Basics of Oil and Gas Treatment - Chapter 1

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Basics of Oil and Gas Treatment - Chapter 1

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Fundamentals of Oil and Gas Processing Book
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Chapter 1 10
Basics of Oil and Gas Treatment 10
1.1 Introduction 10
1.2 Hydrocarbon preparation 10
1.3: Physical properties of Hydrocarbon Gases 11
1.3.1: Hydrocarbon gases 11
1.3.2: Molecular weight and apparent molecular weight 11
1.3.3: Apparent molecular weight of gas mixture 12
1.3.4: Gas Specific Gravity and Density 13
1.3.6: Compressibility Factor (z) 14
1.3.7: Gas density at any condition of Pressure and temperature 18
1.3.8: Gas volume at any condition of Pressure and temperature 19
1.3.9: Velocity of gas, (ft/s) 20
1.3.10: Average pipeline pressure 21
1.3.11: Viscosity of gases 22
1.3.12: The heating value of gases 22
1.4: properties of Hydrocarbon Liquids (Crude Oil) 23
1.4.1: Introduction 23
1.4.2: Crude oil Density and gravity 24
1.4.3: Crude oil Viscosity. 25
1.4.4: Oil-Water Mixture Viscosity 25
1.5: Phase Behavior 27
1.5.1: Introduction 27
1.5.2 System Components 27
1.5.3: Single-Component Systems 28
1.5.4: Multicomponent Systems 31
1.5.5: Prediction of phase envelope 32
1.6: Types of Fluid Flow 42
1.6.1: Reynolds Number 42
Chapter 2 43
Two-phase Oil and Gas Separation 43
2.1 Introduction 43
2.2 Phase Equilibrium 43
2.3: Separation process: 43
2.4: Principles of Physical Separation: 44
2.5: Gravity Separation: 44
2.6: Factors Affecting Separation 46
2.7: Separator categories and nomenclature: 47
2.8: Functional Sections of a Gas-Liquid Separator 47
2.8.1: Inlet Diverter Section 48
2.8.2: Liquid Collection Section 48
2.8.3: Gravity Settling Section 48
2.8.4: Mist Extractor Section 49
2.9: Separator Configurations 49
2.10: Types of Separators 50
2.10.1: Vertical Separators 50
2.10.2: Horizontal Separators 52
2.10.3: Double-Barrel Horizontal Separators 53
2.10.4: Horizontal Separator with a “Boot” or “Water Pot” 54
2.10.5: Filter Separators 54
2.10.6: Scrubbers 56
2.10.7: Slug Catchers 56
2.11: Selection Considerations 57
2.12: Internal Vessel Components 59
2.12.1: Inlet Diverters 59
2.12.2: Wave Breakers 62
2.12.3: Defoaming Plates 62
2.12.4: Vortex Breaker 63
2.12.5: Stilling Well 64
2.12.6: Sand Jets and Drains 64
2.12.7: Mist Extractors 65
2.13: Control Components of Gas–Oil Separators 76
2.14.1: Foamy Crude 77
2.14.2: Paraffin 78
2.14.3: Sand 78
2.14.4: Gas Blowby 78
2.14.5: Liquid Carryover 79
2.14.6: Liquid Slugs 79
2.15: Stage Separation 80
2.15.1: Initial Separation Pressure 80
2.15.2: Stage Separation 81
2.15.3: Selection of Stages 83
2.15.4: Fields with Different Flowing Tubing Pressures 83
2.15.5: Determining Separator Operating Pressures 84
2.15.6: Two-Phase vs. Three-Phase Separators 85
2.16: Separator calculation basics. 85
2.16.1: Liquid Handling and Liquid Retention Time 85
2.16.2: Gas retention time 86
2.16.3: Gas velocity 86
2.16.4: Liquid Re-entrainment 87
2.16.5: Droplet Size (Liquid in gas phase) 88
2.17: Design Principles and sizing of Oil-gas Separator 88
2.17.1: First method Design Theory 89 Slenderness Ratio 95
2.17.2: Second method Design Theory 100
Chapter 3 107
Three-phase Oil and Gas Separation 107
3.1: Introduction 107
3.2: three phase separation equipment’s 108
3.2.1: Horizontal Separators 108
3.2.2: Free-Water Knockout 111
3.2.3: Horizontal Three-Phase Separator with a Liquid “Boot” 111
3.2.4: Vertical Separators 112
3.2.5: Selection Considerations 114
3.3: Internal Vessel components 115
3.3.1: Coalescing Plates 117
3.4: Operating Problems 118
3.4.1: Emulsions 118
3.5: Three-Phase Separator Design Theory 118
3.5.1: Gas Separation 118
3.5.2: Oil–Water Settling 118
3.5.3: Water Droplet Size in Oil 118
3.5.4: Oil Droplet Size in Water 119
3.5.5: Retention Time 119
3.6: Separator Design (first method) 121
3.6.1: Horizontal Three-phase Separator Sizing—Half-Full 121 Retention Time Constraint 121 Settling Water Droplets from Oil Phase 122 Separating Oil Droplets from Water Phase 123
3.6.2: Vertical Separators’ Sizing 124 Gas Capacity Constraint 125 Settling Oil from Water Phase Constraint 125
3.7: Separator Design (second method) 131
Chapter 4 134
Crude oil dehydration 134
4.1: Introduction 134
4.2: Emulsion 134
4.2.1 Energy of Agitation 135
4.2.2 Emulsifying Agents 136
4.2.3: Stability of oil water emulsion 137
4.2.4: Emulsion Treating Theory 139
4.2.5: Demulsifiers 140
4.3: Crude oil treating systems 143
4.3.1: Free-Water Knockouts 143
4.3.2: Gunbarrel tanks with internal and external gas boots 144
4.3.3: Heaters 146
4.4: Emulsion Treating Methods 164
4.4.1: General Considerations 164
4.4.2: Chemical Addition 165
4.5: Heat Required 174
4.5.1: Heat duty 174
4.5.2: Heat Loss 174
4.5.3: Fire Tube Heat Flux 175
4.5.4: Firetube Heat Density 175
4.6: Treater Equipment Sizing 175
4.6.2: Design Procedure 178
4.7: Practical Considerations 184
4.7.1: Gunbarrels with Internal/External Gas Boot 184
4.7.2: Heater-Treaters 184
4.7.3: Electrostatic Heater-Treaters 184
Chapter 5 185
Crude Oil Desalting 185
5.1: Introduction 185
5.1.1: Salt Content 185
5.1.2: Desalting Process 186
5.2: Equipment Description 186
5.2.1: Desalters 186
5.2.2: Mixing Equipment 186
5.3: Process Description 188
5.3.1: Single-Stage Desalting 189
5.3.2: Two-Stage Desalting 189
5.4: Electrostatic Desalting Voltage 189
5.5: Operating Parameters Effects 191
5.6: Design Consideration 191
5.7: Troubleshooting 192
Chapter 6 193
Crude Oil Stabilization and Sweetening 193
6.1: Introduction 193
6-1-1: Crude oil treatment steps 193
6.2: Process Schemes 194
6.2.1: Multi-Stage Separation 194
6.2.2: Oil Heater-Treaters 194
6.2.3: Liquid Hydrocarbon Stabilizer 195
6.2.4: Cold-Feed Stabilizer 197
6.2.5: Stabilizer with Reflux 197
6.3: Stabilization Equipment 199
6.3.1: Stabilizer Tower 199
6.4: Stabilizer Design 205
6.5: Crude Oil Sweetening 206
6.6.1: Stage vaporization with stripping gas. 206
6.6.2: Trayed stabilization with stripping gas. 207
6.6.3: Reboiled trayed stabilization. 208
Chapter 7 209
Fluid Measurements 209
7.1: Gas Measurement 209
7.1.1: Orifice-Meter Measurement 209 Meter Tubes 213
7.1.2: Ultrasonic Measurement 220
7.2: Liquid Measurements 221
7.2.1: Volumetric Measurement Meters (Orifice Meters) 221
7.2.2: Turbine Meters 223
7.2.3: Positive Displacement Meters 224
7.2.4: Turbine and Positive Displacement Meter Selection 224
7.2.5: Mass Measurement Meters 225
Chapter 8 228
Instrumentation and Control 228
8.1: Introduction 228
8.2: Type Selection and Identification 228
8.2.1: Pneumatic Power Supplies 228
8.2.2: Electronic Power Supplies 229
8.3: Sensing Devices 230
8.3.1: Pressure Sensors 230 Bellows (Fig. 8-3) 230
8.3.2: Level Sensors 232
8.3.3: Temperature Sensors 237
8.3.4: Flow Sensors 239
8.4: Signal Transmitters 241
8.4.1: Pneumatic Transmitters 241
8.4.2: Electronic Transmitters 241
8.5: Signal Converters 241
8.5.1: Pneumatic-to-electronic (P/I) 242
8.5.2: Electronic-to-pneumatic (I/P) 242
8.5.3: Isolators 242
8.5.4: Electric signal converters 242
8.5.5: Frequency converters 242
8.6: Recorders and Indicators 242
8.6.1: Recorders 242
8.6.2: Indicators 242
8.7: Control Concepts 243
8.7.1: Control Loops 243
8.8: Control Modes and Controllers 245
8.8.1: Two-Position (on-off) Controllers 245
8.8.2: Proportional Control Mode 245
8.9: Control Valves 246
8.9.1: Control-Valve Bodies 247
8.9.2: Control-Valve Actuators 248
8.9.3: Flow Characteristics and Valve Selection 249
8.9.4: Fundamentals of Control Valve Sizing 250
Chapter 9 256
Process Relief Systems 256
9.1: Introduction 256
9.2: Relief Device Design and Requirements: 256
9.2.1: Blocked Discharge 257
9.2.2: Fire Exposure 257
9.2.3: Tube Rupture 257
9.2.4: Control Valve Failure 257
9.2.5: Thermal Expansion 257
9.2.6: Utility Failure 257
9.3: General discussion 258
9.4: Special Relief System Considerations 260
9.4.1: Pumps and storage equipment 260
9.4.2: Low Temperature Flaring 260
9.5: Relieving Devices 260
9.5.1: Conventional Relief Valves 260
9.5.2: Balanced Relief Valves 262
9.5.3: Pilot Operated Relief Valves 262
9.5.4: Resilient Seat Relief Valves 264
9.5.5: Rupture Disk 265
References. 267


