transportation and storage handout 4

PEE 372: Transportation and Storage Instructor : Dr. Salem S. Al-Marri

Office: Building No. 6 Office No. 149

Petroleum Engineering Technology

College of Technological Studies (CTS)

Public Authority for Applied Education and Training (PAAET)

Email: [email protected]

Course Contents

• Fluid Properties

• ----

• ----

• ----

• ----

• ----

• ----

Course Grade System

Lecture Grade: 100%

The lecture grades will be divided as follows:

Attendance 10%

H.W. & Quizzes 10%

Mid-Terms 40%

Final Exam 40%

Total 100%

Lecture: Section# 1: Thursday 3-5 PM

1

Transportation and Storage

Lecture # 1

Introduction to

Fluid Properties

2

3

Classification of Reservoir Fluid Chemical Properties

PNA and combinations (Paraffinic-Naphthenic-Aromatics)

Resin and asphalthene content

Physical Properties

Phase density (liquid, gas, solid)

Compressibility

Viscosity

Formation volume factor

Gas Oil Ratio

Surface tension

Heat capacity

Thermal conductivity

Pour and Cloud points

4

Classification of Reservoir Fluid Chemical Properties (Compositions)

Organic Components (Hydrocarbon)

PNA and combinations

Paraffinic

All normal paraffins to C10H22

Isobutane

2-Methylbutane

2,3-Dimethylbutane

2-Methylpentane

3-Methylpentane

2-Methylhexane

3-Methylhexane

2-Methylheptane

2,6-Dimethylheptane

2-Methyloctane

5




Naphthenic

Cyclopentane

Cyclohexane

Methylcyclopentane

1,1-Dimethylecyclopentane

Methylcyclohexane

1,3-Dimethylecyclohexane

1,2,4-trimethylecyclohexane

6




Aromatics

Benzene

Toluene

Ethylbenzene

Xylene

1,2,4-Trimethylbenzene

Inorganic Components

Nitrogen

Carbon Dioxide

Hydrogen Sulfide

7

Classification of Reservoir Fluid Physical Properties

Phase density (liquid, gas, solid)

Compressibility

Viscosity


Gas Oil Ratio

Surface tension

Heat capacity

Thermal conductivity

Pour and Cloud points

8


Phase density (liquid)

Specific gravity of a liquid

API gravity

o

m

V

),(

),(

11

11

TP

TP

w

oo

141.5131.5o

o

API

Phase Density (Liquid)

Example : The density of a stock-tank oil at 60 oF is 51.25 lbm/cu ft. Calculate the specific gravity and gravity in oAPI.

Solution

351.25

oo

w

lbm

ft

362.37

lbm

ft

oSecond, calculate gravity in A

0.8217

141.5 141.5131.5 131.5 40.7

0.8

PI:

217

o o

o

API API

10


Phase density (gas)

• From real gas law

g

g

g

g

g

g

g

g

g g g g

g

g g

m

V

PVZ

n RT

mn

M

PV M m PMZ

m RT V ZRT

11


Phase density (gas)

• Gas Specific Gravity

• Assumption Z-factor is very close to one for standard conditions

g

g

PM

Z RTGas DensitySG

Air Density

air

air

PM

Z RT

28.92

28.92

g g

air

g

M M

M

MSG


Example : Calculate the mass and density of methane gas contained at 1000 psia and 68 oF in a cylinder with volume of 3.20 cu ft.

a) Assume that methane is an ideal gas.

b) Assume that methane is an real gas.

Solution:

a) Assume ideal g

,

100

s

0

a

PMVm where Z is equal one for ideal gas

ZRT

psia

m

16.04lbm

lbmol 33.20 ft

10.732psia 3. ft

lbmol . oR(68 460) oR

3 3

9.1

9.12.844

3.20

lbm

m lbm lbm

V ft ft


Example : Calculate the mass and density of methane gas contained at 1000 psia and 68 oF in a cylinder with volume of 3.20 cu ft.

a) Assume that methane is an ideal gas.

b) Assume that methane is an real gas.

Solution:

, 0.89(from McCain fig

a

ure 3

) Assume real gas

.2)

1000

PMVm where Z is equal

ZRT

psia

m

16.04lbm

lbmol 33.20 ft

0.89 10.732psia

3. ft


3 3

10.2

10.23.18

3.20

lbm

m lbm lbm

V ft ft

15

Pseudocritical Properties of Natural Gases

Pseudoreduced Pressure

Pseudoreduced Temperature

pc

prP

PP

pc

prT

TT

16


If only the specific gravity and Mw of of the gases is known then charts are available to estimate these pseudocritical properties (McCain figure 3-10 ).

17


Naturally the degree of accuracy is reduced substantially. We well see methods when compositional information is available, in this case:

cii

N

i

pc PyPc

1

cii

N

i

pc TyTc

1

18


Once Z is evaluated you can find the gas density as

3/ ftlbm

V

Mg

19

Z-factor

chart for

low

reduced

pressures

20


Compressibility

• Definition

Derivative is evaluated at constant T=TA and specified pressure P=PA

1

,

A

g A A

T

VC P T

V P

Isothermal Gas

Compressibility

P1

V2 V1

TB

P2

Vave= (V1+V2)/2

ATave

AAgPP

VV

VTPC

21

211,

TA

PA

21

http://classicpins.com/redwhitandbl.html








22


Compressibility

• Definition

Derivative is evaluated at constant T and specified pressure P

1

,g

T

VC P T

V P

P

Z

ZPP

V

VCg

111

Isothermal Compressibility

Using ideal gas equation

The simplest equation of state is that for ideal gas

23

2

2

1

T

g

nRTPV nRT or V

P

V nRT

P P

nRT PC

V P

nRT

nRT

2

1for ideal gas

PP









Isothermal Compressibility

Using real gas equation

24

2

1

T

T

g

T

ZV nRT

P

ZP Z

V PnRT

P P

V PC

V P nRT

nRT

Z

2

1for Realgas

1 1

T

g

T T

ZP Z

PP

ZC P

Z

Z PZ

P ZP P









Isothermal Compressibility (Cg) of an Ideal Gas

Example 1: The following table gives volumetric data at 150 oF for a natural gas. Determine the coefficient of isothermal compressibility for the is gas at 150 oF and 1000 psia.

