Fluid Flow EPT 09-T-06 April 1998
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Fluid Flow
April 1998
Table of Contents Table of Tables ................................................................ ....................................................................................... ................................................. ............................ .. 7 1.
Scope .............................................. ......................................................................... .................................................. ................................................. ............................ .. 8
2.
References References .................................................. ............................................................................. .................................................. ...................................... ............... 8
3.
4.
2.1.
MEPS–Mobil MEPS–Mobil Engineerin Engineering g Practices Practices.................................................. .................................................................... .................. 8
2.2.
Mobil Tutorials Tutorials ................................................. ............................................................................ .................................................. ......................... .. 9
2.3.
API–American API–American Petroleum Petroleum Institute .................................................... ...................................................................... .................. 9
2.4.
ASME–American ASME–American Society Society of Mechanical Mechanical Engineers Engineers............................................ ............................................ 9
2.5.
ASTM–American ASTM–American Society for Testing Testing & Material Materials s ............................................... ............................................... 9
2.6.
AWS–American Welding Society.................................................. ........................................................................ ...................... 9
2.7.
CFR–U.S. CFR–U.S. Code Code of Federal Federal Regulations Regulations .................................................... ........................................................... ....... 10
Basic Flow Flow Equations Equations and Factors .................................................. ......................................................................... ......................... 10 3.1.
Density ............................................... .......................................................................... .................................................. .................................... ............. 10
3.2.
Fluid Viscosity Viscosity ................................................. ............................................................................ .................................................. ....................... 13
3.3.
Fluid Head and and Bernoulli' Bernoulli's s Law..................................................... ......................................................................... .................... 19
3.4.
Flow Regimes ................................................. ........................................................................... .................................................. ........................ 22
3.5.
Darcy-Weisbach Darcy-Weisbach Equation for Pressure Drop ................................................... ................................................... 25
3.6.
Moody Friction Friction Factor .................................................. ........................................................................... ..................................... ............ 27
3.7.
Effect of of Elevation Elevation Changes.................................................. ......................................................................... ............................ ..... 30
Pressure Drop Drop in Piping Piping ................................................... ........................................................................... ......................................... ................. 35 4.1.
Liquid Flow (General Equation) Equation) ................................................. ........................................................................ ......................... 35
4.2.
Gas Flow ................................................ ........................................................................... .................................................. ................................ ......... 36
4.3.
Two-Phase Two-Phase Flow ................................................. ............................................................................ .............................................. ................... 45
4.4.
Head Loss Loss in Valves Valves and Pipe Fittings Fittings ................................................... .............................................................. ........... 55
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5.
6.
Fluid Flow
April 1998
Choosing Choosing a Line Diameter.............................. Diameter........................................................ ................................................. ................................. .......... 66 5.1.
General .................................................. ............................................................................. .................................................. ................................ ......... 66
5.2.
Erosional Erosional Velocity................................................ ........................................................................... .............................................. ................... 67
5.3.
Liquid Line Sizing Sizing ................................................ ........................................................................... .............................................. ................... 71
5.4.
Gas Line Sizing Sizing ................................................... .............................................................................. .............................................. ................... 72
5.5.
Two-Phase Two-Phase Flow Flow Line Sizing Sizing ..................................................... ............................................................................ ......................... 76
Determining Determining Wall Thickness Thickness ................................................ ........................................................................ ..................................... ............. 80 6.1.
Commonly Available Available Pipe .................................................... ............................................................................ ............................. ..... 80
6.2.
Standards and Requirements Requirements ................................................... .......................................................................... ......................... 83
6.3.
General Hoop Stress Formula .................................................. ......................................................................... ......................... 84
Appendix Appendix A–Nomenclatu A–Nomenclature re ................................................ ........................................................................... .............................................. ................... 86 Appendix Appendix B–Example Problems–Metric Problems–Metric Units ........................................... ............................................................... .................... 90 1.
2.
Pressure Drop .................................................. ............................................................................ ................................................. ................................ ......... 90 1.1.
Liquid Line................................................... ............................................................................. ................................................. ............................ ..... 90
1.2.
Gas Line................................................. ............................................................................ .................................................. ................................ ......... 92
1.3.
Two-Phase Two-Phase Lines ................................................ ........................................................................... .............................................. ................... 96
Choosing Choosing a Line Line Diameter Diameter and Determining Determining Wall Thickness Thickness ................................ ................................ 98 2.1.
Liquid Line................................................... ............................................................................. ................................................. ............................ ..... 98
2.2.
Gas Line................................................. ............................................................................ .................................................. .............................. ....... 101
2.3.
Two-Phase Two-Phase Line .................................................. ............................................................................. ............................................ ................. 103
Appendix Appendix C–Example Problems–Customar Problems–Customary y Units ........................................... ..................................................... .......... 107 1.
Pressure Drop .................................................. ............................................................................ ................................................. .............................. ....... 107 1.1.
Liquid Line................................................... ............................................................................. ................................................. .......................... ... 107
1.2.
Gas Line................................................. ............................................................................ .................................................. .............................. ....... 109
1.3.
Two-Phase Two-Phase Lines ................................................ ........................................................................... ............................................ ................. 113
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Choosing a Line Diameter and Determining Wall Thickness .............................. 116 2.1.
Liquid Line ....................................................................................................... 116
2.2.
Gas Line.......................................................................................................... 118
2.3.
Two-Phase Line .............................................................................................. 121
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Table of Figures Figure 1: Physical Properties of Water (Courtesy of Ingersoll Rand) ......................... 15 Figure 2: Kinematic Viscosity vs. Temperature for Different API Gravity Oils ........... 16 Figure 3: Liquid Viscosity of Pure and Mixed Hydrocarbons Containing Dissolved Gases (Courtesy of GPSA) ............................................................................. 17 Figure 4: Hydrocarbon Gas Viscosity (Courtesy of GPSA).......................................... 19 Figure 5: Friction Factor as a Function of Reynolds Number and Pipe Roughness (Courtesy of API) ............................................................................................. 30 Figure 6: Effect of Elevation Changes on Head ............................................................ 31 Figure 7: Friction Factor vs. Pipe Diameter for Three Correlations ............................ 44 Figure 8: Two-phase Flow Patterns in Horizontal Flow (Source: P. Griffith, "Multiphase Flow in Pipes," JPT, March 1984, pp. 363-367) ....................... 46 Figure 9: Horizontal Multi-phase Flow Map (Source: P. Griffith, "Multiphase Flow in Pipes," JPT, March 1984, pp. 363-367) .......................................................... 48 Figure 10: Two-phase Flow Patterns in Vertical Flow (Source: J.P. Brill, "Multiphase Flow in Wells," JPT, January 1987, pp. 15-21) .............................................. 49 Figure 11: Vertical Multiphase Flow Map (Source: Yaitel, Y., Barhea, D., and Duckler, A.E., "Modeling Flow Pattern Transitions for Steady Upward GasLiquid Flow in Vertical Tubes," AIChE J., May 1980, pp. 345-354.) ............. 50 Figure 12: Resistance Coefficients for Different Types of Pipe Entrances and Exits (Courtesy of Paragon Engineering Services, Inc.) ....................................... 57 Figure 13: Resistance Coefficients for Sudden Enlargements and Contractions (Courtesy of Paragon Engineering Services, Inc.) ....................................... 58 Figure 14: Equivalent Lengths of 90 Degree Bends (Courtesy of Crane Technical Paper 410) ........................................................................................................ 65 Figure 15: Equivalent Length of Miter Bends (Courtesy of Crane Technical Paper 410) ................................................................................................................... 65 Figure 16: Example of a Target Tee ............................................................................... 70 Figure 17: Wear Rate Comparison for Standard Fittings (Source: API OSAPR Project No. 2) ................................................................................................... 70
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Figure 18: Acceptable Pressure Drop for Short Lines (Courtesy of Paragon Engineering Services, Inc.) ............................................................................ 74 Figure 19: General Hoop Stress Free Body Diagram (Courtesy of Paragon Engineering Services, Inc.) ............................................................................ 85
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Table of Tables Table 1: Pipe Roughness ................................................................................................ 29 Table 2: Two Phase Flow Correlations, AGA Multiphase Pipeline Data Bank for GasCondensate Lines (From: Battarra, Mariana, Gentilini and Giaccheta, Oil and Gas Journal, Dec. 30, 1985) .................................................................... 51 Table 3: Resistance Coefficients for Pipe Fittings ....................................................... 56 Table 4: Equivalent Lengths of Valves and Fittings in Feet (Courtesy of GPSA) ...... 61 Table 5: ANSI Pipe Schedules ........................................................................................ 80
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1.
Fluid Flow
April 1998
Scope Piping transports produced fluids from one piece of production equipment to another. Facilities piping, whether in an onshore production facility or an offshore platform, may be required to carry liquids, gas, or two-phase flow. Most facilities piping is made up of short segments, and pressure drop in the piping is minimal. Pressure losses between process components occur primarily in control valves. In these cases, flow velocity and not pressure drop is most important in choosing a line size. However, pressure drop could be critical in sizing lines between vessels operating at or near the same pressure, where elevation changes occur, for long transfer lines between facilities, where back-pressure on wells is critical, and in vent and relief lines. Selection of facilities piping is accomplished in three basic steps. First, determine the allowable pressure drop and flow velocities within the constraints allowed by the process. Second, select the required pipe diameter to meet the process fluid flow velocity and pressure drop requirements. Third, determine the required wall thickness to meet maximum working pressures, corrosion effects, and design code requirements. The design pressure or stress due to thermal expansion, contraction, or bending determines wall thickness and pressure rating class. Pressure rating class may be substantially higher than the operating pressure of the line, since the system shall be designed to contain the maximum pressure (stress) to which it can be subjected under abnormal as well as normal conditions. The selection of a design pressure for a given line is determined by selected distinct locations in the piping system, called "pressure breaks." These are where the maximum pressure the system can be subjected to under abnormal conditions changes. The procedure for determining pressure breaks is discussed in Section 8. The purpose of this Tutorial is to provide the project engineer with information for determining line size, wall thickness, and pressure rating class. Miscellaneous details to be considered in designing a piping system also are discussed.
2.
References The following Mobil Guides and industry publications are referenced herein and shall be considered a part of this EPT. Refer to the latest editions unless otherwise specified.
2.1. MP 16-P-01
MEPS–Mobil Engineering Practices Piping-General Design
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MP 16-P-30A
Piping - Materials and Service Classifications (M&R)
MP 33-P-13
Electrical - MV Motor Control
MP 33-P-23
Electrical - Raceway & Cable Tray Installations
2.2. EPT 09-T-05
2.3.
April 1998
Mobil Tutorials Piping-Code Selection Guide
API–American Petroleum Institute
API RP 14C
Recommended Practice for Analysis, Design, Installation, and Testing of Basic Surface Safety Systems for Offshore Production Platforms Fifth Edition; Errata - 1994
API RP 14E
Recommended Practice for Design and Installation of Offshore Production Platform Piping Systems Fifth Edition
API SPEC 6A
Specification for Wellhead and Christmas Tree Equipment Seventeenth Edition
2.4.
ASME–American Society of Mechanical Engineers 1
ASME B16.5
Pipe Flanges and Flanged Fittings NPS /2 Through NPS 24
ASME B31.1
Power Piping
ASME B31.3
Process Piping
ASME B31.4
Liquid Transportation Systems for Hydrocarbons, Liquid Petroleum Gas, Anhydrous Ammonia, and Alcohols
ASME B31.8
Gas Transmission and Distribution Piping Systems
2.5. ASTM A106
2.6. AWS QC7-93
ASTM–American Society for Testing & Materials Standard Specification for Seamless Carbon Steel Pipe for HighTemperature Service
AWS–American Welding Society Chemical Plant and Petroleum Refinery Piping (Supplement F)
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2.7. 49 CFR 192
3.
Fluid Flow
April 1998
CFR–U.S. Code of Federal Regulations Transportation, Subchapter D - Pipeline Safety, Transportation of Natural and Other Gases by Pipeline: Minimum Federal Safety Standards
Basic Flow Equations and Factors 3.1.
Density 3.1.1. The density of a fluid is an important property in calculating pressure drop. A liquid's density is often specified by giving its specific gravity relative to water at standard conditions: 15.6°C and 101.4 kPa (60°F and 14.7 psia). Thus: Equation 1
Metric : ρ = 1000 (SG ) Customary : ρ = 62.4 (SG ) where : ρ = density of liquid, kg / m
3
(lb / ft ) 3
(SG ) = specific gravity of liquid relative to water
3.1.2. Oil density is often expressed in te rms of API gravity, given in degrees API, which is defined as:
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Equation 2
Metric : API =
o
141.5
(SG at 15.6 C ) o
- 131.5
Customary : API =
o
141.5
(SG at 60 F ) o
- 131.5
3.1.3. The density of a mixture of oil and water can be determined by the volume weighted average of the two densities and is given by: Equation 3
ρ =
ρ wQw
+ ρ oQo
QT
where : ρ = density of liquid, kg / m 3 (lb / ft 3 )
= density of oil, kg / m 3 (lb / ft 3 ) ρ w = density of water, kg / m 3 (lb / ft 3 ) 3 Qw = water flow rate, m / hr ( BPD ) 3 Qo = oil flow rate, m / hr ( BPD ) QT = total liquid flow rate, m 3 / hr ( BPD ) ρ o
3.1.4. Similarly, the specific gravity of an oil and water mixture can be calculated by:
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Equation 4
(SG )m =
(SG )w Qw + (SG )o Qo QT
where :
(SG )m = specific gravity of liquid (SG )o = specific gravity of oil (SG )w = specific gravity of water Qw = water flow rate, m 3 / hr ( BPD ) Qo = oil flow rate, m 3 / hr ( BPD ) QT = total liquid flow rate, m 3 / hr ( BPD )
3.1.5. The density of natural gas at standard conditions of temperature and pressure is determined by its molecular weight. It is often expressed as a specific gravity, which is the ratio of the density of the gas at standard conditions of temperature and pressure to that of air at standard conditions of temperature and pressure. Since the molecular weight of air is 29, the specific gravity of a gas is given by: Equation 5
S =
( MW ) 29
where : S = specific gravity ofgas relative to air
( MW ) = molecular weight of the gas
3.1.6. The density of a gas under specific conditions of temperature and pressure is given by:
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Equation 6
etric : ρ g = 3.48
SP TZ
Customary : ρ g = 2.70
SP TZ
Equation 7
etric : ρ g = 0.1203
( MW ) P TZ
Customary : ρ g = 0.093
( MW ) P TZ
where : ρ g = density of gas, kg / m (lb / ft 3
3
)
P = pressure, kPa ( psia ) T = temperatur e, K ( R ) o
Z = gas compressibility factor S = specific gravity of gas relative to air
( MW ) = molecular weight of the gas
3.2.