Chapter 1

Basics of Oil and Gas Treatment
1.1 Introduction
Oil and gas wells produce a mixture of hydrocarbon gas, condensate or oil, salt water, other gases, including nitrogen, carbon dioxide (CO2), and possibly hydrogen sulfide (H2S), and solids, including sand from the reservoir, dirt, scale, and corrosion products from the tubing. These mixtures are very difficult to handle, meter, or transport. In addition to the difficulty, it is also unsafe and uneconomical to ship or to transport these mixtures to refineries and gas plants for processing. Further, hydrocarbon shipping tankers, oil refineries, and gas plants require certain specifications for the fluids that each receive. Also, environmental constraints exist for the safe and acceptable handling of hydrocarbon fluids and disposal of produced salt water. It is therefore necessary to process the produced fluids in the field to yield products that meet the specifications set by the customer and are safe to handle.
1.2 Hydrocarbon preparation
The goal is to produce oil that meets the purchaser’s specifications that define the maximum allowable amounts of water, salt, and sulfur. In addition to the maximum allowable value of Reid vapor pressure and maximum allowable pour point temperature.
Similarly, the gas must be processed to meet purchaser’s water vapor maximum allowable content (Water dew point), hydrocarbon dew point specifications to limit condensation during transportation, in addition to the maximum allowable content of CO2, H2S, O2, Total Sulfur, Mercaptan, Mercury, and maximum gross heating value.
The produced water must meet the regulatory requirements for disposal in the ocean if the wells are offshore, or to meet reservoir requirements for injection into an underground reservoir to avoid plugging the reservoir.
The specifications for the above requirements may include maximum oil in water content, total suspended solids to avoid formation plugging, bacteria counts, toxicity in case of offshore disposal, and oxygen content. Before discussing the industry or the technology of oil and gas processing it is best to define the characteristic, physical properties and main chemical composition of oil and gas produced.
Figures 1-1 and 1-2, illustrates gas-oil separation plant, and oil flow diagram.

Fig.1-1 .Gas Oil Separation plant function.

Separator may be a slug catcher, free water knock out drum, two phase separator, or gun barrel.
A dehydrator may be a heater treater, separator, or settling tank.
Heat is added upstream or downstream separator depending on crude oil temperature and gas oil ratio.
Crude oil stabilization is usually performed in separation step or during heat addition.
Crude oil sweetening is usually performed upstream or downstream heater treater.
Gas and water are separated and undergoes further treatment processes not in the scope of this book.
Fig.1-2. Crude oil flow Diagram

1.3: Physical properties of Hydrocarbon Gases
1.3.1: Hydrocarbon gases
Most of compounds in crude oil and natural gas consist of molecules made up of hydrogen and carbon, therefore these types of compounds are called hydrocarbon.
The smallest hydrocarbon molecule is Methane (CH4) which consists of one atom of Carbon and four atoms of hydrogen. It may be abbreviated as C1 since it consisted from only one carbon atom. Next compound is Ethane (C2H6) abbreviated as C2, and so on Propane (C3H8), Butane (C4H10)...etc.
Hydrocarbon gases are C1:C4), with the increase of carbon number, liquid volatile hydrocarbon is found (e.g. Pentane C5 is the first liquid hydrocarbon at standard conditions).

1.3.2: Molecular weight and apparent molecular weight
The molecular weight of a compound is the sum of the atomic weight of the various atoms making up that compound. The Mole is the unit of measurements for the amount of substance, the number of moles is defined as follows:

Mole = Weight/(Molecular weight) (Eq. 1-1)

Expressed as n = m/M (Eq. 1-2)
or, in units as, lb-mole = lb/(lb/lb-mole) (Eq. 1-3)


Table 1-1 Physical constants of light hydrocarbons and some inorganic gases. Adapted from GPSA, Engineering Hand Book.

