Natural Gas Production Technology.pdf

VIỆN

DẦU

KHÍ

VIỆT

NAM

TRUNG TÂM ĐÀO TẠO VÀ THÔNG TIN DẦU KHÍ

Natural Gas Production Technology Giảng viên: Trần Vũ Tùng Email: [email protected] - SĐT: 0909803608

HCM, ngày 16-18/8/2017

Introduction  Natural gas is worlds cleanest fossil fuel.  Primarily composed of methane, the lightest of the hydrocarbon compounds.  Because of its high- energy value and its low emission of carbon dioxide (CO2) and nitrogen oxides (NOx) when burned, it has become the fuel of choice for new electric power plants and many new industrial facilities.  Associated gas from oil production in the oilfields can normally be used for: o Energy requirements o Gas re-injection

 Any remaining gas was flared (still present)

Introduction  Why was gas flared? o Not easy to store o Expensive to transport

 Why has natural gas use increased? o Development of pipelines and compressor stations o More environmentally friendly compared to other fossil fuel o Cheaper price (per MSCF)

Introduction  Gas prices are not affected by global concerns (Price is controlled per region)  Development of new technologies with the use of natural gas  Used for hydrogen generation for fuel cells (energy supply)

Introduction  Demand for Natural Gas:  Has increased steady - 100 TCF (2004)  Estimated to be about 162 TCF (2020)  By 2020 (Natural is provide about 28% of word energy demand).Oil will be constant at about 39%  Addition fuel cell technology will increase natural gas demand to 200 TCF (2020)

 Supply of Natural Gas:  Proven Reserve is about 6400 TCF (2005)  Reserves per production ration varies by region (Gas consumption indicator per region)  Most recent reserve data shows: About 30% (Russia), 30% (Middle east), 3.5% (U.S.)

 How can natural gas demand be globally available?

Natural Gas Transportation 1. Pipelines:  Main method for gas transportation

2. LNG (Liquefied Natural Gas)  Factors to affect LNG projects: (Amount of producible natural gas at proposed LNG Location) • LNG plant cost (expensive and associated with gas capacity) i.e. $225 $625 million for 100 MMSCFD • LNG tanker cost (expensive and associated with gas capacity) i.e. one tanker is about $260 million for 3BCF • LNG re-gasification plant cost (expensive)

Natural Gas Transportation 3. GTL (Gas to Liquid)  

Conversion of natural gas into synthetic hydrocarbon liquids (mainly middle distillates) First step- natural gas is reformed and converted to H2 and CO (mixture is called synthetic gas or syn gas) • •

 

Second step- syn gas is blown into a slurry reactor and is converted to liquids (called Fisher –Tropsch process) Benefits of GTL technology: • • •



same process of natural gas conversion into H2 most expensive phase (consumes 50% of total GTL cost)

Help stop gas flaring Syn gas has no or very sulfur content High tech jobs can be made available

One disadvantage is the cost to build a GTL plant (may be about $11 billion per plant)

Natural Gas Transportation 4. Gas Hydrate Pipeline: 

Involves the conversion of natural gas into hydrates and the gas is transported in a slurry form (still I experimental phase)

Natural Gas Sources  Natural gas is found in underground rock formations known as reservoirs.  The reservoir rock is typically a porous sand, sandstone, limestone or dolomite.  The gas in the reservoir may be mixed with oil, or the reservoir may contain only natural gas and natural gas liquids, commonly referred to as condensate.

Natural Gas Production  As production begins from a reservoir, gas, oil or condensate and water are produced simultaneously.  Production equipment located at the surface separates each of the fluids.  The oil and condensate can be stored in aboveground tanks or pumped into a pipeline.  The water is collected for disposal, and the gas can be compressed into a pipeline or re-injected into the reservoir to maintain reservoir pressure.

Natural Gas Properties  Natural gas is colorless.  While delivering the gas as a fuel for household, Local gas utilities add mercaptan for safety purposes-since it allows small leaks at low pressures in the distribution systems to be easily detected.  The specific gravity of natural gas is about 0.6-0.7  The heating value of natural gas is approximately 8900 kcal/ per standard cubic meter. However, it depends upon the actual composition of the gas.

Natural Gas Properties  Since natural gas is primarily methane it is a high-energy, cleanburning fuel. This quality is clearly illustrated in the graph comparing emissions from natural gas, fuel oil and coal.

Components in Reservoir Fluids C1 (Methane)

C2 (Ethane)

C3 (Propane)

n-C4 (Butane)

i-C5 (Isopentane)

n-C5 (Pentane)

i-C4 (Isobutane)

C6

C19

Typical Natural Gas Components Component

Typical Analysis (mole %)

Methane

79.50

Ethane Propane

10.0 3.4

Butane Pentane

1.2 0.25

Hexanes plus

0.1

Carbon Dioxide Nitrogen Specific Gravity

5.5 0.05 0.70

Reservoir Fluid Characterization Pressure path in reservoir

Pressure path in reservoir

Critical 1 point

2 Critical point

Dewpoint line

Black Oil

% Liquid

Volatile oil Pressure

Pressure, psia

 The location of the reservoir temperature with respect to the critical temperature and cricondentherm.

Volatile Oil

Black Oil

% Liquid

 Location of the first-stage separator pressure and temperature with respect to the phase diagram of the reservoir fluid.

33

Separator

Separator Temperature, °F

Temperature

Pressure path in reservoir

Critical point

1

Wet gas

% Liquid Critical point 3

Separator

% Liquid

Pressure

2

Pressure

Pressure

Retrograde gas

Pressure path in reservoir

Pressure path in reservoir 1

1

Dry gas

% Liquid 2

Separator

Temperature

Temperature

Retrograde Gas

Wet Gas

2

Separator Temperature

Dry Gas

Pressure-Temperature Diagrams  Used to visualize the fluids production path from the reservoir to the surface  Used to classify reservoir fluids  Used to develop different strategies to produce oil/gas from reservoir

Phase diagram for a pure component  Saturated liquid and vapor co-exist along the vapor pressure curve  At a given pressure, there is only one temperature where liquid and vapor co-exist.  At a pressure of 1 atm, this is the normal boiling point  The two-phase region terminates at the critical point (where liquid and vapor have identical properties).