Pressure psia

Molar Volume Cu ft/lbmole

700 8.5

800 7.4

900 6.5

1000 5.7

1100 5.0

1200 4.6

1300 4.2

Isothermal Compressibility (Cg) of an Ideal Gas

Solution

26









Isothermal Compressibility (Cg)

3

3

7.1 50.007

800 1100

,

1 1

5.7m

g

mg

ftslope

lbmole

Second compute

VC

V p ft

lbmole

C

3

0.007ft

lbmole

6 11300 10 psipsia

Isothermal Compressibility (Cg) of an Real Gas

Example 2: Compute the coefficient of isothermal compressibility of ethane at 1000 psia and 212 oF.

Isothermal Compressibility (Cg) of an Real Gas

Solution

0.8

0.65

1250 750

0.725

Isothermal Compressibility (Cg) of an Real Gas Solution

Z = 0.72 from figure3-2

1

1 1

o

g

slopeof tangent to 212 F

isotherm at p = 1000 psia, from figure3-2

second calculate C

1413.79 1

0.8 0.650.0003

750 1250

1 1 1 10.0003 0.00141379

1000 0.

0

725

T

g

T

g

Zpsi

p

ZC psi psi

p Z p psia

C

6 1psi

Example The following gas contained in 43560 cu ft. at 9300 psia and 290°F

1) Calculate the molecular weight

2) Calculate the gas specific gravity

3) Calculate the pseudocritical temperature

4) Calculate the pseudocritical pressure

5) Calculate the pseudoreduced temperature

6) Calculate the pseudoreduced pressure

7) Calculate the mass, assume ideal gas

8) Calculate the number of moles, assume ideal gas

9) Calculate the density, assume ideal gas

10) Determine the Z-factor

11) Calculate the mass, assume real gas

12) Calculate the number of moles, assume real gas

13) Calculate the density, assume real gas

14) Calculate the gas formation volume factor

15) Calculate isothermal gas compressibilty

component Composition Mole fraction

Methane 0.850

Ethane 0.090

Propane 0.040

n-Butane 0.020

Total 1.000

Example Solution

component Composition Mole fraction

Mi Tci, R Pci, Psia yiMi yiTci yiPci

Methane 0.850 16.04 343.3 666.40 13.634 291.805 566.44

Ethane 0.090 30.07 549.9 706.50 2.7063 49.491 63.585

Propane 0.040 44.097 666.1 616.00 1.76388 26.644 24.64

n-Butane 0.020 58.123 765.6 550.60 1.16246 15.312 11.012

Total 1.00 19.26664 383.252 665.677

1) Calculate the molecular weight

2) Calculate the gas specific gravity

1

19.26cN

g i ii

lbmM y M

lbmol

19.260.66

29 29

g

g

M

Example 3) Calculate the pseudocritical temperature

4) Calculate the pseudocritical pressure

5) Calculate the pseudoreduced temperature

6) Calculate the pseudoreduced pressure

1

383.25cN

o

pc i cii

T y T R

1

665.67cN

pc i cii

P y P psia

750 o

pr

pc

T RT

T

383.25 oR1.96

9300pr

pc

psiaPP

P

665.67 psia13.97 14

Example 7) Calculate the mass, assume ideal gas

8) Calculate the number of moles, assume ideal gas

9) Calculate the density, assume ideal gas

,

9300

PMVm where Z is equal one for ideal gas

ZRT

psia

m

19.26lbm

lbmol 343560 ft

10.732psia 3. ft


969694.4 lbm

969694.4m lbmn

M

19.26lbm

50330.23lbmol

lbmol

3 3

969694.422.26

43560

m lbm lbm

V ft ft

Example 10) Determine the Z-factor

Z = 1.346 (McCain figure 3-10 ).

1.346

Example 11) Calculate the mass, assume real gas

12) Calculate the number of moles, assume real gas

13) Calculate the density, assume real gas

, 1.346

9300

PMVm where Z is equal for real gas

ZRT

psia

m

19.26lbm

lbmol 343560 ft

1.346 10.732psia

3. ft


720426.7 lbm

720426.7m lbmn

M

19.26lbm

37392.4 lbmol

lbmol

3 3

720426.716.54

43560

m lbm lbm

V ft ft

Example 14) Calculate the gas formation volume factor

15) Calculate isothermal gas compressibility, assume ideal gas

ZT0.0282 0.0282

Pg

psiacuftB

scf

oR

1.346 750 oR

9300 psia

60.003061065

0.003061065

3061.065 10gB

Assume ideal gas

1 1gC

p Z

1

Z

p

1

1

1 10.000107527

0.000107

9300

7

0

52

T

gC

p

ps

sip sia

i

p

38


Viscosity • Viscosity is a measure of the resistance to flow exerted by a

fluid, This is called dynamic viscosity and has units of

centipoise = g-mass / 100 sec cm

• Kinematic viscosity is dynamic viscosity per density, units are in

centistokes = centipoise /g/cc

• Needs of Crude Oil Viscosity

Calculation of two-phase flow

Gas-lift and pipeline design

Calculate oil recovery either from natural depletion or from recovery techniques such as waterflooding and gas-injection processes

Variation of Oil Viscosity O

il V

isco

sit

y

T = constant

Pb

Single Phase Flow

Two Phase Flow

Gas Out of

Solution

39

40



Gas formation volume factor Bg

Reservoir Conditions VR

Standard Conditions VSC

41




Bg=

Rg

SC

VB

V

[res bbl/SCF] or [ft3/SCF]

Gas Formation Volume

Factor

SC

SCSC

P

nRTZ

P

ZnRT

Bg

42










Factor

43

Since Tsc = 520 oR, Psc =14.696 psia, for all practical purposes Zsc = 1, then

14.6

ZT0.02

96

1 520

82P

scg o

sc sc sc sc sc sc

sc c

g

s

ZnRT ZnRTPZ T Z T psiaP PB

Z nRT Z nRT Z T P R P

cuftB

cf

P P

s









Gas Formation Volume Factor Bg

43.47 10 ( )g

TB Z T in K and Pin MPa

P

0.0282 ( )g

TB Z T in R and Pin psia

P

( )g

sc sc sc

P V TZ V Expelled Gas

P V T

44



Factor

Gas Formation Volume Factor

Bg

Pressure45









Oil Formation Volume Factor

46

Volume of Oil + Dissolved gas at Reservoir Pressure & Temperature

Volume of Oil entering Stoc

Reservoir barrels (bbl)