Example 1.1: Methane molecule consists of one carbon atom with atomic weight =12 and 4 hydrogen atoms with atomic weight = 1 each. Molecular weight for Methane (CH4) = (1 × 12) + (4 × 1) = 16 lb/lb-mole. Similarly, Ethane (C2H6) molecular weight = (2 × 12) + (6 × 1) = 30 lb/lb-mole.

Hydrocarbon up to four carbon atoms are gases at room temperature and atmospheric pressure. Reducing the gas temperature and/or increasing the pressure will condense the hydrocarbon gas to a liquid phase. By the increase of carbon atoms in hydrocarbon molecules, consequently the increase in molecular weight, the boiling point increases and a solid hydrocarbon is found at high molecular weight.
Physical constants of light hydrocarbon and some inorganic gases are listed in Table 1-1.

1.3.3: Apparent molecular weight of gas mixture
For compounds, the term molecular weight is used, while, for hydrocarbon mixture the term apparent molecular weight is commonly used. Apparent molecular weight is defined as the sum of the products of the mole fractions of each component times the molecular weight of that component. As shown in Eq. 1-4
MW= ∑▒ Yi (MW)i (Eq. 1-4)
yi =molecular fraction of ith component,
MWi =molecular weight of ith component,
Ʃyi =1.

Example 1.2: Determine the apparent molecular weight for the gas mixture in Table 1-2:

Table 1-2 Gas mixture for Example 1-2

Solution: Using Table 1-1 & Equation 1-4
MW= ∑▒ Yi (MW)i
MW = (Mole Fraction of component 1 × MW of component 1) + (Mole Fraction of component 2 × MW of component 2) + (Mole Fraction of component 3 × MW of component 3) +…etc.
The following table can be made:

Table 1-3 Solution of Example 1-2
The apparent molecular weight is 21.36

1.3.4: Gas Specific Gravity and Density
The density of a gas is defined as the mass per unit volume as follows
Density = mass / volume (Eq. 1-5)

The specific gravity of a gas is the ratio of the density of the gas to the density of air at standard conditions of temperature and pressure.
S = (ρ(gas))/(ρ(air)) (Eq. 1-6)
ρ(gas) ρg = density of gas
ρ(air) ρair = density of air

Both densities must be computed at the same pressure and temperature, usually at standard conditions.
It may be related to the molecular weight by Equation 1-7
S = (MW(gas))/(MW(air)) (Eq. 1-7)
Since molecular weight of air is 28.96 (table 1-1)
Specific gravity of gas S = (MW(gas))/28.96 (Eq. 1-8)

Example 1-3: Determine the specific gravity of the gas mixture in example 1-2.
Apparent molecular weight of gas mixture is 21.36
Gas specific gravity = 21.36/28.96 = 0.7376

Since the gas is a compressible fluid, its density varies with temperature and pressure, calculating the gas density at a certain pressure and temperature will be explained after discussing the general gas law and gas compressibility factor.

1.3.5: General Gas Law
The general (Ideal) Gas equation, or the Perfect Gas Equation, is stated as follows:

PV = nRT (Eq. 1-9)

P = gas pressure, psia
V = gas volume, ft3
n = number of lb moles of gas (mass/molecular weight)
R = universal gas constant, psia ft3/lb mole OR
T = gas temperature, OR (OR = 460 + OF)
The universal gas constant R is equal to 10.73 psia ft3/lb mole OR in field units.

Equation (1-9) is valid up to pressures of about 60 psia. As pressure increases above this level, its accuracy becomes less and the system should be considered a non-ideal gas equation of state.
PV = znRT (Eq. 1-10)

z = gas compressibility factor.

1.3.6: Compressibility Factor (z)
The Compressibility factor, Z is a dimensionless parameter less than 1.00 that represents the deviation of a real gas from an ideal gas. Hence it is also referred to as the gas deviation factor. At low pressures and temperatures Z is nearly equal to 1.00 whereas at higher pressures and temperatures it may range between 0.75 and 0.90. The actual value of Z at any temperature and pressure must be calculated taking into account the composition of the gas and its critical temperature and pressure. Several graphical and analytical methods are available to calculate Z. Among these, the Standing-Katz, and CNGA methods are quite popular. The critical temperature and the critical pressure of a gas are important parameters that affect the compressibility factor and are defined as follows.
The critical temperature of a pure gas is that temperature above which the gas cannot be compressed into a liquid, however much the pressure. The critical pressure is the minimum pressure required at the critical temperature of the gas to compress it into a liquid.
As an example, consider pure methane gas with a critical temperature of 343 0R and critical pressure of 666 psia (Table 1-1).
The reduced temperature of a gas is defined as the ratio of the gas temperature to its critical temperature, both being expressed in absolute units (0R). It is therefore a dimensionless number.
Similarly, the reduced pressure is a dimensionless number defined as the ratio of the absolute pressure of gas to its critical pressure.
Therefore we can state the following:
Tr = T/Tc (Eq. 1-11)
Pr = P/Pc (Eq. 1-12)

P = pressure of gas, psia
T = temperature of gas, 0R
Tr = reduced temperature, dimensionless
Pr = reduced pressure, dimensionless
Tc = critical temperature, 0R
Pc = critical pressure, psia

Example1-4: Using the preceding equations, the reduced temperature and reduced pressure of a sample of methane gas at 70 0F and 1200 psia pressure can be calculated as follows

Tr = (70 +460) / 343 =1.5
Pr = 1200/666 = 1.8

For natural gas mixtures, the terms pseudo-critical temperature and pseudo-critical pressure are used. The calculation methodology will be explained shortly. Similarly we can calculate the pseudo-reduced temperature and pseudo-reduced pressure of a natural gas mixture, knowing its pseudo-critical temperature and pseudo-critical pressure.
The Standing-Katz chart, Fig. 1.3 can be used to determine the compressibility factor of a gas at any temperature and pressure, once the reduced pressure and temperature are calculated knowing the critical properties.
Pseudo-critical properties allow one to evaluate gas mixtures. Equations (1-13) and (1-14) can be used to calculate the pseudo-critical properties for gas mixtures:

P’c = Ʃ yi Pci (Eq. 1-13)

T’c = Ʃ yi Tci (Eq. 1-14)

P’c =pseudo-critical pressure,
T’c =pseudo-critical temperature,
Pci =critical pressure at component i, psia
Tci =critical temperature at component i, 0R
Yi =mole fraction of each component in the mixture,
Ʃ yi =1.

Example 1-5: Calculate the Compressibility factor for the following Gas mixture at 1000F and 800 psig:

Table 1-4 for Example 1-5.

Using Equation 1-11 and 1-12
T`r = (100+460)/464.5 =1.2
P`r = (800+14.7)/659.4 = 1.23
From fig.1-3. Compressibility factor is approximately, z= 0.72

Calculating the compressibility factor for example 1-4, of the gas at 70 0F and 1200 psia, using Standing-Katz chart, fig. 1-3. Z = 0.83 approximately. For ) Tr = 1.5 , Pr = 1.8).

