Chapter 3 Wellbore Performance

Vietnam National University - Ho Chi Minh City
University of Technology
Faculty of Geology & Petroleum Engineering
Department of Drilling - Production Engineering
Course
Petroleum production
analysis and forecast
Trần Nguyễn Thiện Tâm
trantam2512@hcmut.edu.vn
Chapter 3
Wellbore performance
5/20/2023 Petroleum production analysis and forecast 2

References
[1] Michael J. Economides, A. Daniel Hill, Christine Ehlig-Economides, Ding Zhu.
Petroleum Production Systems, 2nd Edition. Prentice Hall, 2013.
[2] Boyun Guo, Xinghui Lou Liu, Xuehao Tan. Petroleum Production Engineering, 2nd
Edition. Gulf Professional Publishing, 2017.
[3] H. Dale Beggs. Production Optimization Using Nodal Analysis. Oil & Gas Consultants
International, 2008.

Contents
❑ Introduction
❑ Single-Phase Flow of an Incompressible, Newtonian Fluid
❑ Single-Phase Flow of a Compressible, Newtonian Fluid
❑ Multiphase Flow in Wells

Introduction
Wellbore flow can be divided into several broad categories, depending on the flow
geometry, the fluid properties, and the flow rate.
First, the flow in a wellbore is either single phase or multiphase; in most production
wells, the flow is multiphase, with at least two phases (e.g., gas and liquid) present.
Some production wells and most injection wells experience single-phase flow.
The flow geometry of interest in the wellbores is usually flow through a circular pipe,
though flow in an annular space, such as between tubing and casing, sometimes occurs.

Introduction
Furthermore, the flow of interest may be in any direction relative to the gravitational
field. The properties of the fluids, both their PVT behavior and their rheological
characteristics, must be considered in describing wellbore flow performance.
Finally, depending on the flow rate and the fluid properties, flow in a wellbore may be
either laminar or turbulent, and this will strongly influence the flow behavior.

Introduction
Wellbore flow performance: predict the pressure as a function of position between the
bottomhole location and the surface.
In addition, the velocity profile and the distribution of the phases in multiphase flow
are sometimes of interest.

Single-Phase Flow of an Incompressible, Newtonian Fluid
▪ Laminar or Turbulent Flow
▪ Pressure-Drop Calculations

Laminar or Turbulent Flow
Single-phase flow can be characterized as being either laminar or turbulent, depending
on the value of a dimensionless group, the Reynolds number, NRe.
Du 
N Re = (3.1)


When the flow is laminar, the fluid moves in distinct laminae, with no fluid motion
transverse to the bulk flow direction. Turbulent flow, on the other hand, is
characterized by eddy currents that cause fluctuating velocity components in all
directions.
Whether the flow is laminar or turbulent will strongly influence the velocity profile in
the pipe, the frictional pressure drop behavior, and the dispersion of solutes contained
in the fluid, among other factors; all of these attributes are considerations at times in
production operations.

The transition from laminar to turbulent flow in circular pipes generally occurs at a
Reynolds number of 2100, though this value can vary somewhat depending on the pipe
roughness, entrance conditions, and other factors (Govier and Aziz, 1977).
To calculate the Reynolds number, all variables must be expressed in consistent units so
that the result is dimensionless.

Example 3.1
Determining the Reynolds Number for Flow in an Injection Well
For the injection of 1.03-specific gravity water (ρ = 64.3 lbm/ft3) in an injection well with
7-in., 32-lb/ft casing, construct a graph of Reynolds number versus volumetric flow rate
(in bbl/d). The viscosity of the water at bottomhole conditions is 0.6cp. At what
volumetric flow rate will the transition from laminar to turbulent flow occur?

Example 3.1 - Method of Solution
The average velocity is simply the volumetric flow rate divided by the cross-sectional
area of flow
q
u= (3.2)
A
For flow in a circular pipe, the cross-sectional area is

A= D2 (3.3)
4
Substituting for u in Equation (3-1).
4q 
N Re = (3.4)
 D

The units must now be converted to a consistent set.

The units must now be converted to a consistent set.
N Re = 26.0q (3.5)
where q is in bbl/d, p in lbm/ft3, D in in., and μ in cp.
For these oilfield units, the Reynolds number can be expressed in general as
1.48q 
N Re = (3.6)
D

We see that the Reynolds number
varies linearly with the volumetric
flow rate for a given pipe size and
fluid (Figure 3-1).
The transition from laminar to
turbulent flow occurs at NRe =
2100, so for this example 2100 =
26.0q and therefore q = 81 bbl/d.
Below about 81 bbl/d, the flow
will be laminar, at rates higher
than 81 bbl/d, flow is turbulent.

As the viscosity increases, the likelihood of
laminar flow also increases.
Figure 3-2 shows the variation of Reynolds number
with flow rate, with pipe size and viscosity as
parameters.

