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Drilling Engineering 1 Course (2nd Ed.)

1. Drillstring Equipment
A. Special Tools
2. Connections; Make–up and Break–out
3. Other Drillstring Equipment

1. Hydraulic Introduction
A. Hydrostatic Pressure
a. Calculation for Incompressible Fluids
b. Calculation for Compressible Fluids
c. Buoyancy
B. Drillstring Design
a. Length of Drill Collars; Neutral Point Calculation

hydrostatic state
By definition,
a static fluid cannot sustain shear stresses, otherwise it
will enter in motion (and will not be static anymore).
The consequence of this is that the state of stress
inside a fluid is such that the normal stresses are
the same in any direction.
This state of stress is called hydrostatic state of stress.
The magnitude of the stress is called pressure.
Spring14 H. AlamiNia Drilling Engineering 1 Course (2nd Ed.) 5

hydrostatic pressure
The hydrostatic pressure
inside a homogeneous fluid
comes from the pressure at
the surface and
the weight of the fluid above
the point in question.
To calculate the pressure at
any point inside a column
of fluid of density ρ
(gas or liquid),
we consider an infinitesimal
element of fluid
with volume dV = dx · dy · dz
as shown in the Figure.
Stress state about a point in a fluid

pressure differential
To remain in equilibrium, the resulting forces acting
in all 3 directions must be zero. Then we have:
The horizontal gradients (dpx/dx and dpy/dy) are zero.
Since in general we will be dealing with depth D, and
since a point at depth D has coordinate z = −D,
the expression for the pressure differential in terms of depth is:

Density and specific weight
Note that ρ
is mass per volume, or specific mass, or density.
Therefore, ρ g
is weight per volume or specific weight,
usually noted as γ.
For a gravitational system of units as the British System,
the acceleration of gravity is equal to 1 G.
Such that: 1 lbm × 1 G = 1 lbf .
That is the magnitude of the force
is numerically equal to the magnitude of the mass.

density dependency on the pressure
For a scientific or non–gravitational system as
the International System (SI),
the acceleration of gravity is equal to 9.80665m/s2,
such that 1 kg × 9.80665m/s2 = 9.80665 N.
To integrate hydrostatic pressure differential,
we need to know how
the density depends on the pressure.
All fluids are compressible, but for some applications,
some fluids can be classified as incompressible.
Liquids, in general are incompressible
up to a considerably high pressures.
Gases are, in general, compressible.

Hydrostatic Pressure for
Incompressible Fluids
For incompressible fluids (liquids in general are in
this class), the density ρ is constant so:
which allows us to calculate the pressure at a point with
depth D1 if we know the pressure p0 at depth of D0.
For pressure in psi, density in lbm/gal, and depth in feet
we have:
For pressure in psi, density in lbm/gal, and depth in
meter we have:

Absolute, gauge and differential
pressure definitions
The zero point of an absolute reference
is the absence of all matter.
There is no pressure at absolute zero.
Absolute pressure may have
the letter "a" after the unit as in "psia".
On the other hand, a gauge pressure measures
pressure relative to the local atmosphere.
Changes in local atmospheric pressure occur
due to weather, altitude, and/or depth.
(standard atmospheric pressure = 14.696 psi)
Gauge pressure indications usually
use a "g" after the unit as in "psig".
For pressure differential,
it is common to use a "d" after the unit.

Complex Fluid Column
If the column of fluid is composed of several
segregated fluids with different densities
(complex fluid column),
we still can use the expression for hydrostatic pressure
above,
considering that at the top of a homogeneous column of
fluid acts a pressure resulting from the fluid above:
ρi is the density of the ith layer of fluid and Ti its thickness.

Equivalent Density
For a complex fluid column,
the equivalent fluid density at a given depth
is the density of a homogeneous fluid that would
cause the same hydrostatic pressure at that depth.
Note that the equivalent density
depends on the depth in consideration.
A general expression for the equivalent density
at depth D = ΣTi is

The density ρ of a compressible fluid
The density ρ of a compressible fluid
(gases in general are compressible) is not constant
and depends on the pressure and temperature.
In order to account for this, we must consider the
equation of state for real gases:
z is the real gas deviation factor of the real gas at
pressure p and temperature T.
Solving for ρ = m/V results in:

Calculating the pressure differential
Considering the expression
for the pressure differential we have:
Since z depends on p,
separating variables and integrating results in:
assuming the temperature is constant in the gas column.