Fluid Viscosity 3.2.1. In determining the pressure drop in a piping system, the viscosity of the fluid flowing at the actual conditions of pressure and temperature in the piping system shall be known. Viscosity is a measure of a fluid's resistance to flow and is expressed in either absolute terms or kinematic terms. The relationship between absolute and kinematic viscosity is given by:
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Equation 8
µ = γρ where : µ = absolute viscosity, Pa - sec (cp ) γ = kinematic viscosity, m 2 / sec (cs ) ρ = density offluid, kg / m 3 (lb / ft 3 )
3.2.2. In the metric system, if absolute viscosity is given in centipoise then kinematic viscosity is in centistokes and the unit of density to use in Equation 8 is gram/cm3. Since water has a density of 1.0 gram/cm3, Equation 8 can be rewritten: Equation 9
Metric : µ = 1000(SG ) γ Customary : µ = (SG ) γ where : µ = absolute viscosity, Pa - sec (cp )
(SG ) = specific gravity of liquid relative to water γ = kinematic viscosity, m 2 / sec (cs )
3.2.3. Figure 1 shows the viscosity of water at various temperatures. The viscosity of oil is highly dependent on temperature and is best determined by measuring the viscosity at two or more temperatures and interpolating to determine the viscosity at any other temperature. When such data are not available, the viscosity can be estimated from Figure 2 if the oil is above its cloud point temperature (the temperature at which wax crystals start to form when crude oil is cooled). Although viscosity is generally a function of API gravity, it is not always true that a heavier crude (lower API gravity) has a
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higher viscosity than a lighter crude. For this reason, this figure shall be used with care.
Figure 1: Physical Properties of Water (Courtesy of Ingersoll Rand)
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Figure 2: Kinematic Viscosity vs. Temperature for Different API Gravity Oils
3.2.4. Figure 2 presents viscosity for "gas free" or stock tank crude oil. Figure 3 can be used to account for the fact that oil at higher pressures has more light hydrocarbon components and so has a higher gravity and lower viscosity than at stock tank conditions. This correction also can be made by using Figure 2 with the API gravity of the oil at the higher pressure rather than its stock tank gravity.
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Figure 3: Liquid Viscosity of Pure and Mixed Hydrocarbons Containing Dissolved Gases (Courtesy of GPSA)
3.2.5. The viscosity of a mixture of oil a nd water is not the weighted average of the two viscosities. Depending on the ratio of water and oil and the violence of mixing (shear rate) in the system, the viscosity of the mixture can be as much as 10 to 20 times that of the oil. The following equation has proven useful in analyzing piping systems:
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Equation 10
µ eff
= (1 + 2.5φ + φ 2 ) µ c
where :
= effective viscosity of the mixture, Pa - sec (cp ) µ c = viscosity of the continuous phase, Pa - sec (cp ) φ = volume fraction of the discontinuous phase µ eff
3.2.6. Normally the breakover between an oil-continuous and a water-continuous phase occurs between 60 and 70 percent water cut.
3.2.7. The viscosity of a natural gas can be determined from Figure 4. For most production facility gas piping applications, a viscosity of 0.012 cp can be assumed.
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Figure 4: Hydrocarbon Gas Viscosity (Courtesy of GPSA)
3.3.
Fluid Head and Bernoulli's Law 3.3.1. The term "head" is commonly used to represent the vertical height of a static column of a liquid corresponding to the mechanical energy contained in the liquid per unit mass. Head also can be considered as the amount of work necessary to move a liquid from its original position to the r equired delivery position. Here, this includes the additional work necessary to overcome the resistance to flow in the line.
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3.3.2. In general, a liquid at any point may have three kinds of head: 1. Static Pressure Head represents the energy contained in the liquid due to its pressure. 2. Potential Head represents the energy contained in the liquid due to its position measured by the vertical height above some plane of reference. 3. Velocity Head represents the kinetic ener gy contained in the liquid due to its velocity.
3.3.3. Bernoulli's Law states that as a fluid flows from one point to another in a piping system the total of static, potential, and velocity head at the upstream point (subscript 1) equals the total of the three heads at the downstream point (subscript 2) plus the friction drop between points 1 and 2. Equation 11
( H SH )1 + ( H PH )1 + ( H VH )1 = ( H SH )2 + ( H PH )2 + ( H VH )2 + H f where : H SH = Static pressure head, m ( ft ) H VH = Velocity head, m ( ft ) H PH = Potential head, m( ft ) H f = Pipe friction loss, m ( ft )
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Equation 12
etric : Z e1
+ 1000
P 1 ρ 1
2
+
V 1
2g
= Z e2 + 1000
2
P 2
+
ρ 2
V 2
2g
+ H f
Customary : Z e1
+
144P 1 ρ 1
2
+
V 1
2g
= Z e2 +
144P 2 ρ 2
2
+
V 2
2g
+ H f
where :
= vertical elevation rise of pipe, m ( ft ) P = pressure, kPa ( psia ) ρ = density of liquid, kg / m 3 (lb / ft 3 ) V = average velocity, m / sec ( ft / sec ) g = acceleration of gravity, 9.81m / sec 2 (32.2ft / sec 2 ) H f = head loss, m ( ft )
Z e
3.3.4. Velocity, as used herein, refers to the average velocity of a fluid at a given cross section, and is determined by the steady state flow equation:
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Equation 13
etric : V =
Q
=
W s
3600A A ρ
Customary : V =
Q A
=
W s A ρ
where : V = average velocity, m / sec ( ft / sec ) Q = rate of flow, m 3 / sec ( ft 3 / sec ) A = cross sectional area of pipe, m 2 ( ft 2 )
= rate of flow, kg / sec (lb / sec ) ρ = density of liquid, kg / m 3 (lb / ft 3 ) W s
3.4.
Flow Regimes 3.4.1. Experiments have demonstrated that there are two basic types of flow in pipe, laminar and turbulent. In laminar flow, fluid particles flow in a straight line, while in turbulent flow the f luid particles move in random patterns transverse to the main flow.
3.4.2. At low velocities, fluid flow is laminar. As the velocity increases, a "critical" point is reached at which the flow regime changes to turbulent flow. This "critical" point varies depending upon the pipe diameter, the fluid density and viscosity, and the velocity of flow.
3.4.3. Reynolds showed that the flow regime can be defined by a dimensionless combination of four variables. This number is referred to as the Reynolds Number (Re), and is given by:
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Equation 14
etric : Re =
ρ DV µ
Customary : Re =
ρ D V µ '
=
1488DV ρ µ
=
124dV ρ µ
=
7738 (SG )dV µ
where : Re = Reynolds number, dimensionl ess
(
ρ = density, kg / m 3 lb / ft 3
)
d = pipe ID, mm (in ) D = pipe ID, m ( ft ) V = average velocity, m / sec ( ft / sec ) µ ' = viscosity, lb / ft - sec [µ (cp ) × 0.000672] µ = absolute viscosity, Pa - sec (cp )
(SG ) = specific gravity of liquid relative to water 3.4.4. At Re < 2000 the flow shall be laminar, and when Re > 4000 the flow shall be turbulent. In the "critical" or "transition" zone, (2000 < Re < 4000), the flow is unstable and could be either laminar or turbulent.
3.4.5. The Reynolds number can be expressed in more convenient terms. For liquids, Equation (14) can be shown to be:
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Equation 15
Metric :
(SG ) Ql
Rel = 353.13
d µ
Customary : Rel = 92.1
(SG ) Ql d µ
where : µ = absolute viscosity, Pa - sec (cp ) d = pipe ID, mm (in )
(SG ) = specific gravity of liquid relative to water Ql = liquid flow rate, m 3 / hr ( BPD )
3.4.6. The Reynolds number for gas flow can be shown to be: Equation 16
Metric : Q g S
Re g = 0.428
d µ
Customary : Re g = 20,000
Q g S d µ
where : Q g = gas flow rate, std m / hr ( MMSCFD ) 3
S = specific gravity of gas relative to air d = pipe ID, mm (in ) µ = absolute viscosity, Pa - sec (cp )
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Fluid Flow
April 1998
Darcy-Weisbach Equation for Pressure Drop 3.5.1. The head loss due to friction is given by the Darcy-Weisbach equation as follows: Equation 17
H f
=
fLV 2 D2g
where : L = length of pipe, m ( ft ) D = pipe ID,m( ft ) f = Moody friction factor V = average velocity, m / sec ( ft / sec ) g = acceleration of gravity, 9.81 m / sec 2 (32.2 ft / sec 2 ) H f
= pipe friction head loss, m ( ft )
3.5.2. Equations 12 and 17 can be used to calculate the pressure at any point in a piping system if the pressure, average flow velocity, pipe inside diameter, and elevation are known at any other point. Conversely, if the pressures, pipe inside diameter, and elevations are known at two points, the flow velocity can be calculated. Neglecting the head differences due to elevation and velocity changes between two points, Equation 12 can be reduced to:
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Equation 18
etric : P 1 - P 2 = ∆ P = 9.81 × 10 ρ H f -3
Customary : P 1 - P 2 = ∆ P =
ρ 144
H f
where : ∆ P = pressure drop, kPa ( psi ) H f
= pipe friction head loss, m ( ft )
ρ = density of liquid, kg / m 3 (lb / ft 3 )
3.5.3. Substituting Equation 17 into Equation 18 and expressing pipe inside diameter in inches: Equation 19
Metric : ∆ P = 0.5
f ρ LV 2 d
Customary : 2
∆ P = 0.0013
f ρ LV d
where : d = pipe ID,mm (in ) f = Moody friction factor ρ = density of liquid, kg / m 3 (lb / ft 3 ) L = length of pipe, m ( ft ) V = average velocity, m / sec ( ft / sec ) ∆ P = pressure drop, kPa ( psi )
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3.6.
Fluid Flow
April 1998
Moody Friction Factor 3.6.1. The Darcy-Weisbach equation can be derived ra tionally by dimensional analysis, except for the friction factor (f), which shall be determined experimentally. Considerable research has been done in reference to pipe roughness and friction factors. The Moody friction factor is generally accepted and used in pressure drop calculations.
3.6.2. Some texts including API RP 14E utilize the "Fanning friction factor," which is one fourth ( 1/4) the value of the Moody friction factor, restating the DarcyWeisbach equations accordingly, where:
f fanning = 1 / 4f
3.6.3. This has been a continual source of confusion in basic e ngineering fluid analysis. This Tutorial uses the Moody friction factor throughout. The reader is strongly cautioned always to note which friction factor (Moody or Fanning) is used in the applicable equations a nd which friction factor diagram is used as a source when calculating pressure drops.
3.6.4. The friction factor for fluids in laminar flow is directly related to the Reynolds Number (Re < 2000), and is expressed:
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Equation 20
etric : f = 64
µ dV ρ
Customary : f =
64 Re
= 0.52
µ dV ρ
where : f = Moody friction factor Re = Reynolds number µ = absolute viscosity, Pa - sec (cp ) d = pipe ID, mm (in ) V = average velocity, m / sec ( ft / sec ) ρ = density of fluid, kg / m (lb / ft 3
3
)
3.6.5. If this quantity is substituted into Equation 19, pressure drop in pounds per square inch for fluids in laminar flow becomes: Equation 21
etric : ∆ P = 32
µ LV d 2
Customary : ∆ P = 0.000676
µ LV d 2
3.6.6. The friction factor for fluids in turbulent flow (Re > 4000) depends on the Reynolds number and the relative roughness of the pipe. Relative roughness is the ratio of pipe absolute roughness, ε ε, to pipe inside diameter. Roughness
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is a measure of the smoothness of the pipe's inner surface. Table 1 shows the absolute roughness, ε ε, for various types of new, clean pipe. For pipe which has been in service for some time it is often recommended that the absolute roughness to be used for ca lculations shall be up to four times as much as the values shown in Table 1.
Table 1: Pipe Roughness Absolute Roughness ( ε ) Type of Pipe (New, clean condition)
(mm)
(ft)
(in)
Unlined Concrete
0.30
0.001-0.01
0.012-0.12
Cast Iron - Uncoated
0.26
0.00085
0.0102
Galvanized Iron
0.15
0.0005
0.006
Carbon Steel
0.046
0.00015
0.0018
Fiberglass Epoxy
0.0076
0.000025
0.0003
Drawn Tubing
0.0015
0.000005
0.00006
3.6.7. The friction factor, f, can be determined from the Moody diagram, Figure 5, or from the Colebrook equation: Equation 22
ε 2.51 = - 2 log 10 + 1 1 3.7D 2 ( f )2 ( ) Re f 1
where : f = Moody friction factor D = pipe ID, m ( ft ) Re = Reynolds number ε = absolute roughness, m ( ft )
3.6.8. The pressure drop between any two points in a piping system can be determined from Equation 21 for laminar flow, or Equation 19 for turbulent flow using the friction factor from Figure 5 or Equation 22.
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Figure 5: Friction Factor as a Function of Reynolds Number and Pipe Roughness (Courtesy of API)
3.7.
Effect of Elevation Changes 3.7.1. In single phase gas or liquid flow the pressure change between two points in the line shall be affected by the relative elevations of those points but not by intermediate elevation changes. This is because the density of the flowing fluid is nearly constant and the pressure increase caused by any decrease in elevation is balanced by the pressure decrease caused by an identical increase in elevation.