Pressure-Drop Calculations
The pressure drop over a distance, z, of single-phase flow in a pipe can be obtained by
solving the mechanical energy balance equation, which in differential form is
dp g udu 2 f f u 2 dL
+ dz + + + dWs = 0 (3.7)
 gc gc gc D
If the fluid is incompressible (ρ = constant), and there is no shaft work device in the
pipeline (a pump, compressor, turbine, etc.), this equation is readily integrated to yield
g  2 f  u 2
L
p = p1 − p2 = z + u +
2 f
(3.8)
gc 2 gc gc D
The three terms on the right-hand side are the potential energy, kinetic energy, and
frictional contributions to the overall pressure drop, or
p = pPE + pKE + pF (3.9)

Pressure-Drop Calculations
• ΔpPE, the Pressure Drop Due to Potential Energy Change
• ΔpKE, the Pressure Drop Due to Kinetic Energy Change
• ΔpF, The Frictional Pressure Drop

ΔpPE, the Pressure Drop Due to Potential Energy Change
ΔpPE accounts for the pressure change due to the
weight of the column of fluid (the hydrostatic
head); it will be zero for flow in a horizontal
pipe. From Equation (3-8). the potential energy
pressure drop is given by
g
p = z (3.10)
gc
Where,
Δz is the difference in elevation between
positions 1 and 2, with z increasing upward.
θ is defined as the angle between
horizontal and the direction of flow. Thus, θ is
+90° for upward, vertical flow, 0° for horizontal
flow, and -90° for downward flow in a vertical
well (Figure 3-4).
ΔpPE, the Pressure Drop Due to Potential Energy Change
For flow in a straight pipe of
length L with flow direction θ,
z = z2 − z1 = L sin  (3.11)

Example 3.2
Calculation of the Potential Energy Pressure Drop
Suppose that 1000 bbl/d of brine (𝛾w = 1.05) is being injected through 2 7/8-in., 8.6-
lbm/ft tubing in a well that is deviated 50° from vertical. Calculate the pressure drop over
1000 ft of tubing due to the potential energy change.

Combining Equations (3-17) and (3-18).
g
p =  L sin  (3.12)
gc
For downward flow in a well deviated 50° from vertical, the flow direction is -40° from
horizontal, so 3 is -40°. Converting to oilfield units, ρ = (1.05) (62.4) lbm/ft3 = 65.5
lbm/ft3 and ΔpPE = -292 psi from Equation (3-19).

For fresh water with 𝛾w = 1 (ρ = 62.4 lbm/ft3), the potential energy pressure drop per
foot of vertical distance is
dp g  lb f   lbm   1 ft 2 
=  = 1   62.4 3   2 
= 0.433 psi/ft (3.13)
dz g c  lbm   ft  144 in 
For a fluid of any other specific gravity,
dp
= 0.433 w (3.14)
dz
where 𝛾w is the specific gravity. Thus,
pPE = 0.433 w z (3.15)
For the example given, 𝛾w = 1.05 and Δz = L sinθ, so ΔpPE = 0.433 𝛾w L sinθ = -292 psi.

ΔpKE, the Pressure Drop Due to Kinetic Energy Change
ΔpKE is the pressure drop resulting from a change in the velocity of the fluid between
positions 1 and 2. It will be zero for an incompressible fluid unless the cross-sectional
area of the pipe is different at the two positions of interest. From Equation (3-8),

pKE = u 2 (3.16)
2 gc
or

pKE =
2 gc
( 2 1)
u 2
− u 2
(3.17)

If the fluid is incompressible, the volumetric flow rate is constant. The velocity then
varies only with the cross-sectional area of the pipe. Thus,
q
u= (3.18)
A
and since A = πD2/4, then
4q
u= (3.19)
D 2
Combining Equations (3-17) and (3-19), the kinetic energy pressure drop due to a pipe
diameter change for an incompressible fluid is
8 q 2  1 1  (3.20)
pKE = 2  4− 4
 g c  D2 D1 

Example 3.3
Calculation of the Kinetic Energy Pressure Drop
Suppose that 2000 bbl/d of oil with a density of 58 lbm/ft3 is flowing through a
horizontal pipeline having a diameter reduction from 4 in. to 2 in., as illustrated in
Figure. Calculate the kinetic energy pressure drop caused by the diameter change.

First, the volumetric flow rate must be converted to ft3/sec: q = 0.130 ft3/sec
From Equation (3.20)
8 q 2  1 1 
pKE = 2  4− 4
 g c  D2 D1 

For oilfield units of bbl/d for flow rate, lbm/ft3 for density, and in. for diameter, the
constants in Equation (3-20) and unit conversions can be combined to yield
 1 1  (3.21)
pKE = 1.53 10  q  4 − 4 
−8 2
 D2 D1 
where q is in bbl/d, ρ in lbm/ft3, and D in in.

ΔpF, The Frictional Pressure Drop
The frictional pressure drop is obtained from the Fanning equation,
2 f f u 2 L
pF = (3.22)
gc D
where ff is the Fanning friction factor.
In laminar flow, the friction factor is a simple function of the Reynolds number,
16 (3.23)
ff =
N Re

In turbulent flow, the friction factor may depend on both the Reynolds number and
the relative pipe roughness, ε.
k
= (3.24)
D
where
k is the length of the protrusions on the pipe wall.
D the pipe diameter.

The relative roughness of some common types of
pipes are given in Figure (Moody, 1944). However, it
should be remembered that the pipe roughness may
change with service, so that the relative roughness is
essentially an empirical parameter that can be
obtained through pressure-drop measurements.

