7b. other abnormal pressure detection methods
TRANSCRIPT
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TAMU - PemexWell Control
Lesson 7BOther Abnormal Pressure
Detection Methods
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Contents
Moores Equation - Drilling rate Gas in the Drilling Fluid
Rock Sample Characteristics
Use of Surge and Swab Pressure todetermine Overbalance
Changes in Drilling Fluid Properties Temperature Indications
Hole Conditions
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Moores Equation
Moore proposed a practical method for
maintaining pore pressure overbalance
while drilling into a transition.
If drilling parameters are kept constant
while drilling into an abnormal pressurezone, the drilling rate will increase.
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Moores Equation
Moore suggests that we increase themud weight sufficiently to keep the
drilling rate from increasing.
The increase in mud weight will then be
a measure of the abnormal pore
pressure.
But how much do we increase the mud
weight?
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Moores
Equation
Transition zoneBegin weighting up
Weight up complete
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Ex.
2.10
?
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Example 2.10
Bit parameters prior to transition were
Bit Weight = 4,700 lbf/in
Rotary Speed = 80 rpm
Transition detected at 9,100 ft and theoperator immediately reduced the bit
weight to 2,900 lbf/in
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Example 2.10
Determine the extrapolated normal
penetration rate at a depth of 9,250 ft
if the bit weight is reduced from itscurrent value of 4,700 to 2,900 lbf.
Use the data in Fig. 2.46 andMoores penetration rate model.
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Solution
The extrapolatednormal penetrationrate at 9,250 is 15.7ft/hr, at 4,700 lbfbitweight.
This would have been
the target rate had thebit weight remainedconstant.
15.7
Fig. 2.46
9,250
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Solution contdThe target penetration rate at the reduced bit
weight of 2,900 lbf is calculated below:
Na
b
Nd
WKR
=
Na
R = 8080700,4 900,27.15
The target rate would revert back to 15.7 ft/hr if
the operator resumes drilling at 4,700 lbf/in.
23.2.' EqsMoore
ft/hr9.7=R(assumes R W)
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Example 2.11 - Fig. 2.46
At 8,300 ft (under normal conditions);
increase the ECD from 9.6 to 10.1 ppg.
In response, the drilling rate decreases from
20.5 ft/hr to 18.5 ft/hr
What is the shale compaction coefficient, c?
2211 loglog RRcc 24.2.Eq
How much should we increase the mud weight? (Moore)
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Solution
ECD changes from 1= 9.6 to 2= 10.1 ppgCalculate c, the shale compaction coefficient
(9.6)clog 20.5 = (10.1)
clog 18.5
(10.1/9.6)c= log 20.5 / log 18.5 = 1.035
c * log 1.052 = log 1.035
c = 0.679
2211 loglog RRcc
Now use Eq. 2.24 to calculate therequired change in mud weight
c
R
R
1
2
112
log
log
2
1
1
2
log
log
R
Rc =
47.1
2
112
log
log
R
R
gal
lb1.10
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Example 2.12
At 9,090 ft the normal
penetration rate is 16.5ft/hr
Actual penetration rate is
18.0 ft/hr, using a mud
weight of 9.6 ppg
Normal MW = 8.3 ppg
16.5
Fig. 2.46
9,090
47.1
2
112log
log
R
R
ppg0.105.16log
0.18log6.9
47.1
2 =
=
p= 8.3 + 0.4 = 8.7 ppg = (10.0 - 9.6) = 0.4 ppg
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Other predictors of
abnormal pressure
Drilling rate is not the only available
predictor of abnormal pressure.
Properties of shale cuttings can be
used:
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Shale density
Transition
OffshoreNigeria
_shaledensityfrom Boatman
n o - g/cm3Density - g/cm3
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Example
2.15
n =2.54
o = 2.44
pp_14,000 = ?
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Solution
At 14,000, n = 2.54 and o = 2.44 g/cm3
so, = 0.1 g/cm3
From Fig. 2.48:
p14,000 = 0.052*14.6*14,000
p14,000 = 10,629 psig
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shale densityfrom Boatman
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Shale density measurement
1. Fill a standard API
mud balance with
shale cuttings (wash
and dry with a towel)until balance reads
8.33 ppg.
2. Fill the cup to top
with water and
record reading
(e.g. 13.3 ppg).