3.7.2. In Figure 6, case A, the elevation head increa ses by H from point 1 to point 2. Neglecting pressure loss due to friction, the pressure drop due to elevation change is given by:
P 1 - P 2 = H
ρ 144
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Figure 6: Effect of Elevation Changes on Head
3.7.3. In case B, the elevation head decrea ses by H from point 1 to point A. Neglecting pressure loss due to friction, the pressure increase from 1 to A is determined from Equation 12:
P A - P B
ρ = - H 144
3.7.4. Similarly, elevation pressure changes due to other segments are:
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PA - PB = 0
ρ 144
P B - P 2 = 2H
3.7.5. The overall pressure change in the pipe due to elevation is obtained by adding the changes for the individual segments:
P 1 - P 2 = P 1 - P A + P A - P B + P B - P 2
P 1 - P 2 = (0 - H + 0 + 2H )
P 1 - P 2 = H
ρ 144
ρ 144
3.7.6. Thus, for single-phase flow, the pressure drop due to elevation changes is determined solely by the elevation change of the end points. Equation 12 can be rewritten as: Equation 23
etric : ∆ P Z ≈ 9.79 (SG ) ∆ Z Customary : ∆ P Z ≈ 0.433 (SG ) ∆ Z where : ∆ P Z = pressure drop due to elevation increase, kPa ( psi ) ∆ Z = total increase in elevation, m ( ft )
(SG ) = specific gravity of liquid relative to water
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3.7.7. In two-phase flow, the density of the fluids in the uphill runs is higher than the density of the fluids in the downhill runs. In downhill lines flow is stratified with liquid flowing faster than gas. The depth of the liquid layer adjusts to the depth where the static head advantage is equal to the pressure drop due to friction, and thus the average density of the mixture approaches that of the gas phase.
3.7.8. The uphill segments at low gas rates are liquid full, and the density of the mixture approaches that of the liquid phase. As a worst case condition, it can be assumed that the downhill segments are filled with gas and the uphill segments are filled with liquid. Referring to Figure 6, case A, assuming fluid flow from left to right and neglecting pressure loss due to friction:
ρ l 144
P 1 - P 2 = H
For Case B:
P 1 - P A
ρ = - H g 144
P A - P B
=0 ρ l 144
P B - P 2 = 2H
Thus:
ρ l (2 ρ - ρ ) l g 144
P 1 - P 2 =
Since ρ l >> ρ g
ρ l 144
P 1 - P 2 ≈ 2H
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3.7.9. Thus, one would expect a higher pressure drop due to elevation change for case B than for case A even though the net change in elevation f rom point 1 to point 2 is the same in both cases.
3.7.10. So, neglecting pressure changes due to any elevation drops, the maximum pressure drop due to elevation changes in two-phase lines can be estimated from: Equation 24
etric : ∆ P Z e Z ≤ 9.79 (SG ) Σ Customary : Z e ∆ P Z ≤ 0.433 (SG ) Σ where : Σ Z e = sum of vertical elevation rises only, m ( ft ) ∆ P Z = pressure drop due to elevation changes, kPa ( psi )
(SG ) = specific gravity of liquid relative to water
3.7.11. With increasing gas flow, the total pressure drop may decrease as liquid is removed from uphill segments. More accurate prediction of the pressure drop due to elevation changes requires c omplete two-phase flow models that are beyond the scope of this manual. There are a number of proprietary computer programs available that take into account fluid property changes and liquid holdup in small line segments; they model pressure drop due to elevation changes in two-phase flow more accurately.
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Pressure Drop in Piping 4.1.
Liquid Flow (General Equation) 4.1.1. For flowing liquids in facility piping, the density is constant throughout the pipe length. Equation 19 can be rewritten to solve for either pressure drop or flow rate for a given length and diameter of pipe as follows: Equation 25
Metric : fL (Ql ) (SG ) 2
∆ P = 6.266 × 10
7
d 5
Customary : ∆ P = (1.15 × 10
-5
)
fL (Ql ) (SG ) 2
d 5
Equation 26
etric : 1
5 2 - 4 ∆ P d Ql = 1.265 × 10 fL (SG )
Customary : 1
∆ P d 5 2 Ql = 295 fL (SG )
4.1.2. The most common use of Equations 25 and 26 is to determine a pipe diameter for a given flow rate and allowable pressure drop. First, however, the Reynolds number shall be calculated to determine the fr iction factor.
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Since the Reynolds number depends on the pipe diameter, the equation cannot be solved directly. One method to overcome this disadvantage is to assume a typical friction factor of 0.025, solve for diameter, compute a Reynolds number, and then compare the assumed friction factor to one rea d from Figure 5. If the two are not sufficiently close, iterate the solution until they converge.
4.1.3. Figure 2.2 in API RP 14E can be used to approximate pressure drop or required pipe diameter. It is based on an assumed friction factor relationship which can be adjusted to some extent for liquid viscosity.
4.2.
Gas Flow 4.2.1.
General Equation 1. The Darcy equation assumes constant density throughout the pipe section between the inlet and outlet points. While this assumption is valid for liquids, it is incorrect for gas pipelines, where density is a function of pressure and temperature. As gas flows through the pipe the drop in pressure due to head loss causes it to expand and, thus, to decrease in density. At the same time, if heat is not added to the system, the gas will cool and tend to increase in density. In control valves, where the change in pressure is nearly instantaneous, (and thus no heat is added to the system), the expansion can be considered adiabatic. In pipe flow, however, the pressure drop is gradual and there is sufficient pipe surface area between the gas and the surrounding medium to add heat to the gas and thus to keep it at constant temperature. In such a case the gas can be considered to undergo an isothermal expansion. 2. On occasion, where the gas temperature is significantly different from ambient, the assumption of isothermal (constant temperature) flow is not valid. In these instances, greater accuracy can be obtained by breaking the line up into short segments that correspond to small temperature changes. 3. The general isothermal equation for the expansion of gas can be given by:
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Equation 27
etric :
( P 1 )2 - ( P 2 )2 (W s ) = fL P 1 P 1 + 2 ln v P 2 D 1.322 × 10 9 g A2
2
Customary :
( P 1 )2 - ( P 2 )2 (W s ) = P 1 fL P 1 v + 2 ln P 2 D 144g A2
2
where :
= rate of flow, kg / sec (lb / sec ) g = acceleration of gravity, 9.81 m / sec 2 (32.2 ft / sec 2 ) A = cross - sectional area of pipe, m 2 ( ft 2 ) v = specific volume of gas at upstream conditions , m 3 / kg ( ft 3 / lb ) f = Moody friction factor l = length of pipe, m ( ft ) D = pipe ID, m ( ft ) P 1 = upstream pressure, kPa ( psia ) P 2 = downstream pressure, kPa ( psia ) W s
4. This equation assumes that: a) No work is performed between points 1 and 2; i.e., there are no compressors or expanders, and no elevation changes. b) The gas is flowing under steady state conditions; i.e., there are no acceleration changes. c) The Moody friction factor, f, is constant as a function of length. There is a small change due to a change in Reynolds number, but this may be neglected. 5. For practical pipeline purposes,
2 ln
P 1 P 2
<<
fl D
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6. Making this assumption and substituting it into Equation 27, one can derive the following equation: Equation 28
etric :
( P 1 ) - ( P 2 ) = 52,430 2
2
S Q g 2 ZTfL d 5
Customary :
( P 1 ) - ( P 2 ) = 25.2 2
2
S Q g 2 ZTfL 5
d
7. The "Z" factor will change slightly from point 1 to point 2, but it is usually assumed to be constant and is chosen for an "average" pressure. 8. Please note that
P 12 - P 22 in Equation 28 is not the same as ( ∆P) . Rearranging Equation 28 and solving for Qg we have: 2
Equation 29
etric : Q g = 4.367 × 10
-3
d ( P 1 - P 2 ) 5
2
2
Z TfL S
Customary :
Q g = 0.199
d ( P 1 - P 2 ) 5
2
2
Z TfL S
9. As was the case for liquid flow, in order to determine a pipe diameter for a given flow rate and pressure drop, it is first necessary to estimate the diameter and then to compute a Reynolds number to determine the friction factor. Once the friction factor is known, a pipe diameter is calculated and compared against the assumed number. If the two are not sufficiently close, the process is iter ated until they converge.
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4.2.2.
April 1998
Small Pressure Drops For small pressure drops, an approximation can be calculated. The following formula can be derived from Equation 28 if P 1 - P2 < 10 percent of P1 and it is assumed that
P 12 - P 22
≈ 2P 1 (∆ P )
Equation 30
etric : S (Q g ) Z T f L 2
∆ P = 26,215
P 1 d 5
Customary : S (Q g ) Z T f L 2
∆ P = 12.6
4.2.3.
P 1 d 5
Weymouth Equation 1. This equation is used for short lengths of pipe where high-pressure drops are likely (turbulent flow). It is based on measurements of compressed air flowing in pipes with inner diameters ranging from 20 to 200 mm (0.8 to 11.8 in), in the range of the Moody diagram where the ε ε/ d curves are horizontal (i.e., high Reynolds number). In this range the Moody friction factor is independent of the Reynolds number and dependent upon the relative roughness. For a given absolute roughness, ε ε, the friction factor is a function of diameter only. For steel pipe the Weymouth data indicate: Equation 31
etric : f =
0.0941 1 3
d
Customary : f =
0.032 1 3
d
2. Substituting this into Equation 29, the Weymouth equation is:
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Equation 32
etric :
P - P Q g = 1.42 × 10 - 2 d 2.67 LSZT 2 1
2 2
1 2
Customary : 1
Q g = 1.11 d
2.67
P - P 2 LSZT 2 1
2 2
3. Assuming a temperature of 290°K (520°R), a compressibility of 1.0, and a specific gravity of 0.6, the Weymouth equation can also be written (This is the form of the equation that is given in the GPSA Engineering Data Book): Equation 33
Metric : 0.5
P 12 - P 22 - 5 T 2.667 b Q = 3.415 × 10 E d P b S Lm T avg Z avg Customary : 0.5
T b P 12 - P 22 2.667 Q = 433.5 E d P b S Lm T avg Z avg where :
= base absolute temperatur e, K ( o R ) P b = base absolute pressure, kPa ( psia ) E = pipeline efficiency factor o T avg = average absolute temperatur e, K ( R ) Z avg = average compressibility S = specific gravity d = pipeline ID, mm (in ) T b
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4. It is important to know what the equation is based on and when it is appropriate to use it. To reiterate, short lengths of pipe with high pressure drops are likely to be in turbulent flow, and thus the assumptions made by Weymouth are appropriate. Industry experience indicates that Weymouth's equation is suitable for most gas piping within the production facility. However, the friction factor used by Weymouth is generally too low for large diameter or low velocity lines, where the flow regime is more properly characterized by the sloped portion of the Moody diagram.
4.2.4.
Panhandle Equation 1. This equation is often used for long, larger diameter pipelines. It assumes that the friction factor can be represented by a straight line of constant negative slope in the moderate Reynolds number region of the Moody diagram. 2. A straight line on the Moody diagram would be expressed: Equation 34
log f = - n log Re + log N
or Equation 35
f =
N
( Re ) n
3. Using this assumption and assuming a constant viscosity for the gas, Equation 29 can be rewritten as the Panhandle equation:
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Equation 36
etric :
P 12 - P 22 -3 Q g = 1.229 × 10 E f 0.961 S ZTLm
0.51
d 2.53
Customary :
P 12 - P 22 Q g = 0.028E f 0.961 S ZTLm
0.51 2.53
d
where : E f
= efficiency factor, dimensionl ess = 1.0 for brand new pipe = 0.95 for good operating conditions = 0.92 for average operating conditions = 0.85 for unfavorabl e operating conditions
4. In practice, the Panhandle equation is commonly used for large diameter long pipelines where the Reynolds number is on the straight-line portion of the Moody diagram. 5. Neither the Weymouth nor the Panhandle equation represents a "conservative" assumption that can always be used to overstate pressure drop. If the Weymouth formula is used and the flow is in a moderate Reynolds number regime, the friction factor will in reality be higher than assumed (because the sloped line portion is higher than the horizontal portion of the Moody curve), and the actual pressure drop will be higher than calculated. If the Panhandle formula is used and the flow is actually in a high Reynolds number regime, the friction factor also will be higher than assumed (because the equation assumes the friction factor continues to decline with increased Reynolds number beyond the horizontal portion of the curve), and the actual pressure drop will again be higher than calculated.
4.2.5.
Spitzglass Equation 1. The Spitzglass equation is used for near-atmospheric pressure lines. It is derived directly from Equation 29 by making the following assumptions: a)
3.6 1 + 0.03d d 100
f = 1 +
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b) Τ = 520 R o
c) P 1 = 15 psi d) Z = 1.0 e)
∆ P < 10 percent of P 1
2. With these assumptions, and expressing pressure drop in terms of inches of water, the Spitzglass Equation can be written: Equation 37
etric :
∆hw d 5 -2 Q g = 3.655 × 10 9.144 S L 1 + + 1.18 × 10 -3 d d Customary :
5 ∆hw d Q g = 0.09 3.6 S L + 1+ 0.03d d where : Q g = gas flow rate, std m 3 / hr ( MMSCFD ) ∆hw = pressure loss, mm of water (in of water ) d = pipe ID, mm (in ) L = Length of pipe, m ( ft ) S = specific gravity of gas relative to air
4.2.6.
Comparison and Recommended Use of Gas Flow Equations 1. The Weymouth and Spitzglass equations both assume that the friction factor is a function of pipe diameter only. Figure 7 compares the friction factors calculated from these equations with the factors indicated by the horizontal line portion of the Moody diagram for two different absolute roughnesses.
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Figure 7: Friction Factor vs. Pipe Diameter for Three Correlations 2. In the small pipe diameter range [75 to 150 mm (3 to 6 in)] , all curves tend to yield identical results. For large diameter pipe [250 mm (10 in), and above], the Spitzglass equation becomes overly conservative. The curve is going in the wrong direction and thus the form of the equation must be wrong. If used, its predictive results are higher pressure drops than actually observed. The Weymouth equation tends to become optimistic with pipe diameters greater than 500 mm (20 in). If used, its predictive results are lower pressure drops than actually observed. Its slope is greater than the general flow equation with ε ε = 0.002 in. These results occur because of the ways the Spitzglass and Weymouth equations approximate the Moody diagram. 3. The empirical gas flow equations use various coefficie nts and exponents to account for efficiency and friction factor. These equations represent the flow condition at which they were derived but may not be accurate under different conditions. Unfortunately, these equations are often used as if they were universally applicable. 4. The following guidelines are recommended in the use of the gas flow equations: a) Use the general gas flow equation for most general usage. If it is inconvenient to use the iterative procedure of the general equation but high accuracy is required, compute the results using both the Weymouth and Panhandle Equations and use the higher calculated pressure drop.