Stock tank barrels (

k tank at Standard Pressure & Temper

STB)

ature o

Units

B

Total Formation Volume Factor

Also called Two-phase formation volume factor

ssbgot RRBBB

Units…

bbl/STB + bbl/SCF * (SCF/STB)

47

Total Formation Volume Factor Bt

Hg

Oil Bob

Hg

Gas

Oil Bo

Bg(Rsb-Rs)

Pb

48

General Shape of Bt Bo,

Bt

Reservoir Pressure

Pb

Reservoir T = constant

Bt=Bo

Bt

Bo

49

50

Classification of Crude Oils Physical Properties

Solution Gas Oil Ratio, Rs

How much gas is dissolved in the oil volume per volume basis

Rs depends upon pressure

Units [= ] SCF gas /STB oil

General Shape of Solution Gas Oil Ratio (Rs)

Rs

Reservoir Pressure

Pb

Reservoir T = constant

51

52

Classification of Reservoirs based on Phase Diagram

Gas Reservoirs (Single Phase)

Gas Condensate Reservoirs (Dew-Point Reservoirs):

Undersaturated Solution-Gas Reservoirs (Bubble-Point Reservoirs):

DRY GAS RESERVOIRS: – GOR > 100,000 SCF/STB

– No liquid produced at surface

– Mostly methane

53

Classification of Reservoirs based on Production and PVT

data









54


data

GAS CONDENSATE RESERVOIRS: GOR between 70,000-100,000 SCF/STB

Density greater than 60 ºAPI

Light in color

C7+ composition < 12.5%

55


data

VOLATILE OIL RESERVOIRS: GOR between1,000-8,000 SCF/STB

Density between 45-60 ºAPI

Oil FVF greater than 2.00 (high shrinkage oils)

Light brown to green in color

C7+ composition > 12.5%

56

BLACK OIL RESERVOIRS: GOR less than 1,000 SCF/STB Density less than 45 ºAPI Reservoir temperatures less than 250 ºF

Oil FVF less than 2.00 (low shrinkage oils)

Dark green to black in color C7

+ composition > 30%


data

Separators

iSTx

)STB/scf( GOR and

y iST

)STB/scf( GOR and

y iSP

iSTx

Wellhead

iSPx

SToilgas

STgas

v

SToilgasSPoil

SPoilgas

SPgas

v

molelbmolelb

molelbf

molelbmolelbmolelb

molelbmolelb

molelbf

ST

SP

57

Separators

)STB/scf( GOR and

y iST

)STB/scf( GOR and

y iSP

iSTx

Wellhead

Usually vented

Note for a black oil have more liquid produced than for a wet gas –

Less oil shrinkage and lower GOR 58

PETE 310

Lecture # 17

Chapter 10 – Properties of Black Oils - Reservoir Fluid Studies

59

Fluid Properties for Reservoir Engineering Processes

The fluid properties of interest to the Reservoir Engineer are those that affect the mobility of fluids within the reservoirs these are used in material balance calculations Properties at surface conditions for

transportation and sales (API, viscosity, oil quality)

60

Fluid Properties for Reservoir Engineering Processes

PVT properties are determined from 5 specific lab procedures

– Flash liberation tests – Differential Liberation Tests – Viscosity Measurements – Separator Tests – Compositional measurements

61

Fluid Properties Determined

Oil Properties – Bubble Point Pressure – Bo

– Rs

– Bt

– Co and mo

Gas properties – z – Bg and mg

Compositions oil & gas

62

Oil Sampling Procedures

Bottom hole or subsurface samples

Separator Samples

SAMPLE BEFORE RESERVOIR PRESSURE DROPS BELOW Pb

63

Flash Vaporization Test

Temperature of Test = Reservoir Temperature

V t1

V t2

V t3

= V

b

V t5

V t4

oil oil oil oil

oil

gas gas

Hg Hg Hg Hg

Hg

P 1 >> P

b P

2 > P

b P

3 = P

b P

4 < P

b P

5 < P

4

1 2 3 4 5

64


Properties determined

– Pb

– Co

65


Example : The data from a flash vaporization on a black oil at 220 oR are given below. Determine the bubble-point pressure and prepare a table of pressure and relative volume for the reservoir fluid study

Pressure, psig Total volume, cc

5000 61.030

4500 61.435

4000 61.866

3500 62.341

3000 62.866

2900 62.974

2800 63.088

2700 63.208

2605 63.455

2591 63.576

2516 64.291

2401 65.532

2253 67.400

2090 69.901

1897 73.655

1698 78.676

1477 86.224

1292 95.050

1040 112.715

830 136.908

640 174.201

472 235.700


Example : Solution

2619.3 psig

63.284 cc


Example : Solution

Pressure, psig Total volume, cc Relative Volume

5000 61.030 0.9644

4500 61.435 0.9708

4000 61.866 0.9776

3500 62.341 0.9851

3000 62.866 0.9934

2900 62.974 0.9951

2800 63.088 0.9969

2700 63.208 0.9988

2619.3 63.284 1.0000 Bubble Point

2605 63.455 1.0027

2591 63.576 1.0046

2516 64.291 1.0159

2401 65.532 1.0355

2253 67.400 1.0650

2090 69.901 1.1046

1897 73.655 1.1639

1698 78.676 1.2432

1477 86.224 1.3625

1292 95.050 1.5020

1040 112.715 1.7811

830 136.908 2.1634

640 174.201 2.7527

472 235.700 3.7245

Lecture 2.1: Continuity and Bernoulli’s Equation

Incompressible Fluid Flow

• Today’s Lecture

– Volume and Mass Flow Rate

– Continuity equation

– Energy in a fluid

– Flow Work

– Bernoulli’s Equation



• Volumetric Flow Rate

– Volume flowing per second along a pipe

• Volume of fluid in element = V = AL

• Velocity of fluid = L/Δt = c

• Volumetric Flow Rate = Volume/Δt = AL/ Δt = Ac

3

sV Ac

m



• Continuity Equation

Density of entry = ρ1 Density of entry = ρ2

Area at entry = A1 Area at entry = A2

Velocity at entry = C1 Velocity at entry = C2

1 1 1m AC 2 2 2m A C




– For incompressible flow, density remains constant

1 2

1

1 1 2AC 2 2

1 1 2 2

1 2

A C

AC A C

V V




– Kinetic Energy (Energy due to velocity as a whole):