Phase diagram for a mixture (at least 2 components)

Critical point

 Saturated liquid and vapor co-exist within the phase envelope  At a given pressure, there is a range of temperatures where liquid and vapor coexist.  The two-phase region can extend to pressures and temperatures that are higher than Pc and Tc

Pressure, psia

Phase Diagrams for different type of fluids

Single Liquid Phase Region

Initial Reservoir State

Critical Cricondenbar Point

Single Gas Phase Region

Two-Phase Region % Liquid

Separator

Temperature, °F

Cricondentherm

Black Oil

Pressure, psia

Black Oil  Pressure path in reservoir Critical Dewpoint line  Point Black Oil

% Liquid

Separator

Temperature, °F

The oil is typically composed of more than 25% C7+. The oil has a relatively low shrinkage when produced.



Typical GOR less than 1000 Scf/bbl. (200 sm3/m3)



The equilibrium gas is relatively dry, and the amount of liquid produced from the reservoir gas is usually negligible (a dry gas BO model is reasonable).

Volatile-Oil Volatile Oil Critical point

Pressure

1 Pressure path in reservoir 2 Volatile oil



Typically have more heavy components than a gas condensate, which makes the fluid more oil-like.



Large shrinkage of oil volume with pressure, due to gas vaporization as the pressure drops below the bubble-point.



Typical GOR range between 1000 to 3000 Scf/Stb (200 to 600 sm3/m3)



The equilibrium gas is relatively rich, and the amount of liquid produced from the reservoir gas is significant (dry-gas BO models 21 cannot be used!).

% Liquid

3 Separator

Temperature, °F

Retrograde Gas Pressure path in reservoir 1

Pressure

Retrograde gas 2

Gas Condensate 

Presence of heavier components (C4C7+) expands the phase envelope, so that the reservoir temperature lies between the critical point and cricondentherm.



Liquid drops out during depletion due to retrograde condensation, further condensation also occurs at surface conditions.



Liquid dropout in the reservoir will generally lead to lower condensate recovery (immobile oil phase) and may result in poor well deliverabilities.



Typical GOR range between 3000 to 50,000 scf/Stb (600 to 10,000 sm3/m3)

Critical point % Liquid

3

Separator

Temperature

Wet Gas Pressure path in reservoir 1

Pressure

Wet gas

Critical % Liquid point 2

Separator

Temperature

Wet Gas  Mainly composed of light HC components (C1, C2, C3 and C4).  Wet gas will not drop out condensate at reservoir conditions, only at surface conditions.  Producing GOR typically higher than 50,000 scf/Stb (10,000 sm3/m3) and will remain constant throughout depletion.

Dry Gas

Pressure

Pressure path in reservoir 1

Dry gas

% Liquid 2

Separator

Temperature

Dry Gas  Mainly composed of C1 and light hydrocarbons, in addition to non-HC components such as N2 and CO2  The gas remains single-phase from the reservoir to surface conditions (no condensate dropout).

Reservoir Fluid Compositions CO2 N2 C1 C2 C3 C4 C5 C6 C7+

GOR, scf/STB CGR, stb/MMscf GOR, m3/m3 CGR, m3/MMm3 API MW- C7+

Dry Gas

Wet Gas

Lean gas condensate

0.1 2.1 86.1 5.9 3.6 1.7 0.5

1.4 0.3 92.5 3.2 1.0 0.5 0.2 0.1 0.8

1.8 3.4 83.5 5.2 1.9 1.1 0.6 0.4 1.7

2.4 0.3 73.2 7.8 3.6 2.2 1.3 1.1 8.2

1.3 0.6 69.4 7.9 4.3 3.0 2.0 1.5 10.0

0.9 0.2 58.8 7.6 4.1 3.0 1.9 1.8 21.8

0.0 0.3 34.6 4.1 1.0 1.3 0.6 1.6 56.4

105,000 10

20,000 50

5,450 180

3,650 275

1,490

300

18701 53

3562 281

971 1030

650 1538

265

53

57 130

55 135

49 184

45 219

38 228

24 274

0

Rich Gas Condensate

Nearcritical oil Volatile Oil Black Oil

Basic Properties of Natural Gas

 Equation-of-State (EOS)  Apparent Molecular Weight of Gas Mixture

 Density of Gas Mixture  Gas Specific Gravity  Z-factor (Gas Compressibility or Gas Deviation Factor)

The Ideal Gas Law pV = nRT p  Absolute Pressure V  Total Fluid Volume n  Number of Fluid Moles T  Temperature R  Universal Gas Constant  8314.472 Pa - m 3 /kmol - K (1998 CODATA)  0.08314472 bara - m 3 /kmol - K

Ideal gases:  No interacting forces between the molecules.  The volume of the molecules is negligible compared to the total gas volume.

 10.73159 psia - ft 3 /lbmol - R

Most gases show ideal behavior at low pressure and temperatures

The Ideal Gas Law pV = nRT Most gases show ideal behavior at low pressure and temperatures. At standard conditions, the volume of 1 mole of gas is independent of its composition.

vg ,sc  379.4 scf / lbmol

vg ,sc  23.69 sm / kmol 3

Apparent MW of a Gas Mixture Normally, petroleum gas is a mixture of various light hydrocarbon (C1-C4). For example: Component

Mole Percent

Molecular Weight (lb/lb-mol)

(1)

(2)

Critical

Critical

Pressure

Temperature

(psia)

(oR)

(3)

(4)

C1

0.85

16.043

666.4

343.00

C2

0.04

30.070

706.5

549.59

C3

0.06

44.097

616.0

665.73

iC4

0.03

58.123

527.9

734.13

nC4

0.02

58.123

550.6

765.29

N

M a   yi M i  20.39 i 1

Density of Gas Mixture Gas density is calculated from the definition of density and the EOS

mg

nM a p pM a g  = = Vg nZRT ZRT

3

(lb/ft )

Gas Specific Gravity The specific gravity is defined as the ratio of the gas density to that of the air

g 

(  g ) sc (  air ) sc

g M a Ma g  = = air M air 28.96

The Real Gas Law

pV  nZRT Quantity

Description

Unit/Value

p

Pressure

psia

V

Volume

ft3

n

Mole Number

lb-mol

Z

Gas Devation Factor

dimensionless

T

Temperature

Rankine

R

Universal Gas constant

10.73 psia.ft3/lb-mole. R

Gas Deviation Factor (Z-factor) Z-factor in the EOS accounts for the difference in the behavior of natural gases in compared with ideal gases.