8.33
Calculate S.G. of shale
cuttings:
S.G. = 8.33/(16.66 - 13.3)
S.G. = 2.48
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Possible Sources
of Gas in aDrilling Fluid
Drilled gas,
Produced gas
Recycled gas
Contamination gas
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Drilled gas, cuttings gas, or liberated gas refers to
gas released from rock cuttings generated by the bit.
Usually small volumes. Increasing MW will not help.
Produced gas refers to gas which enters the wellbore
from the walls of the hole. Increasing MW will reducethe quantity.
Recycled gas is any wellbore gas that remains in the
mud after at least one pass through the surface
equipment.
Contamination gas is gas released from any volatile
hydrocarbons intentionally added to the system (mud
additives).
Possible Sources of Gas in a Drilling Fluid
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Gas in Mud
Connection gas - gas that has enteredthe wellbore when pumps are shut down
to make a connection, can be detected in
a gas trap.
Trip gas - gas that entered the wellbore
during a trip; can also be detected.
Background gas - gas baseline
concentration in the mud usually small.
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Example 2.17
Determine the density of the gas-cutmud returns from a well at a depth of 2 ft
below flowline outlet if:
Clean MW = 12.0 ppg
Flowline MW = 7.0 ppg
Atmospheric press = 14.7 psia
Sample temperature = 100 deg F
Gas gravity = 0.6
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Solution At the Surface
From Eq. 1.22,
g = g * p/(2.77 * Z * T)
g,surface = 0.6*14.7/(2.77*1*560)
g,surface = 0.00569 ppg
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Solution At the Surface
From Eq. 2.26,
gm = m (1 - fg) + g fg
fg = ( m - gm)/ ( m - g)
= (12.0 7.0/(12.0 - 0.00569) = 0.417
This is the gas fraction at the surface, but
fg varies with depth.
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Solution contd
By definition, fg = Vg /(1+Vg)
so, Vg = fg / (1- fg), but, pV = ZnRT
n = fg *p / [ ZRT(1- fg )]
n = 0.000234 lb-moles/gal of mud
This parameter stays constant with depth
provided the downhole gas entry rate
remains constant.
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Assuming the density of the mud-gasmixture does not change appreciably
over two ft of depth.
p2ft = 14.7 + 0.052 * 7.0 * 2
= 15.43 psia
g,2ft = 0.6 * 15.43 / (2.77 * 1 * 560)
= .00597 lbm/gal
Solution 2 ft down
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fg,2ft = [(1 * 0.000234 * 80.275 * 560) / 15.43]
[1+(1 * 0.000234 * 80.275 * 560) / 15.43]
fg,2ft
= 0.405 (down from 0.417 at the surface)
2ft = 12 * (1 - 0.405) + 0.00597 * 0.405
2ft = 7.14 ppg
This is an increase of 0.14 ppg in just 2.
See Fig. 2.51 for plot of entire range
Example 2.17 contd
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Clearly most of the
gas expansion is
near the top of thewellbore.
At 10,000,
MW = 11.9+ ppg.What is the
resulting reduction
in BHP due to the
gas?
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Example 2.18
What is the total change in HSP at the
bottom of the well described in Ex. 2.17?
Average temperature is 150 deg F.
From Eq. 2.28
( ) +
= ssgm
ssg
____
sgreduc
p
ppln
TZf1
TZpfp
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In the annulus, without, gas
BHP = 12 * 10,000 / 19.25
= 6,233.8 psig
BHP = 6,248 psia
Average pressure = (14.7 + 6,248) / 2
= 3,131 psia
From Fig. 1.6, Zavg = 0.868
Example 2.18 contd
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Example 2.18
contd
If pgm = 6,248 psia, then
pred = 60 psi
pgm = 6,248 - 60 = 6,188 psia
EMW = (6,188 14.7) / (0.052 * 10,000)= 11.87 ppg
( ) + 7.14 7.14248,6ln560*1*417.01 610*868.0*7.14*417.0 sreducp
( )
+
=s
sgm
ssg
____
sg
reducp
ppln
TZf1
TZpfp
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Gas cut mud
A second iteration is generally notnecessary if the assumed value for pgm is
reasonably close to the calculated
value.
Furthermore, adding gas to a drillingfluid will increase viscosity, so the
annular friction drop will increase,partially off-setting any reduction in BHPdue to gas.
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Gas cut mud
Another factor that will tend to offset the
reduction in mud density is drilled
cuttings.