Calculating the pressure
The pressure integral can
only be calculated if we
know how z depends on p.
This is normally very
complicated. For short
columns, z can be considered
constant and we can write:
where z1 is the
compressibility factor at
pressure p1 and T. A more
accurate approach is to use
an average value for z given
by:
where z2 is the deviation
factor for p2 calculated using
the expression above. Using
this new average value of the
compressibility factor, a new
pressure p2 is obtained from
and compared with the
previous one.
The process is repeated until
convergence is obtained.

Unit conversion
Values for the universal
gas constant R¯ for
various units are:
It is important to note
that pressure and
temperature must be
given in absolute scales,
as required by the gas
equation of state.
The absolute
temperatures are
normally the Rankine and
the Kelvin scales:

Archimedes principle of buoyancy
Archimedes principle of
buoyancy states that
the buoyant force exerted
on a body fully or partially
immersed in a fluid
is equal in magnitude
(and opposite in direction)
to the weight of the volume
of fluid which is displaced
by that body.
For homogeneous bodies
immersed in
homogeneous fluids,
the net or buoyed weight of
the body
can be calculated from
W is the weight of the body
(in the air),
ρf is the density of the fluid,
ρb is the density of the
body.
The term is called
buoyancy factor.
This expression is valid only
for homogeneous bodies
fully immersed in
homogeneous fluids.

homogeneous bodies vs.
non–homogeneous bodies
For homogeneous bodies,
the geometric center of the
body coincides
with the center of mass.
For non–homogeneous
bodies,
an equivalent density (total
mass/volume) can be used,
but it is important
to keep in mind that
the geometric center (where
the buoyant force applies)
may not coincide with the
center of mass.
In these cases, stable or
instable equilibrium may exist.
If the body is either
not totally submerged, or
submerged in a
inhomogeneous fluid,
the expression above cannot
be used.
This may be complicated for
complex shape bodies.
A more general way to
calculate the buoyed weight,
even for partially immersed
bodies and for complex fluid
column is to calculate
the effect of the hydrostatic
pressure on the body.

drillstring design consideration
The drillstring must be designed
to support (with a safety margin)
all the static and dynamic loadings
that occur during normal and special operations.
also must support some extreme situations like pipe
sticking problems, curved holes, harsh environment, etc.
In addition,
it must be able to provide a suitable conduit for
the drilling fluid flow
without causing excessive frictional pressure drop
both inside and outside the pipe.

apply weight on the bit
One of the purposes of the drillstring
is to apply weight on the bit,
this is obtained by
slacking part of the drillstring weight on the bit.
When this is done,
a portion of the lower end of the drillstring
will be put in compression,
and the upper portion will remain in tension.
Since drill pipe can not be compressed,
a sufficient length of drill collars
and/or heavy weight drill pipes should be used so that
the required weight on bit can be applied
without compressing the drill pipes.

compression
The reason drill pipes
can not be put in compression is that
the moment of inertia of drill pipes are small compared
to its length (a parameter called slenderness =
length/radius of gyration).
On the other hand,
the slenderness of the drill collars are relatively low and
compression is allowed.
The slenderness is an important parameter to determine
the mechanical buckling resistance of a column.
The cause of buckling
is the moment created by compressive forces.

Constituents parameters of buckle the
drillstring
One of the considerations
was that both
the compressive force due
to the weight on bit and
the compressive force due
to the hydrostatic pressure
of the fluid
would contribute to buckle
the drillstring,
since the pressure acting
in the lower area of the
drillstring causes
a substantial compressive
force at the bit,
even with no weight
applied against the rock.
Based on this assumption,
the position of the neutral
point of stress can be
determined,
and a sufficient length of
drill collars must be used
such that
the neutral point lays
in the drill collars
(neutral point occurs
where the axial stress is
equal to zero.)

position of NP (Assuming: pressure
contributes to buckling)
Considering the element of
length x in the lower end of the
column.
FT is the axial force at acting at
the top section of the element,
W is the weight of the element of
length x,
p is the pressure at depth D,
A is the cross section of the
column, and
Fb is the reaction of the force
applied to at the bit.

