PETE 625 Well Control

1
PETE 625Well Control

• Lesson 2
• Gas Behavior and Hydrostatics

2
Contents

• Why Study Well Control?
• Ideal Gases
• Real Gases
• Critical Temperature Pressure
• Pseudo-Critical Temp. and Press.
• Gas Compressibility
• Problems

3
Assignments

• Homework 1
• Ch 1, Problems 1.1-1.10
• due
• Homework 2
• Ch 1, Problems 1.11-1.21
• Read All of Chapter 1

4
(No Transcript)
5
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

6
Why study well control?

• Yet, well control problems, and blowouts occur
• with casualties
• with environmental damage
• at high cost (often in millions/occurrence)

7
Why study well control?

• Most blowouts result from human failure
• Perhaps advanced well control training and
  education can further improve the statistics

8
Why study well control?

• 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

9
Why study well control?

• Well owners, oil field workers, and regulatory
  authorities are becoming increasingly intolerant
  of human error relative to well operations

10
Why study well control?

• The way to prevent failures
• proper training
• responsible engineering and planning
• adequate equipment
• prudently executed operations

11
Why study well control?

• Advanced well control can offer the largest
  impact in the following areas
• proper engineering design of wells, such as
  proper casing setting depths and proper
  materials
• operational planning, and
• the execution of the drilling process

12
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

13
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 the
  surface whenever possible
• must allow gas, if present, to expand

14
Why study well control?

• Different well control methods may result in
  different wellbore pressures
• Accurate pressure predictions require knowledge
  of the influx composition, temperature, and
  pressure
• Influx phase changes can and do occur in the
  process of killing a well

15
Pressure-temperature phase diagram for a pure
substance
C
Pc
Melting Point Curve
Pressure
Gas
Solid
Liquid
Vapor Pressure Curve
Temperature
Tc
Critical Temperature
16
Some Definitions
The critical temperature of a gas is the highest
temperature 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
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
Use absolute units, e.g., oR and psia
17
Physical Properties of Natural Gas Constituents
18
Universal Gas Constant Values

• p V T n R
• psia ft3 oR lbm mole 10.732
• psia gal oR lbm mole 80.275
• psia bbl oR lbm mole 1.911
• kPa m3 oK g mole 0.0083145
• kPa m3 oK kg mole 8.3145

pVZnRT
19
Typical phase diagram for mixtures
Bubble point curve
Dew point curve
20
Ideal Gases

• Boyles Law
• Charles Law
• Ideal Gas Law
• or

21
Problem 1

• A 20 bbl gas influx has entered a well at
  bottomhole 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 65 oF.

22
Solution

• (a) Using Boyles law

V2 4,861 bbl
(243x expansion!)
23
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
(207x expansion!)
24
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?
• (b) Under the specified atmospheric conditions?

25
Solution

• Weighted molecular weight
• MWgas 0.9 16.0 0.1 30.1 (Eq. 1.1)
• 17.41
• Gas Specific gravity

26
Solution

• (a) Under bottomhole conditions the gas density
  (assume Z 1)

27
Solution

• (b) Under the specified atmospheric conditions

28
Properties of H-C gases

• Mol. Specific Critical Critical
• Wt. gravity Temp Press
• oR psia
• Methane, CH4 16.0 0.55 343 668
• Ethane, C2H6 30.1 1.04 550 708
• Propane, C3H8 44.1 1.52 666 616
• n-Butane, C4H10 58.1 2.00 765 551
• Nitrogen, N2 28.0 0.97 227 493
• Carbon Dioxide, CO4 44.0 1.52 548
  1,071
• Hydrogen Sulfide, H2S 34.1 1.18 673
  1,306
• Water, H2O 18.0 0.62 1,166 3,208

29
Real Gases

• The Equation of State, EOS for a non-ideal gas
  is
• pV ZnRT
• The Z-factor, or compressibility factor, is an
  empirical adjustment for the non-ideal behavior
  of a real gas

30
Real Gases

• Z, the compressibility factor, is 1 at
  atmospheric conditions, decreases as the pressure
  increases (min. value 0.25) and then increases
  again, reaching a value of 1 or more at pseudo
  reduced pressures in excess of 9.
• At low temperatures and a pseudo-reduced pressure
  in excess of 25, the value of Z can be as high as
  2.0, or even higher (Fig. 1.7).

31
Problem 3

• Repeat Problem 2 taking into consideration the
  variation in Z-factor with changes in temperature
  and pressure.
• From Problem 2, gg 0.600
• From Fig. 1.5, the Pseudo-critical pressure,
• Ppc 671 psig
• and the pseudo-critical temperature
• Tpc 358 oR

32
671
Tpc (oR) ppc (psia)
358
Fig. 1.5-Gas Specific Gravity (air 1)
33
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

The pseudo-reduced pressure of a gas mixture is
the ratio p/ppc
34
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
35
Problem 3

• From Fig. 2.6, the Z-factors can now be
  determined.
• Under bottomhole conditions,
• Z 0.886
• Under surface conditions,
• Z .995 1

36
Surface
0.995
0.886
Bottomhole
37
Determination of Z-factor

• If a computer is available, Z factors can be
  calculated
• ppr 756.8 - 131gg - 3.6 gg2
• Tpr 169.2 349.5 gg 74 gg2
• Z can be taken from chart or calculated on
  spreadsheet

38
Determination of Z-factor
39
Problem 3
At bottomhole conditions, the density of the gas
is
This is 13 above the value obtained for an ideal
gas
40
Problem 3

• Under surface conditions, with a Z - factor
  near 1, the density is still 0.0059 ppg.
• Note At a pressure of 10,000 psia and
  temperature of 200 OF
• ppr 10,000 / 671 14.9
• Tpr 660 / 358 1.84
• Z 1.41 and rg 2.33 ppg

41
Problem 4
500 psi

• A 12,000 vertical well is shut in with a
  single-phase, 0.6 gravity gas influx on bottom.
  SICP 500 psia. The initial influx height is
  determined to be 400 ft. Mud density 11.5 ppg.
  
• Determine the BHP if BHT 205 deg F

400 ft
42
Solution

• The pressure at the top of the kick is
• p SICP HSPmud
• 500 0.05211.5(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. 2

43
Solution
BHP 7,437 0.052 2.03 400
7,479 psia
44
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.

45
Solution

• Solve by trial and error.
• First assume that Z 1

46
Solution

• Thus, Z 1.132 Then

47
Solution

• Now, Z 1.140 Then

Close enough
48
6,173 psia
12,000 7,472 psia
49
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 120 oF for the case where the hole
  is filled with gas.

50
Solution

• Recall that p 0.052 MW Depth
• so, MW p / (0.052 Depth)

• At TD,
• requiv (7,479-14.4) / (0.05212,000)
• 12.0 ppg

51
Solution

• At 6,000
• p 500 0.05211.56,000 4,048 psia
• requiv (4,048-14.4)/(0.0526,000)
• 13.0 ppg

52
Solution

• What is the equivalent density at 1,000?
• p 500 0.05211.51,000 1,098 psia
• requiv (1,098-14.4)/(0.0521,000)
• 20.8 ppg
• Note how the equivalent density increases as
  depth decreases.

53
Equivalent Mud Weight