Another analytical method of calculating the compressibility factor of a gas is using the CNGA equation as follows:
(Eq. 1-15)
Pavg = Gas pressure, psig. [psig = (psia - 14.7)]
Tf = Gas temperature, 0R
G = Gas gravity (air = 1.00)
The CNGA equation for compressibility factor is valid when the average gas pressure Pavg is greater than 100 psig. For pressures less than 100 psig, compressibility factor is taken as 1.00. It must be noted that the pressure used in the CNGA equation is the gauge pressure, not the absolute pressure.

Example 1-6: Calculate the compressibility factor of a sample of natural gas (gravity = 0.6) at 80 0F and 1000 psig using the CNGA equation.
From the Eq. (1.15), the compressibility factor is
The CNGA method of calculating the compressibility, though approximate, is accurate enough for most gas pipeline hydraulics work and process calculations.

Figure 1-3 Compressibility Factor For lean sweet natural gas (Surface Production Operations).

1.3.7: Gas density at any condition of Pressure and temperature
Once the molecular weight of the gas is known, the density of a gas at any condition of temperature and pressure is given as:

ρg= ((MW)P)/RTZ lb/ft3

Since R=10.73, then
ρg= 0.093 ((MW)P)/TZ lb/ft3 (Eq. 1-16)
ρg = density of gas, lb/ft3,
P =pressure, psia,
T =temperature, 0R,
Z =gas compressibility factor,
MW=gas molecular weight.

Example 1-7: Calculate the pseudo-critical temperature and pressure for the natural gas stream composition given in example 1-2, calculate the compressibility factor, and gas density at 600 psia and 1000F.

Table 1-5 solution of Example 1-7.
From the table MW= 21.36
T`c = 451.5 0R
P`c = 667 psia

From Eq. (1-11) and Eq. (1-12)
Tr = T/T`c = (100+460)/451.5 = 1.24

Pr = P/P`c = 600/667 = 0.9

Compressibility factor z could be calculated from figure 1-3, or from Eq. (1-15)
Value from figure, z = 0.83
From Equation 1-15 z = 0.87
For our further calculation we will use the calculated z value [Eq. (1-15)]
Using eq. (1-16) density of gas
ρg = 0.093 ((21.36)600)/(560 ×0.83) = 2.56 lb/ft3
Comparing ρg at standard condition (z=1)
ρg at standard condition = 0.093 (21.36)14.7/(520 ×1) = 0.056 lb/ft3
We can conclude that density increases with pressure while the volume decreases.

1.3.8: Gas volume at any condition of Pressure and temperature
Volume of a gas is the space occupied by the gas. Gases fill the container that houses the gas. The volume of a gas generally varies with temperature and pressure.
Volume of a gas is measured in cubic feet (ft3).
Gas volume are commonly referred to in "standard" or "normal" units.
Standard conditions commonly refers to gas volumes measured at: 60°F and 14.696 psia
The Gas Processors Association (GPA) SI standard molar volume conditions is 379.49 std ft3/lb mol at 60°F, 14.696 psia.
Therefore, each mole (n) contains about 379.5 cubic feet of gas (ft3)at standard conditions.
Therefore, by knowing the values of mass and density at certain pressure and temperature, the volume occupied by gas can be calculated.

Example 1-8: Calculate the volume of a 10 lb mass of gas (Gravity = 0.6) at 500 psig and 80 0F, assuming the compressibility factor as 0.895. The molecular weight of air may be taken as 29 and the base pressure is 14.7 psia.
The molecular weight of the gas (MW) = 0.6 x 29 = 17.4
Pressure =500+14.7 = 514.7 psia
Temperature = 80+460 = 540 0R
Compressibility factor z= 0.895
The number of lb moles n is calculated using Eq. (1-2). n=m/(MW)
n = 10/17.4
Therefore n= 0.5747 lb mole
Using the real gas Eq. (1-10), PV=nzRT
(514.7) V = 0.895 x 0.5747 x 10.73 x 540. Therefore, V = 5.79 ft3

Example 1-9: Calculate the volume of 1 lb mole of the natural gas stream given in the previous example at 1200F and 1500 psia (compressibility factor Z = 0.811).
Using Eq.(1-10), PV = nzRT
V= 0.811 x 1 x 10.73 x (120+460)/1500. V = 3.37 ft3

Example 1-10: One thousand cubic feet of methane is to be compressed from 60°F and atmospheric pressure to 500 psig and a temperature of 50°F. What volume will it occupy at these conditions?
Moles CH4 (n) = 1000 / 379.5 = 2.64
At final conditions, (Compressibility factor z must be calculated), from equations 1-11 and1-12
Tr = (460 + 50) / 344 = 1.88
Pr = (500 + 14.7) / 673 = 0.765
From Figure 1-3, Z = 0.94
From eqn. 1-10, PV = nzRT
V = ft3
Example 1-11: One pound-mole of C3 H8 (44 lb) is held in a container having a capacity of 31.2 cu ft. The temperature is 280°F. "What is the pressure?
Volume = V = 31.2 ft3
A Trial-and-error solution is necessary because the compressibility factor Z is a function of the unknown pressure. Assume Z = 0.9.
Using Eq. 1-10, PV = nzRT
P ×31.2 = 0.9 × 1.0 × 10.7 × (460 + 280)
P = 229 psia
From table 1-1, eqns. 1-11 and 1-12
Pr = 229 / 616 = 0.37,
Tc = 665ºR
Tr = (460 + 280) / 665 = 1.113
According to Figure 1.3, the value of Z should be about 0.915 rather than 0.9. Thus, recalculate using eq. 1-10, the pressure is 232 rather than 229 psia.

Example 1-12: Calculate the volume of gas (MW=20) will occupy a vessel with diameter 24 in, and 6 ft. length. At pressure 200 psia and temperature 100 0F. (Assume compressibility factor z=0.9), and what will be the volume of gas at 14.7 psia and 60 0F.
Then calculate gas density and mass inside the container at pressure 200 psia and temperature 100 0F.
Volume of vessel = π L r2
V = 3.14 × 6 × (24)2/ (2 ×12)2 ft3
V = 18.8 ft3.
(We divided by 2 to get r from the diameter, and divided by 12 to convert from in. to ft.)
T = 460 + 100 = 560 0R
Using Eq. 1-10, PV=nZRT
n = 18.8 × 200 / (0.9 × 10.73 × 560)
n = 0.7 lb. moles. (Remember gas volume ft3 = 379.5 x n)
Volume of gas at 200 psia and 100 0F= 0.7 * 379.5 = 266 ft3
n of Gas at 14.7 psia and 60 0F ( z=1) = 18.8 × 14.7 / (1 × 10.73 × 520)
n = 0.0495 lb. moles
Volume of gas at 14.7 psia and 60 0F = 0.0495 * 379.5 = 18.8 ft3
From the previous example 1-12, the gas volume will equal to the container volume at standard conditions (14.7 psia and 60 0F).
Gas density is calculated using Eq. 1-16
ρg = 0.093 ((MW)P)/TZ lb/ft3
Density of gas ρg = 0.093 × 20 × 200 / (0.9 × 560) = 0.738 lb/ft3
Mass of gas inside the vessel = Volume × density = 0.738 × 265 = 196 lb mass

1.3.9: Velocity of gas, (ft/s)
The velocity of gas equal the volume flow rate (ft3) per second divided by flow area (ft2).