Z-factor can be expressed as: Z=Z(ppr,Tpr) where

p T p pr  ; Tpr  p pc Tpc

ppr: pseudo-reduced pressure Tpr: pseudo-reduced temperature ppc: pseudo-critical pressure Tpc: pseudo-critical temperature

p pc   yi pci ; Tpc   yiTci i

i

Standing-Katz Chart Step 1: Calculate pseudo-critical pressure and temperature

p pc   yi pci ; Tpc   yiTci i

i

Step 2: Calculate pseudo-reduced pressure and temperature:

p T p pr  ; Tpr  p pc Tpc Step 3: Use Standings-Katz chart to determine Z

Gas Viscosity Most gas viscosities range from 0.01 to 0.03 cP, making them difficult to measure accurately. Accurate determination of gas viscosities has low economic value. Instead, values are normally determined from one of two correlations. The first one in common use today, from Lee et al., is given in equation form as

Gas Viscosity

Gas Formation Volume Factor for a Dry Gas

Bg  (

Vg ,res Vg ,SC

)

psc ZT Bg  ( ) Tsc p

 For a dry gas reservoir, all of the wellstream fluid is produced as gas at standard conditions.

 Gas formation volume factor is defined as the ratio between the gas volume at p and T (reservoir conditions) and the ideal gas volume at standard conditions.

Condensate Gas Ratio (CGR, Rv , rs , rv) Separator gas – Vsg Process Condensate - Vc Wellstream fluid

Vc rv  ( ) Vsg

 For a wet gas or gas condensate, the wellstream fluid will yield both liquid (condensate) and separator gas at the surface.  The CGR is the ratio of the volume of separator gas at standard conditions to the volume of condensate at stock tank conditions. Common units are stb/MMscf, or m3/million sm3.  For a wet gas reservoir, the CGR is more or less constant (well stream composition does not change).  For gas condensates the CGR changes over time, due to liquid drop-out in the reservoir.

Wet gas and separator gas Separator gas – Qsg Process Condensate – Qsg.rv

Wellstream fluid

 Consider a well producing with a CGR rv.  The ‘wet gas’ volume is a hypothetical volume consisting of the separator gas volume and the condensate converted to an equivalent surface gas volume.

 The ‘equivalent surface gas volume’ of condensate assumes that it behaves as an ideal gas at standard conditions. The equivalent volume of condensate is Cog, where

cog  133,000(

c ) Mc

scf/stb

Qwg  Qsg 1  rv cog 

 The ‘wet gas’ rate is given by

Equivalent volume of condensate  The equivalent volume of condensate is

cog  133,000(

c ) Mc

scf/stb

  and M are the condensate specific gravity and molecular weight.  cog is typically about 0.7 to 0.8 Mscf/stb.  cog is the (hypothetical) volume of 1 stb of condensate if it behaved as an ideal gas at standard conditions.  It should not be confused with 1 ‘barrel of oil equivalent’ (BOE) – about 6 Mscf/stb – this is based on equivalent amounts of energy.

FVF for a Wet Gas or Gas Condensate There are two definitions of gas FVF 1.

‘Wet gas’ FVF. Bgw = (gas volume at reservoir conditions) / (wet gas volume at standard P, T)

2.

‘Dry gas’ FVF. Bgd = (gas volume at reservoir conditions) / (separator gas volume at standard P, T)

Bgw

psc ZT ( ) Tsc p

Bgd  Bgw 1  cog rs 

Exercise Given the following real gas composition,

Component C1 C2 CO2 H2S

Mole fraction 0.76 0.07 0.1 0.07

Determine the density of the gas mixture at 1,000 psia and 110 F using Witchert-Aziz correction method.

Dranchuk & Abou-Kassem Correlation FZ  R1  r 

R2

r

 R3  r2  R4  r5  R5  r2 (1  A11  r2 ) exp( A11  r2 )  1  0

 r  0.27 p pr / ( ZTpr ) R1  A1  A2 / Tpr  A3 / Tpr3  A4 / Tpr4  A5 / Tpr5 R2  0.27 p pr / Tpr R3  A6  A7 / Tpr  A8 / Tpr2 R4  A9 ( A7 / Tpr  A8 / Tpr2 ) R5  A10 / Tpr3 A1  0.3265; A2  1.0700; A3  0.5339 A4  0.01569; A5  0.05165; A6  0.5475 A7  0.7361; A8  0.1844; A9  0.1056 A10  0.6134; A11  0.7210

Exercise

Component

yi

Mi

Tci,°R

pci

CO2

0.02

44.01

547.91

1071

N2

0.01

28.01

227.49

493.1

C1

0.85

16.04

343.33

666.4

C2

0.04

30.1

549.92

706.5

C3

0.03

44.1

666.06

616.4

i - C4

0.03

58.1

734.46

527.9

n - C4

0.02

58.1

765.62

550.6

Wichert-Aziz Correction Method Corrected pseudo-critical temperature:

Tpc  Tpc   , o R Corrected pseudo-critical pressure: p pcTpc p pc  , psia Tpc  yH 2 S (1  yH 2 S ) Pseudo-critical temperature adjustment factor   120

 y

H2S

 yCO2



0.9



 yH 2 S  yCO2



1.6

  15  y

0.5 H2S



 yH 2 S 4.0 ,

Sutton Correction Method Step1: Calculate the parameters J and K: 2

0.5     Tc   Tc   o 1 2 J    yi       yi    , R/psia 3 i  pc i  3  i  pc i   Tci K   yi , o R/psia pci i

Step 2: Calculate the adjustment parameters: 2 1   Tc FJ   y 3   pc

 2   Tc    y  C7 3   pc

  

0.5

   C7

 J  0.6081FJ  1.1325 FJ2  14.004 FJ yC  64.434 FJ yC2 7

K

 Tc    pc

  0.3129 yC7  4.8156 yC2 7  27.3751 yC3 7  C7





7

Sutton Correction Method (cont.) Step 3: Adust the parameters J and K

J  J J K  K K Step 4: Calculate the adjusted pseudo-critical terms K2 T pc  J T pc p pc  J

Pseudo Properties of Gas

Petrophysical Properties  Klinkenberg Effect: The permeability of a sample to a gas varies with the molecular weight of the gas and the applied pressure, as a consequence of gas slippage at the pore wall. Klinkenberg[1] determined that liquid permeability (kL) is related to gas permeability (kg) by

Where:  p is the mean flowing pressure  b is a constant for a particular gas in a given rock type

Petrophysical Properties  Non-Darcy Flow: At high fluid velocities, Darcy’s law may not always be accurate. An additional energy loss is often apparent above that predicted from the laminarflow relationship suggested by Darcy’s law.

 where β is called the non-Darcy velocity coefficient, having units of L–1, and u is the volumetric flux (q/A) through the rock.