– Potential Energy (Energy due to the height of the fluid):

– Flow Work Energy (energy required to displace fluid):

– Internal Energy (energy due to temperature of fluid): U • (Normally assume temperature change to be zero, so not normally considered)

21

2kE mC

pE mgz

W pV




– Flow Work

• Force being applied to fluid in plane XX = pressure x area = pA

• Force is moving, so is doing work (work is a form of energy)

• Work done = Force x Distance = pAL

• Volume of element = AL = V


• Bernoulli’s Equation

– Energy in a fluid

• No energy added or extracted from fluid

• Principle of conversion of energy applies

2 2

1 2

1 2

1 1 2 2

1 1

2

1 2

2mC Kinetic Energy mC

mgZ Potential Energy

Energ

mgZ

p V

y at Ener

Flo

gy a

wWork Energy p

t

V



– Incompressible so V is constant: V1 = V2 = V

2 2

1 1 1 1 2 2 2 2

1 1

2 2p V mC mgZ p V mC mgZ constant

2 2

1 1 1 2 2 2

2 2

1 1 1 2 2 2

1 1

2 2

1 1

2 2

m m m mp C gZ p C gZ constant

V V V V

m=

V

p C gZ p C gZ constant



2 2

1 1 1 2 2 2

1 1

2 2p C gZ p C gZ constant

21

2p C gZ Total pressure

Static Pressure Hydrostatic Pressure

Dynamic Pressure



– Conditions

• No heat transfer (adiabatic)

• No work done (no pump or turbines)

• Flow is frictionless (no Temperature change)

• Flow is incompressible (ρ = constant)

– Liquids

– Gases < 100 m/s

2 2

1 1 1 2 2 2

1 1



• Summary

– Volume Flow rate

– Mass Flow rate

– Continuity Equation (conservation of mass/volume)

– Fluid Energy

– Flow Work


V AC

m AC

2 2

1 1 1 2 2 2

1 1



• Example

Water flow through a pipe shown below at the rate

80 L/s. If the pressure at point 1 is 180 kPa, find (a)

the velocity at point 1, (b) the velocity at point 2, and

© the pressure at point 2.


• Example-Solution

a)

b)

c)

1 1 1 2

80 80

80

lt lt

V lt sAC Ct s r

2 2(16 /100)

s

m

31m

1000 lt0.99432

m

s

1 1AC 1 12 2 2

2

ACA C C

A

2

1r

2

1 22

2

0.160.99432 3.9773

0.08

mC

sr

2 2

1 1 1 2 2 2

1180 100

1

02

1

2 2p C gZ p C

a

gZ

kP

kg

m3

220.99432

m

2s

1 N

1kg . m

2s

1

1000

kPa

N2m

3 21000 9.81 0

kg m

m s

2

11000

2p

kg

m3

223.9773

m

2s

1 N

1kg . m

2s

1

1000

kPa

N2m

1000kg

3m9.81

m

2s2 m

1 N

1kg . m

2s

1

1000

kPa

N2m

2

2

0.4943 7.91 19.6180 0

152.

2

965

kPa kP kPa

p

Pa

a

p a

kP

k


• Example

Water enters atypical garden hose of diameter 1.6 cm

with velocity of 3 m/s. Calculate the exit velocity of

water from the garden hose when a nozzle of

diameter 0.5 cm is attached to the end of the hose.

Solution

1 1AC 1 12 2 2

2

ACA C C

A

2

1r

2

1 22

2

0.0083 30.72

0.0025

m mC

s sr


• Example

Oil flows throw a pipe of radius R with speed V.

Some distance down the pipeline, the pipe narrows

to half its original radius. What is the speed of the oil

in the narrow region of the pipe?

Solution:

1 1AC 1 12 2 2

2

ACA C C

A

2R

12

2 1

14

2

4C

C CR

C

Lecture 1.1 flow with Friction & Laminar Flow

Part I: Introduction, Fluid Flow with Friction & Type of Flow

• Today’s Lecture – Fluid flow with friction – Nature of flows: Laminar and Turbulent – Shear Stress and Fluid Viscosity – Laminar Flow

• Flow Velocity • Flow Rate • Pressure drop • Power

– Example


• Fluid Flow with Friction

– Assumption has been no friction

– Bernoulli’s equation:

2

2 2

1 1 1 2 2 2

1

2

1 1

2 2

Total Pressure p C gz constant

p C gz p C gz


• Fluid Flow with Friction

– Now, shear stresses in fluid and with pipe

– Shear stress works against flow


• Fluid Flow with Friction – Bernoulli’s equation:

– How do we determine pressure drop (ΔP)?