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b) Use the Weymouth Equation only for small diameter, short run pipe within the production facility where the Reynolds number is expected to be high. The use of the Weymouth equation for pipe greater than 500 mm (20 in) in diameter or in excess of 4,600 meters (15,000 ft) long is not recommended. c) Use the Panhandle Equation only for large diameter, long run pipelines where the Reynolds number is expected to be moderate. d) Use the Spitzglass equation for low pressure vent lines le ss than 300 mm (12 in) in diameter. e) When using gas flow equations for old pipe, attempt to derive the proper efficiency factor through field tests. Buildup of scale, corrosion, liquids, paraffin, etc., can have a la rge effect on gas flow efficiency.
4.3.
Two-Phase Flow 4.3.1.
General 1. In some single-phase flow conditions, a small volume of gas may be entrained in liquid flow (such as a liquid dump line from a separator), or a small amount of liquid may be carried in the pipe in gas flow (such as gas off a separator). These small amounts usually have a negligible effect on pressure loss and are not considered in single phase flow calculations. However, there are certain flow conditions where sufficient volumes of a second gas or liquid phase exist to produce an appreciable effect on pressure loss. The pressure drop in such lines shall be considered using techniques for two-phase flow. 2. Examples of two-phase flow situations include: a) Fluid coming out of the well bore prior to liquid separation b) Gas and oil that have been metered a nd then recombined for flow in a common line to a central facility 3. Using the best correlations available for pressure drop and liquid hold up, predictions may be in error by ± 20 percent for horizontal flow and ± 50 percent for flow which is slightly inclined.
4.3.2.
Flow Regimes 1. When a gas-liquid mixture enters a horizontal pipeline, the two phases tend to separate with the heavier liquid settling to the bottom. The type of flow pattern depends primarily on the gas a nd liquid flow rates. Figure 8 shows typical flow patterns in horizontal two-phase pipe flow.
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Figure 8: Two-phase Flow Patterns in Horizontal Flow (Source: P. Griffith, "Multiphase Flow in Pipes," JPT, March 1984, pp. 363-367) 2. Horizontal flow regimes can be described as follows: a) Bubble Very low gas-liquid ratios. Gas bubbles rise to the top. b) Elongated Bubble With increasing gas-liquid ratios, bubbles become larger and form gas plugs. c) Stratified Further increases in gas-liquid ratios make the plugs become longer until the gas and liquid are in separate layers. d) Wavy
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As the gas rate increases, the flowing gas causes waves in the flowing liquid. e) Slug At even higher gas rates, the waves touch the top of the pipe, trapping gas slugs between wave crests. The length of these slugs can be several hundred feet long in some cases. f)
Annular Mist At extremely high gas-liquid ratios, the liquid is dispersed into the flowing gas stream.
3. Figure 9 can be used to approximate the type of flow regime expected for any flow condition. In most two-phase lines in the field, slug flow is predominant in level and uphill lines. In downhill lines, stratified flow is predominant.
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Figure 9: Horizontal Multi-phase Flow Map (Source: P. Griffith, "Multiphase Flow in Pipes," JPT, March 1984, pp. 363-367) 4. Two-phase flow patterns in vertical flow are somewhat different than those occurring in horizontal flow. Different flow regimes may occur at different segments of pipe, such as flow in a well tubing where pressure loss causes gas to come out of solution as the fluid moves up the well. Figure 10 shows typical flow regimes in vertical two-phase flow.
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Figure 10: Two-phase Flow Patterns in Vertical Flow (Source: J.P. Brill, "Multiphase Flow in Wells," JPT, January 1987, pp. 15-21) 5. Vertical flow regimes can be described as follows: a) Bubble Small gas-liquid ratio with gas present in small, randomly distributed bubbles. The liquid moves up at a uniform velocity. Gas phase has little effect on pressure gradient. b) Slug Flow The gas phase is more pronounced. Although the liquid phase is still continuous, the gas bubbles coalesce into stable bubbles of the same size and shape, which are nearly the diameter of the pipe. These bubbles are separated by slugs of liquid. Both phases have a significant effect on the pressure gradient. c) Transition Flow or Churn Flow The change from a continuous liquid phase to a continuous gas phase occurs in this region. The gas phase is predominant and the liquid becomes entrained in the gas. The effects of the liquid are still significant. d) Annular-Mist Flow
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The gas phase is continuous and the bulk of the liquid is entrained in and carried by the gas. A film of liquid wets the pipe wall and its effects are secondary. The gas phase is the controlling factor. 6. Normally, flow in oil wells is in the slug or transition flow regime. Flow in gas wells can be in mist flow. Figure 11 can be used to determine the type of regime to be expected.
Figure 11: Vertical Multiphase Flow Map (Source: Yaitel, Y., Barhea, D., and Duckler, A.E., "Modeling Flow Pattern Transitions for Steady Upward GasLiquid Flow in Vertical Tubes," AIChE J., May 1980, pp. 345-354.) 7. In two-phase piping, pressure drop is caused by the friction developed due to the energy transfer between the two phases as well as that between each phase and the pipe wall. Pressure drop calculations shall take into account the additional friction loss due to the energy transfer between phases. 8. The detailed calculation of pressure drops in two-phase pipelines requires an evaluation of phase changes due to pressure and temperature changes, evaluation of liquid holdup using empirical formulas, and evaluation of energy transfer between the phases. These are addressed in the many proprietary computer programs available. It is beyond the
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scope of this manual to evaluate all the equations which have been proposed in the literature or to develop a new computer algorithm. 9. It shall be kept in mind that even under the best conditions small changes from horizontal in piping systems can lead to large errors in calculating pressure drops. Table 2 shows that, although the different correlations analyzed against field data on the average give reasonable results, the standard deviation is large; any one calculation could be as much as 20 to 50 percent in error.
Table 2: Two Phase Flow Correlations, AGA Multiphase Pipeline Data Bank for Gas-Condensate Lines (From: Battarra, Mariana, Gentilini and Giaccheta, Oil and Gas Journal, Dec. 30, 1985) Flow Pattern
Holdup
Down-Hill Recovery
Friction
Mean Error, Percent
Standard Deviation, Percent
BEG2
BEG2
NOCO
BEGO
-10.8
40.5
BEG2
BEG2
NOCO
BEGC
28.1
62.7
MAN1
BEG2
GARE
BEGC
29.5
63.3
MAN1
EATO
FLAN
DUKO
33.7
47.9
MAN1
EATO
FLAN
DUKC
92.5
89.8
MAN1
EATO
FLAN
OLIE
0.5
30.9
BEG2
=
Revised Beggs and Brill
MAN1
=
Mandane
EATO
=
Eaton
NOCO
=
No correction to institute density
GARE
=
Gas recovery only
FLAN
=
Flanigan
BEGO
=
Original Beggs and Brill
BEGC
=
Beggs and Brill with Colbrook
DUKO
=
Original Dukler
DUKC
=
Dukler with Colbrook
OLIE
=
Oliemans 10. The following four correlations have been found to give reasonable results when used within the limitations inherent in their derivation.
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4.3.3.
April 1998
API RP 14E 1. The following formula, presented in the American Petroleum Institute's Recommended Practice API RP 14E is derived from the general equation for isothermal flow assuming that the pressure drop is less than 10 percent of the inlet pressure: Equation 38
Metric : fL(W h )
2
∆ P = 62,561
ρ m d 5
Customary : 3.36 × 10 -6 fL(W h )
2
∆ P =
ρ m d 5
where : ∆ P = pressure drop, kPa ( psi ) L = length of pipe, m ( ft )
= flow rate of liquid and vapor, kg / hr (lb / hr ) ρ m = mixture density, kg / m 3 (lb / ft 3 ) d = pipe ID, mm (in ) f = Moody friction factor W h
2. This equation assumes that there is no energy interchange between the phases, that bubble or mist flow exists so that the fluid can be described by an average mixture density, and that there are no elevation changes. 3. The flow rate of the mixture to use in this equation can be calculated as follows:
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Equation 39
etric : W h
= 1.21 Q g S + 999.7 Ql (SG )
Customary : W h
= 3,180 Q g S + 14.6 Ql (SG )
where :
= flow rate of liquid and vapor, kg / hr (lb / hr ) Q g = gas flow rate, std m 3 / hr ( MMSCFD )
W h
Ql = liquid flow rate, m 3 / hr ( BPD ) S = specific gravity of gas relative to air
(SG ) = specic gravity of liquid relative to water 4. The density of the mixture to use in Equation 38 is given by:
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Equation 40
Metric : ρ m
=
28,814(SG ) P + 34.81R S P 28.82P + 10.0 R T Z
Customary : ρ m
=
12,409 (SG ) P + 2.7R S P 198.7 P + R T Z
where :
= mixture density, kg / m 3 (lb / ft 3 ) P = pressure, kPa ( psia ) (SG ) = specific gravity of the liquid relative to water ρ m
(use the average gravity for the hydrocarbon and water mixture ) S = specific gravity of gas relative to air R = gas / liquid ratio, std m 3 / m 3 ( std ft 3 / bbl ) T = temperatur e, K ( o R ) Z = gas compressibility factor
4.3.4.
AGA Equation The American Gas Association method uses a frictional pre ssure drop calculation originally developed by Dukler and an elevation pressure drop calculation originally developed by Flanigan. This seventeen-step method is an iterative procedure described in the Fluid Flow and Piping Sections of the GPSA Engineering Data Book.
4.3.5.
Beggs and Brill Equation 1. This correlation was developed by two University of Tulsa students, Dale Beggs and James Brill. Their original procedure first appeared in the May 1973 issue of the Journal of Petroleum Technology. Almost all correlations prior to the Beggs and Brill method could predict pressure drop in two-phase flow for vertical or horizontal flow only. Prior to 1973, however, no correlation existed for predicting the pre ssure drop in two-phase flow at any angle of inclination. Beggs and Brill therefore set out to develop such a correlation. 2. The original article presents a specific description of the experimental procedure used to develop this method. Such a description is beyond the
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scope of this Tutorial. The Beggs and Brill method was correlated using small diameter pipe and generally shall be applied to pipelines 200 mm (8 in) or less in diameter.
4.4.
Head Loss in Valves and Pipe Fittings In many piping situations, including those in most production facilities where space is limited, the pressure drop through valves, pipe fittings, and enlargements and contractions is a significant portion of the overall pressure drop in the pipe segment. A pipe flow restriction that changes velocity or direction of the flow stream causes pressure drops greater than that which would normally occur in a straight piece of pipe of the same length. The three most common ways of calculating these pressure drops are by using resistance coefficients for fittings, flow coefficients for valves and equivalent lengths for both valves and fittings.
4.4.1. Resistance Coefficients for Fittings 1. The Darcy-Weisbach equation, Equation 17, can be rewritten as: Equation 41
H f = K r
V 2 2g
where : K r = resistance coefficient, dimensionl ess
= head loss in fitting, m ( ft ) V = average velocity, m / sec ( ft / sec ) 2 2 g = acceleration of gravity, 9.81m / sec (32.2ft / sec )
H f
2. A comparison of Equations 17 and 41 shows that for a straight pipe: Equation 42
K r =
fL D
3. Approximate values of K r are given in Table 3 for various pipe fittings. Figures 12 and 13 show resistance coefficients for sudden contractions and enlargements and for pipe entrances a nd exits.
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Table 3: Resistance Coefficients for Pipe Fittings K r Globe Valve, wide open
10.0
Angle Valve, wide open
5.0
Gate Valve, wide open
0.2
Gate Valve, half open
5.6
Return Bend
2.2
Tee
1.8
90° Elbow
0.9
45°Elbow
0.4
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Figure 12: Resistance Coefficients for Different Types of Pipe Entrances and Exits (Courtesy of Paragon Engineering Services, Inc.)
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Figure 13: Resistance Coefficients for Sudden Enlargements and Contractions (Courtesy of Paragon Engineering Services, Inc.)
4.4.2.
Flow Coefficients for Valves 1. The valve industry generally expresses valve pressure drop characteristics in terms of a flow coefficient, C v. The flow coefficient is measured experimentally for each valve and is defined as the flow of water at 60°F, in gpm, at a pressure drop of 1 psi across the valve. It can be shown from Darcy's equation that Cv can be expressed as follows:
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April 1998
Equation 43
Metric : C v
=
0.0105 d 2
( fl / D )
=
1
0.0105 d 2
( K r )
2
1
2
Customary : C v
=
29.9 d 2
( fl / D )
1
=
29.9 d 2
( K r )
2
1
2
where :
= flow coefficient, m3 / hr ( gpm ) D = fitting ID, m ( ft ) d = fitting ID, mm (in ) L = fitting length, m ( ft ) f = Moody friction factor, dimensionl ess K r = resistance coefficient, dimensionl ess C v
2. For any fitting with a known C v: Equation 44
etric : 2
Q ∆ P = 6.89 l (SG ) C v Customary : 2
Q ∆ P = 8.5 × 10 l (SG ) C v -4
where : Ql = liquid flow rate, m / hr ( BPD ) 3
∆ P = pressure drop, kPa ( psi )
(SG ) = specific gravity of liquid relative to water 3 C v = flow coefficient, m / hr ( gpm )
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4.4.3.
April 1998
Equivalent Length 1. It is often simpler to treat valves and fittings in terms of their equivalent length of pipe. The equivalent length of a valve or fitting is the length of an equivalent section of pipe of the same diameter that gives the same pressure drop as the valve or fitting. Total pressure drop can then be determined by adding all equivalent lengths to the pipe length. The equivalent length, L e, can be determined from K r and Cv as follows: Equation 45
Le =
K r D f
Equation 46
Metric : Le
=
K r d 1000f
Customary : Le
=
K r d 12f
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Equation 47
Metric : Le
=
1.108 × 10 -7 d 5 fC v2
Customary : Le
=
74.5d 5 fC v2
where :
= equivalent length, m ( ft ) K r = resistance coefficient, dimensionl ess D = fitting ID, m ( ft ) d = fitting ID, mm (in ) f = Moody friction factor 3 C v = flow coefficient, m / hr ( gpm ) Le
2. Table 4 summarizes the equivalent lengths of various commonly used valves and fittings. Figure 14 shows equivalent lengths of fabricated bends of different radius. Figure 15 shows equivalent lengths of miter bends.
