Line Sizing Procedure

Fluid Flow
EPT 09-T-06
April 1998
Mobil Oil,1998 1 of 123

EPT 09-T-06 Fluid Flow April 1998
Table of Contents
Table of Tables ................................................................................................................... 7
1. Scope ............................................................................................................................ 8
2. References ................................................................................................................... 8
2.1. MEPSMobil Engineering Practices.................................................................... 8
2.2. Mobil Tutorials ..................................................................................................... 9
2.3. APIAmerican Petroleum Institute ...................................................................... 9
2.4. ASMEAmerican Society of Mechanical Engineers ............................................ 9
2.5. ASTMAmerican Society for Testing & Materials ............................................... 9
2.6. AWSAmerican Welding Society ........................................................................ 9
2.7. CFRU.S. Code of Federal Regulations ........................................................... 10
3. Basic Flow Equations and Factors .......................................................................... 10
3.1. Density .............................................................................................................. 10
3.2. Fluid Viscosity ................................................................................................... 13
3.3. Fluid Head and Bernoulli's Law ......................................................................... 19
3.4. Flow Regimes ................................................................................................... 22
3.5. Darcy-Weisbach Equation for Pressure Drop ................................................... 25
3.6. Moody Friction Factor ....................................................................................... 27
3.7. Effect of Elevation Changes .............................................................................. 30
4. Pressure Drop in Piping ............................................................................................ 35
4.1. Liquid Flow (General Equation) ......................................................................... 35
4.2. Gas Flow ........................................................................................................... 36
4.3. Two-Phase Flow ............................................................................................... 45
4.4. Head Loss in Valves and Pipe Fittings .............................................................. 55

5. Choosing a Line Diameter......................................................................................... 66
5.1. General ............................................................................................................. 66
5.2. Erosional Velocity.............................................................................................. 67
5.3. Liquid Line Sizing .............................................................................................. 71
5.4. Gas Line Sizing ................................................................................................. 72
5.5. Two-Phase Flow Line Sizing ............................................................................. 76
6. Determining Wall Thickness ..................................................................................... 80
6.1. Commonly Available Pipe ................................................................................. 80
6.2. Standards and Requirements ........................................................................... 83
6.3. General Hoop Stress Formula .......................................................................... 84
Appendix ANomenclature .............................................................................................. 86
Appendix BExample ProblemsMetric Units ............................................................... 90
1. Pressure Drop ............................................................................................................ 90
1.1. Liquid Line ......................................................................................................... 90
1.2. Gas Line ............................................................................................................ 92
1.3. Two-Phase Lines .............................................................................................. 96
2. Choosing a Line Diameter and Determining Wall Thickness ................................ 98
2.1. Liquid Line ......................................................................................................... 98
2.2. Gas Line .......................................................................................................... 101
2.3. Two-Phase Line .............................................................................................. 103
Appendix CExample ProblemsCustomary Units ..................................................... 107
1. Pressure Drop .......................................................................................................... 107
1.1. Liquid Line ....................................................................................................... 107
1.2. Gas Line .......................................................................................................... 109
1.3. Two-Phase Lines ............................................................................................ 113

2. 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

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 Gas-
Liquid 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

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

Table of Tables
Table 1: Pipe Roughness ................................................................................................ 29
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) .................................................................... 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

1. 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. MEPSMobil Engineering Practices
MP 16-P-01 Piping-General Design

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. Mobil Tutorials
EPT 09-T-05 Piping-Code Selection Guide
2.3. APIAmerican 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. ASMEAmerican Society of Mechanical Engineers
ASME B16.5 Pipe Flanges and Flanged Fittings NPS 1/2 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. ASTMAmerican Society for Testing & Materials
ASTM A106 Standard Specification for Seamless Carbon Steel Pipe for High-
Temperature Service
2.6. AWSAmerican Welding Society
AWS QC7-93 Chemical Plant and Petroleum Refinery Piping (Supplement F)

2.7. CFRU.S. Code of Federal Regulations
49 CFR 192 Transportation, Subchapter D - Pipeline Safety, Transportation of Natural

and Other Gases by Pipeline: Minimum Federal Safety Standards
3. 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: 15.6C and 101.4 kPa (60F and 14.7 psia).
Thus:
Equation 1
Metric :
= 1000 (SG )
Customary :
= 62.4 (SG )
where :
= 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 terms of API gravity, given in degrees API,
which is defined as:

Equation 2
Metric :
141.5
o
API =
(
SG at 15.6 o C )
- 131.5
Customary :
141.5
o
API =
(
SG at 60 o F )
- 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
wQw + oQo
=
QT
where :
o = density of oil, kg / m 3 (lb / ft 3 )
w = density of water, kg / m 3 (lb / ft 3 )
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 )
3.1.4.
Similarly, the specific gravity of an oil and water mixture can be calculated
by:

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:

Equation 6
Metric :
SP
g = 3.48
TZ
Customary :
SP
g = 2.70
TZ
Equation 7
Metric :
g = 0.1203
(MW )P
TZ
Customary :
g = 0.093
(MW )P
TZ
where :
g = density of gas, kg / m 3 (lb / ft 3 )
P = pressure, kPa ( psia )
T = temperature, 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:

Equation 8
where :
= absolute viscosity, Pa - sec (cp )
= kinematic viscosity, m 2 / sec (cs )
= 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 :
= 1000 (SG )
Customary :
= (SG )
where :
= 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

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)

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.

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 and 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:

Equation 10
eff = (1 + 2.5 + 2 ) c
where :
eff = effective viscosity of the mixture, Pa - sec (cp )
c = viscosity of the continuous phase, Pa - sec (cp )
= volume fraction of the discontinuous phase
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.

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 required delivery
position. Here, this includes the additional work necessary to overcome the
resistance to flow in the line.