2 2

1 1 1 2 2 2

1 1

2 2p C gz p C gz P

Lecture 1.1 flow with Friction &

Laminar Flow

• Nature of Flow

– ΔP due to shear stress

– Vary depending on nature of flow

– Two types (in general) – Laminar

– Turbulent


Laminar Flow

• Laminar Flow:

– Layers of adjacent fluid slide over each other

– Streamlines are straight

– Flow near wall slower than centre

– Example: honey falling off spoon


Laminar Flow

• Turbulent Flow:

– Particle paths irregular and chaotic

– Large scale mixing

– Flow in radial direction

– Example: smoke billowing from chimney


Laminar Flow • Transition Flow:

– Transition between laminar and turbulent flow related to fluid density, fluid viscosity, flow velocity and pipe diameter

– The Reynolds number is as the following:

– Where • Re: Reynolds number

• Ρ: Fluid density

• c: Flow velocity

• D: Pipe diameter

• : Fluid viscosity

cDRe

m


Laminar Flow

– The Reynolds number non-dimensional number:

cDRe

m

Reynolds Number Flow

Less than approx. 2000 Laminar

2000-4000 Critical Region

Above 4000 Partial or fully turbulent


Laminar Flow

• Fluid Viscosity & Flow Velocity

• Fluid Viscosity – Relative movement of sliding fluids layers

– Shear Stress

– = dynamic viscosity

– As increases, shear stress increase • Hi viscosity: honey, toothpaste

• Low viscosity: water, alcohol

• Generally, viscosity decreases with increase in temperature

Difference in speed (between layers)

Distance between layers m


Laminar Flow

• Shear Stress & Fluid Viscosity

– = dynamic viscosity

– = kinematic viscosity

• Relationship:

• Reynolds number

cD

Re

m

Lecture 1.2 Laminar Flow-Flow

Velocity

• Element of Fluid

– Pipe radius R

– Element radius r

– Length dx

– Force on LHS: pA

– Force RHS: -(p+dp)A

– Shear stresses:

– Velocity in pipe is: 22

14

R dp rC

dx Rm

Lecture 1.2 Laminar Flow-Flow

Velocity

• Velocity:

– Quadratic equation

– Parabolic velocity profile

– If r = R

• Velocity = 0

– If r = 0

• Velocity = Cmax

22

14

R dp rC

dx Rm

Pipe walls

Flo

w D

irec

tio

n

Cen

ter

lin

e o

f p

ipe

Cmix

0 0

Turbulent and Laminar Flow Profile

http://www.google.com.kw/url?sa=i&source=images&cd=&cad=rja&docid=hRBQ_BJlkm1U5M&tbnid=gTRe63Azz95hMM:&ved=0CAUQjRw&url=http://www.pumpfundamentals.com/pump_glossary.htm&ei=yYY8UdG9IseBtAamyIHQDA&psig=AFQjCNENCECHGM5FZt8-e52gRJ0VPTdd4Q&ust=1363007537597776

Lecture 1.3 Fluid Flow with Friction & Laminar Flow

• Part 3: Volumetric Flow Rate, Pressure Drop, Power & Summary

– Volumetric Flow Rate

• Cannot just use

• Velocity not constant across cross section

V AC

4

8

dpV R

dx

m

V

Lecture 1.3 Fluid Flow with Friction & Laminar Flow

• Fluid element:

• Consider pipe:

dp P

dx L

Lecture 1.3 Laminar Flow-Pressure

Drop

• Using:

• Velocity:

• Flow rate: 4 4

8 8

dpV R

dx

PV R

L

mm

dp P

dx L

22 22

14

14

R dp r R P rC

L RC

dx R mm


Drop

• From Flow rate:

• Rearranging:

• Bernoulli’s Equation:

4

8V

PV R

L

m

4

8 LVP

R

m

2 2

1 1 1 2 2 2

1 1

2 2p C gz p C gz P


Drop • Average Flow Velocity

• Pressure Drop

• Power

• Consider force applied to fluid

• Power = Force x Velocity – Force = pA

– Velocity = C

Power pAC pV Power Loss pV

Lecture 1.3 Laminar Flow-Pressure Drop

Laminar Flow: Example • Fluid flow with friction:

– Friction produces pressure drop


• Nature of flow determined by Reynolds number:

– Re < 2000 Laminar

– Re < 2000 Turbulant

• Flow Velocity:

– Parabolic profile

2 2

1 1 1 2 2 2

1 1

2 2p C gz p C gz P

cD cDRe

m

22

14

R P rC

L Rm


Summary of Today’s Lecture • Flow rate:

• Pressure Drop:

• Power Consumed:

Power consumed pV

4

8 LVP

R

m

4

8

PV R

L

m


Summary of Today’s Lecture • Example

Oil of density 900 kg/m3 and viscosity 0.17 Pas is pumped through a

75 mm diameter pipe 750 m long at the rate of 2.7 kg/s. if the critical

Reynolds number is 2300, show that the critical velocity is not

exceeded and calculate the pressure required at the pump and the

power required. The pipe is horizontal.


Summary of Today’s Lecture • Solution

Mass flow is given, we need velocity in order to find the Re and

volumetric flow in order to find the Δp and power

3

2

900 , 0.17 . , 75 , 750

2.7 Re 2300 ? ?

2.7

4

kgPa s D mm l m

m

kgm P power

s

kg

m mm Ac c

AD

m

900

s

kg

3m

20.0754

m

0.679

900

Re

m

s

kg

cD

m

m

30.679

m

s0.075 m

0.17 .Pa s

.Pa s

kg

m . s

The flow is l

2

a

69.

ar

6

min





3900 , 0.17 . , 75 , 750

2.7 Re 2300 ? ?

2.7

kgPa s D mm L m

m

kgm P power

s

kg

mV

m

900

s

kg

3

3

4

0.003

8 0.17 .8

m

s

m

Pa sLV

pR

m

750 m3

0.003m

4

40.075

2

s

m

.

.

kg

m s

Pa s

2492795.5

.

492795.5

kg

m s

kgp

2.m s

1

1

N

kg

2

.m

s

2

2

492795.5

492795.5

N

m

Np

m

2

1

100000

bar

N

m

9

4.93

4. 3

bar

p bar





3

2

900 , 0.17 . , 75 , 750

2.7 Re 2300 ? ?

492795.5

kgPa s D mm L m

m

kgm P power

s

Npower pV

m

m

3

0.003m

.