Example 1-13: Calculate the gas velocity for gas flow rate 100 MMscfd through 24 in. internal diameter gas pipe, the gas specific gravity is 0.7, pressure 500 psia, Temperature 100 0F, and assume compressibility factor 0.85.
Solution: Using Eq. 1-10, PV=nzRT, and remember that n= V (ft3)/379.5).
n = 100 × 106/379.5
Gas volume at operating conditions V= 100 × 106 × 0.85 × 10.73 × 560 / (379.5 × 500)
= 2,695,000 ft3/day
Gas flow rate cubic foot per second = 2,695,000 / (24×60×60) = 31.2 ft3/sec
Area of flow = π r2 = 3.14 × 12 × 12 / (144) = 3.14 ft2
(144 to convert r2 from in. to ft2.)
Velocity of gas will be 31.2/3.14 = 9.9 ft/s
The gas velocity may be calculated directly from the following equation:
Velocity = 6 ZTQ/(100,000×Pd2) ft/s. Eq 1-17
Where Q = Flow rate scfd, d = diameter in inches.

The maximum recommended velocity of dry gas in pipes is 100 ft/s, (60 ft/s for wet gas), and to be less than the erosional velocity which is defined as:
Erosional velocity: The erosional velocity represents the upper limit of gas velocity in a pipeline. As the gas velocity increases, vibration and noise result. Higher velocities also cause erosion of the pipe wall over a long time period. The erosional velocity Vmax may be calculated approximately as follows:

Vmax = 100 √(2&ZRT/29GP) Eq 1-18

Where G= gas sp. Gt (air=1), P = pressure psia
For Example 1-12, the erosional velocity Vmax is:
Vmax = 100 √(2&0.85×10.73×560/(29×0.7× 500)) Vmax = 70.9 ft/s.

1.3.10: Average pipeline pressure
The gas compressibility factor Z used in the General Flow equation is based upon the flowing temperature and the average pipe pressure. The average pressure may be approximated as the arithmetic average
Pavg = (P1+P2)/2 of the upstream and downstream pressures P1 and P2. However, a more accurate average pipe pressure is usually calculated as follows
Pavg = 2/3 (P1+P2 - (P1× P2)/(P1+ P2)) Eq 1-19
P1, P2, Pavg = pressure, psia

Example 1-14: A natural gas pipeline with internal diameter 19 in. transports natural gas (Sp. Gr.= 0.65) at a flow rate of 200 MMscfd. Calculate the gas velocity at inlet and outlet of the pipe, assuming isothermal flow. The inlet temperature of 70 0F, inlet pressure is 1200 psig, and outlet pressure is 900 psig. Use compressibility factor of 0.95. Also, calculate the erosional velocity for this pipeline.
Using Eq. 1-17, the gas velocity at inlet of the pipe:
Velocity = 6 × 0.95× 530×200,000,000/(100,000×1214.7×192) ft/s.
Velocity = 13.8 ft/s.
The gas velocity at outlet of the pipe:
Velocity = 6 × 0.95× 530×200,000,000/(100,000×914.7×192) ft/s.
Velocity = 18.3 ft/s.
Finally, the erosional velocity can be calculated using Eq. 1-18
Vmax = 100 √(2&0.95 ×10.73×530/29×0.65×1214.7)
Vmax = 48.6 ft/s.

The above example may be solved by calculating the gas density at inlet and outlet of the pipe, then calculating the operational flow rate, divide it by pipe cross sectional area to get the velocity as follows:
Gas molecular weight = 0.65 × 28.96 = 18.8
Using Eq. 1-10, PV = nzRT
Calculating n = 200,000,000 / 379.5
Flow rate under operating conditions =
Gas volume V (= flow rate Q) = 200,000,000 × 0.95 × 10.73 × 530 / (379.5 ×1214.7)
Q = 2,347,000 ft3 per day at operating conditions. Q = 27.16 ft3/s.
Pipe cross sectional area = π r2 = 3.14 × 19 × 19 /(4× 144) = 1.97 ft2
Velocity of gas at the inlet = 27.16/1.97 = 13.8 ft/s.

1.3.11: Viscosity of gases
Viscosity of a fluid relates to the resistance to flow of the fluid. Higher the viscosity, more difficult it is to flow. Viscosity is a number that represents the drag forces caused by the attractive forces in adjacent fluid layers. It might be considered as the internal friction between molecules, separate from that between the fluid and the pipe wall.
The viscosity of a gas is very small compared to that of a liquid. For example, a typical crude oil may have a viscosity of 10 centipoise (cp), whereas a sample of natural gas has a viscosity of 0.0019 cp.
Viscosity may be referred to as absolute or dynamic viscosity measured in cp or kinematic viscosity measured in centistokes (cSt). Other units of viscosity are lb/ft-sec for dynamic viscosity and ft2/s for kinematic viscosity.
Fluid viscosity changes with temperature. Liquid viscosity decreases with increasing temperature, whereas gas viscosity decreases initially with increasing temperature and then increases with further increasing temperature.

Table 1- 6 Viscosity conversion factors

Figure 1-4 can be used to estimate the viscosity of a hydrocarbon gas at various conditions of temperature and pressure if the specific gravity of the gas at standard conditions is known. It is useful when the gas composition is not known. It does not make corrections for H2S, CO2, and N2. It is useful for determining viscosities at high pressure.

1.3.12: The heating value of gases
The heating value of a gas is expressed in Btu/ft3. It represents the quantity of heat in Btu (British Thermal Unit) generated by the complete combustion of one cubic foot of the gas with air at constant pressure at a fixed temperature of 60 0F.
Hydrogen in the fuel burns to water and when the flue gases are cooled to 60°F, the physical state — either vapor or liquid — of this water must be assumed. So the latent heat of vaporization of the water may or may not be considered to be part of the heating value. The result is two definitions for the heating value. The higher or gross heating value, HHV, includes the heat of condensation and the lower or net heating value, LHV, assumes the water remains in the vapor state.
For gas mixture the heating value is calculated as follows:
H = Ʃ xi Hi Eq. 1-20

Example 1-15: Calculate the heating value of gas mixture of Example 1-2
Table 1-7 Solution of Example 1-15

From table 1-7 the Gross calorific value HHV = 1246 Btu/ft3

The higher, ideal, dry heating value of sweet natural gas at 60°F and 760 mm Hg may be calculated with the following equation:
HHV=1568.72 × SG – 2524.88 × XCO2 – 1658.37 × XN2 +141.05 Eq. 1- 21

Applying Eq 1-21 for Example 1-15
The apparent molecular weight= 21.36
Gas Specific gravity = 21.36/28.96 = 0.738
HHV = 1568.72 × 0.738 – 2524.88 × 0.015 – 1658.37 × 0.01 +141.05 = 1244 Btu/ft3

1.4: properties of Hydrocarbon Liquids (Crude Oil)
1.4.1: Introduction
Crude oils are complex mixtures of a vast number of hydrocarbon compounds. Properties of crude petroleum vary appreciably and depend mainly on the origin.
Liquid hydrocarbons are started from Pentane C5 (Natural gasoline) up to solid hydrocarbon (C20) which has a melting point 100 0F. Heavier hydrocarbons (Paraffin and Asphalteen have higher melting points and may be soluble or dispersed in the liquid crude oil depending on solution temperature.)
Crude oil properties depends on its composition which is deferent and variable from crude to another.
Figure 1-4 Hydrocarbon gas viscosity.