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 energy 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 )

Equation 12
Metric :
P1 V12 P V2
Z e1 + 1000 + = Z e2 + 1000 2 + 2 + H f
1 2g 2 2g
Customary :
144P1 V12 144P2 V22
Z e1 + + = Z e2 + + +Hf
1 2g 2 2g
where :
Z e = vertical elevation rise of pipe, m ( ft )
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 )
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:

Equation 13
Metric :
Q W
V= = s
3600A A
Customary :
Q Ws
V= =
A A
where :
(
Q = rate of flow, m 3 / sec ft 3 / sec )
A = cross sectional area of pipe, m 2 ft 2 ( )
Ws = rate of flow, kg / sec (lb / sec )
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 fluid 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:

Equation 14
Metric :
DV
Re =

Customary :
D V 1488DV 124dV 7738 (SG )dV
Re = = = =
'
where :
Re = Reynolds number, dimensionless
= density, kg / m 3 (lb / ft 3 )
d = pipe ID, mm (in )
D = pipe ID, m ( ft )
' = viscosity, lb / ft - sec [ (cp ) 0.000672]
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:

Equation 15
Metric :
Rel = 353.13
(SG ) Ql
d
Customary :
Rel = 92.1
(SG ) Ql
d
where :
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 :
Qg S
Reg = 0.428
d
Customary :
Qg S
Reg = 20,000
d
where :
Qg = gas flow rate, std m 3 / hr (MMSCFD )

3.5. 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
fLV 2
Hf =
D2g
where :
L = length of pipe, m ( ft )
D = pipe ID, m( ft )
f = Moody friction factor
(
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:

Equation 18
Metric :
P1 - P2 = P = 9.81 10 -3 H f
Customary :

P1 - P2 = P = Hf
144
where :
P = pressure drop, kPa ( psi )
H f = pipe friction head loss, m ( ft )
3.5.3.
Substituting Equation 17 into Equation 18 and expressing pipe inside
diameter in inches:
Equation 19
Metric :
fLV 2
P = 0.5
d
Customary :
fLV 2
P = 0.0013
d
where :

3.6. Moody Friction Factor
3.6.1.
The Darcy-Weisbach equation can be derived rationally 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 Darcy-
Weisbach equations accordingly, where:
f fanning = 1 / 4f
3.6.3.
This has been a continual source of confusion in basic engineering 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 and 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:

Equation 20
Metric :

f = 64
dV
Customary :
64
f = = 0.52
Re dV
where :
Re = Reynolds number
= density of fluid, kg / m 3 (lb / ft 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
Metric :
LV
P = 32
d2
Customary :
LV
P = 0.000676
d2
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, , to pipe inside diameter. Roughness

is a measure of the smoothness of the pipe's inner surface. Table 1 shows

the absolute roughness, , 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 calculations shall be up to four times as
much as the values shown in Table 1.
Table 1: Pipe Roughness
Absolute Roughness ()
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

= - 2 log 10
1 2.51
+
1 3.7D 1
( f )2 Re( f )2
where :
Re = Reynolds number
= 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.

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 increases by H from point 1 to point
2. Neglecting pressure loss due to friction, the pressure drop due to
elevation change is given by:

P1 - P2 = H
144

Figure 6: Effect of Elevation Changes on Head
3.7.3.
In case B, the elevation head decreases 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:

PA - PB = - H
144
3.7.4.
Similarly, elevation pressure changes due to other segments are:

PA - PB = 0

PB - P2 = 2H
144
3.7.5.
The overall pressure change in the pipe due to elevation is obtained by
adding the changes for the individual segments:
P1 - P2 = P1 - PA + PA - PB + PB - P2

P1 - P2 = (0 - H + 0 + 2H )
144

P1 - P2 = H
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
Metric :
PZ 9.79 (SG ) Z
Customary :
PZ 0.433 (SG ) Z
where :
PZ = pressure drop due to elevation increase, kPa ( psi )
Z = total increase in elevation, m ( ft )

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:

P1 - P2 = H l
144
For Case B:

P1 - PA = - H g
144
PA - PB = 0

PB - P2 = 2H l
144
Thus:

P1 - P2 = l (2 l - g )
144
Since l >> g

P1 - P2 2H l
144

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 from 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
Metric :
PZ 9.79 (SG ) Z e
Customary :
PZ 0.433 (SG ) Z e
where :
Z e = sum of vertical elevation rises only, m ( ft )
PZ = pressure drop due to elevation changes, kPa ( psi )
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 complete 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.

4. 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 = 6.266 107
d5
Customary :
fL (Ql ) (SG )
2
P = (1.15 10 -5 )
d5
Equation 26
Metric :
1
P d 5 2
Ql = 1.265 10 -4
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 friction factor.

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 read
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:

Equation 27
Metric :
g A2 ( P1 ) - ( P2 )
(Ws )2 = 1.322 10
9 2 2

fL P1 P1
v + 2 ln
D P2
Customary :
144g A2 (P1 )2 - (P2 )2
(Ws ) 2
=
fL P1 P1
v + 2 ln
D P2
where :
Ws = rate of flow, kg / sec (lb / sec )
(
A = cross - sectional area of pipe, m 2 ft 2 ( )
(
v = specific volume of gas at upstream conditions, m 3 / kg ft 3 / lb )
l = length of pipe, m ( ft )
P1 = upstream pressure, kPa ( psia )
P2 = downstream pressure, kPa ( psia )
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,
P1 fl
2 ln <<
P2 D

6. Making this assumption and substituting it into Equation 27, one can
derive the following equation:
Equation 28
Metric :
S Qg2 ZTfL
(P1 )2 - (P2 )2 = 52,430
d5
Customary :
S Qg2 ZTfL
(P1 ) - (P2 )
2 2
= 25.2
d5
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
P12 - P22
in Equation 28 is not the same as (P) 2. Rearranging Equation 28 and

solving for Qg we have:
Equation 29
Metric :
(
d 5 P12 - P22
Qg = 4.367 10 -3 )

Z TfL S
Customary :
(
d 5 P12 - P22
Qg = 0.199
)

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 iterated until they converge.