1478.4

.1478.4

N m

s s

N mpower

s

1

.1

W

N m

s

1478.4

1478.4

power W

W

Introduction and Fluids at Rest

Fluids Lecture 1.1 Hydrostatic Pressure


– Phases

– Types of fluids

– Density

– Pressure

• Hydrostatis Pressure

• Measurement of atmospheric pressure

• Measurements of liquid pressure

• Measurement of gas pressure



• Phases

– A substance can exist as

• A solid

• A liquid

• A gas

– e.g. Water: ice; water; steam



• Phases

– Solids

• Maintains shape

• Can withstand tensile (pulling), compressive and shear

sliding stress

– Fluids

• Includes liquids and gases

• Little resistance to a permanent change in shape

• Take on the shape of container



• Phases

– Fluids

• Two types of fluids (generally)

– Incompressible Fluids: Liquids

» Molecules arranged so a given mass of fluid will retain

virtually the same volume irrespective of pressure

• Compressible Fluids: Gases and Vapors

– Free molecular arrangement, so tend to fill any vessel

– Pressure changes produce considerable change in volume



• Phases

– Density

• Density is mass per unit volume

• Density of water:

• Relative density

3

( )

( )

mass kg mDensity

Volume m V

31000w

kg

m

ll

w



• Pressure – Closed cylinder

– The pressure is the force exerted per unit area

– Neglecting the weight of fluid:

– Units: • N/m2 or Pascal (Pa); 1 N/m2 = 1 Pa

• Bar: 1 bar = 105 N/m2

2

( )

( )

Force N Fpressure p

Area m A



• Pressure in a Fluid – At rest:

– Pressure in fluid acts equally in all directions

– Pressure exerted on a surface is normal to the surface

– At any two points in the same horizontal plane, pressure is equal



• Hydrostatic Pressure – Column of liquid of height hp

– Pressure in fluid due to the weight of the fluid above:

p = ρghp – Hydrostatic equation, and hp = pressure head

p

p

p

p

Volumeof liquid =cross - sectional area×height Ah

Mass of liquid =density volume Ah

Weight of liquid =mass g Ah g

Ah gPressure on base = weight of liquid/arae of base

A


Fluids Lecture 1.1 Atmospheric Pressure

• Atmospheric Pressure patmos

– Column of air: 1 m2 cross sectional area

– Weight, on average: 101300 N

– Atmospheric pressure is nearly 1bar

2

2

5

2

101300

101.300

1.013 10 1.013

atmos

Np

m

kN

m

Nbar

m


Absolute and Gauge Pressure

• Absolute Pressure

– Vessel completely empty

– Pressure in a vacuum is absolute zero of pressure

• Gauge pressure

– Subjected to patmos, so only indicates pressures

that differ from patmos

Atmospheric pressure = zero gauge pressure

Absolute pressure = gauge pressure + atmospheric pressure

abs g atmosp p p


Pressure Measurement

• Measurement: Barometer

– Used to measure absolute pressure of the atmosphere

• Pressure in the same horizontal plane is equal

• Mercury will rise up until pressure due to

weight will balance patmos

• At level xx:

p = patmos

where

p = ρghb

hb = barometric head

ρ = density of mercury = 13600 kg/m3


Piezometer Tube

• Measurement of liquid pressure

– Gauge pressure at xx:

pgauge = ρghp

– Absolute pressure at xx:

p = patmos + ρghp

hp = pressure head, measured in meters


U-Tube Manometer

• Gas pressure measurement

– Gauge pressure at xx:

• When p > patmos :

– Gauge pressure of gas:

pgauge = ρghp

– Absolute pressure of gas:

p = patmos + ρghp

hp = pressure head, measured in meters

Gas at

Pressure

p

patmos

hp


Lecture Summary


– Density, ρ: mass per unit volume; kg/m3

– Pressure, p: force/area; N/m3

– Hydrostatic pressure: p = ρghp where hp is pressure head

– Atmospheric pressure: 101300 N/m3 (avg)

– Measurement of patmos: barameter

– Measurement of liquid pressure: Piezometer

– Measurement of gas pressure: U-tube Manometer

Course Content

1. Fluid Properties Overview

2. Flow in Horizontal Pipeline o Liquid Phase

o Gas Phase

o Multiphase

Dr. Almarri

Eng. Aldousari

Petroleum Transportation & Storage

Engineering 124

• Oil and gas properties are very important in the pipeline

flow calculations

• API, Oil specific gravity, and Oil formation volume factor

• Gas density, gas formation volume factor, and gas specific

gravity.

Dr. Almarri

Eng. Aldousari


Engineering 125

Fluid Properties Overview

• Specific gravity of a liquid

• API gravity

Dr. Almarri

Eng. Aldousari


Engineering 126

),(

),(

11

11

TP

TP

w

oo

141.5131.5o

o

API

Crude Oil Properties

• Phase density (gas)

Dr. Almarri

Eng. Aldousari


Engineering 127

g

g

g

g

g

g

g

g

g g g g

g

g g

m

V

PVZ

n RT

mn

M

PV M m PMZ

m RT V ZRT

Dr. Almarri

Eng. Aldousari


Engineering 128

g

g

PM

Z RTGas DensitySG

Air Density

air

air

PM

Z RT

28.92

28.92

g g

air

g

M M

M

MSG

Dr. Almarri

Eng. Aldousari


Engineering 129

Pb

Oil

Vis

cosi

ty

T = constant

Single Phase Flow

Two Phase Flow

Gas Out of Solution


Dr. Almarri

Eng. Aldousari


Engineering 130

Reservoir Conditions VR

Standard Conditions VSC

Dr. Almarri

Eng. Aldousari


Engineering 131

Rg

SC

VB

V

Bg=

Dr. Almarri

Eng. Aldousari


Engineering 132


SC

SCSC

P

nRTZ

P

ZnRT

Bg

Dr. Almarri

Eng. Aldousari


Engineering 133

Since Tsc = 520 oR, Psc =14.696 psia, for all practical purposes Zsc = 1, then

14.6

ZT0.02

96

1 520

82P

scg o

sc sc sc sc sc sc

sc c

g

s

ZnRT ZnRTPZ T Z T psiaP PB

Z nRT Z nRT Z T P R P

cuftB

cf

P P

s

Oil Formation Volume Factor

Dr. Almarri

Eng. Aldousari


Engineering 134

Volume of Oil + Dissolved gas at Reservoir Pressure & Temperature

Volume of Oil entering Stoc

Reservoir barrels (bbl)

Stock tank barrels (

k tank at Standard Pressure & Temper

STB)

ature o

Units

B

1. Introduction

• We will talk about the transport of fluids from the

wellhead to the facility where processing of the fluids

begins.