At a moderate to high drilling rate, the
quantity of cuttings present in the mud
at any time, may be significant.
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Gas in mud
Gas in mud is monitored as the mudexits the flowline. A gas trap is placed
to sample the gas before the mud
passes over the shale shaker.
The gas concentration is recorded in
arbitrary gas units.
Look for relative changes.
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Gas detection unit
Gas detectorlocated in
the shale shakers
possum belly.
BBG = Background gas
This is the baseline gas
concentration in the mud,
and is usually in the orderof a few gas units.
CG = Connection Gas
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CG = constant
BGG = constant
Overbalanced
CG increases
BGG increases
Underbalanced
CG increases
BGG constant
?
CG increases
BGG increases
Transition zone
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Measuring Surge Pressure
Swab pressure is
hard to measure, but
surge is not.
Run one stand of pipein hole at constant
velocity.
Repeat at different
velocities.
Plot surge pressure
vs. pipe velocity.
Mud Level
Flowline
Closed Safety Valve
Pressure Recorder Sub
Drillpipe
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Measuring Surge Pressure
By assuming
surge = swab,
we can predict
the swabpressure at
different pipe
pulling speeds.
Surge/S
wabPressur e
,psi
Pipe Velocity, ft/sec
E l
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Example
67 sec/std
59 sec/std
48 sec/std
452 min-units
1,036 min-units
2,132 min-units
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Example
Estimate the pore pressure at TD if
MW = 11.7 ppg
The length of each stand is 90 ft.
V1 = 90 ft / 48 sec = 1.88 ft/sec
V2 = 90 / 59 = 1.53 ft/sec
V3 = 90 / 67 = 1.34 ft/sec
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Example
From Figure:
p1 = 405 psi
p2 = 300 psi
p3 = 242 psi
Surge/S
wab
Pressur e
,psi
Pipe Velocity, ft/sec
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Example
From Figure, plotof gas units vs.swab pressure,
when line isextrapolated tozero velocity (zerogas), overbalance
is found to be 197psi
197
0
Pressure, psi
GasUnits
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Example
With an overbalance of 197 psi:
Pore pressure = MW - (overbalance)
0.052 * TD
Pore pressure = 11.7 - (197 / 0.052 * 13,600)
pp = 11.4 ppg.
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Changes in drilling fluid properties
Gas in mud
reduced density
increased viscosity
Salt water inflow chloride content
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Salt water inflow
Chloride content
Flocculation of sodium bentonite clay increases yield point
increases gel strength
increases water loss
poor filter cake
pH change
Changes in drilling fluid properties
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Drilled rock salt can have similar effect
CO2 and H2S may reduce pH
H2S is very poisonous and is corrosive
Raise pH and precipitate out any solublesulfides using scavengers.
Changes in drilling fluid properties
Temperature and abnormal press
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Temperature and abnormal press.
Geothermaltemperature
vs. depth
Undercompacted rockLower thermal conductivity
Rock conducts heat better
than pore fluid
Poor conductivityrequires higher
temperature
gradient to maintain
constant heat flux.
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Temperature indicators
Temperature gradient tends to increasewithin a pore pressure transition
Rock grains have a much higherthermal conductivity than pore fluids
Well planning predictions may beassisted by downhole temperaturemeasurements in offset wells
T t i di t
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Temperature indicatorsNote that wellbore
circulation of fluids will
distort the true
temperature profile.
The drilling fluid
temperature increases asthe fluid moves down the
drillpipe.
As fluid enters the annulus
its temperature increasesfor a short while.
Higher up the annulus
temperature decreases
Fl li t t f N th S ll
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Flowline temperature from a North Sea well
Predictable increase in
temperature of mud returns as
depth increases
A deviation from
the normal
temperature trendmay signal
abnormal pore
pressure
Important tool if
no shales are
present
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Hole Conditions
Drilling torque when rotating pipe, and
drag during trips or connections, result
from friction between the drillstring or bitand the walls of the hole.
Torque and drag (T&D) will generally
increase with depth, gradually.
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Hole Conditions
A sudden increase in T&D may becaused by hole instability.
Circulate bottoms up and observesamples.
If abnormal pressure caused anincrease in T&D, the rock samples willhelp to tell the story.
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Sample Shale Cuttings
Normally pressured shales Abnormally pressured shales