4.2.2. Small Pressure Drops

For small pressure drops, an approximation can be calculated. The
following formula can be derived from Equation 28 if P1 - P2 < 10 percent of
P1 and it is assumed that
P12 - P22 2P1 (P )
Equation 30
Metric :
S (Qg ) Z T f L
2
P = 26,215
P1 d 5
Customary :
S (Qg ) Z T f L
2
P = 12.6
P1 d 5
4.2.3. 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 /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, , the
friction factor is a function of diameter only. For steel pipe the
Weymouth data indicate:
Equation 31
Metric :
0.0941
f = 1
d3
Customary :
0.032
f = 1
d3
2. Substituting this into Equation 29, the Weymouth equation is:

Equation 32
Metric :
1
P 2 - P22 2
Qg = 1.42 10 - 2 d 2.67 1
LSZT
Customary :
1
P 2 - P22 2
Qg = 1.11 d 2.67 1
LSZT
3. Assuming a temperature of 290K (520R), 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
T P12 - P22
Q = 3.415 10 -5 b E d 2.667
Pb S Lm Tavg Z avg
Customary :
0.5
T P12 - P22
Q = 433.5 b E d 2.667
Pb S Lm Tavg Z avg
where :
Tb = base absolute temperature, K ( R)
o
Pb = base absolute pressure, kPa ( psia )

E = pipeline efficiency factor
Tavg = average absolute temperature, K ( R)
o
Z avg = average compressibility

S = specific gravity
d = pipeline ID, mm (in )

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
N
f =
( )
Re n
3. Using this assumption and assuming a constant viscosity for the gas,
Equation 29 can be rewritten as the Panhandle equation:

Equation 36
Metric :
0.51
P12 - P22
Qg = 1.229 10 E f 0.961
-3
d 2.53
S ZTLm
Customary :
0.51
P12 - P22
Qg = 0.028E f 0.961 d 2.53
S ZTLm
where :
E f = efficiency factor, dimensionless
= 1.0 for brand new pipe
= 0.95 for good operating conditions
= 0.92 for average operating conditions
= 0.85 for unfavorable 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:
3.6 1
a) f =1+ + 0.03d
d 100

b) = 520 o R
c) P1 = 15 psi
d) Z = 1.0
e) P < 10 percent of P1
2. With these assumptions, and expressing pressure drop in terms of inches
of water, the Spitzglass Equation can be written:
Equation 37
Metric :

hw d 5
Qg = 3.655 10 - 2
S L 1 + 9.144 + 1.18 10 -3 d

d
Customary :

hw d 5
Qg = 0.09
S L 1 + 3.6 + 0.03d

d
where :
hw = pressure loss, mm of water (in of water )
L = Length of pipe, m ( ft )
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.

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 = 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 coefficients 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.

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 less 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 large 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 and 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 and liquid flow rates.
Figure 8 shows typical flow patterns in horizontal two-phase pipe flow.

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

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.

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.

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

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 Gas-
Liquid 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

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 Holdup Down-Hill Friction Mean Error, Standard

Pattern Recovery Percent 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.

4.3.3. 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 (Wh )
2
P = 62,561
md 5
Customary :
3.36 10 -6 fL (Wh )
2
P =
md 5
where :
Wh = flow rate of liquid and vapor, kg / hr (lb / hr )
m = mixture density, kg / m 3 (lb / ft 3 )
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:

Equation 39
Metric :
Wh = 1.21 Qg S + 999.7 Ql (SG )
Customary :
Wh = 3,180 Qg S + 14.6 Ql (SG )
where :
Wh = flow rate of liquid and vapor, kg / hr (lb / hr )
(SG ) = specic gravity of liquid relative to water
4. The density of the mixture to use in Equation 38 is given by:

Equation 40
Metric :
28,814 (SG ) P + 34.81R S P
m =
28.82P + 10.0 R T Z
Customary :
12,409 (SG ) P + 2.7R S P
m =
198.7 P + R T Z
where :
m = mixture density, kg / m 3 (lb / ft 3 )
(SG ) = specific gravity of the liquid relative to water
(use the average gravity for the hydrocarbon and water mixture )
(
R = gas / liquid ratio, std m 3 / m 3 std ft 3 / bbl )
T = temperatur e, K ( R)
o
4.3.4. AGA Equation

The American Gas Association method uses a frictional pressure 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 pressure 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

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
V2
H f = Kr
2g
where :
K r = resistance coefficient, dimensionless
H f = head loss in fitting, m ( ft )
(
g = acceleration of gravity, 9.81m / sec 2 32.2ft / sec 2 )
2. A comparison of Equations 17 and 41 shows that for a straight pipe:

Equation 42
fL
Kr =
D
3. Approximate values of Kr 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 and exits.

Table 3: Resistance Coefficients for Pipe Fittings
Kr
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
45Elbow 0.4

Figure 12: Resistance Coefficients for Different Types of Pipe Entrances and
Exits (Courtesy of Paragon Engineering Services, Inc.)