• For oil production:

The facility is typically a two or three phase separator.

• For gas production:

The facility is a gas plant which is a compressor

station or a transport pipeline.

Dr. Almarri

Eng. Aldousari


Engineering 135

2. Flow in Horizontal Pipeline

• We will study the flow in horizontal pipes for gas, liquid,

and multiphase.

• Our main objective is to predict the pressure drop due to

friction.

• The pressure drop due to potential energy is zero in

horizontal pipelines.

Dr. Almarri

Eng. Aldousari


Engineering 136

2.1 Single-Phase Flow: Liquid

• The potential energy pressure drop is zero

• Incompressible fluid and the pipe diameter is constant thus; the kinetic energy pressure drop is zero.

• The mechanical energy balance simplifies to

∆𝑃 = 𝑃1 − 𝑃2 =2𝑓𝑓𝜌𝑣

2𝐿

𝑔𝑐𝐷

• The friction factor is calculated from Moody diagram.

𝑣 =𝑞

𝐴

A=𝜋

4𝐷2

𝑣 =4𝑞

𝜋𝐷2

𝑣: Velocity, ft/sec

A: Pipe cross-sectional area, 𝑓𝑡2

D: Pipe diameter, ft

Dr. Almarri

Eng. Aldousari


Engineering 137

• Single phase flow can be characterized as being:

1. Laminar

2. Turbulent

• Depending on the value of a dimensionless group known as the Reynolds number, Nre.

𝑁𝑅𝑒 =1.48 𝑞𝜌

𝐷𝜇

q: flow rate STB/d

𝜌: density, lbm/ft3

D: Diameter, in

𝜇: Viscosity, cp

If Nre >= 2100 then the flow is turbulent.

If Nre< 2100 then the flow is laminar.


Dr. Almarri

Eng. Aldousari


Engineering 138

Example

• 1000 bbl/d is supplied to the wellhead through a 3000 ft

long, 1 ½ in I.D. flow line from a central pumping station.

The relative roughness of the pipe is 0.004. If the pressure

at the wellhead is 100 psia. What is the pressure at the

pumping station. The water has a specific gravity of 1.05

and a viscosity of 1.2 cp. Fresh water density is 62.4 lbm/ft3


Dr. Almarri

Eng. Aldousari


Engineering 139

Solution

• Solution

2 3

3

2 2 2

1000 3000 1.5 0.004

100 1.05 1.2 62.4

1.05 62.4 65.52

3.142861.5

4 4

fw

fw

bblq L ft D in

d

lbmP psia cp

ft

lbm

ft

SOLUTI

A D in

ON

m

2

2

1

144

ft

in 20.01227

1000

ft

bbl

qV

A

d

35.615 ft

2

1bbl

1d

2

24 60 60

0.01227

s

ft

5.298

ft

s

Solution

2 3

1000 3000 1.5 0.004

100 1.05 1.2 62.4

65.52

fw

Re

bblq L ft D in

d

l

SOLUTION

bmP psia cp

ft

lbm

VDN

m

m

3ft5.298

ft

s

1.5

12ft

1.2 cp

46.72 10lbm

ft . s

1cp

0.89811.1098

2

0.89811.1098

53808

1 5.0452 7.1494log log

3.7065 2.8257

10.0076

0.004 5.0452 0.004 7.1494log log

3.7065 53808 2.8257 53808

Re ReN Nf

f

Solution

• Solution

2 3

2

3

2

1 2

1000 3000 1.5 0.004

100 1.05 1.2 62.4

65.52 0.01227 5.298

53808 0.0076

2*0.0076*65.522

fw

Re

c

bblq L ft D in

d

lbmP psia cp

ft

lbm ftA f

SOLU

t Vft s

N f

lbm

f V LP P P

TION

g D

m

3ft

2

2*(5.298)ft

2s*3000 ft

1*

12

32.174

ft

in

lbm . ft

2.flb s

2

1 2

144.8

*1.5

100 144.8 244.8

flb

inin

P P P psi

0.00764

Solution

• 𝛾 =𝜌

𝜌𝑓𝑤 ≫ ρ = 𝛾𝜌𝑓𝑤 = 1.05 62.4

• 𝑁𝑅𝑒 =1.48 1,000 65.52

1.5 1.2= 53,872

• f = 0.0076 from Moody chart.