The Fanning friction factor
is most commonly obtained
from the Moody friction
factor chart (Figure; Moody,
1944). This chart was
generated from the
Colebrook-White equation,
1   1.2613 
= −4 log  + 
ff  3.7065 N Re f f 
 
(3.25)

The Colebrook-White equation is implicit in ff, requiring an iterative procedure, such as
the Newton-Raphson method, for solution. An explicit equation for the friction factor
with similar accuracy to the Colebrook-White equation (Gregory and Fogarasi, 1985) is
the Chen equation (Chen, 1979):
1 
  5.0452   1.1098
 7.149 
0.8981
 
= −4 log  − log  +   (3.26)
ff  3.7065 N Re  2.8257  N Re   

Example 3.4
Calculating the Frictional Pressure Drop
Calculate the frictional pressure drop for the 1000 bbl/d of brine injection described in
Example 3-2. The brine has a viscosity of 1.2 cp, and the pipe relative roughness is 0.001.

First, the Reynolds number must be calculated to determine if the flow is laminar or
turbulent.
1.48q 
N Re =
D
Note that the oilfield units used here are obviously not consistent; however, the constant
1.48 converts the units to a consistent set. The Reynolds number is well above 2100, so
the flow is turbulent.

Either the Moody diagram or the Chen equation can be used to determine the friction
factor. Using the Chen equation,
1 
  5.0452   1.1098
 7.149 
0.8981
 
= −4 log  − log  +  
ff  3.7065 N Re  2.8257  N Re   
Now using Equation (3-31), and noting that 2 7/8-in., 8.6-lbm/ft tubing has an I.D. of
2.259 in.,
q 4q
u= =
A  D2
then,
2 f f u 2 L
pF =
gc D

Single-Phase Flow of a Compressible, Newtonian Fluid
To calculate the pressure drop in a gas well, the compressibility of the fluid must be
considered. When the fluid is compressible, the fluid density and fluid velocity vary
along the pipe, and these variations must be included when integrating the
mechanical energy balance equation.
To derive an equation for the pressure drop in a gas well, we begin with the mechanical
energy balance, Equation (3-14). With no shaft work device and neglecting for the time
being any kinetic energy changes, the equation simplifies to
dp g 2 f f u 2 dL
+ dz + =0 (3.27)
 gc gc D
Since dz is sinθdL, the last two terms can be combined as
dp  g 2 f u 2

+  sin  + f
 dL = 0 (3.28)
  g c g c D 

From the real gas law, the density is expressed as
MWp
= (3.29)
ZRT
or, in terms of gas gravity,
28.97 g p
= (3.30)
ZRT
The velocity can be written in terms of the volumetric flow rate at standard conditions,
q,
4  T   psc 
u= qZ     (3.31)
D 2
 Tsc   p 

Then, substituting for ρ and u from Equations (3-30) and (3-31), Equation (3-28)
becomes
ZRT  g 32 f f  T
2
  psc   
dp +  sin  + 2    qZ   dL = 0 (3.32)
28.97 g p  gc  gc D5  Tsc  p   
 
This equation still contains three variables that are functions of position: Z, the
compressibility factor, temperature, and pressure.
To solve Equation (3-32) rigorously, the temperature profiles can be provided and the
compressibility factor replaced by a function of temperature and pressure using an
equation of state. This approach will likely require numerical integration.

Alternatively, single, average values of temperature and compressibility factor over
the segment of pipe of interest can be assumed.
If the temperature varies linearly between upstream position 1 and downstream
position 2, the average temperature can be estimated as the mean temperature (T1 +
T2)/2 or the log-mean temperature (Bradley. 1987), given by
T2 − T1
T1m = (3.33)
ln(T2 / T1 )

An estimate of the average compressibility factor, 𝑍,ҧ can be obtained as a function of
average temperature, 𝑇,ത and the known pressure, p1.
Once the pressure, p2, has been calculated, 𝑍ҧ can be checked using 𝑇ത and the mean
pressure, (p1 + p2)/2. If the new estimate differs significantly, the pressure calculation
can be repeated using a new estimate of 𝑍.ҧ

Using average values of Z and T, Equation (3-32) can be integrated for nonhorizontal
flow to yield
 ZTqpsc
32 f f  s
p =e p + 2
2 s 2
  (e − 1) (3.34)
2
 g c D sin   Tsc
15

where s is defined as
−(2)(28.97) g ( g / g c ) sin  L
s= (3.35)
ZRT

For the special case of horizontal flow, sin θ and s are zero; integration of Equation (3-
32) gives
(64)(28.97) g f f ZT  qpsc 
2
p −p =
2 2
  L (3.36)
1 2
 gc D R
2 5
 Tsc 
To complete the calculation, the friction factor must be obtained from the Reynolds
number and the pipe roughness. Since the product, ρq, is a constant for flow of a
compressible fluid, NRe can be calculated based on standard conditions as
4(28.97) g qpsc
N Re = (3.37)
 D  RTsc
The viscosity should be evaluated at the average temperature and pressure.

The constants and conversion factors for oilfield units for Equations (3.48) through (3-
51) can be combined to give:
For vertical flow or inclined flow:
f f ( ZTq ) 2
p22 = e s p12 + 2.685 10−3 (e s − 1) (3.38)
D sin 
5
where
−0.0375 g sin  L
s= (3.39)
ZT

For horizontal flow:
−4
 g f f ZTq L2
p − p = 1.007 10
2
1
2
2 5
(3.40)
D
 gq
N Re = 20.09 (3.41)
D

Frequently, in production operations, the unknown pressure may be the upstream
pressure, p1.
For example, in a gas production well, in calculating the bottomhole pressure from the
surface pressure, the upstream pressure is the unknown. Rearranging Equation (3-38)
to solve for p1, we have
f f ( ZTq ) 2
p12 = e − s p22 − 2.658 10−3 (1 − e − s ) (3.42)
D sin 
5

Equations (7-51) through (7-56) are the working equations for computing the pressure
drop in gas wells. Remember that these equations are based on the use of an average
temperature, compressibility factor, and viscosity over the pipe segment of interest.
The longer the flow distance, the larger will be the error due to this approximation. It is
advantageous to divide the well into multiple segments and calculate the pressure drop
for each segment if the length (well measured depth) is large. We have also neglected
changes in kinetic energy to develop these equations, even though we know that
velocity will be changing throughout the pipe. The kinetic energy pressure drop can
be checked after using these equations to estimate the pressure drop and corrections
made, if necessary.