Gas Metering  Gas Metering is used to monitor the amount of gas that is delivered by gas producers from gas wells or by gas pipelines to end users  Gas meters placed on pipeline are used to monitor and measure the amount of gas moving through a pipeline  Types of Gas Meters:  Head meters (i.e. orifice meters): A typical Head meter is an Orifice meter  Turbine meters  Positive displacement meters  Variable area meters  Vortex meters  Sonic meters

Gas Meters 1. Orifice Meters  Advantages: Simple to use, Ease of installation and maintenance, Measurement accuracy within +/- 2%  Limitation: Not suitable for erosive or corrosive gases and in the presence of solids or liquid droplets

2. Turbine meters 

Used mostly for liquid flow

3. Positive Displacement meters   

Measurement accuracy within +/- 1% Expensive and needs significant maintenance More ideal for liquid measurement

4. Variable area meters   

Simple and inexpensive meter Used for low gas flow-rates Measurement accuracy within +/- 10%

Gas Meters 5. Vortex Meter 



Advantages: can be applied wide flow rate ranges, improved reliability, reduced maintenance requirement, low installation cost for pipeline diameters less than 6 inches, can be applicable for liquid and gas measurement, low permanent pressure loss compared to orifice meters Limitations: costly installation for large pipe diameter, not suitable for use for very low gas densities (i.e. hydrogen) or high fluid velocities

6. Sonic Meter  

Advantages: Nonintrusive measurements (no permanent pressure loss), high degree of accuracy +/- 0.5% Limitations: complex and difficult instrumentation and measurement, presence of solids and liquids can cause measurement errors

Fully Developed Single-Phase Pipe Flow  Consider a pipe with a constant diameter, d:  Cross sectional area – A = pd2/4  Perimeter – Sw = pd  Inclination angle – q (Positive angles indicate upward flow. Negative angles indicate downward flow. Zero angle indicates horizontal flow.)

Geometry z

dz

θ z

 gz   g z   g sinq   g z   g sinq 

q

 g

Forces Acting on Control Volume



Take a small section of the pipe as control volume W is mass flow rate W = ρVA



Acceleration and forces acting on control volume:



  w S w dz

 W V  dV 

 AP  dP  WV q

AP

 Adzg sinq

Force Balance  For steady state flow (no change of momentum in control volume), all forces must be balanced on the control volume PA  ( P  dP ) A   V 2 A   V V  dV A   w Sw dz   g sinq  Adz  0

Sw dP dV   w   g sinq    V dz A dz

Sw 4 p d 4   2 A pd d

dP 4 dV    g sinq    w   V dz d dz

Three Parts of Pressure Gradient

dP 4 dV    g sinq    w   V dz d dz Gravitational Part: Due to elevation differences – main part in tubing strings

Frictional Part: Due to friction between fluid and pipe wall – main part in flowlines

Accelerational Part: Due to acceleration of fluids – Important only for high speed gas flow

Pressure Gradient  The acceleration term can be neglected for most cases  Shear stress at pipe wall is proportional to velocity gradient of flow in normal direction of pipe wall and fluid viscosity – Newton’s Law

dv  w   dr

r  rp

dP 4  dv    g sinq   dz d dr

r  rp

Velocity Profile  Under certain flow conditions (low velocity, high viscosity, small diameter), the velocity profile inside the pipe is parabolic obtained from the Newton’s Law  Under those conditions, the layers of fluids slide over each other  This flow regime is called laminar flow  If tracking particles are injected into the pipe, they will follow straight lines

Stream lines

v( r )

Laminar Pipe Flow  Velocity profile of laminar pipe flow

Pfriction 1  v r   R2  r 2  L 4 Average velocity (by integration of above equation)

Pfrictiond 2 V 32 L

dP dz

32  V  2 d friction

L, Pfriction

Laminar Pipe Flow...  Pressure gradient

dP dz

128  q    g sinq   4 p d total

psi/ft

dP dz

cp

 0.433  sinq   7.9628  10 total

6

bbl/d

q d4 in

Turbulent Flow  Transition from laminar to turbulent happens at about Re=2100  Turbulent pipe flow  Flow structures irregular  No analytical expression for velocity profile  Momentum transfer by both shear and Reynolds stress Turbulent flow

Turbulent Flow…  The transition between laminar and turbulent flow is very important in determining the pressure gradient in the pipe.  Notice that in order to calculate the pressure gradient we must know the derivative of the velocity profile at the wall.  The transition between laminar and turbulent flow is governed by the fluid viscosity, pipe diameter and fluid flowrate or average velocity.

v( r ) Particles Path

Turbulent Flow…  The frictional pressure gradient can not be obtained using the Newton’s Law since the flow is no longer laminar.  If we know the tubulent velocity profile we can obtain a new expression for the turbulent frictional pressure gradient.  Experimental correlations have been obtained to describe the velocity profile in turbulent flow. They can be used to estimate the frictional pressure gradient, but their accuracy is not good.

dP dz

dP dz

dP  dz total

dP  dz gravity

friction

   g sinq  gravity

dP dz

4  dv  d dr friction

r  rw

Velocity Profiles 

Turbulent and laminar velocity profiles

r rw

0.8

60  r v  V  1  49  rw

0.6

Turbulent

1

  

1 7

0.4 0.2

  r v  2V  1     rw 

0 -0.2

Laminar

-0.4 -0.6

Average Velocity V

-0.8 -1 0

0.5

1

1.5

2

2.5

v V

  

2

   

Velocity Profiles...  This turbulent velocity profile based on experimental data gives a reasonable aproximation of the actual velocity profile the pipe.  Note that the velocity profile should be flat at the middle of the pipe and this is not followed by the velocity profile correlation.  Also this equation is not accurate enough to give a reasonable aproximation for the velocity gradient at the wall. This information is crucial to determine the frictional pressure gradients.  There are other correlations based on experimental data that better describes the flow field close to the wall. Those correlations can be used to obtain the velocity gradient and then the friction losses in turbulent flow.  The frictional losses in laminar flow can be obtained from Newton’s Law while the turbulent frictional losses can only be obtained with experimental data either to directly correlate the friction losses or to obtain a velocity profile that enables the calculation of the losses.

