part 2 gas behavior and hydrostatics

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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• Most blowouts result from human failure

• Perhaps advanced well control training and

education can further improve the statistics


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• 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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• The way to prevent failures:

– proper training

– responsible engineering and planning

– adequate equipment

– prudently executed operations


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• 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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• 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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• 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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• 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

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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

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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

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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

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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

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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

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• 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

part 2 gas behavior and hydrostatics

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