1.4.2: Crude oil Density and gravity
Density is defined as the mass of a unit volume of material at a specified temperature. It has the dimensions of grams per cubic centimeter or lb/ft3.
Another general property, which is more widely, is the specific gravity. It is the ratio of the density of oil to the density of water and is dependent on two temperatures, those at which the densities of the oil sample and the water are measured. When the water temperature is 60 0F. The standard temperatures for specific gravity in the petroleum industry is 15/15 0C and 60/60 0F.
Although density and specific gravity are used extensively in the oil industry, the API gravity is considered the preferred property. It is expressed by the following relationship:

0API = 141.5/(Sp.Gr @ 60 Deg F) - 131.5 Eq. 1-22

1.4.3: Crude oil Viscosity.
The best way to determine the viscosity of a crude oil at any temperature is by measurement. If the viscosity is known at only one temperature, Figure 1-5 can be used to determine the viscosity at another temperature by striking a line parallel to that for crudes “A,” “C,” and “D.” Care must be taken to assure that the crude does not have its pour point within the temperature range of interest. If it does, its temperature-viscosity relationship may be as shown for crude “B.”
Solid phase high-molecular-weight hydrocarbons, otherwise known as paraffins, can dramatically affect the viscosity of the crude sample. The cloud point is the temperature at which paraffins first become visible in a crude sample. The effect of the cloud point on the temperature viscosity curve is shown for crude “B” in Figure 1-5. This change in the temperature-viscosity relationship can lead to significant errors in estimation. Therefore, care should be taken when one estimates viscosities near the cloud point.
The pour point is defined as the lowest temperature (5 0F) at which the oil will flow.
The lower the pour point, the lower the paraffin content of the oil.

Figure 1-5, typical viscosity-temperature curves for crude oils. (Courtesy of ASTM D-341.)
(Light crude oil (300–400API), Intermediate crude oil (200–300), & Heavy crude oil (less than 200 API)

In the absence of any laboratory data, correlations exist that relate viscosity and temperature, given the oil gravity. The following equation relating viscosity, gravity, and temperature was developed by Beggs and Robinson after observing 460 oil systems:

µ = 10x -1 Eq. 1-23
µ = oil viscosity, cp,
T = oil temperature, 0F,
x = y (T)−1.163,
y = 10z
z = 3.0324 – 0.02023G,
G= oil gravity, API@ 60 0F.
Figure 1-6 is a graphical representation of another correlation.

1.4.4: Oil-Water Mixture Viscosity
The viscosity of produced water depends on the amount of dissolved solids in the water as well as the temperature, but for most practical situations, it varies from 1.5 to 2 centipoise at 500F, 0.7 to 1 centipoise at 1000F, and 0.4 to 0.6 centipoise at 1500F.
When an emulsion of oil and water is formed, the viscosity of the mixture may be substantially higher than either the viscosity of the oil or that of the water taken by themselves. The modified Vand’s equation allows one to determine the effective viscosity of an oil-water mixture and is written in the form
µeff = (1+2.5 ϕ +10 ϕ2) µc Eq. 1- 24
µeff = effective viscosity, cp
µc = viscosity of the continuous phase (Oil), cp
Φ = volume fraction of the discontinuous phase (Water).

Figure 1-6, Oil viscosity vs. gravity and temp. (Courtesy of Paragon Eng. Services, Inc.)

1.5: Phase Behavior
1.5.1: Introduction
Before studying the separation of gases and liquids, we need to understand the relationship between the phases. Phase defines any homogeneous and physically distinct part of a system that is separated from other parts of the system by definite bounding surfaces:
The matter has three phases, the simplest example is water.
• Solid (ice),
• Liquid (liquid water),
• Vapor (water vapor).
Solids have a definite shape and are hard to the touch. They are composed of molecules with very low energy that stay in one place even though they vibrate. Liquids have a definite volume but no definite shape. Liquids assume the shape of the container but will not necessarily fill that container. Liquid molecules possess more energy than a solid (allows movement from place to another). By virtue of the energy, there is more space between molecules, and liquids are less dense than solids. Vapors do not have a definite volume or shape and will fill a container in which they are placed. Vapor molecules possess more energy than liquids (very active) and are less dense than liquids.
Our primary concern in this section is the difference in energy level between phases.
Energy is added to melt a solid to form a liquid. Additional energy will cause the liquid to vaporize. One needs to know the phase or phases that exist at given conditions of pressure, volume, and temperature so as to determine the corresponding energy level, to do this we need to study the phase diagram or phase behavior, but first we have to separate components into three classifications:
• Pure substance (single-component systems),
• Two substances,
• Multicomponent.
Phase diagrams illustrate the phase that a particular substance will take under specified conditions of pressure, temperature, and volume.

1.5.2 System Components
Natural gas systems are composed primarily of the lighter alkane series of hydrocarbons, with methane (CH4) and ethane (C2H6) comprising 80% to 90% of the volume of a typical mixture. Methane and ethane exist as gases at atmospheric conditions.
Propane (C3H8), butane (n-C4H10 and i-C4H10), and heavier hydrocarbons may be extracted from the gas system and liquefied for transportation and storage. These are the primary components of liquefied petroleum gas, or LPG.
The intermediate-weight hydrocarbons (pentane through decane) exist as volatile liquids at atmospheric conditions. These components are commonly referred to as pentanes-plus, condensate, natural gasoline, and natural gas liquids (NGL).
Natural gas systems can also contain non-hydrocarbon constituents, including hydrogen sulfide (H2S), carbon dioxide (CO2), nitrogen (N2), and water vapor. These constituents may occur naturally in gas reservoirs, or they may enter the system as contaminants during production, processing, and transportation. In addition, operators may intentionally add odorants, tracers (such as helium), or other components.
Dry, or lean, natural gas systems have high concentrations of the lighter hydrocarbons (methane and ethane), while wet, or rich, gas systems have higher concentrations of the intermediate-weight hydrocarbons. Lean gases burn with a low air-to-gas ratio and display a colorless to blue or yellow flame, whereas rich gases require comparatively higher amounts of air for combustion and burn with an orange flame. Intermediate-weight hydrocarbons may condense from rich gases upon cooling.

Table 1-8 shows typical compositions for a lean gas and a rich gas.

Table 1-8 typical composition of Lean and Rich gases.