Friction Factor  Fanning definition

1 2  w  f Fanning  V  2  pd 2 1 2 P   wpdL  f Fanning  V pdL 4 2  dP  dz

1 2 4   f Fanning  V  2 d friction

Friction Factor...  Moody definition

1 2 L P  f Moody V  2 d f Moody  1 dP 2    V  dz friction d 2 

Friction Factor...  Friction factor is function of Reynolds Number for laminar flow

f  f  Re 

Vd Re  

 Friction factor is function of Reynolds Number and pipe wall roughness for turbulent flow

  f  f  Re,  d 

Relative Laminar

Critical Zone Transition

Roughness

Complete Turbulent

0.10 0.09 0.08

 d



3500  d Re

0.05

0.07

Moody Friction Factor

0.06 0.02

0.05

0.015

0.04

0.010 0.006

0.03

f Moody 

64 Re 0.002

0.02

0.001 0.0006 0.0002

Smooth Pipe 0.0001 0.00005

0.01

102

103

104

105 Reynolds Number

106

107

108

Moody Friction Factor  Several regions exists in this graph  Laminar Region • This is the region for Reynolds Number smaller than 2100. • The flow is laminar and the friction factor can be obtained using Newton Law of Viscosity 64

f Moody 

 Critical Region

Re

• For Reynolds Number between 2100 and 4000 the flow can be either laminar or turbulent depending on several factors such as changes in cross section, direction of flow, obstructions, vibrations. This region is called critical region.

 Transition Region • For Reynolds Number above 4000 the flow is turbulent. The friction factor depends on the Reynolds number and also on the rougheness of the wall.

 Complete Turbulent • Above certain Reynolds Numbers, the flow becomes fully turbulent and the friction factor is only function of the roughness of the wall. This region is called Complete Turbulent regime and the criteria for it to occur is:

 d



3500 Re

Moody Friction Factor...  The pipe wall roughness depends on the material and the quality of the finishing of the internal surface as well as past operating conditions.  This is just a reference Material



( ft )

Riveted Steel

0.003 - 0.03

Concrete

0.001 – 0.01

Cast Iron

0.00085

Galvanized Iron

0.0005

Asphalted Cast Iron

0.0004

Commercial Steel

0.00015

Drawn Tubing

0.000005

Friction Factor

 Laminar pipe flow  Fanning:

16 f  , Re  Moody:

64 f  Re

Vd Re  

Moody Friction Factor...

 Turbulent pipe flow  Smooth pipe:

0.184 f  , 0.2 Re

Vd Re  

Moody Friction Factor...  Turbulent friction factors for rough pipe  Colebrook-White (iterative)

  1 2.52   2.0 log   3.7 d Re f f 

   

 Haaland (1983) (explicit)

  6.9     f   1.8 log   Re  3.7 d     

1.11

     

2

Moody Friction Factor...  Churchill 1977 developped a correlation to be used to estimate the friction factor across the whole range of Reynolds Number.

f Moody

 8 12  1     8    3  Re   2   A  B  

    1   A  2.457 ln 0.9   7         0.27     Re   d  

1

12

16

 37530  B   Re 

16

Moody Friction Factor...  Another interpretation of Churchill formulas is given below. They are equivalent to the original formulation.

f Moody

          12   f laminar        1         f critical   

f laminar

64  Re

 Re  f critical  8   37530 

 8        

     1  8 1/ 8      1    f turbulent    and transition    

f turbulent and transition

2



12

          

1

12

8     1  2.457 ln 0.9   7      0 . 27   Re  d   

       

2

Summary of Single Phase Flow in Pipes  Flow regime is determined by Reynolds Number V d Re   

The pressure gradient in the pipe is given by:

f Moody  V 2 dP    g sinq   dz d 2 

The moody friction factor can be estimated using correlations such as:  8  1     8    3  Re   2   A  B   12

f Moody

1

12

    1  A   2.457 ln 0.9   7         0.27    Re    d  

 37530  B  Re  

16

16

Summary of Single Phase Flow in Pipes... 

Reynolds number with variables in field units lb/ft3

ft/s in

 Vd Re  124  cp 

Reynolds number expressed in terms of volumetric flowrate (and in field units) of liquid lb/ft3

bbl/d

q Re  1.4775 d c p

bbl/d

in

q Re  92.2 d c p

in

Summary of Single Phase Flow in Pipes... 

The pressure gradient in the pipe is given by:

dP  q2    g sinq   8 f Moody 2 5 dz p d 

To account for the proper units we have:

psi/ft

bbl/d

2 dP  q  0.433  sinq   1.1471  10  5 f Moody 5 dz d

in

Single Phase Flow in Pipes

Frictional Pressure Gradient (psi/ft)

0.05



  1 cp   0.00015 ft

0.04

1.2

d  2 in

0.03

1.0 0.8

0.02

0.01

0.00 0

500

1000 Flowrate (bpd)

1500

2000

Single Phase Flow in Pipes...

Frictional Pressure Gradient (psi/ft)

0.05

 1



  0.00015 ft

0.04

5.0

d  2 in

0.03

2.0 1.0

0.02

0.01

0.00 0

500

1000 Flowrate (bpd)

1500

2000

Single Phase Flow in Pipes... 0.0015

Frictional Pressure Gradient (psi/ft)

0.05

  1 cp

 1

0.04

d  2 in

0.03

0.00015 0



0.02

0.01

0.00 0

500

1000 Flowrate (bpd)

1500

2000

Single Phase Flow in Pipes... 1.75

Frictional Pressure Gradient (psi/ft)

0.05

  1 cp  1

0.04

  0.00015 ft

0.03

2.00

0.02 2.25

d

0.01

0.00 0

500

1000 Flowrate (bpd)

1500

2000

Single Phase Flow in Pipes... 

Pressure change from separator to upstream (opposite to flow direction)

 dP  P  Psep     dz  dz    q2  5  Psep    0.433  sinq   1.1471  10 f Moody 5 dz d   

Bottomhole flowing pressure



Pwfo  Psep  0.433 ZTubing  Z Flowline



Tubing Flowline  f L f LFlowline  Tubing Moody 5 2  Moody   1.1471  10  q  5 5   dTubing d Flowline  

Single Phase Flow in Pipes...  

Pipeline diameter has a significant effect on the economics A large diameter pipeline will:

 Reduce the frictional pressure losses in the system  Require smaller pumping pressure  Require smaller pump and motor

 Reduce capital investment on pumps and motors  Consume less horsepower during operation  Require higher capital investment on pipeline  Require higher operational costs for the pipeline

Single Phase Flow in Pipes... 

Horse power needed to transport single phase fluid through pipeline with a pressure increase Po – Pi

HP 





1.7014  10 q Po  Pi  0.433  Z pipeline  1.9516  10 5

 10

f Moody

 q3 d5

Lpipeline

 motor pump

 motor and pump are the efficiencies of motor and pump 

This equation is very important for optimizing pipeline and pumping system design

Single Phase Flow in Pipes... 