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, Cv. The flow coefficient is
measured experimentally for each valve and is defined as the flow of
water at 60F, 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:

Equation 43
Metric :
0.0105 d 2 0.0105 d 2
Cv = 1
= 1
( fl / D )2 (K r )2
Customary :
29.9 d 2 29.9 d 2
Cv = 1
= 1
( fl / D )2 (K r )2
where :
Cv = flow coefficient, m 3 / 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
2. For any fitting with a known Cv:

Equation 44
Metric :
2
Q
P = 6.89 l (SG )
Cv
Customary :
2
Q
P = 8.5 10 l (SG )
-4
Cv
where :
Cv = flow coefficient, m / hr ( gpm ) 3

4.4.3. 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, Le, can be determined from Kr and Cv as follows:
Equation 45
Kr D
Le =
f
Equation 46
Metric :
Kr d
Le =
1000f
Customary :
Krd
Le =
12f

Equation 47
Metric :
1.108 10 -7 d 5
Le =
fCv2
Customary :
74.5d 5
Le =
fCv2
where :
Le = equivalent length, m ( ft )
K r = resistance coefficient, dimensionless
D = fitting ID, m ( ft )
d = fitting ID, mm (in )
Cv = flow coefficient, m 3 / hr ( gpm )
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 Globe Valve or Ball Angle Valve Swing Check Valve Plug Valve, Gate
Size (in) Check Valve 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

Nominal Pipe Globe Valve or Ball Angle Valve Swing Check Valve Plug Valve, Gate
Size (in) Check Valve 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

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

Table 4: (Concluded)
Enlargement Contraction
Sudden Standard Reducer Sudden Standard Reducer
Equivalent length in terms of small diameter
d/D = 1/4 d/D = 1/2 d/D = 3/4 d/D = 1/2 d/D = 3/4 d/D = 1/4 d/D = 1/2 d/D = 3/4 d/D = 1/2 d/D = 3/4
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.

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)

5. 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 across
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.

5.2. 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
Metric :
Ce
Ve = 1.22 1
( )2
Customary :
Ce
Ve = 1
( )2
where :
Ve = erosional flow velocity, m / sec ( ft / sec )
Ce = empirical constant, dimensionl ess
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 Ce

= 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 Ce = 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

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 )
2
vol = 9.806 10 -3
gPh
Customary :
12KW (V p )
2
vol =
gPh
where :
vol = volume of metal eroded, mm 3 in 3 ( )
V p = particle velocity, m / sec ( ft / sec )
Ph = penetratio n hardness of the material, kPa ( psi )
= a value between 0.5 and 1.0 depending upon the
impingemen t angle of the particle
K = erosive wear coefficient, dimensionl ess
W = total weight of impinging solid particles, kg (lb )
(
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

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 "Ce"
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.

Figure 16: Example of a Target Tee
Figure 17: Wear Rate Comparison for Standard Fittings (Source: API OSAPR
Project No. 2)

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 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 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 erosion-
corrosion is anticipated. (See MP 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 will be
eroded until a new equilibrium velocity is reached and 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
velocity of 0.9 m/sec (3 ft/sec) is normally recommended.
5.3.4.
Liquid velocity can be determined from the following equation:

Equation 50
Metric :
Ql
V = 353.68
d2
Customary :
Ql
V = 0.012
d2
where :
Ql = liquid flow rate, m 3 / hr ( BPD )
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. 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. 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 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.

5.4.3.
By using the following equation, Figure 18 can be used to choose a pipe
diameter directly:
Equation 51
Metric :
S Tf Qg2
d 5 = 8.186 10 5
P (P / 100 ft )
Customary :
1260 S Tf Qg2
d =
5
P (P / 100 ft )
where :
o
f = Moody friction factor, dimensionl ess

P / 100ft = desired pressure drop per 100 ft from Figure 18, kPa ( psi )

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 CO2 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:

Equation 52
Metric :
1
T Z 2
Ve = 0.644 Ce
S P
Customary :
1
T Z 2
Ve = 0.6 Ce
S P
where :
Ve = erosional velocity, m / sec ( ft / sec )
o

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:

Equation 53
Metric :
Qg T Z
Vg = 122.7
d 2P
Customary :
Qg T Z
Vg = 60
d 2P
where :
o

Vg = gas velocity, m / sec ( ft / sec )
Z = gas compressib ility 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 facility,
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.

5.5.3.
A minimum flow velocity of 3 to 4.5 m/sec (10 to 15 ft/sec) is recommended
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 CO2
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
Metric :
Ce
Ve = 1.22 1
( m )2
Customary :
Ce
Ve = 1
( m )2
where :
Ve = erosional flow velocity, m / sec ( ft / sec )
m = mixture density, kg / m 3 (lb / ft 3 ), from Equation 40
5.5.4.
It can be shown that the minimum cross-sectional area of pipe for a
maximum allowable velocity can be expressed as:

Equation 55
Metric :
ZRT
9.35 + 3.24 P
a = 29.69 Ql
Vmax

Customary :
ZRT
9.35 + 21.25P
a= Ql
1,000 Vmax

where :
a = minimum required cross - sectional area, mm 2 in 2 ( )
Vmax = maximum allowable velocity, m / sec ( ft / sec )
(
R = gas / liquid ratio, std m 3 / m 3 std ft 3 / bbl )
o
5.5.5.
This can be solved for pipe inside diameter:

Equation 56
Metric :
1
ZRT 2
11.9 + 4.13 Ql
P
d = 5.448
Vmax

Customary :
1
ZRT 2
11.9 + Ql
16.7 P
d =
1,000 Vmax

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 m from Equation 40.

2. Determine the erosional velocity, Ve, from Equation 54.
3. For the design, use the smaller of Ve 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 there is enough
driving force available.