Example 3.5
Calculation of the Bottomhole Flowing Pressure in a Gas Well
Suppose that 2 MMSCF/d of natural gas is being produced through 10,000 ft of 2 7/8-in.
tubing in a vertical well. At the surface, the temperature is 150°F and the pressure is 800
psia; the bottomhole temperature is 200°F. The gas has the composition given in
Example 4-3. and the relative roughness of the tubing is 0.0006 (a common value used
for new tubing).
Calculate the bottomhole flowing pressure directly from the surface pressure. Repeat the
calculation, but this time dividing the well into two equal segments. Show that the
kinetic energy pressure drop is negligible.

Example 3.5 - Summary
q = 2 MMSCF/d
L = 10,000 ft
D = 2 7/8-in
T2 = 150°F
p2 = 800 psia
T1 = 200°F
ε = 0.0006

From Example 4-3. Tpc is 374oR, ppc is 717 psia, and 𝛾g is 0.709. Using the mean
temperature, 175°F, the pseudoreduced temperature is Tpr = (175 + 460)/374 = 1.70;
and using the known pressure at the surface to approximate the average pressure, ppr =
800/717 = 1.12.
From Figure 4-1, 𝑍ҧ = 0.935. Following Example 4-4, the gas viscosity is estimated: from
Figure 4-4, μ1 atm = 0.012 cp; from Figure 4-5, μ/μ1 atm = 1.07, and therefore μ = (0.012
cp)(1.07) = 0.013 cp

The Reynolds number is, from Equation
 gq (0.709)(2000)
N Re = 20.09 = 20.09 = 9.70 105
D (2.259)(0.013)
and ε = 0.0006, so, from the Moody diagram (Figure 7-7), ff = 0.0044. Since the flow
direction is vertical upward, θ = +90°.
Now, using Equation
−0.0375 g sin  L −0.0375(0.709)(sin 90o )(10, 000)
s= = = −0.448
ZT (0.935)(635)
The bottomhole pressure is calculated from Equation:
2
f ( ZTq )
p12 = e − s p22 − 2.658 10−3 f 5 (1 − e − s )
D sin 
2
(0.0044)[(0.935)(635)(2000)]
e0.448 (800)2 − 2.658 10−3 5 o
(1 − e 0.448
) p1 = pwf = 1078 psi
2.259 sin 90
The well is next divided into two equal segments and the calculation of bottomhole
pressure repeated.
The first segment is from the surface to a depth of 5000 ft.
For this segment, 𝑇ത is 162.5°F, Tpr is 1.66, and ppr is 1.12 as before. From Figure 4-1, 𝑍ҧ =
0.93. The viscosity is essentially the same as before, 0.0131cp. Thus, the Reynolds
number and friction factor will be the same as in the previous calculation.
−0.0375 g sin  L −0.0375(0.709)(sin 90o )(5, 000)
s= = = −0.2296
ZT (0.935)(622.5)
f f ( ZTq ) 2
p12 = e − s p22 − 2.658 10−3 (1 − e − s )
D5 sin 
2
−3 (0.0044)[(0.935)(622.5)(2000)]
e 0.2296
(800) − 2.658 10
2
5 o
(1 − e 0.2296
)
2.259 sin 90
p5000 = 935 psi
For the second segment, from a depth of 5000 ft to the bottomhole depth of 10,000 ft,
we use 𝑇ത is 187.5°F and p = 935 psia. Thus, Tpr = 1.73, ppr = 1.30, and from Figure 4-1,
𝑍ҧ = 0.935. Viscosity is again 0.0131 cp. So, for this segment,
−0.0375 g sin  L −0.0375(0.709)(sin 90o )(5, 000)
s= = = −0.2196
ZT (0.935)(647.5)
f f ( ZTq ) 2
p12 = e − s p22 − 2.658 10−3 (1 − e − s )
D5 sin 
2
(0.0044)[(0.935)(647.5)(2000)]
e0.2196 (935) 2 − 2.658 10−3 5 o
(1 − e 0.2196
)
2.259 sin 90
p1 = pwf = 1078 psia
Since neither temperature nor pressure varied greatly throughout the well, little
error resulted from using average T and Z for the entire well.
It is not likely that kinetic energy changes are significant, also because of the small
changes in temperature and pressure, but this can be checked.
The kinetic energy pressure drop in this well can be estimated by
 
pKE =
2 gc
u =
2
2 gc
( 2 1)
u 2
− u 2
This calculation is approximate, since an average density is being used. The velocities at
points 1 and 2 are
Z1 ( psc / p1 )(T1 / Tsc )q Z 2 ( psc / p2 )(T2 / Tsc )q
u1 = u2 =
A A
and the average density is
28.97 g p
=
ZRT

At position 2 (surface), T is 150°F, Tpr = 1.63, p = 800 psia, ppr = 1.12, and Z = 0.925;
while at 1 (bottomhole), T = 200°F, Tpr = 1.76, p = 1078 psia, ppr = 1.50, and Z = 0.93. For
2 7/8-in., 8.6-lbm/ft tubing, the I.D. is 2.259 in., so the cross-sectional area is 0.0278 ft2.
To calculate average densities, the average pressure, 939 psia, average temperature,
175°F, and average compressibility factor, 0.93, are used. We then calculate ΔpKE = 0.06
psia.
The kinetic energy pressure drop is negligible compared with the potential energy and
frictional contributions to the overall pressure drop.