contributes to buckling) (Cont.)
Substituting in the expression
above we have for FT:
 𝐹 𝑇 = 𝑊 − 𝑝𝐴 − 𝐹𝑏 = 𝜌𝑠 𝑔𝐴𝑥
− 𝜌 𝑓 𝑔𝐷𝐴 − 𝐹𝑏 = 𝑔𝐴(𝜌𝑠 𝑥
− 𝜌 𝑓 𝐷) − 𝐹𝑏
The stress at x is calculated
 dividing FT by the area A:
The neutral line is the position
xn where σ = 0, that is:
Solving for xn we get:
 wc = ρs g A is the linear weight
of the drill collar.
 Note that even if the weight on
bit is zero (drillstring hanging off
the bottom), this expression
tells us that the neutral point is
quite above the bit, and that the
lower end of the drillstring will
be under compression.

(pressure contributes to
buckle the drillstring)
Calculate the position of the neutral point for
a column 10,000 ft long
hanging off bottom and submerged
in a borehole filled
with a 9.3 ppg (pound per gallon) fluid.
What is the length of drill collar with 147 lb/ft
is required to apply 100,000 lbf on the bit,
assuming that, for safety,
only 85% of the total length of DC is compressed.
Calculate the suspended weight of the drillstring
using nominal 19.5 lb/ft DP (actual 22.28 lb/ft).

(pressure contributes to
buckle the drillstring)
Off bottom:
On bottom:
Since this length should be 85% of the total,
the minimum length of DC’s is
The hook load (buoyed weight) of the drillstring is:

doesn't contributes to buckling)
The fact, however, is that
the neutral point position
calculated as above is
calculated assuming that
the forces created by
hydrostatic pressure
will contribute to
the buckling of the column.
But this is not true.

doesn't contributes to buckling) (Cont)
If we recall the Archimedes’s effect,
the resultant force due to hydrostatic pressures
is equal in magnitude and opposite in direction
to the weight of the displaced fluid.
But that is not all.
In addition to that,
the resulting moment of the hydrostatic pressure
must be ZERO, otherwise,
the fluid would rotate about its center of mass.
• This means that
the hydrostatic pressure acting on an immersed column
can not contribute to buckle it.
The easiest (and correct) way to figure out the required length
to apply a weight on bit is exactly doing this, that is,
calculating the length of a column whose buoyed weight is equal to
the required weight on bit.

Considering the diagram we have:
wDC is the linear weight of the drill collar.
Note that the denominator corresponds
to the buoyed linear weight of the drill collar.

(pressure doesn’t
Calculate the position of the neutral point for
a column 10,000 ft long
hanging off bottom and submerged
in a borehole filled
with a 9.3 ppg (pound per gallon) fluid.
What is the length of drill collar with 147 lb/ft
is required to apply 100,000 lbf on the bit,
assuming that, for safety,
only 85% of the total length of DC is compressed.
Calculate the suspended weight of the drillstring
using nominal 19.5 lb/ft DP (actual 22.28 lb/ft).

(pressure doesn’t
Off bottom:
The neutral point position is given by
For the off bottom condition the weight on bit Fb = 0
and the neutral point is at xn = 0 ft.
For a weight on bit of
100,000 lbf we have:
Since this length should be 85% of the total,
the minimum length of DC’s is
The hook load (buoyed weight) of the drillstring is:

The length calculated with previous expression
determines the neutral point of buckling.
Above this point the drillstring will not buckle.
Bellow this point, which depends on the weight on bit,
the drillstring might be buckled or not, and this will
depend, among other factors, on the weight on bit and
on the polar moment of inertia of the drill collars.
A more general expression for the neutral point
position includes the possibility that the fluids
inside and outside the pipe have different densities:

1. Jorge H.B. Sampaio Jr. “Drilling Engineering
Fundamentals.” Master of Petroleum
Engineering. Curtin University of Technology,
2007. Chapter 4 and 5

1. Drillstring design
A. Maximum Tensile Force
a. tapered drill pipe
B. Maximum Torque
C. Internal (Burst) and External (Collapse) Pressures

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