6. 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 from the actual pipe outside diameter.
Table 5: ANSI Pipe Schedules
Nominal Pipe O. D. Wall Thickness Weight/Ft Schedule Class

Size
3
/4 in 1.050 .113 1.131 40 STD.
.154 1.474 80 XH
.218 1.937 160
.308 2.441 XXH
1 in 1.315 .133 1.679 40 STD.
.179 2.172 80 XH
.250 2.844 160
.358 3.659 XXH
1 1/2 in 1.900 .145 2.718 40 STD.
.200 3.631 80 XH
.281 4.859 160
.400 6.408 XXH
2 in 2.375 .154 3.653 40 STD.
.218 5.022 80 XH
.343 7.444 160
.436 9.029 XXH
3 in 3.500 .216 7.576 40 STD.
.300 10.25 80 XH
.437 14.32 160


Size
.600 18.58 XXH

4 in 4.50 .237 10.79 40 STD.
.281 12.66 60
.337 14.98 80 XH
.437 19.01 120
.531 22.51 160
.674 27.54 XXH
6 in 6.625 .280 18.97 40 STD.
.432 28.57 80 XH
.562 36.39 120
.718 45.30 160
.864 53.16 XXH
8 in 8.625 .250 22.36 20
.277 24.70 30
.322 28.55 40 STD.
.406 35.64 60
.500 43.39 80 XH
.593 50.87 100
.718 60.63 120
.812 67.76 140
.906 74.69 160
.875 72.42 XXH
10 in 10.75 .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


Size
12 in 12.75 .250 33.38 20

.330 43.77 30
.375 49.56 STD.
.406 53.53 40
.500 65.42 XH
.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
14 in 14.0 .250 36.71 10
.312 45.68 20
.375 54.57 30 STD.
.437 63.37 40
.500 72.09 XH
.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
16 in 16.0 .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


Size
1.593 245.1 160
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

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 for 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 accordance
with these standards, it is necessary 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
2 tL = Pi (d o - 2t )L
where :
= hoop stress in pipe wall, kPa ( psi )
t = pipe wall thickness, mm (in )
Pi = internal design pipepressu re, kPa ( psig )
d o = pipe OD, mm (in )
L = pipe length, m ( ft )

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
Pi d o
t=
2( + Pi )

Appendix ANomenclature
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
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
Kr = resistance coefficient, dimensionless

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
(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

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
Ze = vertical elevation rise of pipe, m, ft
= density of liquid, kg/m, lb/ft
g = density of gas, kg/m, lb/ft
m = mixture density, dg/m
o = density of oil, kg/m, lb/ft
w = density of water, kg/m, lb/ft
= absolute viscosity, Pa-sec, centipoise (cp)

' = viscosity, kg/m sec, lb/ft-sec ( (cp) x 0.000672)
c = viscosity of the continuous phase, Pa-sec, centipoise (cp)
eff = effective viscosity of the mixture, Pa-sec, centipoise (cp)
= kinematic viscosity, m/sec, centistoke
= a value between 0.5 and 1.0, depending on the impingement angle of the particle
= volume fraction of the discontinuous phase
= 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
Ze = sum of vertical elevation rises of pipe, m, ft
= hoop stress in pipe wall, kPa, psi

Appendix BExample ProblemsMetric Units
1. Pressure Drop
1.1. Liquid Line
1.1.1. Given:
Flow Rates: Condensate = 5.30 m/hr

Water = 1.52 m/hr
Specific Gravity: Condensate = 0.87
Water = 1.05
Viscosity = 0.003 Pa-sec
Length = 2130 m
Inlet Pressure = 6200 kPa (abs)
Temperature = 27C
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:

(Equation 4)
(SG )m = Qo (SG )o + QW (SG )W

QT
(SG )m = (5.30 )(0.87 ) + (1.52 )(1.05 )

5.30 + 1.52
= 0.91
2. Calculate the Reynolds number from the formula:

(Equation 15)
353.13 (SG ) Q
Re =
d
Re =
(353.13 )(0.91)(6.82 )
d (0.003 )
730,530
Re =
d
3. Calculate the pressure drop using the general equation:

(Equation 25)
fLQl2 (SG )m
P = 6.266 10 7
d5
P=
(6.266 10 ) (2130 )(6.82 ) (0.91) f
7 2
d5
P=
(5.649 10 ) f 12
d5
4. Assume an absolute roughness () of 0.045 mm.
MEPBusinessSectorName Mobil Oil, 1997 91 of 123

Diameter
50.8 mm (2 in) 101.6 mm (4 in)
Re 14.4 x 10 7.2 x 10
/d 0.0009 0.00045
f, Moody friction factor (from chart) 0.029 0.034
P 484 kPa 17.7 kPa
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 = 27C
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:

(Equation 16)
0.428 Qg S
Re =
d
Re =
(0.428 )(27,100 )(0.85 )
(1.3 10 ) d
-5
760,000,000
=
d
c) Assume an absolute roughness () of 0.045 mm.

d) Calculate the pressure drop from the general equation:
(Equation 28)
S Qg2 Z T f L
P - P = 52,430
1
2
2
2
d5
2
P -P 2
= 52,430
(0.85 )(27,100 ) (0.67 )(300 )(2,130 ) f
2
1 2
d5
1.40 10 19 f
P -P =
1
2
2
2
d5
Diameter
101.6 mm (4 in) 152.4 mm (6 in)
Re 7.5 x 106 5.0 x 106
/d 0.00045 0.0003
P12 - P22 2.121 x 107 2.554 x 106
P2 4303 kPa (abs) 6097 kPa (abs)
P 2000 kPa 206 kPa
2. The approximate equation where it is assumed that P < 10 percent of

P1:

(Equation 30)
S Qg2 Z T f L
P = 26,215
P1d 5
P = 26,215
(0.85 )(27,100 )2 (0.67 )(300 )(2,130 ) f
6303 d 5
1.11 10 15 f
P =
d5
Diameter
101.6 mm (4 in) 152.4 mm (6 in)
P 1684 kPa 222 kPa
3. Panhandle Equation
(Equation 36)
0.51
P1 - P 2
Qg = 1.229 10 E f 0.9612 2
-3
d 2.53
S Z T Lm
2,130
Lm = = 2.13 km
1,000
E = 0.95 (assumed )
(
27,100 = 1.229 10 -3 (0.95 ) )
(6303)2 - P22
0.51

d 2.53
(0.85 )
0.961
(0.67 )(300 )(2.13)
1.015 10 17
P = 3.973 10 -
2
2 7
d 4.96

Diameter
101.6 mm (4 in) 152.4 mm (6 in)
P 969 kPa 121 kPa
4. Weymouth Equation
(Equation 32)
1
P12 - P22 2
Qg = 1.42 10 - 2 d 2.67
L S ZT
a) Rearranging to solve for P1 or P2:
Qg2 L S Z T
P -P =
2 2
1 2
(1.42 10 )
-2 2
d 5.34
b) Note: We could just as easily have solved for P1, knowing P2 !

