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NATURAL GAS ENGINEERING
CHAPTER 5 GAS WELL PERFORMANCE
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CONTENTS
5.1 Gas Well Performance
5.2 Static Bottom-hole Pressure(static BHP)
5.3 Flowing Bottom-hole Pressure(flowing BHP)
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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) usingdifferent methods
Determine flowing bottom-hole pressure(flowing BHP) using
different methods
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Gas Well Performance
Figure (5.1) Gas Production Schematic
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Gas Well Performance
Referring to Fig.(5.1), ability of a gas reservoir to produce
for a given set of reservoir conditions dependsdirectly on theflowing bottom-hole pressure,Pwf.
The ability of reservoir to deliver a certain quantity of gasdepends on
the inflow performance relationship (IPR)
flowingbottom-hole pressure (FBHP)
Flowing bottom-hole pressure dependson
Separator pressure
Configuration of the piping system
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Gas Well Performance
These conditions can be expressed as:
(8.1)
(8.2)
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7/417Figure (5.2) Deliverability test plot
The static or flowing pressure at the formation must beknownin order to predictthe productivity or absolute open flow
potential (AOF) 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
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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)
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Basic Energy Equation
Figure (5.3) Flow in pipe (After Aziz.)
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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 wsis zero.
Reduced form of the mechanical energy equation may be
written as:
OR
cg
udu
2
(8.5)
(8.6)
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Basic Energy Equation
All equations now in use for gas flow and static head
calculationsare various forms of this Equation.The density of a gas ( ) at a point in a vertical pipe at
pressurepand temperature Tmay be calculated as:
(8.7)
g
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Fig.(5.4) Compressibility
factor for natural gases
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13/4113Fig.(5.5) Moody Friction Factor Chart
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Basic Energy Equation
The velocity of gas flow ugat a cross section of a vertical pipe
is
(8.8)
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Basic Energy Equation
General vertical flow equation assuming a constant average
temperaturein the interval of interest is
(8.10)
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Basic Energy Equation
Sukker & Cornell, and Poettmann assumed gas deviation
factor varies only 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.
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StaticBottom-Hole Pressure
Average Temperature and Deviat ion Factor Method
The Equation is:
(8.20)
l ( )
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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.(Use Average Temperature & Deviation Factor
Method).
Solution
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First trial
Second trial
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Exercise 1
1. Calculate the static bottom-hole pressure of a gas
well having a depth of 8570 ft. The gas gravity is 0.63and the pressure at the wellhead is 2800 psia. The
average temperature of the flow string is 124oF.Use
average Temperature and Deviation Factor method.
=672 psia, =358 R
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.
=672 psia, =358 R
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Cullender and Smith Method
This is a more realistic approach that gas deviation factor is a
functionof both temperature and pressure.
Define
(8.25)
(8.26)
C ll d d S ith M th d
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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
WhereItsis evaluated atH= 0,ImsatZ/2 andIwsatZ.
(8.30)
(8.31)
(8.32)
(8.27)
(8.29)
C ll d d S ith M th d
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Cullender and Smith MethodCalculation procedure
First: to solve for an intermediate temperature and pressurecondition at the mid point of the vertical column;
Second: Repeat the calculations for bottom-holecondition.
-A value of Its is first calculated from Eqn 8.27 at surfaceconditions.
-Then, Imsis assumed(Its=Imsat first approximation) and pmsis
calculated for the mid point conditions.
-Using this value of Ims , a new value of Ims is computed.
-The new value of Imsis then used to recalculate pms.
-This procedure is repeated until successive calculations of pms
are within the desired accuracy (usually within 1 psi difference).
Cullender and Smith Method
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Cullender and Smith Method
-The Cullender and Smith method is the mostaccurate method for calculating bottom-hole
pressures.
-This method is generally applicable to shallow
and deep wells, sour gases, and digital
computations.
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E l ( )
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Example (2)
Solution
(a) Determine the value ofzat wellhead conditions and computeIts.
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(b) CalculateImsfor intermediate conditions at a depth of 5790/2or 2895 ft, assuming a straight line temperature gradient. As a
first approximation, assume
Ims =Its= 178Then, from Eqn 8.30,
(8.30)
(8.27)
(8.30)
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(c) CalculateIwsat 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 arerepeated withPms=2477 psia.
This is a check of the pressure at 2895 ft.
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Repeating the calculation,
(d) Finally, using Eqn8.32,
QUIZZE i 2
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QUIZZExercise 2
1.Calculate static bottom-hole pressure using the following data
given:
Depth of the well=7900 ft.
Gas gravity = 0.65
Pressure at the wellhead = 2800 psia.
Temperature at well head=74oF
Average temperature of flow string=117F
Ppc=672psia
Tpc=358R
Take initial pressure 3100 psia for your trial and errorcalculation.
QUIZZ
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QUIZZ
Home Work(2)
1.Calculate static bottom-hole pressure by
using thefollowing data given in example 2.
Take tubing head pressure to be 3400 psia.Use Cullender and Smith method.
Flowing Bottom Hole Pressure
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FlowingBottom-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 energylosses resulting from friction.
As kinetic energy changeis 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 wellheadobservations.
The only assumptions made so far are single-phasegas flow
and negligible kinetic energy change.
(8.33)
Average Temperature And Average
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Assumptions in the average temperature and average gasdeviation 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 smalland may be neglected
4. Constant temperatureat some average value
5. Constant gas deviation factor at some average value6. Constant friction factor over the length of the conduit
Average Temperature And Average
Gas Deviation Factor Method
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(8.40)
If Fanning friction factor is used, use thefollowing equation.
Moody friction factor = 4 * Fanning friction factor
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Equation8.39is to be applied when MoodyFriction factoris used.
Equation8.40is to be applied when Fanning
Friction factor is used.
Example (3)
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Example (3)
Calculate the flowing bottom hole pressureof a gas well from
the following surface measurements: Use Average temperature
and Deviation Factor method.
SolutionUsing Eqn 8.39,
First trial Guess P = 2500 psia
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First trial Guess,Pwf = 2500 psia
At 1.0 atm and 121.5oF.
Viscosity at average pressure:
The Reynolds number is given by
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The Reynold s number is given by
From the Moody friction factor chart, or by applying Jain Eqn
Then, Pwf= 2543 psia
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Second trial
There is no appreciable change in z for this trial; (first zvalue=0.825, and 0.825 again in second trial) ,so, first trial is
sufficiently accurate.
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