An application example

 Single phase flow in well tubing strings and surface    

flowlines Production at a certain flowrate from an aquifer Assume that the aquifer is very large and IPR does not change with time Flow rate is constant and does not change with time Tools: Linear IPR Single phase flow pressure losses in pipes

Single Phase Flow in Pipes... Ps

Flowline

Separator 

Tubing

The flowrate q this well can produce using only the aquifer pressure is called natural equilibrium flowrate. The total pressure difference between the reservoir and the separator is the pressure loss in the system

Pr  Ps  PAquifer q   PTubing q   PFlowlineq 

Flowrate

Pwf

Aquifer

Pr

Single Phase Flow in Pipes... Pr  Ps  PAquifer q   PTubing q   PFlowlineq  Pr  PAquifer q   Ps  PTubing q   PFlowlineq 

Pwfi q  

Pwfo q 

The natural equilibrium flowrate is the solution to the following equation

Pwfi q   Pwfo q 

Single Phase Flow in Pipes... 

The inflow bottomhole pressure as a function of the flowrate is known as the IPR and for an aquifer it is

q P q   Pr  J i wf



The outflow bottomhole pressure as a function of the flowrate is known as the OPR and it is the sum of the separator pressure and the pressure losses in the tubing and in the flowline:

Pwfo q   Ps  PTubing q   PFlowlineq  PTubing q   0.433  ZTubing  1.1471  10

5

f

Tubing Moody

 q2 d

5 Tubing

Flowline PFlowlineq   0.433Z Flowline  1.1471  10  5 f Moody

LTubing

 q2

d

5 Flowline

LFlowline

Single Phase Flow in Pipes... Pwfo q   Ps  PTubing q   PFlowlineq 



Pwfo q   Ps  0.433  ZTubing  Z Flowline



Tubing Flowline   f L f Moody LFlowline 5 2  Moody Tubing   1.1471  10  q  5 5   d d Tubing Flowline   

The natural equilibrium flowrate can be obtained by solving:





q Pr   Ps  0.433  ZTubing  Z Flowline J Tubing Flowline   f L f Moody LFlowline 5 2  Moody Tubing   1.1471  10  q  5 5   d d Tubing Flowline  

Single Phase Flow in Pipes... Tubing Flowline   f L f Moody LFlowline 5 2  Moody Tubing  1.1471  10  q  5 5   d d Tubing Flowline   q   Pr  Ps  0.433  ZTubing  Z Flowline  0 J









This equation can not be solved directly since the Reynolds number and the friction factors are also functions of the flowrate Two ways of solving this equation:

 By iteration

 Graphically

Single Phase Flow in Pipes... 







We must first guess an initial value for the equilibrium flowrate Using this guessed flowrate to calculate the Reynolds number and the friction factors in the tubing and flowline Now we are ready to solve the equation for a new estimate of the equilibrium flowrate The new flowrate is compared with the previous one. If they are close enough the solution is reached. If not repeat the process with the new estimate of the flowrate

Iteration Method... 

Solution procedure 1. 2.

Give a estimation of flowrate, q0 Calculate Reynolds number

q Re  92.2 d 3.

Calculate the Moody friction factor (Churchill’s method)

f Moody

Iteration Method... 4.

Calculate the new flowrate using the OPR Tubing Flowline   f L f Moody LFlowline 5 2  Moody Tubing  1.1471  10  q  5 5   dTubing d Flowline   q   Pr  Ps  0.433  ZTubing  Z Flowline  0 J



5.



If the new flowrate is close to the previous flowrate, stop calculation. Otherwise repeat from the second step

Iteration Method... 

The solution in step 4 is:

 b  b 2  4 ac q 2a Tubing Flowline   f L f Moody LFlowline  5  Moody Tubing  a  1.1471  10   5 5   d d Tubing Flowline  



c  Ps  Pr  0.433  ZTubing  Z Flowline 1 b J



Iteration Method... Stop

Guess q0

Yes

Calculate Re No

|qi+1-qi|<e ?

Calculate fMoody

Cal. Pwf from IPR

Cal. q from OPR

Single Phase Flow in Pipes... 

Example: Calculate the equilibrium flowrate for the following water well:

 Pr   

   

- 5000 psig J - 2 bpd/psi Psep - 100 psig  - 0.000015 ft  - 1 cp  -1 Tubing - 2” - 5000 ft - Vertical Flowline - 2” - 10000 ft - Horizontal

Single Phase Flow in Pipes... Flowline Iteration 1

2 3 4

5 6

qi 5000

Re

f

Flowline Moody

Tubing Re

f

Tubing Moody

qi+1

Single Phase Flow in Pipes... Flowline Iteration

qi

1

5000

2

3436

3 4

5 6

Tubing

Re

Flowline f Moody

Re

Tubing f Moody

230500

0.03800

231078

0.03800

qi+1

3436

Single Phase Flow in Pipes... Flowline Iteration

q

Tubing

Re

Flowline f Moody

Re

Tubing f Moody

q

1

5000

230500

0.03800

231078

0.03800

3436

2

3436

158392

0.04256

158789

0.04256

3379

3

3379

155782

0.04275

156173

0.04275

3377

4

3377

155661

0.04275

156051

0.04275

3376

5

3376

155655

0.04276

156045

0.04276

3376

6

3376

155655

0.04276

156045

0.04276

3376

Single Phase Flow in Pipes... 

The other solution is the graphical method:

Bottom Hole Flowing Pressure (psi)

5000

OPR

4500





4000

Pwfo ( q )  Ps  0.433 ZTubing  Z Flowline

3500

Tubing Flowline  f L f LFlowline  Moody Tubing Moody 5 2   1.1471 10  q  5 5   dTubing d Flowline   IPR

3000 2500 2000 1500

Pwfi ( q )  Pr 

1000

q J

500 0 0

2000

4000

6000

Flowrate (bpd)

8000

10000

Single Phase Flow in Pipes... 

As an example let’s examine the equilibrium flowrates for different flowline and tubing diameters:  Flowlines of 1 ½”, 2”, 2 ½”, 3”  Tubings of 2 3/8”, 2 7/8”, 3 ½”

Single Phase Flow in Pipes...