Multiphase Flow in Wells
Multiphase flow - the simultaneous flow of two or more phases of fluid - will occur in
almost all oil production wells, in many gas production wells, and in some types of
injection wells.
In an oil well, whenever the pressure drops below the bubble point, gas will evolve,
and from that point to the surface, gas–liquid flow will occur. Thus, even in a well
producing from an undersaturated reservoir, unless the surface pressure is above the
bubble point, two-phase flow will occur in the wellbore and/or tubing. Many oil wells
also produce significant amounts of water, resulting in oil–water flow or oil–water–gas
three-phase flow.
Two-phase flow behavior depends strongly on the distribution of the phases in the pipe,
which in turn depends on the direction of flow relative to the gravitational field.

Multiphase Flow in Wells
▪ Holdup Behavior
▪ Two-Phase Flow Regimes
▪ Two-Phase Pressure Gradient Models

Holdup Behavior
In two-phase flow, the amount of the pipe
occupied by a phase is often different from its
proportion of the total volumetric flow rate. As
an example of a typical two-phase flow situation,
consider the upward flow of two phases, α and
β, where α is less dense than β, as shown in
Figure.

Holdup Behavior
Typically, in upward two-phase flow, the lighter
phase (α) will be moving faster than the denser
phase (β). Because of this fact, called the holdup
phenomenon, the in-situ volume fraction of the
denser phase will be greater than the input
volume fraction of the denser phase - that is,
the denser phase is “held up” in the pipe
relative to the lighter phase.

Holdup Behavior
This relationship is quantified by defining a
parameter called holdup, y, as
V
y = (3.43)
V
where
Vβ = volume of denser phase in pipe segment and
V = volume of pipe segment.

Holdup Behavior
The holdup, yβ, can also be defined in terms of a
local holdup, yβl, as
A
1
y =  y l dA (3.44)
A0
The local holdup, yβl, is a time-averaged quantity
- that is, yβl is the fraction of the time a given
location in the pipe is occupied by phase β

Holdup Behavior
The holdup of the lighter phase, yα, is defined
identically to yβ as
V
y = (3.45)
V
or, because the pipe is completely occupied by
the two phases,
y = 1 − y (3.46)
In gas–liquid flow, the holdup of the gas phase,
yα, is sometimes called the void fraction.

Holdup Behavior
Another parameter used in describing two-phase
flow is the input fraction of each phase, λ,
defined as
q
 = (3.47)
q + q
and
 = 1 −  (3.48)
where qα and qβ are the volumetric flow rates of
the two phases. The input volume fractions, λα
and λβ, are also referred to as the “no-slip
holdups.”

Holdup Behavior
Another measure of the holdup phenomenon
that is commonly used in production log
interpretation is the “slip velocity,” us. Slip
velocity is defined as the difference between the
average velocities of the two phases. Thus,
us = u − u (3.49)
where 𝑢ത α and 𝑢ത𝛽 the average in-situ velocities
of the two phases.
q
 = (3.50)
q + q
 = 1 −  (3.51)

Holdup Behavior
Slip velocity is not an independent property
from holdup, but is simply another way to
represent the holdup phenomenon. In order to
show the relationship between holdup and slip
velocity, we introduce the definition of
superficial velocity, usα or usβ, defined as
q
us = (3.52)
A
and
q
us = (3.53)
A

Holdup Behavior
The superficial velocity of a phase would be the
average velocity of the phase if that phase
filled the entire pipe - that is, if it were single-
phase flow. In two-phase flow, the superficial
velocity is not a real velocity that physically
occurs, but simply a convenient parameter.

Holdup Behavior
The average in-situ velocities 𝑢ത α and 𝑢ത𝛽 are
related to the superficial velocities and the
holdup by
us
us = (3.54)
y
and
us
us = (3.55)
y

Holdup Behavior
Substituting these expressions into the equation
defining slip velocity (7-74) yields
1  q q 
us =  −  (3.56)

A  1 − y y 
Correlations for holdup are generally used in
two-phase pressure gradient calculations; the
slip velocity is usually used to represent holdup
behavior in production log interpretation.

Example 3.6
Relationship between Holdup and Slip Velocity
If the slip velocity for a gas–liquid flow is 60 ft/min and the superficial velocity of each
phase is also 60 ft/min, what is the holdup of each phase?

Example 3.6 - Summary
us = usg = usl = 60 ft/min

From Equation (7-79), since superficial velocity of a phase is q / A,
1  q q  usg usl
us =  − = −
A  1 − y y  1 − yl yl
Solving for yl, a quadratic equation is obtained:
us yl2 − (us − usg − usl ) yl − usl = 0
For us = usg = usl = 60 ft/min, using Solver gives yl = 0.62.
The holdup of the gas phase is then yg = 1 – yl = 0.38. The holdup of the liquid is greater
than the input fraction (0.5), as is typical in upward gas–liquid flow.