Table 4: Equivalent Lengths of Valves and Fittings in Feet (Courtesy of GPSA) Nominal Pipe Size (in)
Globe Valve or Ball Check Valve
Angle Valve
Swing Check Valve
Plug Valve, Gate Valve or Ball Valve
1 1 /2
55
26
13
1
2
70
33
17
2
2 1 /2
80
40
20
2
3
100
50
25
2
4
130
65
32
3
6
200
100
48
4
8
260
125
64
6
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Nominal Pipe Size (in)
Globe Valve or Ball Check Valve
Angle Valve
Swing Check Valve
Plug Valve, Gate Valve or Ball Valve
10
330
160
80
7
12
400
190
95
9
14
450
210
105
10
16
500
240
120
11
18
550
280
140
12
20
650
300
155
14
22
688
335
170
15
24
750
370
185
16
30
21
36
25
42
30
48
35
54
40
60
45
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Table 4: (Continued) 45 Ell
Short Radius Ell
Long Radius Ell
Branch of Tee
Run of Tee
Weld
Thread
Weld
Thread
Weld
Thread
Weld
Thread
Weld
Thread
1
2
3
5
2
3
8
9
2
3
2
3
4
5
3
4
10
11
3
4
2
5
3
12
3
2
6
4
14
4
3
7
5
19
5
4
11
8
28
8
6
15
9
37
9
7
18
12
47
12
9
22
14
55
14
10
26
16
62
16
11
29
18
72
18
12
33
20
82
20
14
36
23
90
23
15
40
25
100
25
16
44
27
110
27
21
55
40
140
40
25
66
47
170
47
30
77
55
200
55
35
88
65
220
65
40
99
70
250
70
45
110
80
260
70
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Table 4: (Concluded) Enlargement Sudden
Contraction
Standard Reducer
Sudden
Standard Reducer
Equivalent length in terms of small diameter 1
1
d/D = /4 d/D = /2
3
1
3
1
1
3
d/D = /4
d/D = /2
d/D = /4
d/D = /4
d/D = /2
d/D = /4
1
3
d/D = /2 d/D = /4
5
3
1
4
1
3
2
1
1
7
4
1
5
1
3
3
1
1
8
5
2
6
2
4
3
2
2
10
6
2
8
2
5
4
2
2
12
8
3
10
3
6
5
3
3
18
12
4
14
4
9
7
4
4
1
25
16
5
19
5
12
9
5
5
2
31
20
7
24
7
15
12
6
6
2
37
24
8
28
8
18
14
7
7
2
42
26
9
20
16
8
47
30
10
24
18
9
53
35
11
26
20
10
60
38
13
30
23
11
65
42
14
32
25
12
70
46
15
35
27
13
Notes:
1. Source of data is GPSA Data book, 1981 Revision. 2. d is inside diameter of smaller outlet. D is inside diameter of larger outlet.
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Figure 14: Equivalent Lengths of 90 Degree Bends (Courtesy of Crane Technical Paper 410)
Figure 15: Equivalent Length of Miter Bends (Courtesy of Crane Technical Paper 410)
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5.
Fluid Flow
April 1998
Choosing a Line Diameter 5.1.
General 5.1.1. When choosing a line size, it is necessary to examine both pressure drop and flow velocity for anticipated maximum and minimum flow rates expected during the life of the facility. In certain cases, it may be advisable to add surge factors to the anticipated flow rates to insure there is sufficient pressure available to force the fluid through the piping system. The following surge factors are sometimes used: 1. 20 percent Facility handling primary production. 2. 30 percent Facility handling primary production from wells not located adjacent to the facility. 3. 40 percent Facility handling primary production from wells not located adjacent to the facility where there are large elevation changes. 4. 50 percent Facility handling gas lifted production.
5.1.2. The line diameter shall be large enough that the pressure available shall drive the fluid through the line from one point to another. Thus, the operating pressures at the various process points of the facility shall be known. Normally, pressure drop is not a governing criterion in production facility piping system design since most of the pressure drop occurs acr oss control valves. The pressure drop in the line is relatively small compared to the pressure available in the system.
5.1.3. Line diameters also shall be sized for maximum and minimum flow velocities. The fluid shall be kept below some maximum velocity to prevent such problems as erosion, noise, and water hammer. The fluid also shall be kept above some minimum velocity to minimize surging and to keep the lines swept clean of sand and other solids or liquids.
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April 1998
Erosional Velocity 5.2.1. Liquid erosion occurs when liquid droplets impact the wall with enough force to erode either the base metal itself or the products of corrosion (erosion-corrosion). The higher the velocity of flow, the greater the tendency for erosion to occur. Experiments in liquid flow systems indicate that erosion of the products of corrosion occurs when the velocity of flow exceeds the value given by: Equation 48
etric : V e
= 1.22
C e 1
( ρ )2
Customary : V e
=
C e 1
( ρ )2
where :
= erosional flow velocity, m / sec ( ft / sec ) ρ = density of liquid, kg / m 3 (lb / ft 3 ) C e = empirical constant, dimensionl ess V e
5.2.2. Various values have been proposed for "Ce." Prior to 1990, API RP 14E suggested a value of 100 for continuous service and 125 for non-continuous service. Analysis of field data indicates that constants higher than 100 can be used if corrosion is controlled. In 1990 API RP 14E was rewritten as follows: "Industry experience to date indicates that for solids-free fluids, values of C e = 100 for continuous service and Ce = 125 for intermittent service are conservative. For solids-free fluids where corrosion is not anticipated or when corrosion is controlled by inhibition or by employing corrosion resistant alloys, values of C e = 150 to 200 may be used for continuous service; values up to 250 have been used successfully for intermittent service. If solids production is anticipated, fluid velocities shall be
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significantly reduced. Different values of 'Ce' may be used where specific application studies have shown them to be appropriate. "Where solids and/or corrosive contaminants are present or where 'Ce' values higher than 100 for continuous service are used, periodic surveys to assess pipe wall thickness shall be considered. The design of any piping system where solids are anticipated shall consider the installation of sand probes, cushion flow tees, and a minimum of three feet of straight piping downstream of choke outlets."
5.2.3. Erosion of the pipe material itself can occur if solids are present in the fluid. There is no minimum velocity below which this erosion will not occur. One equation proposed to evaluate the erosion of metal is: Equation 49
Metric : KW (V p ) β 2
vol = 9.806 × 10
-3
gP h
Customary : 12KW (V p ) β 2
vol =
gP h
where : vol = volume of metal eroded, mm (in 3
3
)
= particle velocity, m / sec ( ft / sec ) P h = penetration hardness of the material, kPa ( psi ) β = a value between 0.5 and 1.0 depending upon the V p
impingement angle of the particle K = erosive wear coefficient, dimensionl ess W = total weight of impinging solid particles, kg (lb ) g = acceleration of gravity, 9.81 m / sec (32.2 ft / sec 2
2
)
5.2.4. The form of this equation indicates that there is no threshold velocity at which erosion starts. Rather, erosion occurs even at small velocities, and the amount of erosion increases with the square of the velocity. It can be seen
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from Equation 49 that the velocity for a given erosion rate is a function of 1/W. Since the percent of solids impinging on any surface is inversely proportional to the density of the fluid, the erosional velocity can be expected to be proportional to the fluid density. This is contrary to the form of Equation 48. Thus, it is not correct to use Equation 48 with a low "C e" value when solids are present.
5.2.5. The rate of erosion depends on both the concentration of solids in the flow stream and the way in which these particles impinge on the wall. At an ell, one would expect centrifugal force to cause a high percentage of the particles to impinge on the wall in a concentrated area. It can be shown that with a solids concentration of 4.5 kg/month (10 lb/month) in the flow stream the velocity for a .25 mm/year (10 mil/year) erosion rate in an ell can be as low as 1.5 m/sec (5 ft/sec). At higher concentrations the erosional velocity would be even lower. For this reason, where sand production is anticipated, it is usually recommended that right angle turns in the pipe be accomplished with very long radius fabricated bends or target tees. Figure 16 shows a target tee and Figure 17 shows the greater life that can be expected by use of a target tee instead of a long radius ell.
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April 1998
Figure 16: Example of a Target Tee
Figure 17: Wear Rate Comparison for Standard Fittings (Source: (Source: API OSAPR Project No. 2)
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5.2.6. Where sand production is expected, piping shall be inspected periodically for loss of wall thickness at the outside of all direction changes.
5.3.
Liquid Line Sizing 5.3.1. In sizing a liquid line, the two factors that have the greatest effect are the velocity of the fluid and the pressure drop in the pipe. When considering the pressure drop, it is necessary to take take into account the equivalent lengths of valves and fittings, as well as elevation changes.
5.3.2. The maximum velocity used in sizing liquid lines depends on service conditions, pipe materials, and economics. For example, API RP 14E recommends that the maximum velocity not exceed 4.5 m/sec (15 ft/sec). Most companies, however, specify values for cement lined pipe [2.4 to 3.0 m/sec (8 to 10 ft/sec)], fiberglass f iberglass pipe [3.7 to 4.5 m/sec (12 to 15 ft/sec )], or where erosion-corrosion is a problem [3.0 to 4.5 m/sec (10 to 15 ft/sec)]. Even lower velocities may be used for cement lined pipe or where erosioncorrosion is anticipated. anticipated. (See MP 16-P-01.) 16-P-01.)
5.3.3. Liquid lines are normally sized to maintain a velocity sufficient to keep solid particles from depositing in the line. If sand is transported in a pipe, it is deposited on the bottom until an equilibrium flow velocity over the bed is reached. At this point, sand grains are being eroded from the bed at the same rate as they are being deposited. If the flow rate is increased, the bed bed will be eroded until a new equilibrium velocity is reached a nd the bed is once again stabilized. If the flow rate is decreased, sand is deposited until a new equilibrium velocity is established. In most practical cases, a velocity of 0.9 to 1.2 m/sec (3 to 4 ft/sec) is sufficient to keep from building a sufficiently high bed to affect pressure drop calculations. For this reason, a minimum minimum velocity of 0.9 m/sec (3 ft/sec) is normally recommended.
5.3.4. Liquid velocity can be determined from the following equation:
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April 1998
Equation 50
etric : V = 353.68
Ql d 2
Customary : V = 0.012
Ql d 2
where : V = average velocity, m / sec ( ft / sec ) Ql = liquid flow rate, m / hr ( BPD ) 3
d = pipe ID, mm (in )
5.3.5. Figure 2.1 in API RP 14E shows the liquid flow velocity in ft/sec as a function of liquid flow rate in bbl/day for different pipe sizes.
5.4.
Gas Line Sizing 5.4.1. As with liquid line sizing, the two factors that have a pronounced effect on gas line size are the velocity of the gas and the pressure drop. drop. The pressure drop is usually the governing factor in long gas gathering and transmission systems, or in relief/vent piping. The pressure drop also may be important where it necessitates increased compressor horsepower.
5.4.2. In a typical production facility the gas lines are short and the pressure drop does not govern sizing. sizing. For some lines, the pressure lost due to friction shall be recovered by recompressing the gas. In such cases, it is possible to strike an economic balance between the cost of a larger pipe to minimize the pressure drop and the cost of additional compression. Figure 18 is an approximation that attempts to strike this balance by showing acceptable pressure drop versus operating pressure. In most production production facility lines, Figure 18 has little significance since the bulk of the pressure loss is due to a pressure control device, and the size and operating pressure of the compressor are not affected by the incremental pressure drop in the line.
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5.4.3. By using the following equation, Figure 18 can be used to choose a pipe diameter directly: Equation 51
etric : 5
d
= 8.186 × 10
S Tf Q g 2
5
P (∆ P / 100 ft )
Customary : 5
d
=
1260 S Tf Q g 2 P (∆ P / 100 ft )
where : d = pipe ID, mm (in ) S = specific gravity of gas relative to air T = temperatur e, K ( o R ) f = Moody friction factor, dimensionl ess Q g = gas flow rate, std m / hr ( MMSCFD ) 3
P = pressure, kPa ( psia ) ∆ P / 100ft = desired pressure drop per 100 ft from Figure 18, kPa ( psi )
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Figure 18: Acceptable Pressure Drop for Short Lines (Courtesy of Paragon Engineering Services, Inc.)
5.4.4. As in liquid lines, the flow velocity in gas lines shall be kept between some maximum and minimum value. It is recommended that a minimum velocity of 3 to 4.5 m/sec (10 to 15 ft/sec) be maintained to minimize liquid settling out in low spots. Gas velocities are normally kept below 18 to 24 m/sec (60 to 80 ft/sec) to minimize noise and to allow for corrosion inhibition. In systems with CO2 present in amounts as low as 1 to 2 percent, some operators limit the velocity to less than 9 to 15 m/sec (30 to 50 ft/sec). Field experience indicates that it is difficult to inhibit CO 2 corrosion at higher velocities.
5.4.5. Although the erosional criterion was derived for two-phase flow, it shall be verified that this criterion is still met as the liquid flow rate approaches zero. Erosional velocity due to small amounts of liquid in the gas can be calculated from Equation 48 as:
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April 1998
Equation 52
etric : 1
V e
T Z 2 = 0.644 C e S P
Customary : 1
V e
T Z 2 = 0.6 C e S P
where :
= erosional velocity, m / sec ( ft / sec ) C e = empirical constant, dimensionl ess o T = temperatur e, K ( R ) S = specific gravity of gas relative to air P = pressure, kPa ( psia ) Z = gas compressibility factor V e
5.4.6. For most instances, with pressures less than 7,000 to 14,000 kPa (1,000 to 2,000 psi), the erosional velocity shall be greater than 18 m/sec (60 ft/sec) and thus the erosional criteria shall not govern. At high pressures, it may be necessary to check for erosional velocity before sizing lines for 18 m/sec (60 ft/sec) maximum velocity.
5.4.7. Actual gas velocity can be determined by:
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April 1998
Equation 53
etric : V g = 122.7
Q g T Z 2
d P
Customary : V g = 60
Q g T Z d 2 P
where : Q g = gas flow rate, std m 3 / hr ( MMSCFD ) T = temperatur e, K ( R ) o
d = pipe ID, mm (in ) P = pressure, kPa ( psia ) V g = gas velocity, m / sec ( ft / sec ) Z = gas compressibility factor
5.5.
