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1 GAS FIELD ENGINEERING Gas Well Performance

Chapter 6_Gas Well Performance

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Chapter 6_Gas Well Performance

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Page 1: Chapter 6_Gas Well Performance

1

GAS FIELD ENGINEERING

Gas Well Performance

Page 2: Chapter 6_Gas Well Performance

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CONTENTS

6.1 Gas Well Performance

6.2 Static Bottom-hole Pressure(static BHP)

6.3 Flowing Bottom-hole Pressure(flowing BHP)

Page 3: Chapter 6_Gas Well Performance

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LESSON LEARNING OUTCOME

At the end of the session, students should be able to:

• Determine static bottom-hole pressure(static BHP) using

different methods

• Determine flowing bottom-hole pressure(flowing BHP) using

different methods

Page 4: Chapter 6_Gas Well Performance

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Gas Well Performance

Figure (6.1) Gas Production Schematic

Page 5: Chapter 6_Gas Well Performance

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Gas Well Performance

• Referring to Fig.(6.1), ability of a gas reservoir to produce for a given set of reservoir conditions depends directly on the flowing bottom-hole pressure, Pwf.

• The ability of reservoir to deliver a certain quantity of gas depends on

• the inflow performance relationship

• flowing bottom-hole pressure

• Flowing bottom-hole pressure depends on

• Separator pressure

• Configuration of the piping system

Page 6: Chapter 6_Gas Well Performance

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Gas Well Performance

• These conditions can be expressed as:

(8.1)

(8.2)

Page 7: Chapter 6_Gas Well Performance

7 Figure (6.2) Deliverability test plot

•The static or flowing pressure at the formation must be

known in order to predict the productivity or absolute open flow

potential of gas wells.

•Preferred method is a bottom-hole pressure gauge (down-

hole pressure gauge).

•However, Static BHP or Flowing BHP can be estimated from

wellhead data (gas specific gravity, well head pressure, well head

temperature, formation temperature, and well depth.)

Static and Flowing Bottom-Hole Pressures

Page 8: Chapter 6_Gas Well Performance

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Basic Energy Equation

In the case of steady-state flow, energy balance can be

expressed as follows:

OR

(8.3)

(8.4)

Page 9: Chapter 6_Gas Well Performance

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Basic Energy Equation

Figure (6.3) Flow in pipe (After Aziz.)

Page 10: Chapter 6_Gas Well Performance

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Basic Energy Equation

• Second term ( ) kinetic energy is neglected in pipeline flow

calculations.

• If no mechanical work is done on the gas (compression) or by

the gas (expansion through a turbine), the term ws is zero.

• Reduced form of the mechanical energy equation may be

written as:

OR

cg

udu

2

(8.5)

(8.6)

Page 11: Chapter 6_Gas Well Performance

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Basic Energy Equation

•All equations now in use for gas flow and static head

calculations are various forms of this Equation.

•The density of a gas( )at a point in a vertical pipe at pressure

p and temperature T may be written as:

(8.7)

g

Page 12: Chapter 6_Gas Well Performance

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Fig.(6.4) Compressibility

factor for natural gases

for Ppr (0 to 10)

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Fig.(6.5) Compressibility

factor for natural gases

for Ppr (9 to 20)

Page 14: Chapter 6_Gas Well Performance

14 Fig.(6.6) Moody Friction Factor Chart

Page 15: Chapter 6_Gas Well Performance

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Basic Energy Equation

• The velocity of gas flow ug at a cross section of a vertical pipe

is

(8.8)

Page 16: Chapter 6_Gas Well Performance

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Basic Energy Equation

• General vertical flow equation assuming a constant average

temperature in the interval of interest is

(8.10)

Page 17: Chapter 6_Gas Well Performance

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Basic Energy Equation

•Sukker & Cornell and Poettmann assumed gas deviation

factor varies with pressure. But accurate in relatively shallow

wells.

•A more realistic approach is that of Cullender & Smith.

•They treated gas deviation factor as a function of both

temperature and pressure.

Page 18: Chapter 6_Gas Well Performance

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Static Bottom-Hole Pressure

Average Temperature and Deviation Factor Method

The Equation is:

(8.20)

Page 19: Chapter 6_Gas Well Performance

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Example (1)

Calculate the static bottom-hole pressure of a gas well having a

depth of 5790 ft. The gas gravity is 0.60 and the pressure at the

wellhead is 2300 psia. The average temperature of the flow

string is 117oF.

Solution

Page 20: Chapter 6_Gas Well Performance

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First trial

Second trial

Page 21: Chapter 6_Gas Well Performance

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QUIZZ 4

1. Calculate the static bottom-hole pressure of a gas

well having a depth of 8570 ft. The gas gravity is 0.63

and the pressure at the wellhead is 2800 psia. The

average temperature of the flow string is 124oF.Use

average Temperature and Deviation factor method.

Pc=672,Tc=358

2. Calculate the static bottom-hole pressure of a gas

well having a depth of 9230 ft. The gas gravity is 0.66

and the pressure at the wellhead is 3100 psia. The

average temperature of the flow string is 119oF. .Use

average Temperature and Deviation factor method.

Pc=672, Tc=358

Page 22: Chapter 6_Gas Well Performance

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Cullender and Smith Method

This is a more realistic approach that gas deviation factor is a

function of both temperature and pressure.