Two-Phase Flow Regimes
The manner in which the two phases are distributed in the pipe significantly affects
other aspects of two-phase flow, such as slippage between phases and the pressure
gradient.
The “flow regime” or flow pattern is a qualitative description of the phase
distribution.

In gas-liquid, vertical, upward flow, four flow regimes are now generally agreed upon
in the two-phase flow literature: bubble, slug, chum, and annular flow.
A brief description of these flow regimes is as follows.
1. Bubble flow: Dispersed bubbles of gas in a continuous liquid phase.
2. Slug flow: At higher gas rates, the bubbles coalesce into larger bubbles, called
Taylor bubbles, that eventually fill the entire pipe cross section. Between the large gas
bubbles are slugs of liquid that contain smaller bubbles of gas entrained in the liquid.
3. Chum flow: With a further increase in gas rate, the larger gas bubbles become
unstable and collapse, resulting in chum flow, a highly turbulent flow pattern with
both phases dispersed. Chum flow is characterized by oscillatory, up-and-down motion
of the liquid.
4. Annular flow: At very high gas rates, gas becomes the continuous phase, with
liquid flowing in an annulus coating the surface of the pipe and with liquid droplets
entrained in the gas phase.

The flow regime in gas-liquid vertical
flow can be predicted with a flow
regime map, a plot relating flow
regime to flow rates of each phase, fluid
properties, and pipe size. One such map
that is used for flow regime
discrimination in some pressure-drop
correlations is that of Duns and Ros
(1963), shown in Figure.

The Duns and Ros map correlates flow
regime with two dimensionless
numbers, the liquid and gas velocity
numbers, Nvl and Nvg, defined as
l
N vl = usl 4 (3.57)
g
and
l
N vg = usg 4 (3.58)
g

where
ρl is liquid density,
g is the acceleration of gravity, and
σ is the interfacial tension of the
liquid-gas system.
This flow pattern map does account for
some fluid properties; note, however,
that for a given gas-liquid system, the
only variables in the dimensionless
groups are the superficial velocities of
the phrases.

Duns and Ros defined three distinct
regions on their map, but also included
a transition region where the flow
changes from a liquid-continuous to a
gas-continuous system.
Region I contains bubble and low-
velocity slug flow,
Region II is high-velocity slug and
chum flow, and
Region III contains the annular flow
pattern.

A flow regime map that is based on a
theoretical analysis of the flow regime
transitions is that of Taitel, Bamea,
and Dukler (1980). This map must be
generated for particular gas and
liquid properties and for a particular
pipe size; a Taitel-Dukler map for air-
water flow in a 2-in. I.D. pipe is shown
in Figure.

This map identifies five possible flow
regimes: bubble, dispersed bubble (a
bubble regime in which the bubbles are
small enough that no slippage occurs),
slug, chum, and annular.

The slug/chum transition is significantly
different than that of other flow regime
maps in that chum flow is thought to be
an entry phenomenon leading to slug
flow in the Taitel-Dukler theory.

The D lines show how many pipe
diameters from the pipe entrance chum
flow is expected to occur before slug flow
develops. For example, if the flow
conditions fell on the D line labeled LE/D
= 100, for a distance of 100 pipe
diameters from the pipe entrance, chum
flow is predicted to occur; beyond this
distance slug flow is the predicted flow
regime.

Example 3.7
Predicting Two-Phase Flow Regime
200 bbl/d of water and 10,000 fit3/day of ah are flowing in a 2-in. vertical pipe. The
water density is 62.4 lbm/ft3 and the surface tension is 74 dynes/cm. Predict the flow
regime that will occur using the Duns-Ros and the Taitel-Dukler flow regime maps.

First, the superficial velocities are calculated as
ql (200 bbl/d)(5.615 ft 3 /bbl)(1 d/86,400 s)
usl = = 2
= 0.6 ft/s = 0.18 m/s
A 0.02182 ft
qg (10, 000 ft 3 /d)(1 d/86,400 s)
usg = = 2
= 5.3 ft/s = 1.62 m/s
A 0.02182 ft

For the Duns and Ros map, the liquid and gas velocity numbers must be calculated. For
units of ft/s for superficial velocity, lbm/ft3 for density, and dynes/cm for surface
tension, these are
l
N vl = usl 4
g
l
N vg = usg 4
g
Using the physical properties and flow rates given, we find Nvl = 1.11 and Nvg = 9.8.

Referring to Figure. the flow conditions
fall in region 2; the predicted flow
regime is high-velocity slug or chum
flow.

Using the Taitel-Dukler map, the flow
regime is also predicted to be slug or
chum, with LE/D of about 150. Thus, the
Taitel-Dukler map predicts that chum
flow will occur for the first 150 pipe
diameters from the entrance; beyond
this position, slug flow is predicted.

Two-Phase Pressure Gradient Models
A differential form of the mechanical energy balance equation is
dp  dp   dp   dp  (3.59)
=   +  + 
dz  dz  PE  dz  KE  dz  F
In most two-phase flow correlations, the potential energy pressure gradient is based on
the in-situ average density, ρത
 dp  g (3.60)
  =  sin 
 dz  PE g c
where
 = (1 − yl )  g + yl l (3.61)
Various definitions of the two-phase average velocity, viscosity, and friction factor are
used in the different correlations to calculate the kinetic energy and frictional pressure
gradients.