1

P2 = (6303 ) -
2 ( )
27,100 2 (2,130 )(0.85 )(0.67 )(300 ) 2

( 2
1.42 10 -2 d 5.34 )
1
2 1.325 10
18 2
P2 = (6303 ) -
d 5.34
Diameter
101.6 mm (4 in) 152.4 mm (6 in)
P 2523 kPa 236 kPa
5. Summary

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. Given:

Water = 1.52 m/hr
Gas = 27,100 std m/hr
Water = 1.05
Gas = 0.85
Viscosity: Condensate and Water = 0.003 Pa-sec
Gas = 1.3 X 10-5 Pa-sec
Inlet Pressure = 6200 kPa
Temperature = 27C
Compressibility Factor = 0.67
Length = 2,130 m
Absolute Roughness = 0.045 mm
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

(Equation 38)
62,561 fLWh2
P =
md 5
where :
Wh = 1.21 Qg S + 999.7 Ql (SG )m
28,814 (SG )m P + 34.81 R S P

m =
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
(SG )m = (5.30 )(0.87 ) + (1.52 )(1.05 )

6.82
(SG )m = 0.91
b) Calculate the rate of flow of liquid and vapor:

(Equation 39)
Wh = (1.21)(27,100 )(0.85 ) + (999.7 )(6.82 )(0.91)
Wh = 34,077 kg / hr
c) Calculate the mixture density:

(Equation 40)
m =
(28,814 )(0.91)(6303) + (34.81)(27,100 / 6.82 )(0.85 )(6303)
(28.82 )(6303) + (27,100 / 6.82 )(300 )(0.67 )
m = 110.95 kg / m 3
d) Assume f = 0.0204 for rough pipe.

e) Calculate the change in pressure:
(Equation 38)
P =
(62,561)(0.0204 )(2,130 )(34,077 )2
(110.95 ) (d 5 )
Diameter
101.6 mm (4 in) 152.4 mm (6 in) 203.2 mm (8 in)
P = 2628 kPa 346 kPa 82 kPa
2. Choosing a Line Diameter and Determining Wall

Thickness
2.1. Liquid Line
2.1.1. Given:

Water = 1.52 m/hr
Water = 1.05
Viscosity = 0.003 Pa-sec
Length = 2130 m


Temperature = 27C
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:
Vmax = 4.5 m / sec
Vmin = 0.9 m / sec
pressure drop = 6200 - 1035 = 5165 kPa
1. Calculate the diameter from Equation (50):
Ql
V = 353.68
d2
V=
(353.68 )(6.82 )
d2
2412
V= 2
d
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

Equation (59)
Pd o 100
t = tc + tth +
2(SE + PY ) 100 - TOL

t = 0.05 + 0 +
(1480 )(2.375 ) 100
(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)
Pi d o
t=
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.

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 strength.
2.2. Gas Line
2.2.1. Given:
Flow Rates: Gas = 27,100 std m/hr

Viscosity = 1.3 X 10-5 Pa-sec
Gravity: Gas = 0.85
Length = 2130 m
Inlet Pressure = 6200 kPa (g)
Temperature = 27C
Compressibility factor = 0.67
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:
Vmax = 18 m / sec
Vmin = 3 to 4.5 m / sec
Pressure drop = 6200 - 5515 = 685 kPa
At a pressure this low, erosional velocity is not important.

Qg TZ
Vg = 122.7
d 2P
Vg =
(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)
Pi d o 100
t = tc + tth +
2 (S t E L + PiY ) 100 - TOL

t = 0.05 + 0 +
(1480 )(6.675 ) 100
(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)
Pi d o
t=
2(FE LTSY )
1. Outside facility F = 0.72

t=
(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. Given:

Water = 1.52 m/hr
Gas = 27,100 std m/hr
Water = 1.05
Gas = 0.85
Viscosity: Condensate and Water = 0.003 Pa-sec
Gas = 1.3 X 10-5 Pa-sec
Temperature = 27C
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.

2.3.3. Solution:
Vmax = 18m / sec orVe
Vmin = 3 to 4.5 m / sec
Pressure drop = 690 kPa
1. Erosional Velocity
Equation (54)
1.22 C
Ve = 1
( m )2
Critical condition occurs at the lowest pressure, but for computing Ve

use 6200 kPa to be conservative.
m = 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

348.23
d= 1
2
V

(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)
Pi d o 100
t = tc + tth +
2(S t E L + Pi Y ) 100 - Tol

t = 0.05 + 0 +
(1480 )(8.625 ) 100
(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)

Pi d o
t=
2 (FE L TS y )
t=
(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.
t=
(1480 )(8.625 )(25.4 )
2 (0.6 )(1)(1)(35,000 )
t = 7.72 mm (0.304 in )
Use standard weight.

Appendix CExample ProblemsCustomary Units
1. Pressure Drop
1.1. Liquid Line
1.1.1. Given:
Flow Rates: Condensate = 800 BPD

Water = 230 BPD
Water = 1.05
Viscosity = 3 cp
Length = 7,000 ft
Inlet Pressure = 900 psia
Temperature = 80F
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:

Equation (4)
(SG )m = Qo (SG )o + QW (SG )W

QT
(SG )m = (800 )(.87 ) + (230 )(1.05 )

1,030
= 0.91
2. Calculate the Reynolds number from the formula:

Equation (15)
92.1 (SG ) Ql
Re =
d
Re =
(92.1)(0.91)(1,030 )
3d
28,775
Re =
d
3. Calculate the pressure drop using the general equation:

Equation (25)
fLQl2 (SG )m
P = 11.5 10 -6
d5
P=
( 11.5 10 ) (7000 )(1,030 ) (0.91) f
-6 2
d5
77716 f
P=
d5
4. Assume an absolute roughness () of 0.00015 feet.