Define

(8.25)

(8.26)

Page 23: Chapter 6_Gas Well Performance

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Cullender and Smith Method

Which, for the static case, reduces to

For the upper half,

For the lower half,

Static bottom-hole pressure at depth Z in the well is finally given by

Where Its is evaluated at H = 0, Ims at Z/2 and Iws at Z.

(8.30)

(8.31)

(8.32)

(8.27)

(8.29)

Page 24: Chapter 6_Gas Well Performance

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Cullender and Smith Method Calculation procedure

First: to solve for an intermediate temperature and pressure

condition at the mid point of the vertical column;

Second: Repeat the calculations for bottom-hole condition.

-A value of Its is first calculated from Eqn 8.27 at surface

conditions.

-Then, Ims is assumed(Its=Ims at first approximation) and pms is

calculated for the mid point conditions.

-Using this value of Ims , a new value of Ims is computed.

-The new value of Ims is then used to recalculate pms .

-This procedure is repeated until successive calculations of pms

are within the desired accuracy (usually within 1 psi difference).

Page 25: Chapter 6_Gas Well Performance

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Cullender and Smith Method

-The Cullender and Smith method is the

most accurate method for calculating

bottom-hole pressures.

-This method is generally applicable to

shallow and deep wells, sour gases, and

digital computations.

Page 26: Chapter 6_Gas Well Performance

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Example (2)

Calculate the static bottom-hole pressure for the gas well of

Example 1 using the Cullender and Smith method.

Solution

(a) Determine the value of z at wellhead conditions and compute

Its.

depth of the well=5790 ft.,

gas gravity = 0.60

pressure at the wellhead = 2300 psia.

Temperature at well head=74oF

Average temperature of flow string=117°F

Ppc =672psia

Tpc=358°R

Page 27: Chapter 6_Gas Well Performance

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Example (2)

Solution

(a) Determine the value of z at wellhead conditions and compute

Its.

Page 28: Chapter 6_Gas Well Performance

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(b) Calculate Its for intermediate conditions at a depth of 5790/2 or

2895 ft, assuming a straight line temperature gradient. As a

first approximation, assume

Ims = Its = 178

Then, from Eqn 8.30,

(8.30)

(8.27)

(8.30)

Page 29: Chapter 6_Gas Well Performance

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(c) Calculate Iws at bottom-hole conditions assuming, for the first

trial, Iws = Ims = 191. Then, from Eqn 8.31,

Since the two values of Pms are not equal, calculations are

repeated with Pms=2477 psia.

This is a check of the pressure at 2895 ft.

Page 30: Chapter 6_Gas Well Performance

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Repeating the calculation,

(d) Finally, using Eqn 8.32,

Page 31: Chapter 6_Gas Well Performance

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QUIZZ

E

1. Calculate static bottom-hole pressure

by using the same data given in exercise.

Take tubing head pressure to be 3340

psia.

Page 32: Chapter 6_Gas Well Performance

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Flowing Bottom-Hole Pressure

Flowing bottom-hole pressure of a gas well is the sum of the

flowing wellhead pressure, the pressure exerted by the weight

of the gas column, the kinetic energy change, and the energy

losses resulting from friction.

As kinetic energy change is very small, it is assumed zero.

For the situation of no heat loss from gas to surroundings and

no work performed by the system.

This equation is the basis for all methods of calculating flowing

bottom-hole pressures from wellhead observations.

The only assumptions made so far are single-phase gas flow

and negligible kinetic energy change.

(8.33)

Page 33: Chapter 6_Gas Well Performance

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Assumptions in the average temperature and average gas

deviation factor method are:

1. Steady-state flow

2. Single-phase gas flow, although it may be used for condensate

flow if proper adjustments are made in the flow rate, gas

gravity and Z-factor

3. Change in kinetic energy is small and may be neglected

4. Constant temperature at some average value

5. Constant gas deviation factor at some average value

6. Constant friction factor over the length of the conduit

Average Temperature And Average

Gas Deviation Factor Method

Page 34: Chapter 6_Gas Well Performance

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(8.39)

Equation for Average Temperature and Deviation

Factor method is

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(8.40)

If Fanning friction factor is used, use the

following equation.

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Equation 8.39 is to be applied when

Moody Friction factor is used.

Equation 8.40 is to be applied when

Fanning Friction factor is used.

Moody friction factor= 4* Fanning friction factor

Page 37: Chapter 6_Gas Well Performance

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Example (3)

Calculate the sandface pressure of a flowing gas well from the

following surface measurements: Use Average temperature and

Deviation Factor method.

Solution

Using Eqn 8.39,

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First trial Guess, Pwf = 2500 psia

At 1.0 atm and 121.5oF.

Viscosity at average pressure:

Page 39: Chapter 6_Gas Well Performance

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The Reynold’s number is given by

From the Moody friction factor chart,

OR

Pwf = 2543 psia

Page 40: Chapter 6_Gas Well Performance

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Second trial

There is no appreciable change in z for this trial; so, first trial is

sufficiently accurate.

Page 41: Chapter 6_Gas Well Performance

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QUIZZ QUIZZ 5

1. Calculate the sand-face pressure of a flowing gas well

from the following surface measurements:

q= 12 MMscfd D=4 in.

γg = 0.62 Depth = 8400 ft. (bottom of casing)

Twf = 160°F Ttf = 83°F

Ptf = 2755 psia e= 0.0006 in

viscosity at average pressure= 0.0167 cp

length of tubing= 8350 ft. Pc= 672 Tc=358

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Q & A

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Thank You