There are many different correlations that have been developed to calculate gas-liquid
pressure gradients, ranging from simple empirical models to complex deterministic
models.

▪ The Modified Hagedorn and Brown Method
▪ Flow Regimes other than Bubble Flow: The Original Hagedorn-Brown Correlation
▪ Bubble Flow: The Griffith Correlation
▪ The Beggs and Brill Method
▪ The Gray Correlation

The Modified Hagedorn and Brown Method
The modified Hagedorn and Brown method (mH-B) is an empirical two-phase flow
correlation based on the original work of Hagedorn and Brown (1965).
The heart of the Hagedorn-Brown method is a correlation for liquid holdup; the
modifications of the original method include:
• using the no-slip holdup when the original empirical correlation predicts a liquid
holdup value less than the no-slip holdup and
• the use of the Griffith correlation (Griffith and Wallis. 1961) for the bubble flow
regime.

The Modified Hagedorn and Brown Method
These correlations are selected based on the flow regime as follows. Bubble flow exists if
λg < LB, where
 um2 
LB = 1.071 − 0.2218   (3.62)
D
and LB ≥ 0.13.
Thus, if the calculated value of LB is less than 0.13, LB is set to 0.13. If the flow regime
is found to be bubble flow, the Griffith correlation is used; otherwise, the original
Hagedorn-Brown correlation is used.

Flow Regimes other than Bubble Flow: The Original Hagedorn-Brown Correlation
The form of the mechanical energy balance equation used in the Hagedorn-Brown
correlation is dp g 2 f  um2 (um2 / 2 g c )
= + + (3.63)
dz gc gc D z
which can be expressed in oilfield units as
dp fm 2 (um2 / 2 g c )
144 =+ + (3.64)
where dz (7.413 10 D ) 
10 5
z
f is the friction factor,
𝑚ሶ is the total mass flow rate (lbm/d),
ρത is the insitu average density [Equation (7-90)] (lbm/ft3),
D is the diameter (ft),
um is the mixture velocity (ft/sec), and
the pressure gradient is in psi/ft.

The mixture velocity used in H-B is the sum of the superficial velocities,
um = usl + usg (3.65)
To calculate the pressure gradient with Equation (3-93). the liquid holdup is obtained
from a correlation and the friction factor is based on a mixture Reynolds number. The
liquid holdup, and hence, the average density, is obtained from a series of charts using
the following dimensionless numbers.

In field units, these are
• Liquid velocity number
N vl = 1.938usl 4 l /  (3.66)
• Gas velocity number
N vg = 1.938usg 4 l /  (3.67)
• Pipe diameter number
N D = 120.872 D l /  (3.68)
• Liquid viscosity number
1
N L = 0.15726l 4 (3.69)
 l 3
where superficial velocities are in ft/sec, density is in lbm/ft3, surface tension in

dynes/cm, viscosity in cp, and diameter in ft.

The holdup is obtained from Figures 7-12 through 7-14 or calculated from equations
that fit the correlation curves presented on the charts (Brown, 1977).
First, CNL is read from Figure 7-12 or calculated by
(3.70)

Then the group
(3.71)
is calculated; from Figure 7-13,

we get yl/ψ, or it is calculated by
(3.72)

Here p is the absolute pressure at
the location where pressure
gradient is wanted, and pa is
atmospheric pressure. Finally,
compute
(3.73)
and read ψ from Figure 7-14 or

calculate it with
(3.74)

The liquid holdup is then
(3.75)
The mixture density is then calculated from

(3.76)
The frictional pressure gradient is based on a Fanning friction factor using a mixture
Reynolds number, defined as
(3.77)
where mass flow rate, 𝑚ሶ is in lbm/day, D is in ft, and viscosities are in cp. The friction
factor is then obtained from the Moody diagram orcalculated with the Chen equation for
the calculated Reynolds number and the pipe relative roughness.

The kinetic energy pressure drop will in most instances be negligible; it is calculated
from the difference in velocity over a finite distance of pipe, Δz.

Bubble Flow: The Griffith Correlation
The Griffith correlation uses a different holdup correlation, bases the frictional pressure
gradient on the in-situ average liquid velocity, and neglects the kinetic energy pressure
gradient. For this correlation,
(3.78)
where ul is the in-situ average liquid velocity, defined as

(3.79)

Bubble Flow: The Griffith Correlation
For field units, Equation (7-111) is
(3.80)
where ml is the mass flow of the liquid only. The liquid holdup is
(3.81)
where us = 0.8 ft/sec. The Reynolds number used to obtain the friction factor is based on
the in-situ average liquid velocity,
(3.82)

Example 3.8
Pressure Gradient Calculation Using the Modified Hagedorn and Brown Method
Suppose that 2000 bbl/d of oil (ρ = 0.8 g/cm3, μ = 2 cp) and 1 MMSCF/d of gas of the
same composition as in Example 7-5 are flowing in 2 7/8-in. tubing. The surface tubing
pressure is 800 psia and the temperature is 175°F.The oil–gas surface tension is 30
dynes/cm, and the pipe relative roughness is 0.0006. Calculate the pressure gradient at
the top of the tubing, neglecting any kinetic energy contribution to the pressure
gradient,

From Example 3-5, we have μg = 0.0131 cp and Z = 0.935. Converting volumetric flow
rates to superficial velocities with A = (π/4)(2.259/12)2 = 0.0278 ft2
The gas superficial velocity can be calculated from the volumetric flow rate at standard
conditions with Equation (7-45)

The mixture velocity is
and the input fraction of gas is
First, we check whether the flow regime is bubble flow. Using Equation
but LB must be ≥ 0.13, so LB = 0.13. Since λg (0.65) is greater than LB, the flow regime is
not bubble flow and we proceed with the Hagedorn-Brown correlation.