Diameter
2 in 4 in
Re 14.4 x 10 7.2 x 10
/d 0.0009 0.00045
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 = 80F
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)

20,100 Qg S
Re =
d
Re =
(20,100 )(23)(.85 ) =
30,227,000
(0.013) d d
c) Assume an absolute roughness () of 0.00015 ft.

d) Calculate the pressure drop from the general equation:
(Equation 28)
S Qg2 Z T f L
P - P = 25.2
1
2
2
2
d5
P12 - P22 = 25.2

(0.85 )(23 )2 (0.67 )(540 )(7,000 ) f
d5
2.87 10 10 f
P12 - P22 =
d5
Diameter
4 in 6 in
Re 7.6 x 106 5.0 x 106
/d 0.00045 0.0003
2 2 5
P 1-P 2 4.596 x 10 5.536 x 104
P2 614 psia 883 psia
P 301 psi 32 psi
2. The approximate equation where it is assumed that P < 10 percent of

P1:

(Equation 30)
S Qg2 Z T f L
P = 12.6
P1d 5
P = 12.6
(0.85 )(23)2 (0.67 )(540 )(7,000 ) f
915 d 5
1.59 107 f
P =
d5
Diameter
4 in 6 in
P 251 psi 30 psi
3. Panhandle Equation

(Equation 36)
0.51
P21 - P22
Qg = 0.028E f 0.961 d 2.53
S Z T Lm
7,000
Lm = = 1.33 miles
5,280
E = 0.95 (assumed )
(915 )2 - P22
0.51

23 = (0.028 )(0.95 ) d 2.53
(0.85 )
0.961
(0.67 )(540 )(1.33)
1.96
8.37 10 5 - P22 23 1
=
(0.028 )(0.95 )
4.96
412 d
2.35 10 8
P22 = 8.37 10 5 -
d 4.96
Diameter
4 in 6 in
P2 771 psia 897 psia
P 144 psia 18 psia
4. Weymouth Equation
(Equation 32)
1
P12 - P22 2
Qg = 1.11 d 2.67
L S ZT
a) Rearranging to solve for P1 or P2:

Qg2 L S Z T
P -P =
1
2
2
2
1.112 d 5.34
b) Note: We could just as easily have solved for P1, knowing P2!
1

P2 = (915 ) -
2 ( )
23 2 (7,000 )(0.85 )(0.67 )(540 ) 2

1.23 d 5.34
1
925.84 10 6 2
P2 = (915 ) -
2

d 5.34
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. Given:

Water = 230 BPD
Gas = 23 MMSCFD
Water = 1.05

Gas = 0.85
Viscosity: Condensate and Water = 3 cp
Gas = 0.013 cp
Inlet Pressure = 900 psi
Temperature = 80F
Length = 7,000 ft
Absolute Roughness = 0.00015 ft
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 -6 fLWh2
P =
md 5
where :
Wh = 3,180 Qg S + 14.6 Ql (SG )m
12,409 (SG )m P + 2.7 R S P

=
198.7 P + RTZ
1. Calculate the specific gravity of the liquid:

(Equation 4)
(SG )m = Qo (SG )o + Qw (SG )w

QT
(SG )m = (800 )(0.87 ) + (230 )(1.05 )

1,030
(SG )m = 0.91
2. Calculate the rate of flow of liquid and vapor:

(Equation 39)
Wh = (3,180 )(23 )(0.85 ) + (14.6 )(1,030 )(0.91)
Wh = 75,854 lb / hr
3. Calculate the mixture density:

(Equation 40)
m =
(12,409 )(0.91)(915 ) + (12.7 )(22,330 )(0.85 )(915 )
(198.7 )(915 ) + (22,330 )(540 )(0.67 )
m = 6.93 lb / ft 3
4. Assume f = 0.0204 for rough pipe.

5. Calculate the change in pressure:
(Equation 38)
P =
(3.36 10 ) (0.0204 )(7,000 )(75,854 )
-6 2
(6.93) (d )
5
Diameter
4 in 6 in 8 in
P = 389 psi 51 psi 12 psi

2. Choosing a Line Diameter and Determining Wall

Thickness
2.1. Liquid Line
2.1.1. Given:

Water = 230 BPD
Water = 1.05
Viscosity = 3 cp
Length = 7,000 ft
Temperature = 80F
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:
Vmax = 15 ft / sec
Vmin = 3 ft / sec
pressure drop = 900 - 150 = 750 psi

Ql
V = 0.012
d2
V=
(0.012 )(1030 )
d2
12.36
V=
d2
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)
Pd o 100
t = tc + tth +
2(SE + PY ) 100 - TOL

t = 0.05 + 0 +
(1480 )(2.375 ) 100
2 [(20,000 )(1) + (1480 )(0.4 )] 100 - 1.5
t = 0.155 in
Can use standard weight pipe.
b) ASME B31.4

Equation (60)
Pi d o
t=
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 strength.
2.2. Gas Line
2.2.1. Given:
Flow Rate: Gas = 23 MMSCFD

Viscosity = 0.013 cp
Gravity: Gas = 0.85
Length = 7,000 ft
Inlet Pressure = 900 psi

Temperature = 80F
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:
Vmax = 60 ft / sec
Vmin = 10 to 15 ft / sec
Pressure drop = 900 - 800 = 100 psi
At a pressure this low, erosional velocity is not important.
QgTZ
Vg = 60
d 2P
Vg =
(60 )(23)(540 )(0.67 )
d 2 (815 )
612.62
=
d2
V I.D.
10 ft/sec 7.83 in
15 ft/sec 6.39 in
60 ft/sec 3.20 in
2. Pressure Drop

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)
Pi d o 100
t = tc + tth +
2 (S t E L + PiY ) 100 - TOL

t = 0.05 + 0 +
(1480 )(6.675 ) 100
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)
Pi d o
t=
2(FE L TS y )
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.

t=
(1480 )(6.675 )
2 (0.6 )(1)(1)(35,000 )
t = 0.235 in
Use standard weight pipe.