Two-Phase Flow Line Sizing 5.5.1. Typically, flow lines from wells, production manifolds, and two-phase gas/liquid pipelines are sized as two-phase lines. Gas outlets from separators or other process equipment contain small amounts of liquids but are not considered two-phase lines. Similarly, liquid outlets from separators or other process equipment are usually considered single-phase liquid lines, even though gas evolves due to both the pressure decrease across a liquid control valve and the pressure loss in the line. The amount of gas evolved in liquid outlet lines rarely will be sufficient to affect a pressure loss calculation based on an assumption of liquid flow. A relatively large pressure drop is needed to evolve enough gas to affect this calculation.
5.5.2. Since most two-phase lines operate at high pressure within the f acility, pressure drop usually is not a governing criterion in selecting a diameter. However, pressure drop may have to be considered in some long flow lines from wells and in most two-phase gas/liquid pipelines.
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April 1998
5.5.3. A minimum flow velocity of 3 to 4.5 m/sec (10 to 15 ft/sec) is re commended to keep liquids moving in the line and to minimize slugging of separator or other process equipment. This is very important in long lines with elevation changes. The maximum allowable velocity is equal to 18 m/sec (60 ft/sec) for noise, 9 to 15 m/sec (30 to 50 ft/sec) if it is necessary to inhibit for CO 2 corrosion, or the erosional velocity, whichever is least. For two-phase flow, the general erosional velocity equation, Equation 48, is usually expressed as: Equation 54
etric : V e
C e
= 1.22
1
( ρ m )2
Customary : V e
=
C e 1
( ρ m )2
where :
= erosional flow velocity, m / sec ( ft / sec ) C e = empirical constant, dimensionl ess ρ m = mixture density, kg / m 3 (lb / ft 3 ) , from Equation 40 V e
5.5.4. It can be shown that the minimum cross-sectional area of pipe for a maximum allowable velocity can be expressed as:
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April 1998
Equation 55
etric :
9.35 + 3.24 ZRT P Q a = 29.69 l V max Customary :
9.35 + ZRT 21.25P Q a= l 1,000 V max where : a = minimum required cross - sectional area, mm (in 2
2
)
Ql = liquid flow rate, m 3 / hr ( BPD ) V max
= maximum allowable velocity, m / sec ( ft / sec )
Z = gas compressibility factor R = gas / liquid ratio, std m / m ( std ft / bbl ) 3
3
3
P = pressure, kPa ( psia ) T = temperatur e, K ( R ) o
5.5.5. This can be solved for pipe inside diameter:
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April 1998
Equation 56
etric :
ZRT + 11.9 4.13 Ql P d = 5.448 V max
1 2
Customary : 1
ZRT 2 + 11.9 Ql 16.7 P d = 1,000 V max
5.5.6. Figure 2.5 in API RP 14E is a chart developed to minimize the calculation procedure. Care shall be taken when utilizing this chart, as it is based on the assumptions listed. It is better to use Equations (54), (40), and (56) directly as follows: 1. Determine ρ ρ ρ ρ from Equation 40. m
2. Determine the erosional velocity, Ve, from Equation 54. 3. For the design, use the smaller of V e or that velocity required by the noise or CO2 inhibition criteria. 4. Determine the minimum ID from Equation 56. 5. Check pressure drop, if applicable, to make certain ther e is enough driving force available.
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6.
Fluid Flow
April 1998
Determining Wall Thickness 6.1.
Commonly Available Pipe 6.1.1. Pipe comes in standard diameters and wall thicknesses as shown in Table 5 for customary units. In customary units the pipe is designated by a nominal size which is usually different fr om the actual pipe outside diameter.
Table 5: ANSI Pipe Schedules Nominal Pipe Size 3
/4 in
1 in
1 1/2 in
2 in
3 in
O. D.
Wall Thickness
Weight/Ft
Schedule
Class
1.050
.113
1.131
40
STD.
.154
1.474
80
XH
.218
1.937
160
.308
2.441
.133
1.679
40
STD.
.179
2.172
80
XH
.250
2.844
160
.358
3.659
.145
2.718
40
STD.
.200
3.631
80
XH
.281
4.859
160
.400
6.408
.154
3.653
40
STD.
.218
5.022
80
XH
.343
7.444
160
.436
9.029
.216
7.576
40
STD.
.300
10.25
80
XH
1.315
1.900
2.375
3.500
.437
14.32
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XXH
XXH
XXH
XXH
160
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Nominal Pipe Size
4 in
6 in
8 in
10 in
O. D.
4.50
6.625
8.625
10.75
Wall Thickness
April 1998
Weight/Ft
Schedule
Class
.600
18.58
.237
10.79
40
.281
12.66
60
.337
14.98
80
.437
19.01
120
.531
22.51
160
.674
27.54
.280
18.97
40
STD.
.432
28.57
80
XH
.562
36.39
120
.718
45.30
160
.864
53.16
.250
22.36
20
.277
24.70
30
.322
28.55
40
.406
35.64
60
.500
43.39
80
.593
50.87
100
.718
60.63
120
.812
67.76
140
.906
74.69
160
.875
72.42
.250
28.04
20
.307
34.24
30
.365
40.48
40
STD.
.500
54.74
60
XH
.593
64.33
80
.718
76.93
100
.843
89.20
120
1.000
104.1
140
1.125
115.7
160
© Mobil Oil,1998
XXH STD.
XH
XXH
XXH
STD.
XH
XXH
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Fluid Flow
Nominal Pipe Size
12 in
14 in
16 in
O. D.
12.75
14.0
16.0
Wall Thickness
April 1998
Weight/Ft
Schedule
Class
.250
33.38
20
.330
43.77
30
.375
49.56
.406
53.53
.500
65.42
.562
73.16
60
.687
88.51
80
.843
107.2
100
1.000
125.5
120
1.125
139.7
140
1.312
160.3
160
.250
36.71
10
.312
45.68
20
.375
54.57
30
.437
63.37
40
.500
72.09
.593
84.91
60
.750
106.1
80
.937
130.7
100
1.093
150.7
120
1.250
170.2
140
1.406
189.1
160
.250
42.05
10
.312
52.36
20
.375
62.58
30
STD.
.500
82.77
40
XH
.656
107.5
60
.843
136.5
80
1.031
164.8
100
1218
192.3
120
1.437
223.5
140
© Mobil Oil,1998
STD. 40 XH
STD.
XH
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Nominal Pipe Size
Fluid Flow
O. D.
Wall Thickness
1.593
April 1998
Weight/Ft
245.1
Schedule
Class
160
6.1.2. Pipe wall thickness can be given by actual thickness, weight per foot, schedule, or class. The most commonly available pipe wall thicknesses are standard, XH, and XXH. Wall thicknesses corresponding to different schedules are the next most commonly available. See MP 16-P-01for proper use of pipe schedules and classes for various pressure ratings in different facility piping applications.
6.2.
Standards and Requirements 6.2.1. After selecting the appropriate inside diameter, it is necessary to choose a pipe with sufficient wall thickness to withstand the internal pressure.
6.2.2. There are different standards used throughout the world in calculating the required wall thickness of a pipe. The following is a list of the standards used in the United States. These are the most common used in oil production facility design and are similar to national standards which exist in other parts of the world. 1. ASME B31.1 - Power Piping This standard deals with steam and power fluids. It is required by the U.S. Coast Guard on all mobile offshore drilling units. 2. ASME B31.3 - Chemical Plant and Petroleum Refinery Piping This standard is required by the U.S. Minerals Management Service for offshore platforms in federal waters. It is also used extensively for offshore facilities in state waters, and for both onshore and offshore facilities in other parts of the world. 3. ASME B31.4 - Liquid Transportation Systems for Hydrocarbons, Liquid Petroleum Gas, Anhydrous Ammonia, and Alcohols This standard is used in onshore oil production facilities. 4. ASME B31.8 - Gas Transmission and Distribution Piping Systems
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April 1998
This standard is often used for gas lines in onshore production facilities and for the transport or distribution of gas. In general, the U.S. Department of Transportation has adopted this standard f or gas pipelines, although it has modified some sections.
6.2.3. ASME B31.1 and ASME B31.3 use the same equation to calculate the required wall thickness, although the allowable material stress at elevated temperatures differs between the two codes. ASME B31.4 is actually a subset of ASME B31.8 as it relates to calculating wall thickness. Therefore, from a wall thickness standpoint, only ASME B31.3 and ASME B31.8 are in common use. In general, but not always, ASME B31.3 is the more conservative in calculating required wall thickness (see EPT 09-T-05).
6.3.
General Hoop Stress Formula 6.3.1. Before discussing the determination of pipe wall thickness in acc ordance with these standards, it is necessar y to introduce the concept of hoop stress. Figure 19 is a free body diagram of a length of pipe that was cut in half. The hoop stress in the pipe is considered a uniform stress over the thickness of the wall, for a thin wall cylinder. Therefore, the force equilibrium equation can be expressed as: Equation 57
2σ tL = P i (d o - 2t ) L where : σ = hoop stress in pipe wall, kPa ( psi ) t = pipe wall thickness, mm (in ) P i = internal design pipepressure, kPa ( psig )
= pipe OD, mm (in ) L = pipe length, m ( ft ) d o
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April 1998
Figure 19: General Hoop Stress Free Body Diagram (Courtesy of Paragon Engineering Services, Inc.)
6.3.2. Rearranging and solving for required wall thickness, the equation reduces to: Equation 58
t =
P i d o 2(σ + P i )
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Appendix A–Nomenclature A = cross sectional area of pipe, m�, ft� a = minimum required cross-sectional area, mm�, in� Cc = check valve constant, dimensionless Ce = empirical constant, dimensionless Cv = flow coefficient, m�/hr, gpm D = pipe or fitting ID, m, ft d = pipe or fitting ID, mm, in do = pipe OD, mm, in E = pipeline efficiency factor, dimensionless Ef = efficiency factor, dimensionless EL = longitudinal weld joint factor, dimensionless F = design factor 49 CFR 192, ASME B31.8, dimensionless f = Moody friction factor g = acceleration of gravity, 9.81 m/sec�, 32.2 ft/sec� H = elevation head, m, ft Hf = pipe friction head loss, m, ft Hf = head loss in fitting, m, ft HPH = potential head, m, ft HSH = static pressure head, m, ft HVH = velocity head, m, ft K = erosive wear coefficient, dimensionless K r = resistance coefficient, dimensionless
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April 1998
L = length of pipe, or fitting m, ft Le = equivalent length, m, ft Lf = fitting length, m, ft Lm = length of pipe, km, miles (MW) = molecular weight of gas P = pressure, kPa, psia Ph = penetration hardness of material, kPa, psi Pi = internal design pipe pressure, kPa, psig P1 = upstream pressure, kPa, psia P2 = downstream pressure, kPa, psia Q = flow rate, m�/hr, ft�/sec Ql = liquid flow rate, m�/hr, BPD Qg = gas flow rate, std m�/hr, MMSCFD Qo = oil flow rate, m�/hr, BPD QSTD = gas flow rate, std m�/hr, SCFD QT = total liquid flow rate, m�/hr, BPD QW = water flow rate, m�/hr, BPD R = gas/liquid ratio, std m�/m�, std ft�/bbl Re = Reynolds number, dimensionless S = specific gravity of gas relative to air (SG) = specific gravity of liquid relative to water (SG) o = specific gravity of oil relative to water (SG) W = specific gravity of water (SG) m = specific gravity of mixture relative to water St = allowable stress for pipe material, kPa, psi SY = minimum yield strength of pipe material, kPa, psi
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Fluid Flow
April 1998
T = temperature, K, °R T = temperature, derating factor, dimensionless Tol = pipe manufacturer's allowed tolerance t = required pipe wall thickness, mm, in tc = corrosion allowance, mm, in tth = thread of groove depth, mm, in V = average velocity, m/sec, ft/sec Ve = erosional flow velocity, m/sec, ft/sec Vg = gas velocity, m/sec, ft/sec Vmax = maximum allowable velocity, m/sec, ft/sec Vmin = minimum allowable velocity, m/sec, ft/sec VP = particle velocity, m/sec, ft/sec vol = volume of metal eroded, mm�, in� v = specific volume of gas at upstream conditions, m�/kg, ft�/lb Ws = flow rate, kg/sec, lb/sec Wh = flow rate of liquid and vapor, kg/hr, lb/hr W = total weight of impinging solid particles, kg, lb Y = coefficient Z = gas compressibility factor Ze = vertical elevation rise of pipe, m, ft ρ ρ ρ = density of liquid, kg/m�, lb/ft� ρ ρ ρg = density of gas, kg/m�, lb/ft� ρ ρ ρ = mixture density, dg/m m
ρ ρ ρ = density of oil, kg/m�, lb/ft� o
ρ ρ ρ = density of water, kg/m�, lb/ft� w
µ = absolute viscosity, Pa-sec, centipoise (cp)
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April 1998
µ' = viscosity, kg/m sec, lb/ft-sec ( µ (cp) x 0.000672) µc = viscosity of the continuous phase, Pa-sec, centipoise (cp) µeff = effective viscosity of the mixture, Pa-sec, centipoise (cp) γ = kinematic viscosity, m�/sec, centistoke β = a value between 0.5 and 1.0, depending on the impingement angle of the particle φ φ = volume fraction of the discontinuous phase
ε = absolute roughness, m, ft ∆hW = pressure loss, mm of water, in of water ∆P = pressure drop, kPa, psi ∆PZ = pressure drop due to elevation changes, kPa, psi ∆Z = total increase in elevation, m, ft ΣZe = sum of vertical elevation rises of pipe, m, ft σ = hoop stress in pipe wall, kPa, psi
© Mobil Oil,1998
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Appendix B–Example Problems–Metric Units
1.
Pressure Drop 1.1.
Liquid Line 1.1.1.
Flow Rates:
Given: Condensate
=
5.30 m�/hr
Water
=
1.52 m�/hr
Condensate
=
0.87
Water
=
1.05
Viscosity
=
0.003 Pa-sec
Length
=
2130 m
Inlet Pressure
=
6200 kPa (abs)
Temperature
=
27°C
Specific Gravity:
1.1.2.
Problem: Solve for pressure drop in a 50.8 mm (2 in) and 101.6 mm (4 in) ID line using the general equation
1.1.3.
