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8/19/2019 Part 2 Gas Behavior and Hydrostatics
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8/19/2019 Part 2 Gas Behavior and Hydrostatics
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Contents
• Why Study Well Control?
• Types of influx
• Ideal Gases
• Real Gases
• Critical Temperature & Pressure
• Pseudo-Critical Temp. and Press.
• Gas Compressibility
• Problems
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4
Why study well control?
• Well control fundamentals are quite well
known and understood
• Individuals involved in drilling operations
have, in general, received well control training
• Yet, well control problems, and blowouts
occur – with casualties
– with environmental damage
– at high cost (often in $millions/occurrence)
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5
Why study well control?
• Most blowouts result from human failure
• Perhaps advanced well control training and
education can further improve the statistics
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Why study well control?
• Well owners, oil field workers, and regulatoryauthorities are becoming increasingly
intolerant of human error relative to well
operations
• At times unconventional well control
procedures are necessary in order to avoid
blowouts
• We can all learn from the mistakes made in
the past to help avoid problems in the future
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Why study well control?
• The way to prevent failures:
– proper training
– responsible engineering and planning
– adequate equipment
– prudently executed operations
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Why study well control?
• Advanced well control can offer the largest
impact in the following areas:
– proper engineering design of wells, such asproper casing setting depths and proper materials
– operational planning, and
– the execution of the drilling process
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Why study well control?
• Costs may be higher in the short term, but
future profits will not be spent cleaning up
and litigating past mistakes
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Why study well control?
• Influx into wellbore may be gas, oil, and/or
water
• All well control methods:
– maintain a constant BHP
– consider the behavior of gas under changing
wellbore conditions
– are designed to move gas up a wellbore to thesurface – whenever possible
– must allow gas, if present, to expand
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Why study well control?
• Different well control methods may result in
different wellbore pressures
•Accurate pressure predictions requireknowledge of the influx composition,
temperature, and pressure
• Influx phase changes can and do occur in theprocess of killing a well
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Pressure-temperature phase diagram
for a pure substance
Temperature
Pr
e
s
su
r
e
Solid Liquid Gas
C
Tc
Pc Melting Point Curve
Vapor Pressure Curve
Critical
Temperature
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Some Definitions
The reduced pressure of a pure gas is the ratio of the
gas pressure to the critical pressure of the gas, p/pc
The reduced temperature of a pure gas is the ratio of the
gas temperature to its critical temperature, T/Tc
The critical temperature of a gas is the highesttemperature at which a fluid can exist as a liquid or vapor.
Above this temperature the fluid is a gas, at any pressure.
The critical pressure is the pressure needed to
condense a vapor at its critical temperature
Use absolute units, e.g.,oR and psia
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Types of Influx• Influx into wellbore may be gas, oil, and/or water
• All well control methods:
• Maintain a constant BHP
• Consider the behavior of gas under changing wellbore conditions
• Are designed to move gas up a wellbore to the surface – whenever
possible• Must allow gas, if present, to expand
• Different well control methods may result in different wellbore
pressures
•Accurate pressure predictions require knowledge of the influxcomposition, temperature, and pressure
• Influx phase changes can and do occur in the process of killing a well
14
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Pressure-temperature Phase Diagram for A
Pure Substance
15
Temperature
P r e s s u r e
Solid Liquid Gas
C
Tc
Pc Melting Point Curve
Vapor Pressure Curve
Critical
Temperature
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Some Definitions
16
• The reduced pressure of a pure gas is the ratio of the
gas pressure to the critical pressure of the gas, p/pc
• The reduced temperature of a pure gas is the ratio of
the gas temperature to its critical temperature, T/Tc
• The critical temperature of a gas is the highest
temperature at which a fluid can exist as a liquid orvapor. Above this temperature the fluid is a gas, at any
pressure.