2.3. Two-Phase Line
2.3.1. Given:

Water = 230 BPD
Gas = 23 MMSCFD
Water = 1.05
Gas = 0.85
Viscosity: Condensate and Water = 3 cp
Gas = 0.013 cp
Temperature = 80F
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:
Vmax = 60 ft / sec or Ve
Vmin = 10 to 15 ft / sec
Pressure drop = 100 psi
1. Erosional Velocity
Equation (54)
C
Ve = 1
( m )2

Critical condition occurs at the lowest pressure, but for computing Ve

use 900 psi to be conservative.
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):
1
ZRT 2
11.9 + 16.7 P Ql
d =
1000 V

1

11.9 +
(0.67 )(22,330 )(540 ) 1030 2

(16.7 )(815 )
d =
1,000 V

24.97
d= 1
2
V
(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

45.58 ft/sec 3.69 in

53.18 ft/sec 3.42 in
3. Pressure Drop
From Example 1.3, the pressure drop in a 4 in and 6 in line is
unacceptable; use an 8 in line.
a) AWS QC7-93
Equation (59)
Pi d o 100
t = tc + tth +
2(S t E L + Pi Y ) 100 - Tol

t = 0.05 + 0 +
(1480 )(8.625 ) 100
2 [20,000 (1) + (1480 )(0.4 )] 100 - 12.5
t = 0.411 in
Use 8 in Sch. 80 or XH. Could use 0.438 in wall if available.
b) ASME B31.8
Equation (60)

Pi d o
t=
2 (FE L TS y )
t=
(1480 )(8.625 )
2 (0.72 )(1)(1)(35,000 )
t = 0.253 in
Use 8 in 0.277 in wall, could use 0.322 in wall (standard) if
available.
t=
(1480 )(8.625 )
2 (0.6 )(1)(1)(35,000 )
t = 0.304 in
Use 8 in standard.

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Fluid Flow

Mobil Oil,1998 1 of 123

Table of Tables ................................................................................................................... 7

2.1. MEPSMobil Engineering Practices.................................................................... 8

2.2. Mobil Tutorials ..................................................................................................... 9

2.3. APIAmerican Petroleum Institute ...................................................................... 9

2.4. ASMEAmerican Society of Mechanical Engineers ............................................ 9

2.5. ASTMAmerican Society for Testing & Materials ............................................... 9

2.6. AWSAmerican Welding Society ........................................................................ 9

2.7. CFRU.S. Code of Federal Regulations ........................................................... 10

3. Basic Flow Equations and Factors .......................................................................... 10

3.1. Density .............................................................................................................. 10

3.2. Fluid Viscosity ................................................................................................... 13

3.3. Fluid Head and Bernoulli's Law ......................................................................... 19

3.4. Flow Regimes ................................................................................................... 22

3.5. Darcy-Weisbach Equation for Pressure Drop ................................................... 25

3.6. Moody Friction Factor ....................................................................................... 27

3.7. Effect of Elevation Changes .............................................................................. 30

4. Pressure Drop in Piping ............................................................................................ 35

4.1. Liquid Flow (General Equation) ......................................................................... 35

4.2. Gas Flow ........................................................................................................... 36

4.3. Two-Phase Flow ............................................................................................... 45

4.4. Head Loss in Valves and Pipe Fittings .............................................................. 55

Mobil Oil,1998 2 of 123

5. Choosing a Line Diameter......................................................................................... 66

5.1. General ............................................................................................................. 66

5.2. Erosional Velocity.............................................................................................. 67

5.3. Liquid Line Sizing .............................................................................................. 71

5.4. Gas Line Sizing ................................................................................................. 72

5.5. Two-Phase Flow Line Sizing ............................................................................. 76

6. Determining Wall Thickness ..................................................................................... 80

6.1. Commonly Available Pipe ................................................................................. 80

6.2. Standards and Requirements ........................................................................... 83

6.3. General Hoop Stress Formula .......................................................................... 84

Appendix ANomenclature .............................................................................................. 86

Appendix BExample ProblemsMetric Units ............................................................... 90

1. Pressure Drop ............................................................................................................ 90

1.1. Liquid Line ......................................................................................................... 90

1.2. Gas Line ............................................................................................................ 92

1.3. Two-Phase Lines .............................................................................................. 96

2. Choosing a Line Diameter and Determining Wall Thickness ................................ 98

2.1. Liquid Line ......................................................................................................... 98

2.2. Gas Line .......................................................................................................... 101

2.3. Two-Phase Line .............................................................................................. 103

Appendix CExample ProblemsCustomary Units ..................................................... 107

1. Pressure Drop .......................................................................................................... 107

1.1. Liquid Line ....................................................................................................... 107

1.2. Gas Line .......................................................................................................... 109

1.3. Two-Phase Lines ............................................................................................ 113

Mobil Oil,1998 3 of 123

2. 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

Mobil Oil,1998 4 of 123

Figure 3: Liquid Viscosity of Pure and Mixed Hydrocarbons Containing Dissolved

Figure 4: Hydrocarbon Gas Viscosity (Courtesy of GPSA).......................................... 19

Figure 5: Friction Factor as a Function of Reynolds Number and Pipe Roughness

Figure 6: Effect of Elevation Changes on Head ............................................................ 31

Figure 8: Two-phase Flow Patterns in Horizontal Flow (Source: P. Griffith,