Solution: General Equation 1. Calculate the specific gravity of the combined liquid from the following formula:
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Fluid Flow
April 1998
(Equation 4)
(SG )m =
+ QW (SG )W
Qo (SG )o
QT
(5.30 )(0.87 ) + (1.52 )(1.05 ) (SG )m = 5.30 + 1.52
= 0.91
2. Calculate the Reynolds number from the formula: (Equation 15)
Re =
Re =
Re =
353.13 (SG ) Q d µ
(353.13)(0.91 )(6.82 ) d (0.003) 730,530 d
3. Calculate the pressure drop using the general equation: (Equation 25)
∆ P = 6.266 × 10
7
fLQl 2 (SG )m d 5
(6.266 × 10 ) (2130 )(6.82) (0.91) f ∆ P = 2
7
5
d
(5.649 × 10 ) f ∆ P = 12
d 5
4. Assume an absolute roughness (ε ε ε) of 0.045 mm.
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April 1998
Diameter 50.8 mm (2 in)
101.6 mm (4 in)
Re
14.4 x 10�
7.2 x 10�
ε/d
0.0009
0.00045
f, Moody friction factor (from chart)
0.029
0.034
484 kPa
17.7 kPa
∆P
1.2.
Gas Line 1.2.1.
Given:
Flow Rate:
Gas
=
27,100 std m�/hr
Specific Gravity:
Gas
=
0.85
Length
=
2130 m
Inlet Pressure
=
6200 kPa (g)
Temperature
=
27°C
Z
=
0.67
1.2.2.
Problem: Solve for pressure drop in a 101.6 mm (4 in) and 152.4 mm (6 in) I. D. line using 1. The general equation 2. The approximate assumption of ∆P < 10 percent P1 3. The Panhandle Equation 4. The Weymouth Equation
1.2.3.
Solution: 1. General Equation a) Determine the gas viscosity from Figure 4.
Viscosity = 1.3 × 10 -5 Pa - sec (0.013 cp )
b) Calculate the Reynolds Number from the formula:
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April 1998
(Equation 16)
Re =
Re =
0.428 Q g S d µ
(0.428 )(27,100 )(0.85 )
(1.3 × 10 ) d -5
=
760,000,000 d
c) Assume an absolute roughness (ε ε ε) of 0.045 mm. d) Calculate the pressure drop from the general equation: (Equation 28) 2 1
2 2
2 1
2 2
2 1
2 2
P - P
P - P
P - P
= 52,430
= 52,430
=
S Q g 2 Z T f L d 5
(0.85 )(27,100 )2 (0.67 )(300 )(2,130 ) f d 5
1.40 × 10 19 f d 5
Diameter 101.6 mm (4 in)
152.4 mm (6 in)
Re
7.5 x 106
5.0 x 106
ε/d
0.00045
0.0003
f, Moody friction factor (from chart)
0.0164
0.015
2.121 x 107
2.554 x 106
P2
4303 kPa (abs)
6097 kPa (abs)
∆P
2000 kPa
206 kPa
P12 - P22
2. The approximate equation where it is a ssumed that ∆P < 10 percent of P1:
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April 1998
(Equation 30)
S Q g 2 Z T f L
∆ P = 26,215
∆ P = 26,215
∆ P =
P 1d 5
(0.85 )(27,100 )2 (0.67 )(300 )(2,130 ) f 6303 d 5
1.11 × 10 15 f d 5
Diameter
∆P
101.6 mm (4 in)
152.4 mm (6 in)
1684 kPa
222 kPa
3. Panhandle Equation (Equation 36)
P 21 - P 22 Q g = 1.229 × 10 E f 0.961 S Z T Lm -3
Lm
=
2,130 1,000
0.51
d 2.53
= 2.13 km
E = 0.95 (assumed ) 27,100 = (1.229 × 10 - 3 ) (0.95 ) ×
(6303)2 - P 22 0.961 0.85 0.67 300 2.13 ( ) ( )( ) ( ) 2 2
P
«MEPBusinessSectorName»
= 3.973 × 10 7
0.51
d 2.53
1.015 × 1017 d 4.96
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Fluid Flow
April 1998
Diameter 101.6 mm (4 in)
152.4 mm (6 in)
P2
5334 kPa (abs)
6182 kPa (abs)
∆P
969 kPa
121 kPa
4. Weymouth Equation (Equation 32)
P - P Q g = 1.42 × 10 - 2 d 2.67 L S ZT 2 1
2 2
1 2
a) Rearranging to solve for P1 or P2: 2 1
2 2
P - P
=
Q g 2 L S Z T
(1.42 × 10 )
-2 2
d 5.34
b) Note: We could just as easily have solved for P1, knowing P2 !
P 2 = (6303) 2
(27,100 ) (2,130 )(0.85 )(0.67 )(300 ) (1.42 × 10 ) d 2
P 2 = (6303) 2
-2 2
1.325 × 10
18
5.34
d
1 2
5.34
1 2
Diameter 101.6 mm (4 in)
152.4 mm (6 in)
P2
3780 kPa (abs)
6067 kPa (abs)
∆P
2523 kPa
236 kPa
5. Summary
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April 1998
Diameter 101.6 mm (4 in)
152.4 mm (6 in)
General Equation:
∆P = 2000 kPa
206 kPa (abs)
Approximate Equation:
∆P = 1684 kPa
222 kPa
Panhandle Equation:
∆P = 969 kPa
121 kPa
Weymouth Equation:
∆P = 2523 kPa
236 kPa
1.3.
Two-Phase Lines 1.3.1.
Flow Rates:
Given:
Condensate
=
5.30 m�/hr
Water
=
1.52 m�/hr
Gas
=
27,100 std m�/hr
Condensate
=
0.87
Water
=
1.05
Gas
=
0.85
Condensate and Water
=
0.003 Pa-sec
Gas
=
1.3 X 10-5 Pa-sec
Inlet Pressure
=
6200 kPa
Temperature
=
27°C
Compressibility Factor
=
0.67
Length
=
2,130 m
Absolute Roughness
=
0.045 mm
Specific Gravity:
Viscosity:
1.3.2.
Problem: Solve for the pressure drop in 101.6 mm (4 in), 152.4 mm (6 in), and 203.2 mm (8 in) I.D. lines using the API RP 14E method.
1.3.3.
Solution: 1. API RP 14E
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April 1998
(Equation 38)
∆ P =
62,561 fLW h2 ρ m d 5
where : W h
= 1.21 Q g S + 999.7 Ql (SG )m
ρ m
=
28,814 (SG )m P + 34.81 R S P 28.82 P + 10.0 RTZ
a) Calculate the specific gravity of the liquid: (Equation 34)
(SG )m =
Qo (SG )o
+ Qw (SG )w QT
(5.30 ) (0.87 ) + (1.52 ) (1.05 ) (SG )m = 6.82
(SG )m = 0.91 b) Calculate the rate of flow of liquid and vapor: (Equation 39)
W h
= (1.21) (27,100 ) (0.85 ) + (999.7 ) (6.82 ) (0.91 )
W h
= 34,077 kg / hr
c) Calculate the mixture density:
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April 1998
(Equation 40)
(28,814 ) (0.91) (6303 ) + (34.81 ) (27,100 / 6.82 ) (0.85 ) (6303 ) (28.82 ) (6303 ) + (27,100 / 6.82 ) (300 ) (0.67 )
ρ m
=
ρ m
= 110.95 kg / m 3
d) Assume f = 0.0204 for rough pipe. e) Calculate the change in pressure: (Equation 38)
(62,561) (0.0204 ) (2,130 ) (34,077 )2 ∆ P = = (110.95 ) (d 5 )
Diameter
∆P =
2.
101.6 mm (4 in)
152.4 mm (6 in)
203.2 mm (8 in)
2628 kPa
346 kPa
82 kPa
Choosing a Line Diameter and Determining Wall Wall Thickness 2.1.
Liquid Line 2.1.1.
Flow Rates:
Given:
Condensate
=
5.30 m�/hr
Water
=
1.52 m�/hr
Condensate
=
0.87
Water
=
1.05
Viscosity
=
0.003 Pa-sec
Length
=
2130 m
Specific Gravity:
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April 1998
Inlet Pressure
=
6200 kPa
Temperature
=
27°C
Liquid flows to a low pressure separator operating at 1035 kPa. The line is rated for 10,200 kPa (1480 psi).
2.1.2.
Problem: Choose a line size and determine the wall thickness using AWS QC7-93, ASME B31.4, and ASME B31.8.
2.1.3.
Solution:
= 4.5 m / sec V min = 0.9 m / sec pressure drop = 6200 - 1035 = 5165 kPa V max
1. Calculate the diameter from Equation (50):
V = 353.68 V = V =
Ql
d 2 (353.68 )(6.82 ) d 2 2412 d 2
V
I.D.
0.9 m/sec
51.8 mm
4.5 m/sec
23.2 mm 2. Pressure Drop From Example 1.1 the pressure drop in the 50.8 mm (2 in) line is acceptable. For flexibility and mechanical strength it is better to use a 50.8 mm (2 in) line rather than a 25.4 or 38.1 mm (1 or 1 1/2 in) line. a) AWS QC7-93
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April 1998
Equation (59)
100 Pd o + t th + 2(SE + PY ) 100 - TOL
t = t c
t = 0.05 + 0 +
100 (1480 )(2.375 ) (25.4 ) 2 [(20,000 )(1) + (1480 )(.4 )] 100 - 1.5
t = 3.94 mm (0.155 in.)
Can use standard weight pipe. b) ASME B31.4 Equation (60)
t =
P i d o
2 ( FE L TS y )
F = 0.72
t =
(1480 )(2.375 )(25.4 ) 2(0.72 )(1)(1)(35,000 )
t = 1.77 mm (0.0697 in )
Would use a standard weight pipe for mechanical strength. c) ASME B31.8 Use the same equation as B31.4. Outside the facility use a design factor (F) of 0.72 wall thickness, same as above. Within the facility use a factor (F) of 0.60.
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April 1998
for F = 0.6
t =
(1480 )(2.375 )(25.4 ) 2 (0.6 )(1)(1)(35,000 )
t = 2.126 mm (0.0837 in )
Use a standard weight pipe for mechanical str ength.
2.2.
Gas Line 2.2.1.
Given:
Flow Rates:
Gas
=
27,100 std m�/hr
=
1.3 X 10-5 Pa-sec
=
0.85
Length
=
2130 m
Inlet Pressure
=
6200 kPa (g)
Temperature
=
27°C
Compressibility factor
=
0.67
Viscosity Gravity:
Gas
Gas flows to a dehydrator which operates at 5515 kPa. The line is rated for 10,200 kPa (1480 psi).
2.2.2.
Problem: Choose a line size and determine the wall thickness using AWS QC7-93 and ASME B31.8.
2.2.3.
Solution:
= 18 m / sec V min = 3 to 4.5 m / sec Pressure drop = 6200 - 5515 = 685 kPa V max
At a pressure this low, erosional velocity is not important. 1. Calculate the diameter from Equation (53):
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V g = 122.7
V g =
April 1998
Q g TZ d 2 P
(122.7 )(27,100 )(300 )(0.67 ) d 2 (5618 )
V
I.D.
3 m/sec
199.1 mm
4.5 m/sec
162.6 mm
18 m/sec
81.3 mm 2. Pressure Drop From example 1.2, the pressure drop in the 101.6 mm (4 in) line is not acceptable, but it is acceptable in the 152.4 mm (6 in) line. This also gives reasonable velocities between 4.5 and 18 m/sec. a) AWS QC7-93 Equation (59)
t = t c
+ t th +
100 2 (S t E L + P iY ) 100 - TOL
t = 0.05 + 0 +
P i d o
100 (1480 )(6.675 ) (25.4 ) 2 ((20,000 )(1) + (1480 )(0.4 )) 100 - 12.5
t = 8.41 mm (0.331 in )
Use XH, could use 9.5 mm (0.375 in) wall if available. b) ASME B31.8 Equation (60)
t =
P i d o 2( FE LTS Y )
1. Outside facility F = 0.72
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t =
April 1998
(1480 )(6.675 )(25.4 ) 2 (0.72 )(1)(1)(35,000 )
t = 4.98 mm (0.196 in ) Use standard weight pipe. Could use 5.56 mm (0.219 in) wall if available. 2. Inside facility F = 0.6
t =
(1480 )(6.675 )(25.4 ) 2 (0.6 )(1)(1)(35,000 )
t = 5.97 mm (0.235 in ) Use standard weight pipe.
2.3.
Two-Phase Line 2.3.1.
Flow Rates:
Given:
Condensate
=
5.30 m�/hr
Water
=
1.52 m�/hr
Gas
=
27,100 std m�/hr
Condensate
=
0.87
Water
=
1.05
Gas
=
0.85
Condensate and Water
=
0.003 Pa-sec
Gas
=
1.3 X 10-5 Pa-sec
Inlet Pressure
=
6200 kPa
Temperature
=
27°C
Compressibility Factor
=
0.67
Specific Gravity:
Viscosity:
Fluid flows to a separator, which operates at 5515 kPa. The line is rated for 10,200 kPa (1480 psi).
2.3.2.
Problem: Choose a line size and determine the wall thickness using B31.3 and B31.8.
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2.3.3.
April 1998
Solution: V max
= 18m / sec orV e
V min
= 3 to 4.5 m / sec
Pressure drop = 690 kPa
1. Erosional Velocity Equation (54)
V e
=
1.22 C 1
( ρ m )2
Critical condition occurs at the lowest pressure, but for computing V e use 6200 kPa to be conservative.
ρm = 110.95 kg/m3 from example 1.3 C
Ve
80
9.27 m/sec
100
11.58 m/sec
120
13.90 m/sec
140
16.22 m/sec 2. Minimum I. D. Calculate the inside diameter from Equation (56): 1
ZRT 2 11.9 4.13 Ql + P = d = 5.448 V d =
348.23 1
V 2
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(Note: Velocities are from previous step.) V
I.D.
