gas viscosity is that it is mostly independent from pressure. Temperature, however, is the most influential parameter on gas viscosity. Sutherland’s law or formula is an expression of how a gas’s viscosity is dependent on temperature. This law was discovered by William Sutherland, an Australian physicist, in 1893. As a rule, increasing the temperature of a gas results in faster movement of the molecules and an increase in the viscosity of the gas. However, the effect of temperature on a liquid is the opposite: the viscosity of a liquid is decreased by increasing the temperature. Eq. (1.12), known as Sutherland’s formula, shows how viscosity can be changed by modifying the temperature of a gas:

Equation 1.12: Effect of gas temperature on gas viscosity (Sutherland’s formula)

si53_e    (1.12)

Where:

μ: Gas viscosity at temperature T using the Kelvin scale

μ0: Gas viscosity in centipoise at a reference temperature of T0 in Kelvin (K)

T: Temperature in Kelvin (K)

T0: Reference temperature in Kelvin (K)

S: Sutherland’s constant temperature (equal to 110.4 K)

For the conversion of temperatures in Celsius (°C) to Kelvin (K), Eq. (1.13) should be used:

Equation 1.13: Conversion of temperatures from Celsius to Kelvin

si54_e    (1.13)

Example 1.3

The viscosity of air at a reference temperature of 0 ºC is 1.716 ×  si55_e . Given that Sutherland’s constant temperature is 110.4 K, by what percentage does the viscosity of air increase when the temperature is increased to 100 ºC.

Answer

The reference temperature is 0 °C, which is equal to 273.15 K. The viscosity at a temperature of 100 °C, which is equal to 373.15 K, can be calculated by using Eq. (1.12) (Sutherland’s formula).

si56_e

This calculation shows that the viscosity of the gas is increased by 0.454 * 10− 5. Thus, the percentage increase in the viscosity of the gas is calculated as follows:

Gas viscosity increases by increasing the temperature to 100 °C = si57_e = 26.46%

Natural gas is a mixture of different gases with various viscosities, molecular weights, and mole fractions. In some cases, natural gas viscosity can be calculated using the viscosities, mole fractions, and molecular weights of the gases that compose it, as in Eq. (1.14):

Equation 1.14: Gas viscosity calculation using viscosities, molecular weights, and mole fractions of the constituent gases

si58_e    (1.14)

Where:

μg: Viscosity of the gas mixture at the desired temperature and atmospheric pressure

yi: Mole fraction of the/th composition

μi: Viscosity of the/th component of the gas mixture at the desired temperature and atmospheric pressure

Mgi: Molecular weight of the/th component of the gas mixture

N: Total number of components of the gas mixture

For Eq. (1.14), it is important to know the viscosity of each gas in order to calculate the viscosity of the gas mixture. But in some cases, no information is available about the viscosity of the different gases in natural gas. In such cases, following steps should be followed to calculate the viscosity. The first step is to calculate the viscosity of the gas at one atmospheric pressure, represented by μga in Fig. 1.6, using the temperature of the gas as well as the molecular weight or gravity of the gas. It is essential that the obtained viscosity of the natural gas at one atmospheric pressure from the chart be corrected for certain non-hydrocarbon elements or compounds in the gas, such as nitrogen (N2), carbon dioxide (Co2), and hydrogen sulfide (H2S). It is possible to obtain the viscosity of nitrogen, carbon dioxide, and hydrogen sulfide from the small charts provided in the corners of Fig. 1.6. The corrected viscosities for each of these non-hydrocarbon elements or compounds can be estimated using the mole percentage of non-hydrocarbon elements and the specific gravity of the natural gas. The mole percentage is calculated by multiplying the mole fraction by 100. The values of the viscosities of nitrogen, carbon dioxide, and hydrogen sulfide should be added to the estimated viscosity of the natural gas. The presence of each of the non-hydrocarbons increases the viscosity of the hydrocarbon mixture or natural gas.

Fig. 1.6

Fig. 1.6 Estimated viscosity of natural gas at 1 atmospheric pressure, parameter μ ga . (Courtesy: Carr, N. L., Kobayashi, R., & Burrows, D. B. (1954). Viscosity of hydrocarbon gases under pressure. Journal of Petroleum Technology, 6(10), 47–55. https://doi.org/10.2118/297-G.)

Example 1.4

Table 1.3 displays the chemical composition of natural gas, including the molar weight and mole fraction of each compound or element. Assuming the temperature of the gas to be 200 °F, what would be the viscosity of natural gas at 1 atmospheric pressure?

Table 1.3

Answer

The first step is to calculate the molar weight of the gas by using the mole fractions and molar weight of each component as follows:

si59_e

It is now possible to obtain the viscosity of the natural gas, parameter μga, at 1 atmospheric pressure from Fig. 1.6 by using the molecular weight of the gas, which is equal to 20.29, and the temperature, which is equal to 200 °F. The obtained viscosity, parameter μga, is approximately 0.012 CP. But a correction is needed because the nitrogen content is equal to a 15% mole fraction. In order to find the viscosity of nitrogen as per the small charts provided in Fig. 1.6, it is essential to know the specific gravity of the main gas, which is equal to 0.7 according to the following calculations:

si60_e

The estimated viscosity of the nitrogen is approximately equal to 0.0012 CP, given a 15% mole fraction and a specific gravity of the main gas of 0.7. The viscosity of the nitrogen should be added to the viscosity of the natural gas, which is 0.012 CP. Thus, the viscosity of the natural gas, parameter μga, after adding the nitrogen correction factor is 0.0132