NATURAL GAS ENGINEERING
CHAPTER 5 GAS WELL PERFORMANCE
CONTENTS
5.1
Gas Well Performance
5.2
Static Bottom-hole Pressure(static BHP)
5.3
Flowing Bottom-hole Pressure(flowi Pressure(flowing ng BHP)
CONTENTS
5.1
Gas Well Performance
5.2
Static Bottom-hole Pressure(static BHP)
5.3
Flowing Bottom-hole Pressure(flowi Pressure(flowing ng BHP)
LESSON LEARNING OUTCOME
At the end of of the session, session, students should should be able to: • Determine static bottom-hole pressure(static BHP) using different methods • Determine flowing bottom-hole pressure(flowing BHP) using different methods
Gas Well Performance
Gas Well Performance • Referring to Fig.(5.1), ability of a gas reservoir to produce for a given set of reservoir conditions depends directly on the flowing bottom-hole pressure, P wf . • The ability of reservoir to deliver a certain quantity of gas depends on • the inflow performance relationship (IPR) • flowing bottom-hole pressure (FBHP) • Flowing bottom-hole pressure depends on • Separator pressure • Configuration of the piping system
Gas Well Performance • These conditions can be expressed as:
(8.1) (8.2)
Static and Flowing Bottom-Hole Pressures •The static or flowing pressure at the formation must be known in order to predict the productivity or absolute open flow potential (AOF) of gas wells. •Preferred method is a bottom-hole pressure gauge (downhole 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.)
Basic Energy Equation In the case of steady-state flow, energy balance can be expressed as follows: (8.3)
OR (8.4)
Basic Energy Equation
Basic Energy Equation udu
• Second term ( ) kinetic energy is neglected in pipeline flow calculations. 2 g c • 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:
(8.5)
OR (8.6)
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 ( g ) at a point in a vertical pipe at pressure p and temperature T may be calculated as:
(8.7)
Fig.(5.4) Compressibility factor for natural gases
Basic Energy Equation • The velocity of gas flow ug at a cross section of a vertical pipe is
(8.8)
Basic Energy Equation • General vertical flow equation assuming a constant average temperature in the interval of interest is
(8.10)
Basic Energy Equation
•Sukker & Cornell, and Poettmann assumed gas deviation factor varies o n l y 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 b o t h temperature and pressure.
Static Bottom-Hole Pressure A v e r ag e Te m p e r at u r e a n d D ev i a t i o n F a c t o r M et h o d
The Equation is: (8.20)
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
First trial
Second trial
Exercise 1 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 124 oF.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 119 oF. .Use average Temperature and Deviation Factor method. =672 psia, =358 °R
Cullender and Smith Method This
is a more realistic approach that gas deviation factor is a function of both temperature and pressure. (8.25)
Define
(8.26)
Cullender and Smith Method Which, for the static case, reduces to
(8.27) (8.29)
For the upper half ,
(8.30)
For the lower half ,
(8.31)
Static bottom-hole pressure at depth Z in the well is finally given by
(8.32)
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 I t s is first calculated from Eqn 8.27 at surface conditions. -Then, I m s is assumed( I t s =I m s at first approximation) and p m s is calculated for the mid point conditions. -Using this value of I m s , a new value of I m s is computed. -The new value of I m s is then used to recalculate p m s . -This procedure is repeated until successive calculations of p m s are within the desired accuracy (usually within 1 psi difference).
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.
Example (2) Calculate the static bottom-hole pressure for the gas well of Example 1 using the Cullender and Smith method.
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 pc =672psia P pc =358°R T
Example (2) (a) Determine the value of z at wellhead conditions and compute I ts. Solution
(b) Calculate I ms for intermediate conditions at a depth of 5790/2 or 2895 ft, assuming a straight line temperature gradient. As a first approximation, assume I ms = I ts = 178 Then, from Eqn 8.30, (8.30)
(8.27) (8.30)
Since the two values of P ms are not equal, calculations are repeated with P ms=2477 psia.
This is a check of the pressure at 2895 ft. (c) Calculate I ws at bottom-hole conditions assuming, for the first trial, I ws = I ms = 191. Then, from Eqn 8.31,
Repeating the calculation,
(d) Finally, using Eqn 8.32,
QUIZZ Exercise
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=117°F pc =672psia P pc =358°R T
Take initial pressure calculation.
3100 psia for your trial and error
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
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. (8.33)
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
Average Temperature And Average Gas Deviation Factor Method 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
Equation for Average Temperature and Deviation Factor method (8.39)
If Fanning friction factor is used, use the following equation.
(8.40)
Moody friction factor = 4 * Fanning friction factor
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.
Example (3) Calculate the flowing bottom hole pressure of a gas well from the following surface measurements: Use Average temperature and Deviation Factor method.
Solution Using Eqn 8.39,
First trial
Guess, P wf = 2500 psia
At 1.0 atm and 121.5oF.
Viscosity at average pressure:
The Reynold’s number is given by
From the Moody friction factor chart, or by applying Jain Eqn
Then,
P = 2543 psia
Second trial
There is no appreciable change in z for this trial; (first z value=0.825, and 0.825 again in second trial) ,so, first trial is sufficiently accurate.