Bottom Hole Flowing Pressure (psi)

1 ½”

Flowline

5000

2”

2 1/2”

3”

4500 4000 3500 3000 2500 2000 1500 Tubing 2 3/8”

1000 500 0 0

2000

4000

6000

Flowrate (bpd)

8000

10000

Single Phase Flow in Pipes... 1 ½”

Flowline

2”

Bottom Hole Flowing Pressure (psi)

5000

2 1/2”

4500

3”

4000 3500 3000 2500 2000 1500 Tubing 2 7/8”

1000 500 0 0

2000

4000

6000

Flowrate (bpd)

8000

10000

Single Phase Flow in Pipes... Flowline

1 ½”

Bottom Hole Flowing Pressure (psi)

5000

2” 2 1/2”

4500 4000

3”

3500 3000 2500 2000 1500 Tubing 3 1/2”

1000 500 0 0

2000

4000

6000

Flowrate (bpd)

8000

10000

Single Phase Flow in Pipes... Flowline

Bottom Hole Flowing Pressure (psi)

5000

2”

1 ½”

2 1/2”

4500 4000

3”

3500 3000 2500 2000 1500

Tubing 2 3/8” Tubing 2 7/8”

1000

Tubing 3 1/2”

500 0 0

2000

4000

6000

Flowrate (bpd)

8000

10000

Single Phase Flow in Pipes... 

The natural equilibrium flowrates for different tubing and flowline sizes are:

Natural Equilibrium Flowrate (bpd) Flowline Tubing

1 1/2”

2”

2 ½”

3”

2 3/8”

2307

3376

3925

4156

2 7/8”

2372

3605

4319

4649

3 ½”

2398

3705

4507

4894

Single Phase Flow in Pipes...

Natural Equilibrium Flowrate (bpd)

6000

4 1/2” 3 1/2” 2 7/8”

5000

2 3/8”

4000

Tubing String

3000 2000 1000 0 0

1

2

3

Flowline Diameter (in)

4

5

Single Phase Flow in Pipes... 

 





The last example will serve also to illustrate an important aspect of oil production. This is the production control. Usually for naturally flowing wells (and sometimes under certain conditions for artificially lifted wells) we must exert some control over production conditions to limit the value of the natural flow equilibrium flowrate. The way used to reduce the natural equilibrium flowrate is by inclusion of a localized pressured drop in the system. The device used to create this localized pressured drop is called choke. Localized pressure drop can also be caused by:  Restrictions  Wax, Scale, etc…  Subsurface Safety Valves  etc…

Single Phase Flow in Pipes...

Bottom Hole Flowing Pressure (psi)

5000 4500

Operational Range with Choke

4000

Pchoke

3500 3000 2500 2000 1500

Natural Equilibrium Flowrate

1000 500 0 0

2000

4000

6000

Flowrate (bpd)

8000

10000

Darcy’s Experiment 

In 1856 Darcy (1803 – 1858) performed experiments for the design of sand filters for water purification in France

Sand

P + dP

A

q Viscous Fluid

P

 dx

Darcy’s Law 

After several experiments Darcy concluded that:

A dP q is proportion al to  dx kA dP q  dx where k is defined as the permeability of the porous media  In 1933 it was proposed to measure permeability in a unit called Darcy

Darcy’s Law… 

The unit of k can be obtained by:

q dx k A dP 3

2

2

L MLt 1 t L k   2 2 2 L  L2 t t L L ML 

The permeability k has the unit of area

Darcy’s Law… 

The definition of 1 Darcy is the permeability of a porous media that will allow the flow of 1 cm3/s of a fluid with 1 cp viscosity when the pressure gradient is 1 atm/cm and the flow area is 1 cm2

q dx k A dP cm 3 cp cm  13 2 1 Darcy  1  9.869  10 m 2 s cm atm

Darcy’s Law… 

The value of 1 Darcy by definition is then:

1 D  9.869  10 13 m 2 

How big is 1 Darcy?  Human hair thickness is about 60 m = 6x10-5

m. Its cross sectional area is

Ahair 

pd

2



p 6  10



5 2

4 4  28.27  10  10 m 2  2865 Darcy

Darcy’s Law…

kA dP q  dx q – Liquid flow rate a – Cross sectional area  – Liquid viscosity dP/dx – pressure gradient

Darcy’s Law 

Example of application of Darcy’s Law  Obtain relationship between the steady

state flow rate and the pressure at the perforations for a well in a reservoir with a radius re 

Assumption:  Pressure at the edge of the reservoir, re, is

constant and equal to Pr  Fluid is incompressible

P Pr

Pwf rw dr

h

A

r

re

q

r

Darcy’s Law…

kA dP q  dx

A  2p r h

2p k h dP q r  dr

Darcy’s Law…

2p k h dP q r  dr 



This is an ordinary differential equation (ODE) that can be solved to obtain pressure as a function of the radial distance To properly solve this ODE we must impose a boundary condition

Steady State Radial Liquid Flow 

Steady-State Radial Horizontal Liquid Flow



For liquids, the product of Boμo is approximately constant over a fairly wide pressure range so that for practical purposes

Steady State Radial Gas Flow 

Steady-State Radial Horizontal Gas Flow

where real-gas potential Δm is defined as:

Steady State Radial Gas Flow

A typical plot of p/μz vs. p. Twice the area under the curve between any two pressures represents the real-gas potential difference. Note that at high pressures, p/μz is approximately constant. Also, although it is not readily apparent from the plot, at low pressures the product μz is approximately constant.

Typical result of values for Δm are usually determined by numerically integration from gas viscosity and zfactor data from measurements or correlations.

Real Gas Potential Form of The Compressible Flow Equation 1   mp    i c i mp  r  r r  r  k t This form of the compressible flow equation is especially used at intermediate pressures (1000-4000psia [6895 -27580 kPa]) where the variation of z with pressure is strongly non-linear.

Pressure-Squared Form At low pressures (<1000 psia [6895 kPa]), the z product of a hydrocarbon gas remains constant. In this low pressure range, the real gas potential transformation simplifies to: p

p 1 2 mp    dp  p z 2 z 0

1   p 2   c p 2  r   r r  r  k t

Pressure Form At high pressures (>4000 psia [27580 kPa]), the variation of z with p exhibits a linear relationship. In other words, the p/z representing the slope of this linear relationship remains constant

p

p p mp    dp  z z 0

1   p   c p r  r r  r  k t

Skin

pwf, ideal

pwf, real

ps

Non-Darcy Flow In gas wells, there may be a significant non-Darcy component of flow that results in an additional potential difference that depends on the square of the flow rate. The non-Darcy effect appears in well-deliverability equations as a flow-rate-dependent skin,

where S is the fixed or "mechanical" skin, and S′ represents the total apparent skin including non-Darcy effects. D is called the non-Darcy-flow coefficient, having units of D/Mscf or d/std m3. Although the non-Darcy coefficient may be calculated from laboratory measurements of the nonDarcy coefficient, it is typically determined in the field from well tests.