1.5.3: Single-Component Systems
A pure component of a natural gas system exhibits a characteristic phase behavior, as shown in Fig. 1-7. Depending on the component’s pressure and temperature, it may exist as a vapor, a liquid, or some equilibrium combination of vapor and liquid

Figure 1-7 P-T Diagram for pure component

Lines HD, HC, and FH are the equilibrium lines - combinations of pressure and temperature at which the adjoining phases are in equilibrium. At equilibrium, one can change phase, by simply adding or removing energy from the system. Point H, the triple point, is the only combination of pressure and temperature at which all three phases can exist together.
Along line FH no liquid phase is ever present and solid sublimes to vapor. The use of "dry ice" for cooling is an example of this. Line HD is the equilibrium line between solid and liquid. Ice water at 0°C [32°F] and atmospheric pressure occurs on this line. Line HD can have a positive or negative slope depending on whether the liquid expands or contracts on freezing. The energy change occurring along line HD is called the heat of fusion. At any P and T along this line the system can be all solid, all liquid or a mixture of the two depending on the energy level.
This line could be called the solid-liquid saturation or solid-liquid equilibrium line.

Line HC is the saturation or equilibrium curve between vapor and liquid. It starts at the triple point and terminates at the critical point "C." The pressure and temperature conditions at this latter point are known as critical temperature (Tc) and critical pressure (Pc).
At this point the properties of the liquid and vapor phases become identical. For a pure substance the critical point can be defined as that point above which liquid cannot exist as a unique separate phase. Above (Pc), and (Tc), the system is often times referred to as a dense fluid to distinguish it from normal vapor and liquid.
Line HC is often referred to as the vapor pressure curve. Such vapor pressure curves are available from many sources. Line HC is also the bubble point and dew point curve for the pure substance.
The vapor pressure line in Figure 1-8 divides the liquid region from the vapor region.

In Figure 1-7, consider a process starting at pressure P1, and proceeding at constant pressure.
From "m" to "n" the system is entirely solid. The system is all liquid for the segment o-b. At "b" the system is a saturated liquid - any further addition of energy will cause vaporization. At "d," the system is in the saturated vapor state. At temperatures above "d," it is a superheated vapor.
Line HC is thus known by many names - equilibrium, saturated, bubble point, dew point and vapor pressure. For a pure substance these words all mean the same thing.

At the pressure and temperature represented by HC the system may be all saturated liquid, all saturated vapor or a mixture of vapor and liquid.
The rectangle "bfghd" illustrates another important phase property that is confirmed experimentally.
Suppose we place a liquid in a windowed cell at condition "b" and light it so it is easily visible.
We then increase pressure at constant temperature (isothermally). As we proceed toward point “f” the color will begin to fade. At some point, the color disappears completely. The cell now contains what looks like a vapor, but no bubble of vapor was ever seen to form.
At “ f ” (above the critical) the system is in a fourth phase that cannot be described by the senses. It is usually called dense phase fluid, or simply fluid. The word "fluid" refers to anything that will flow and applies equally well to gas and liquid.
This fluid at "f' looks like a gas but possesses different properties from regular gas found to the right of line HC and below the critical pressure. It is denser than regular gas but is more compressible than a regular liquid. “Properties of the liquid and vapor phases become identical”.
Table 1-1 lists Critical pressures and critical temperatures, along with molecular weights, of some pure components present in many natural gas systems.
Figure 1- 8 shows vapor pressure line for light hydrocarbons, where the left part of any component line, represents its liquid phase while the right part represents its gas phase.

Figure 1-8 Vapor pressure for light hydrocarbons.

1.5.4: Multicomponent Systems
In reality, natural gas systems are not pure substances. Rather, they are mixtures of various components, with phase behavior characteristics that differ from those of a single-component system. Instead of having a vapor pressure curve, a mixture exhibits a phase envelope, as shown in Figure 1-9.
Figure 1-9 typical phase envelop of hydrocarbon mixture.

The phase envelope (curve BCD in Figure 1-9) separates the liquid and gas phases. The area within this envelope is called the two-phase region and represents the pressure and temperature ranges at which liquid and gas exist in equilibrium.
The upper line of the two-phase region (curve BC) is the bubble-point line. This line indicates where the first bubble of vapor appears when the pressure of the liquid phase mixture is lowered at constant temperature, or when the temperature increases at constant pressure.
The lower section of the phase envelope (curve CD) is the dewpoint line. When the pressure of a mixture in the gaseous phase is decreased at constant temperature, or when the temperature is lowered at constant pressure, the first drop of liquid forms on this line. The bubble-point line and the dewpoint line meet at the critical point (C).
The highest pressure in the two-phase region is called the cricondenbar, while the highest temperature in the two-phase region is called the cricondentherm.

Figure 1-10, is another example of phase envelope, where:
Cricondenbar - maximum pressure at which liquid and vapor may exist (Point N).
Cricondentherm - maximum temperature at which liquid and vapor may coexist in equilibrium (Point M).
Retrograde Region - that area inside phase envelope where condensation of liquid occurs by lowering pressure or increasing temperature (opposite of normal behavior).
Quality Lines - those lines showing constant percentages which intersect at the critical point (C) and are essentially parallel to the bubble point and dew point curves. The bubble point curve represents 0% vapor and the dew point curve 100% vapor.
Line ABDE represents a typical isothermal retrograde condensation process occurring in a condensate reservoir. Point A represents the single phase fluid outside the phase envelope. As pressure is lowered, Point B is reached where condensation begins. As pressure is lowered further, more liquid forms because of the change in the slope of the quality lines. As the process continues outside the retrograde area, less and less liquid forms until the dewpoint is reached (Point E). Below E no liquid forms.

Figure 1-10 shows another phase envelope for hydrocarbon mixture.

1.5.5: Prediction of phase envelope
The location of the bubblepoint and dewpoint lines may be calculated using vapor-liquid equilibrium (VLE) methods. For most naturally occurring systems above about [2000 psia], the validity of the standard calculation becomes questionable.
The application of K-values to calculate phase quantities and compositions proceeds as follows.
For any stream (F) with mole fractions of components (Z1+Z2+Z3,.., etc.) entering a vessel at certain pressure and temperature, the stream will be divided into Vapor stream(V) with mole fractions of components (Y1+Y2+Y3,.., etc.), and into a liquid phase (L) with mole fractions of components (X1+X2+X3,.., etc.).
Component balance:
Fzi = Vyi + Lxi Eq. 1-25

Figure 1-11 flash separation for hydrocarbon mixture.
zi = mol fraction of any component in total feed stream to separation vessel
yi = mol fraction of any component in the vapor phase
xi = mol fraction of any component in the liquid phase
Ki = equilibrium vaporization ratio (equilibrium constant) = yi/xi
F = total mols of feed
V = total mols of vapor
L = total mols of liquid

If we set F = 1.0 so that L and V are now liquid and vapor-to-feed ratios
then zi = Vyi + Lxi
Since yi = Kixi
So, zi = V Ki xi + L xi
xi = zi / ( L + V Ki) Eq. 1-26
Since the summation of liquid fractions must equal one, we can write the following equation.

∑ xi = ∑ zi / ( L + V Ki) = 1 Eq. 1-27

The equation serves as the objective function in an interactive calculation to determine the quantity of L or V. The calculation procedure is as follows:
1. Determine K values of each component at the temperature and pressure of the system.
2. Assume a value of L (remember, V = 1 - L)
3. Solve the equation Eq. 1-27. If ∑xi ≠ 1.0 assume a new value of L and repeat step 2.
4. When ∑ xi = 1.00, the phase quantities L and V are known as well as the liquid phase composition. Vapor phase compositions may be calculated by remembering that yi = Kixi,

The foregoing calculations is known as a flash calculation and is used to predict the equilibrium quantities and compositions of two phase systems.