3.0 m/sec
201.0 mm
4.5 m/sec
164.2 mm
9.27 m/sec
114.4 mm
11.58 m/sec
102.3 mm
13.90 m/sec
93.4 mm
16.22 m/sec
86.5 mm
3. Pressure Drop From Example 1.3, the pressure drop in a 101.6 mm (4 in) and 152.4 mm (6 in) line is unacceptable; use a 203.2 mm (8 in) line. a) AWS QC7-93 Equation (59)
t = t c
+ t th +
100 2(S t E L + P i Y ) 100 - Tol
t = 0.05 + 0 +
P i d o
100 (1480 )(8.625 ) (25.4 ) 2 [20,000 (1) + (1480 )(0.4 )] 100 - 12.5
t = 10.44 mm (0.411 in )
Use Sch. 80 or XH. Could use 11.33 mm (0.438 in) wall if available. b) ASME B31.8 Equation (60) 1. Outside facility F = 0.72
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t =
t =
April 1998
P i d o
2 ( FE L TS y )
(1480 )(8.625 )(25.4 ) 2 (0.72 )(1)(1)(35,000 )
t = 6.43 mm (0.253 in ) Use 7.04 mm (0.277 in) wall, could use 8.18 mm (0.322 in) wall standard weight if available. 2. Inside facility F = 0.6
t =
(1480 )(8.625 )(25.4 ) 2 (0.6 )(1)(1)(35,000 )
t = 7.72 mm (0.304 in ) Use standard weight.
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Appendix C–Example Problems–Customary Units
1.
Pressure Drop 1.1.
Liquid Line 1.1.1.
Flow Rates:
Given: Condensate
=
800 BPD
Water
=
230 BPD
Condensate
=
0.87
Water
=
1.05
Viscosity
=
3 cp
Length
=
7,000 ft
Inlet Pressure
=
900 psia
Temperature
=
80°F
Specific Gravity:
1.1.2.
Problem: Solve for pressure drop in a 2 in and 4 in I.D. line using the general equation.
1.1.3.
Solution: 1. Calculate the specific gravity of the combined liquid form the following formula:
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Equation (4)
(SG )m =
Qo (SG )o
+ QW (SG )W QT
(800 )(.87 ) + (230 )(1.05 ) (SG )m = 1,030
= 0.91
2. Calculate the Reynolds number from the formula: Equation (15)
92.1 (SG ) Ql
Re =
Re =
Re =
d µ
(92.1)(0.91)(1,030 ) 3d 28,775 d
3. Calculate the pressure drop using the general equation: Equation (25)
∆ P = 11.5 × 10
-6
fLQl 2 (SG )m d 5
( 11.5 × 10 ) (7000 )(1,030 ) (0.91) f ∆ P = 2
-6
d 5
∆ P =
77716 f 5
d
4. Assume an absolute roughness (ε ε ε) of 0.00015 feet.
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Diameter 2 in
4 in
Re
14.4 x 10�
7.2 x 10�
ε ε/ d
0.0009
0.00045
f, Moody friction factor (from chart)
0.029
0.034
∆P
70 psi
2.6 psi
1.2.
Gas Line 1.2.1.
Given:
Flow Rate:
Gas
=
23 MMSCFD
Specific Gravity:
Gas
=
0.85
Length
=
7,000 ft
Inlet Pressure
=
900 psig
Temperature
=
80°F
Z
=
0.67
1.2.2.
Problem: Solve for pressure drop in a 4 in and 6 in I.D. line using: 1. The general equation 2. The approximate assumption of ∆P < 10 percent P1 3. The Panhandle Equation 4. The Weymouth Equation
1.2.3.
Solution: 1. General Equation a) Determine the gas viscosity from Figure 4. Viscosity = 0.013 cp b) Calculate the Reynolds number from the formula: (Equation 16)
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Re =
Re =
April 1998
20,100 Q g S d µ
(20,100 )(23 )(.85 ) 30,227,000 = (0.013 ) d d
c) Assume an absolute roughness (ε ε ε) of 0.00015 ft. d) Calculate the pressure drop from the general equation: (Equation 28) 2 1
2 2
2 1
2 2
2 1
2 2
P - P
P - P
P - P
= 25.2
= 25.2
=
S Q g 2 Z T f L 5
d
(0.85 )(23 )2 (0.67 ) (540 )(7,000 ) f d 5
2.87 × 10 10 f d 5
Diameter 4 in
6 in
Re
7.6 x 106
5.0 x 106
ε ε/ d
0.00045
0.0003
f, Moody friction factor (from chart)
0.0164
0.015
4.596 x 105
5.536 x 104
P2
614 psia
883 psia
∆P
301 psi
32 psi
P21 - P22
2. The approximate equation where it is a ssumed that ∆P < 10 percent of P1:
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(Equation 30)
∆ P = 12.6
∆ P = 12.6
∆ P =
S Q g 2 Z T f L P 1d 5
(0.85 )(23 )2 (0.67 )(540 )(7,000 ) f 915 d 5
1.59 × 107 f d 5
Diameter
∆P
4 in
6 in
251 psi
30 psi
3. Panhandle Equation
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(Equation 36)
P 21 - P 22 Q g = 0.028E f 0.961 S Z T Lm Lm
=
7,000 5,280
0.51 2.53
d
= 1.33 miles
E = 0.95 (assumed )
(915 )2 - P 22 23 = (0.028 )(0.95 ) 0.961 ( ) ( )( ) ( ) 0.85 0.67 540 1.33 8.37 × 10 5 - P 22 412
2 2
P
23 = (0.028 )(0.95 )
= 8.37 × 10 5
1.96
0.51
d 2.53
1 4.96
d
2.35 × 10 8 d 4.96
Diameter 4 in
6 in
P2
771 psia
897 psia
∆P
144 psia
18 psia
4. Weymouth Equation (Equation 32) 1
P - P 2 Q g = 1.11 d 2.67 L S ZT 2 1
2 2
a) Rearranging to solve for P1 or P2:
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2 1
2 2
P - P
=
April 1998
Q g 2 L S Z T 1.112 d 5.34
b) Note: We could just as easily have solved for P1, knowing P 2!
(23 ) (7,000 ) (0.85 ) (0.67 ) (540 ) 2
P 2 = (915 ) 2
2
1.23 d 5.34
1
1
925.84 × 10 2 2 P 2 = (915 ) d 5.34 6
Diameter 4 in
6 in
P2
522 psi
879 psi
∆P
393 psi
36 psi
5. Summary Diameter 4 in
6 in
General Equation:
∆P =
301 psi
32 psi
Approximate Equation:
∆P =
251 psi
30 psi
Panhandle Equation:
∆P =
144 psi
18 psi
Weymouth Equation:
∆P =
393 psi
36 psi
1.3.
Two-Phase Lines 1.3.1.
Flow Rates:
Specific Gravity:
Given:
Condensate
=
800 BPD
Water
=
230 BPD
Gas
=
23 MMSCFD
Condensate
=
0.87
Water
=
1.05
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Gas
=
0.85
Condensate and Water
=
3 cp
Gas
=
0.013 cp
Inlet Pressure
=
900 psi
Temperature
=
80°F
Compressibility Factor
=
0.67
Length
=
7,000 ft
Absolute Roughness
=
0.00015 ft
Viscosity:
1.3.2.
Problem: Solve for the pressure drop in 4 in, 6 in, and 8 in I.D. lines using the API RP 14E method:
1.3.3.
Solution: API RP 14E (Equation 38)
3.4 × 10 fLW h -6
∆ P =
2
ρ m d 5
where : W h
= 3,180 Q g S + 14.6 Ql (SG )m
ρ =
12,409 (SG )m P + 2.7 R S P 198.7 P + RTZ
1. Calculate the specific gravity of the liquid:
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April 1998
(Equation 4)
(SG )m =
+ Qw (SG )w
Qo (SG )o
QT
(800 ) (0.87 ) + (230 ) (1.05 ) (SG )m = 1,030
(SG )m = 0.91 2. Calculate the rate of flow of liquid and vapor: (Equation 39)
W h
= (3,180 ) (23 ) (0.85 ) + (14.6 ) (1,030 ) (0.91 )
W h
= 75,854 lb / hr
3. Calculate the mixture density: (Equation 40)
(12,409 ) (0.91) (915 ) + (12.7 ) (22,330 ) (0.85 ) (915 ) (198.7 ) (915 ) + (22,330 ) (540 )( ) (0.67 )
ρ m
=
ρ m
= 6.93 lb / ft 3
4. Assume f = 0.0204 for rough pipe. 5. Calculate the change in pressure: (Equation 38)
(3.36 × 10 ) (0.0204 ) (7,000 ) (75,854 ) ∆ P = (6.93) (d )
2
-6
5
Diameter
∆P =
4 in
6 in
8 in
389 psi
51 psi
12 psi
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Fluid Flow
April 1998
Choosing a Line Diameter and Determining Wall Thickness 2.1.
Liquid Line 2.1.1.
Flow Rates:
Given:
Condensate
=
800 BPD
Water
=
230 BPD
Condensate
=
0.87
Water
=
1.05
Viscosity
=
3 cp
Length
=
7,000 ft
Inlet Pressure
=
900 psig
Temperature
=
80°F
Specific Gravity:
Liquid flows to a low pressure separator operating at 150 psi. The line is rated for 1480 psi.
2.1.2.
Problem: Choose a line size and determine the wall thickness using AWS QC7-93, ASME B31.4, and ASME B31.8.
2.1.3.
Solution:
= 15 ft / sec V min = 3 ft / sec pressure drop = 900 - 150 = 750 psi V max
1. Calculate the diameter from Equation (50):
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V = 0.012
V =
V =
April 1998
Ql d 2
(0.012 )(1030 ) d 2 12.36 d 2
V
I.D.
3 ft/s
2.03 in
15 ft/s
0.91 in 2. Pressure Drop From Example 1.1 the pressure drop in the 2 in line is acceptable. For flexibility and mechanical strength it is better to use a 2 in line rather than a 1 or 1 1/2 in line. a) AWS QC7-93 Equation (59)
100 Pd o + t th + 2(SE + PY ) 100 - TOL
t = t c
t = 0.05 + 0 +
100 (1480 )(2.375 ) 2 [(20,000 )(1) + (1480 )(0.4 )] 100 - 1.5
t = 0.155 in
Can use standard weight pipe. b) ASME B31.4
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Equation (60)
t =
P i d o
2 ( FE L TS y )
F = 0.72
t =
(1480 )(2.375 ) 2 (0.72 )(1)(1)(35,000 )
t = 0.0697 in
Would use a standard weight pipe for mechanical strength. c) ASME B31.8 Use the same equation as ASME B31.4. Outside the facility use a design factor (F) of 0.72 wall thickness, same as above. Within the facility use a factor (F) of 0.60.
for F = 0.6
t =
(1480 )(2.375 ) 2 (0.6 )(1)(1)(35,000 )
t = 0.0837 in
Use a standard weight pipe for mechanical str ength.
2.2.
Gas Line 2.2.1.
Flow Rate:
Given: Gas
=
23 MMSCFD
=
0.013 cp
=
0.85
Length
=
7,000 ft
Inlet Pressure
=
900 psi
Viscosity Gravity:
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Gas
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April 1998
Temperature
=
80°F
Compressibility factor
=
0.67
Gas flows to a dehydrator which operates at 800 psi. The line is rated for 1480 psi.
2.2.2.
Problem: Choose a line size and determine the wall thickness using AWS QC7-93 and ASME B31.8.
2.2.3.
Solution:
= 60 ft / sec V min = 10 to 15 ft / sec Pressure drop = 900 - 800 = 100 psi V max
At a pressure this low, erosional velocity is not important. 1. Calculate the diameter from Equation (53):
V g = 60
V g =
Q g TZ d 2 P
(60 )(23 )(540 )(0.67 ) d 2 (815 ) =
612.62 d 2
V
I.D.
10 ft/sec
7.83 in
15 ft/sec
6.39 in
60 ft/sec
3.20 in 2. Pressure Drop
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From example 1.2, the pressure drop in the 4 in line is not acceptable, but it is acceptable in the 6 in line. This also gives reasonable velocities between 15 and 60 ft/sec. a) AWS QC7-93 Equation (59)
t = t c
+ t th +
100 2 (S t E L + P iY ) 100 - TOL
t = 0.05 + 0 +
P i d o
100 (1480 )(6.675 ) 2 ((20,000 )(1) + (1480 )(0.4 )) 100 - 12.5
t = 0.331 in
Use 6 in XH, could use 6 in 0.375 wall if available. b) ASME B31.8 Equation (60)
t =
P i d o
2( FE L TS y )
1. Outside facility F = 0.72
t =
(1480 )(6.675 ) 2 (0.72 )(1)(1)(35,000 )
t = 0.196 in Use standard weight pipe. Could use 0.219 in wall if available. 2. Inside facility F = 0.6
t =
(1480 )(6.675 ) 2 (0.6 )(1)(1)(35,000 )
t = 0.235 in Use standard weight pipe.
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2.3.
Fluid Flow
Two-Phase Line 2.3.1.
Flow Rates:
April 1998
Given:
Condensate
=
800 BPD
Water
=
230 BPD
Gas
=
23 MMSCFD
Condensate
=
0.87
Water
=
1.05
Gas
=
0.85
Condensate and Water
=
3 cp
Gas
=
0.013 cp
Inlet Pressure
=
900 psig
Temperature
=
80°F
Compressibility Factor
=
0.67
Specific Gravity:
Viscosity:
Fluid flows to a separator, which operates at 800 psi. The line is rated for 1480 psi.
2.3.2.
Problem: Choose a line size and determine the wall thickness using B31.3 and B31.8.
2.3.3.
Solution: V max
= 60 ft / sec or V e
V min
= 10 to 15 ft / sec
Pressure drop = 100 psi
1. Erosional Velocity Equation (54)
V e
«MEPBusinessSectorName»
=
C 1
( ρ m )2
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April 1998
Critical condition occurs at the lowest pressure, but for computing V e use 900 psi to be conservative.
ρ m = 6.93 lb / ft 3 from example 3
C
Ve
80
30.38 ft/sec
100
37.98 ft/sec
120
45.58 ft/sec
140
53.18 ft/sec 2. Minimum I.D. Calculate the inside diameter from Equation (56):
ZRT 11.9 + 16.7 P Ql d = 1000 V
1 2
1
2 (0.67 )(22,330 )(540 ) + 11.9 1030 ( )( ) 16.7 815 d = 1,000 V d =
24.97 1
V 2
(Note: Velocities are from previous step.) V
I.D. min
10.00 ft/sec
7.89 in
15.00 ft/sec
6.44 in
30.38 ft/sec
4.53 in
37.98 ft/sec
4.05 in
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