• The critical pressure is the pressure needed to
condense a vapor at its critical temperature
Use absolute units, e.g.,oR and psia
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Physical Properties of Natural Gas
Constituents
17
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Typical Phase Diagram for Mixtures
18
Bubble point curve
Dew point curve
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Ideal Gases
• Boyle’s Law:
• Charles Law:
• Ideal Gas Law:
or
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1 1 2 2
1 2
constant p V p V
T T
1 2
1 2
constant p pT T
1 1 2 2 constant pV p V
nRTpV
constantT
constantV
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Universal Gas Constant Values
p V T n R
psia ft3 °R lbm mole 10.732
psia gal °R lbm mole 80.275
psia bbl °R lbm mole 1.911
kPa m3 °K g mole 0.0083145
kPa m3 °K kg mole 8.3145
20
pV=ZnRT
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Problem 1
• A 20 bbl gas influx has entered a well atbottom hole pressure of 3,500 psia.
Determine the gas volume when the kick
exits the well.
a. Assume atmospheric pressure of 14.4 psia
and no change in the gas temperature.
b. Assume initial gas temperature of 150 oF
and surface temperature of 65oF.
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Solution
(a) Using Boyle’s law:
2
112p
VpV
22
2211 VpVp
psia4.14
bbl20*psia3,500V2
V2 = 4,861 bbl (243x expansion!)
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Solution
(b) Using the Ideal Gas law:
V2 = 4,148 bbl
Note: If a real change in temperature is ignored (in this
example) the predicted volume is high by approx. 17%
2
22
1
11
TVp
TVp
460)(150*14.4
)46065(*20*500,3
Tp
TVpV
12
2112
(207x expansion!)
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Problem 2
• What is the density of the gas from the
previous example if it contains 90% methane
and 10% ethane.
a. Under bottomhole conditions?
a. Under the specified atmospheric conditions?
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Solution
• Weighted molecular weight:
MWgas = 0.9 * 16.0 + 0.1 * 30.1 = 17.41
• Gas Specific gravity:a
g
gMW
MW
25
600.029
41.17 g
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Solution
(a) Under bottomhole conditions the gas density
(assume Z = 1):
26
)460150(28.80*1
)500,3)(600.0(2929,
ZRT
pgbottomg
ppg24.1bottom,g
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Solution
(b) Under the specified atmosphericconditions:
ZRT
pgg
29
27
)46065(28.80*1
)4.14)(600.0(29,
surfaceg
ppg00594.0surface,g
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Properties of H-C gases
Gas Mol.
Wt.
Specific
gravity
Critical
Temp F
Critical
Press psia
Methane, CH4 16.0 0.55 343 668
Ethane, C2H6 30.1 1.04 550 708Propane, C3H8 44.1 1.52 666 616
n-Butane, C4H10 58.1 2.00 765 551
Nitrogen, N2 28,0 0.97 227 439
Carbon Dioxide, CO4 44.0 1.25 448 1071
Hydrogen Sulfide, H2S 34.1 1.18 673 1306
Water, H2O 18.0 0.62 1166 3208
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Real Gases
• The Equation of State (EOS) for a non-ideal gas is:
pV = ZnRT
• The Z-factor, or compressibility factor, is anempirical adjustment for the non-ideal behavior
of a real gas
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Real Gases
• Z, the compressibility factor, is 1 at atmosphericconditions, decreases as the pressure increases (min.value ~ 0.25) and then increases again, reaching avalue of 1 or more at pseudo reduced pressures in
excess of 9.
• At low temperatures and a pseudo-reduced pressurein excess of 25, the value of Z can be as high as 2.0,or even higher .
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Problem 3
•Repeat Problem 2 taking into considerationthe variation in Z-factor with changes in
temperature and pressure.
• From Problem 2, g = 0.600
• From Fig. 1.5, the Pseudo-critical pressure,
Ppc
= 671 psig
• and the pseudo-critical temperature
Tpc = 358oR
32
Fi 1 5
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33Gas Specific Gravity (air = 1)
T p c
( o R )
p p c (
p s i a )
671
358
Fig. 1.5
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Problem 3
• The psuedo-reduced pressure,
ppr = p / ppc
At Bottomhole conditions,ppr = 3,500 / 671 = 5.22
At the surface,
ppr = 14.7 / 671 = .022
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The pseudo-reduced pressure of a gas mixture is the ratio p/ppc
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Problem 3
• The psuedo-reduced Temperature,
Tpr = T / Tpc
• At Bottomhole conditions,
Tpr = 610 / 358 = 1.70
• At the surface,
Tpr = 525 / 358 = 1.47
The pseudo-reduced temperature is the ratio T/Tpc
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Problem 3
• The Z-factors can now be determined.