We next compute the dimensionless numbers, Nvl, Nvg, ND, and NL. Using Equations (7-
99) through (7-102), we find Nvl = 10.28, Nvg = 19.20, ND = 29.35, NL = 9.26 × 10–3. Now,
we determine liquid holdup, yl, from Figures 7-12 through 7-14 or Equations 7-103
through 7-107. From Figure 7-12 or Equation 7-103, CNL = 0.0022. Then
and, from Figure 7-13 or Equation 7-105, yl/ψ = 0.46. Finally, we calculate
and from Figure 7-14 or Equation 7-107, ψ = 1.0. Note than ψ will generally be 1.0 for
low-viscosity liquids. The liquid holdup is thus 0.46. This is compared with the input
liquid fraction, λl, which in this case is 0.35. If yl is less than λl, yl is set to λl.

Next, we calculate the two-phase Reynolds number using Equation (7-110). The mass
flow rate is
The gas density is calculated from Equation (7-44)
so
and

From Figure 7-7 or Equation (7-35), f = 0.0046. The in-situ average density is
Finally, from Equation (7-93).

The Beggs and Brill Method
The Beggs and Bill (1973) correlation differs significantly from that of Hagedorn and
Brown in that the Beggs and Brill correlation is applicable to any pipe inclination and
flow direction.
This method is based on the flow regime that would occur if the pipe were horizontal;
corrections are then made to account for the change in holdup behavior with
inclination.
It should be kept in mind that the flow regime determined as part of this correlation is
the flow regime that would occur if the pipe were perfectly horizontal and is probably
not the actual flow regime that occurs at any other angle.
The Beggs and Brill method is the recommended technique for any wellbore that is not
near vertical.

The Beggs and Brill method uses the general mechanical energy balance and the in-situ
average density to calculate the pressure gradient and is based on the following
parameters:

The horizontal flow regimes used as correlating parameters in the
Beggs-Brill method are segregated, transition, intermittent, and
distributed. The flow regime transitions are given by the following.
Segregated flow exists if
Transition flow occurs when
Intermittent flow exists when
Distributed flow occurs if

The same equations are used to calculate the liquid holdup, and hence the average
density, for the segregated, intermittent, and distributed flow regimes. These are
with the constraint that ylo ≥ λl and
where

where a, b, c, d, e, f, and g depend on the flow regime and are given in Table 7-2. C must
be ≥ 0

If the flow regime is transition flow, the liquid holdup is calculated using both the
segregated and intermittent equations and interpolated using the following:
where
and

The frictional pressure gradient is calculated from
where
and

The no-slip friction factor, fn, is based on the actual pipe relative roughness and the
Reynolds number,
for ρm in lbm/ft3, um in ft/s, D in in., and μm in cp, and where

The two-phase friction factor, ftp, is then
where
and
Since S is unbounded in the interval 1 < x < 1.2, for this interval

The kinetic energy contribution to the pressure gradient is accounted for with a
parameter Ek as follows:
where

In comparisons with extensive measurements with natural gas and water flowing in
inclined schedule 40 2-inch I.D. pipe, Payne et al. [1979] found that the Beggs and Brill
correlation underpredicted friction factors and overpredicted liquid holdup.
To correct these errors, Payne et al. suggest that the friction factor be calculated
incorporating pipe roughness (the original correlation assumed smooth pipe), and
found the following holdup corrections improved the correlation.
Denoting the liquid holdup calculated by the original correlation as ylo, the corrected
liquid holdup is
or

Example 3.9
Pressure Gradient Calculation Using the Beggs and Brill Method
Repeat Example 3-8, using the Beggs and Brill method.
Suppose that 2000 bbl/d of oil (ρ = 0.8 g/cm3, μ = 2 cp) and 1 MMSCF/d of gas of the
same composition as in Example 7-5 are flowing in 2 7/8-in. tubing. The surface tubing
pressure is 800 psia and the temperature is 175°F. The oil–gas surface tension is 30
dynes/cm, and the pipe relative roughness is 0.0006. Calculate the pressure gradient at
the top of the tubing, neglecting any kinetic energy contribution to the pressure
gradient,

The Gray Correlation
The Gray correlation was developed specifically for wet gas wells and is commonly used
for gas wells producing free water and/or condensate with the gas.
This correlation empirically calculates liquid holdup to compute the potential energy
gradient and empirically calculates an effective pipe roughness to determine the
frictional pressure gradient.

First, three dimensionless numbers are calculated:
where

The liquid holdup correlation is
where
The potential energy pressure gradient is then calculated using the in-situ average
density.

Example 3.10
Pressure Gradient Calculation Using the Gray Method
An Appendix C gas well is producing 2 MMscf/day of gas with 50 bbl of water produced
per MMscf of gas. The surface tubing pressure is 200 psia and the temperature is 100°F.
The gas-water surface tension is 60 dynes/cm, and the pipe relative roughness is 0.0006.
Calculate the pressure gradient at the top of the tubing, neglecting any kinetic energy
contribution to the pressure gradient, At this location, the water density is 65 lbm/ft3
and the viscosity is 0.6 cp.

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Vietnam National University - Ho Chi Minh City

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