Transient Flow

Deliverability Testing Stabilized-Deliverability Test. A stabilized deliverability, sometimes called a four-point or backpressure test, is conducted by producing a well at four rates. In this test, each rate is run long enough to reach stabilized conditions. To be theoretically valid, "stabilized" should mean to pseudosteady state, although in practice tests are sometimes run only until little variation in well flowing pressure is observed.

Deliverability Testing The most widely used method of presenting such data was first suggested by Rawlins and Schellhardt. This method is not based on the pseudosteady-stateflow equations but is based on an empirical observation that Δp2 vs. qg plotted on a log-log graph typically lies on a straight line. The equation for a straight line on a loglog graph is

Deliverability Testing Isochronal Tests. Another type of test often run on gas wells is the isochronal test. The difference between an isochronal test and a stabilized-deliverability test is that the flow periods are not run long enough to reach stabilized flow. This is done to shorten testing time and to conserve gas, particularly where no pipeline is available. Note that although the shut-in times are sufficiently long to approach initial reservoir pressure, the producing times are not long enough to reach pseudosteady state.

Deliverability Testing Lines are drawn through a Δp2-vs.-qg or Δψ/qsc-vs.-qsc plot at common producing times; that is, during each flowing period, pressures are read at fixed times since the initiation of flow. For a given producing time, a line through the data should have a constant slope. This slope should be the same no matter what the producing time because the transient drainage radius does not depend on flow rate. Thus, from each flow period, data are plotted for identical producing times.

Deliverability Testing A modified isochronal test is also sometimes run in which the shut-in times are also shortened. This type of test also works well if the value of pressure at the end of the last shut-in period is used in place of the average reservoir pressure.

Gas-Well Deliverability Relationships In general, the most theoretically valid deliverability equation should be used:

Solving for qg:

In some circumstances, it may be desirable to use either the C/n equation,

or a productivity index (PI) equation,

Gas In-place using Volumetric Method

Exercise 1

Gas in-place (cont.)

Gas Reserve

Gas Recovery Factor

Exercise 2

Exercise 2 (Cont.)

Exercise 2 (Cont.)

Exercise 3

Volumetric Gas Reservoirs  Volumetric method is an alternative to the MBE methods  The derivations are based on the important assumption that the only reservoir drive mechanism is simply the gas expansions  The gas expansion is a very efficient drive mechanism for the natural gas reservoirs  Even though gas saturations at the abandonment can be quite high, ultimate recoveries from the volumetric gas reservoirs could be as high as 80-90% of the original gas in-place

Volumetric Gas Reservoirs  Volumetric method is an alternative to the MBE methods  The derivations are based on the important assumption that the only reservoir drive mechanism is simply the gas expansions  The gas expansion is a very efficient drive mechanism for the natural gas reservoirs  Even though gas saturations at the abandonment can be quite high, ultimate recoveries from the volumetric gas reservoirs could be as high as 80-90% of the original gas in-place

Gas Reservoirs w/ Water Influx  Natural water influx from an aquifer occurs when the pressure at the reservoir/aquifer boundary is reduced following gas production from the reservoir  Consequently, the PV occupied by the gas decreases by an amount equal to the net volume of water entering the reservoir and remaining un-produced  If the initial gas saturation Swi and the residual gas saturation Sgr at abandonment are known, then, we can still use the volumetric equations to calculate the gas reserves. We now have Gp=G-Ga as:

Gas Reservoirs w/ Water Influx  or

 where the recovery factor RF is now given as:

 These equations are derived under the implicit assumption that the gas sweep efficiency for water is 100%; i.e., water efficiently displaces the gas. However experimental studies suggest that significant gas volumes can be bypassed and trapped by an advancing water front

Gas Reservoirs w/ Water Influx  To account for the un-swept portions of the reservoir, we introduce a volumetric sweep efficiency factor, EV, into the volumetric equation as follows:  In the above equation, note the reservoir volume occupied by the trapped gas due to inefficiency of displacement during the water encroachment. Now, similarly, we can re-write this equation as:

 Which can be re-arranged to yield

Gas Reservoirs w/ Water Influx  The recovery factor becomes:

 Because gas is often bypassed and trapped by the encroaching water, RF values for gas reservoirs with water drive can be significantly lower (<0.7) than for volumetric reservoirs produced by simple gas expansion (0.8-0.9)

Gas Reservoirs under Water Drive  Production Strategies: As water invades the reservoir, the pressure is maintained at a higher level than if there were no water encroachment. This leads to higher abandonment pressures for water-drive gas reservoirs. Because the main mechanism of recovery is gas expansion, RF values are significantly lower

Material Balance Method for Gas Reservoir

Volumetric Dry Gas Reservoirs

 Tank-type model:

Material Balance for Gas Reservoirs  Rearranging the above equation

At least 3 years of production history

p/z

pi/zi

Slope = -pi/(ziG)

Gp

G (initial Gas in place)

Material Balance for Gas Reservoirs  The equation suggests that, if the reservoir is volumetric, a plot of p/z versus Gp should be a straight line, from which we can estimate both original gas in place, G, and gas reserve at some abandonment conditions.

Exercise 1

Exercise 1

Dry-gas Reservoir w/ Water influx

 Usually G and We are unknown.

Exercise 2

Exercise 2 (Cont.)

Exercise 2 (Cont.)

Exercise 3

Exercise 3 (Cont.)

Volumetric Geopressured Gas Reservoirs  MBE for abnormal high pressured (geopressured) reservoir

Volumetric Geopressured Gas Reservoirs

System Performance      

Tubing and choke sizes. Amount of entrained liquids (condensate and water). Accumulation of sand or debris in the wellbore. Flowline pressure drops. Compression ratios across compressors. Pressure losses in separators.

Reservoir Deliverability The intersection points of the well- and equipment-performance curves can be used to construct a relationship between average reservoir p/z and maximum well deliverability. This curve represents the rate that the combined well and equipment design is capable of delivering at any particular average reservoir pressure. Note that this curve can be used to determine the abandonment p/z by knowing the economic limit (i.e., the minimum economic rate). Another method would be to first determine reservoir deliverability constants:

Production Forecast

Example production rate-vs.time curve for a gas reservoir.

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