• Under bottomhole conditions,
Z = 0.886
• Under surface conditions,
Z = .995 ~ 1
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Bottomhole
Surface
0.995
0.886
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Determination of Z-factor
• If a computer is available, Z factors can be
calculated:
ppr = 756.8 - 131g - 3.6 g2
Tpr = 169.2 + 349.5 g – 74 g2
• Z can be taken from chart or calculated onspreadsheet
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Determination of Z-factor
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Problem 3
)610)(28.80)(886.0()500,3)(6.0(29
29,
ZRTPgbottomg
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At bottomhole conditions, the density of
the gas is:
This is 13% above the value obtained
for an ideal gas
ppgbottomg 4.1,
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Problem 3
•Under surface conditions, with a Z - factornear 1, the density is still ~ 0.0059 ppg.
• Note: At a pressure of 10,000 psia andtemperature of 200 OF
ppr = 10,000 / 671 = 14.9
Tpr = 660 / 358 = 1.84Z = 1.41 and g = 2.33 ppg
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Problem 4
• A 12,000’ vertical well is shut in
with a single-phase, 0.6 gravity gas
influx on bottom.
• SICP = 500 psia. The initial influxheight is determined to be 400 ft.
Mud density = 11.5 ppg.
• Determine the BHP if
BHT = 205 deg F
42
500 psi
400 ft
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Solution
• The pressure at the top of the kick is:
p = SICP + HSPmud
= 500 + 0.052*11.5*(12,000-400)
p = 7,437 psia
• pr = p/ppr = 7,437/671 = 11.08, and
• Tr = T/Tpr = 665/358 = 1.86
• Z = 1.195 from Fig
43
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Solution
)665()27.80()195.1(
)437,7()6.0(29
ZRT
p29 gg
44
BHP = 7,437 + 0.052 * 2.03 * 400
= 7,479 psia
ppgbottomg 03.2,
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Problem 5
• Consider the same well. What would the
SICP be if all the drilling fluid had been
unloaded from the hole prior to shut-in?
• Assume BHP = 7,479 psia as calculated in
problem 4. Also assume that the average
wellbore temperature is 160 deg F.
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Solution
• Solve by trial and error.
• First assume that Z = 1
46
73.1358/620T
05.10671/2
009,6479,7p
psia009,6p
p243.1ep479,7
epp
pr
pr
o
o
)620)(0.1(3.53
)0000,12(6.0
o
ZT3.53
)DD(
o
0g
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Solution
Now, Z = 1.140, Then:
48
)620)(14.1(3.53)0000,12(6.0
oep479,7
Close enough
psia173,6po
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6,173 psia
12,000’
7,472 psia
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Problem 6
• For the same well, determine the equivalent
density at depths of 6,000’ and 12,000’.
• Assume the average temperature from the
surface to 6,000’ is 120oF for the case where
the hole is filled with gas.
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Solution
• At TD,
equiv = (7,479-14.4) / (0.052*12,000)
= 12.0 ppg
51
• Recall that: p = 0.052 * MW * Depth
so, MW = p / (0.052 * Depth)
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Solution
• At 6,000’
p = 500 + 0.052*11.5*6,000 = 4,048 psia
equiv = (4,048-14.4)/(0.052*6,000)
= 13.0 ppg
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Solution
• What is the equivalent density at 1,000’?
p = 500 + 0.052*11.5*1,000 = 1,098 psia
equiv = (1,098-14.4)/(0.052*1,000)
= 20.8 ppg
• Note how the equivalent density increases asdepth decreases.
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End