Special cases of a flash calculation include bubble point (V = 0, L = 1) and dew point (V = 1, L = 0), calculations. Equations for bubblepoint, and dewpoint are as follows:

Bubblepoint condition:

∑ Ki xi = 1.0 Eq. 1-28

Dewpoint condition:

∑ yi/Ki = 1.0 Eq. 1-29

Flash calculation are usually made by computer software, but knowing the basic of calculations is important in understanding the gas-liquid separation process.

Example 1- 16: Calculate the bubblepoint and dewpoint temperature at 250 psia of the following hydrocarbon mixture. Then calculate the amount of vapor and liquid and the composition of the two phases if these feed entered a vessel @ 250 psia and 150 0F.

Table 1-9 hydrocarbon component for example 1-16.

Bubblepoint calculation : To calculate the bubblepoint temperature at certain pressure, (All the components are in liquid phase xi = 1).
From eq. 1-28, the bubblepoint will be reached when ∑ Ki xi ≅ 1
Solution Steps:
Assume a temperature value (100 0F), for example.
From the K chart of each compound, find the K value at the system pressure and assumed temperature.
Multiply mole fraction xi of each component by its equilibrium value taken from the table Ki.
Take the sum ∑ Ki xi , if it’s less than 1, choose higher temperature, (1500F for example), and repeat as in the table.
If, ∑ Ki xi is higher than 1, choose a lower temperature.
Repeat till ∑ Ki xi ≅ 1.
Table 1-10 bubblepoint calculation for example 1-16.

We assumed two values of temperature , we found the first value (100 0F) is lower than the bubble point since ∑ Ki xi < 1.00 , and the second value (150 0F) is higher than the bubble point since ∑ Ki xi > 1.00 , the bubble point will be between the two values where ∑ Ki xi ≅ 1.
The Ki values in previous table where collected from “ Design operation and maintenance of gas plants - John Campbell Co.” , since it’s hard to obtain Ki numbers at temperature rather than the pre-drawn temperature lines in K-Charts.
The Values of Ki can be extracted from individual component charts (figures 1-12 to 1-16) (Methane K-chart, Ethane K-chart ….etc.), or can be extracted from “DePriester” chart, fig 1-17.

Dewpoint calculation: To calculate the dewpoint temperature at certain pressure, (All the components are in gas phase yi = 1). From eq. 1-29, the dewpoint will be reached when ∑ yi /Ki ≅ 1
Solution Steps:
Assume a temperature value (150 0F), for example.
From the K chart of each compound, find the K value at the system pressure and assumed temperature.
Divide mole fraction Yi of each component by its equilibrium value taken from the table Ki.
Take the sum ∑ Yi /Ki , if it’s higher than 1, choose higher temperature, (2000F for example), and repeat as in the table.
If, ∑ Yi /Ki is less than 1, choose a lower temperature.
Repeat till ∑ Yi /Ki ≅ 1.


Table 1-11 dewpoint calculation for example 1-16.

We assumed two values of temperature , we found the first value (150 0F) is lower than the dewpoint since ∑ yi /Ki > 1.00 , and the second value (200 0F) is higher than the bubble point since ∑ yi /Ki value is < 1.00 , the dewpoint will be between the two values where ∑ yi /Ki ≅ 1 .
The Ki values in previous table where collected from “ Design operation and maintenance of gas plants - John Campbell Co.” , since it’s hard to obtain Ki numbers at temperature rather than the pre-drawn temperature lines in K-Charts.
The Values of Ki can be extracted from individual component charts (Methane K-chart, Ethane K-chart ….etc.), or can be extracted from “DePriester” chart, fig 1-17.

Flash calculations:
Different values of “L” will be assumed (remember, V = 1 - L), and accordingly Xi will be calculated till we obtain ∑ xi = 1.00.
Ki from chart at 250 psia and 150 0F
Using Eq. 1 -26 xi = zi / ( L + V Ki)

Table 1-12 flash calculations for example 1-16.

The assumed value of L=0.5, found to be lower than the correct value, and the assumed value of L= 0.75 found to be higher than the correct value.
The correct value must be in between the two previous assumed values, and found to be 0.649.
Flash calculations usually performed by computer software, for manual calculations, some K value charts are included in this chapter for the illustration of manual calculations for the previous example. (Figures 1-12 to 1-16)
Other K-values are included in Chapter 25 “Equilibrium Ratio (K) Data” in the “GPSA Engineering Data Book”, or Appendix 5A Volume 1 “Gas conditioning and Processing – The Basic Principles,” Campbell Petroleum Series. In the other hand, the DePriester Chart Figure 1-17, may be used for all hydrocarbon components.

K Value charts:

Figure 1-12 Equilibrium ratio (K) for Methane.
Figure 1-13 Equilibrium ratio (K) for Ethane.

Figure 1-14 Equilibrium ratio (K) for Propane.

Figure 1-15 Equilibrium ratio (K) for i-Butane.
Figure 1-16 Equilibrium ratio (K) for n-Butane.

Figure 1-17 the DePriester (K) Chart for hydrocarbon components.

1.6: Types of Fluid Flow
When a fluid moves through a pipe, two distinct types of flow are possible, laminar and turbulent. Laminar flow occurs in fluids moving with small average velocities and turbulent flow becomes apparent as the velocity is increased above a critical velocity. In laminar flow, the fluid particles move along the length of the pipe in a very orderly fashion, with little or no sideways motion across the width of the pipe.
Turbulent flow is characterized by random, disorganized motion of the particles, from side to side across the pipe as well as along its length. The two types of fluid flow are described by different sets of equations. In general, for most practical situations, the flow will be turbulent.

Figure.1-18. Laminar and turbulent flow in pipes.
1.6.1: Reynolds Number
A useful factor in determining which type of flow is involved is the Reynolds number. This is the ratio of the dynamic forces of mass flow to the shear resistance due to fluid viscosity and is given by the following formula.
Re = VDρ/ µ Eq. 1-30

Re = Reynolds number, dimensionless
V = average gas velocity, ft/s
D = pipe inside diameter, ft
ρ = gas density, lb/ft3
μ = gas viscosity, lb/ft-s (1 cp = 0.000672 lb/ft-s)

The flow is considered to be laminar flow when the Reynolds number is below 2000.
Turbulent flow is said to exist when the Reynolds number is greater than 4000. When the Reynolds numbers is between 2000 and 4000, the flow is called critical flow, or undefined flow.
Re <= 2000 Flow is laminar
Re > 4000 Flow is turbulent
2000 < Re <= 4000 Flow is critical flow
In terms of the more commonly used units in the gas pipeline industry, the following formula for Reynolds number is more appropriate:
Re = 1.35 x 10-5 (GQ/µd) Eq. 1-31
G = gas specific gravity (air = 1)
Q = gas flow rate, standard ft3/day (scfd)
d = pipe inside diameter, in.
μ = gas viscosity, lb/ft-s (1 cp = 0.000672 lb/ft-s) 
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