• 𝑣 =𝑞

𝐴

• We know 5.615 𝑓𝑡3 = 1 𝑏𝑏𝑙

• 𝑣 =1000

𝜋

4

1.5

12

2 𝑏𝑏𝑙

𝑑 5.615 𝑓𝑡3

𝑏𝑏𝑙

1 𝑑

86400 𝑠𝑒𝑐 = 5.3 𝑓𝑡/𝑠𝑒𝑐

Dr. Almarri

Eng. Aldousari


Engineering 144

Lecture 3

Dr. Almarri

Eng. Aldousari 145


Engineering

• The mechanical energy balance is

•𝑑𝑝

𝜌+

𝑣𝑑𝑢

𝑔𝑐+

2𝑓𝑓𝑣2𝑑𝐿

𝑔𝑐𝐷= 0

• 𝑝12 − 𝑝2

2 = 4.195 × 10−6 𝛾𝑔𝑍 𝑇 𝑞2

𝐷4

24𝑓𝑓𝐿

𝐷+ 𝑙𝑛

𝑝1

𝑝2

• P1 and p2 are in psi, T is in oR, q is MSCF/d, D is in in and

L is in ft

• 𝑁𝑅𝑒 = 20.09𝛾𝑔𝑞

𝐷𝜇

2.2 Single-Phase Flow: Gas

Dr. Almarri

Eng. Aldousari


Engineering 146

Example

• Gas production from a low-pressure gas well (wellhead

pressure = 100 psia) is to be transported through 1,000 ft of

a 3 in. I.D. (Ԑ= 0.001) to a compressor station, ehre the inlet

pressure must be at least 20 psia. The gas has a specific

gravity of 0.7, a temperature of 100 oF and an average

viscosity of 0.012 cp. What is the maximum flow rate

possible through this gas line? (Hint: when the pressure is

low z= 1). Final answer = 10,800 MSCF/d

2.2 Single-Phase Flow: Gas

Dr. Almarri

Eng. Aldousari


Engineering 147

Solution

• Solution

1 2

2 2 2

100 1000 3 0.001

100 20 0.7 0.012

3.142863

4 4

o

g

T F L ft D in

P psia P psia c

SOL

p

A D in

UTION

m

2

2

1

144

ft

in 2

Re

0.04911

500

0.7*1000020.09 20.09 195319.4

3*0.012

g

ft

MSCFAssume q

d

qN

D

m

Solution

1 2

2

Re

0.89811.1098

100 1000 3 0.001 1

100 20 0.7 0.012

0.04911 500 195319.4

1

5.0452 7.1494log log

3.7065 2.8257

o

g

Re Re

T F L ft D in Z

P psia P psia cp

MSCFA ft Assume q N

SOLUTION

d

f

N N

m

2

2

0.89811.1098

10.00527761

0.001 5.0452 0.001 7.1494log log

3.7065 195319.4 2.8257 195319.4

f

Solution

• Solution

1 2

2

Re

2 2 4

1 2

7

100 1000 3 0.001 1

100 20 0.7 0.012

0.04911 500 195319.4 0.00527761

(4.195 10 ) 24

o

g

g

T F L ft D in Z

P psia P psia cp

MSCFA ft Assume q N f

d

q is calculated again by the following :

P P Dq

LZT f

D

SOLUTION

m

1

2

2 2 4

7

ln

100 20 310386.80102

1000 100(4.195 10 ) 0.7 1 (460 100) 24 0.00528 ln

3 20

P

P

q

Solution

1 2

2

Re

100 1000 3 0.001 1

100 20 0.7 0.012

0.04911 500 195319.4 0.00527761

10386.80102

By Trial and Error we get the following:

o

g

i

T F L ft D in Z

P psia P psia cp

MSCFA ft Assume

SOL

q N f

UTION

d

q

m

q Nre f

500 195319.4 0.005278

10386.8 4057488 0.004927

10735.62 4193749 0.004927

10736.31 4194020 0.004927

10736.31 4194020 0.004927

2.3 Two Phase Flow

Flow Regimes

• Flow regime does not affect the pressure drop as

significantly in horizontal flow as it does in vertical flow,

because there is no potential energy contribution.

• Most important, the occurance of slug flow nesessitates

designing separators or sometimes special pieces of

equipment. (Slug catchers) to handle the large volume of

liquid containes in a slug.

Dr. Almarri

Eng. Aldousari


Engineering 153

• Segregated flow: two phases are separate

• Intermittent flow: gas and liquid are alternating.

• Distibutive flow: one phase is dispersed in the other

Dr. Almarri

Eng. Aldousari


Engineering 154

Segregated flow

• Stratified smooth flow: consists of liquid flowing along the

bottom of the pipe and gas flowing along the top of the

pipe with smooth interface between the phases, occires at

tye low rates od both phases.

• At higher gas rates, the interface becomes wavy, and

stratified wavy flow results.

Dr. Almarri

Eng. Aldousari


Engineering 155

• Annular flow: occurs at high gas rates and relativley high

liquid rates and consists of an annulus of liquid coating the

wall of the pipe and a central core of gas flow, with liquid

droplets entrained in the gas.

Dr. Almarri

Eng. Aldousari


Engineering 156

Intermittent flow

• Slug flow: consists of large liquid slugs alternating with

high velocity bubbles of gas that fill almost the entire pipe.

• Plug flow: large gas bubbles flow along the top of the pipe.

Dr. Almarri

Eng. Aldousari


Engineering 157

Distributive flow

• Bubble flow: gas bubble concentrated on the upper side of

the pipe.

• Mist flow: occurs at high gas rates and low liquid rates and

consists of gas with liquid droplets entrained.

Dr. Almarri

Eng. Aldousari


Engineering 158

Determination of flow type

• Flow regime in horizontal flow are predicted with flow

regime maps one of the first of these is Baker (1953). We

will study

• 1- Baker (1953)

• 2- Mandhane (1974)

• Beggs and Brill (1973)

Dr. Almarri

Eng. Aldousari


Engineering 159

Baker flow regime map

• The axes for this plot are 𝐺𝐿𝜆∅ = 𝐺𝑔, where 𝐺𝐿 and 𝐺𝑔 are

the mass fluxes of liquid and gas, 𝑙𝑏𝑚/ℎ𝑟 − 𝑓𝑡2

• 𝜆 =𝜌𝑔

0.075

𝜌𝐿

62.4

12

• ∅ =73

𝜎𝐿𝜇𝐿

62.4

𝜌𝐿

21

3

• Where densites in lbm/ft3, viscosity is in cp, and 𝜎𝐿 is in

dunes/cm

Dr. Almarri

Eng. Aldousari


Engineering 160

Mandhane flow regime map

• This map uses the gas and liquid superficial velocities.

Dr. Almarri

Eng. Aldousari


Engineering 161

• The superficial velocity of a phase would be the average

velocity of the phase if that phase filled the entire pipe;

that is, if it were single phase flow. It is not a real velocity

that physically occurs.

• 𝑣𝑠𝐿 =𝑞𝐿

𝐴

• 𝑣𝑠𝑔 =𝑞𝐿𝑔

𝐴

Dr. Almarri

Eng. Aldousari


Engineering 162

• The Begges and Brill Correlation

• Is based on a horizontal flow regime map that divides the

domain into the tree flow regime categories, segregated,

intermittent, and distributed. This map needs mixture

Froud number as

• 𝑁𝑓𝑟 =𝑣𝑚2

𝑔𝐷

• Vm: vsl + vsg

• Vm: mixture velocity, ft/s

• g: gravity constant 32.17 ft/sec2

• D: diameter, ft

Dr. Almarri

Eng. Aldousari


Engineering 163

• Example

• Using the baker, Mandhane, and Beggs and Brill flow

regime maps, determine the flow regime for the flow of

2000 bbl/d of oil and 1 MM SCF/d of gas at 800 psia and

175 oF in a 2.5 in I.D. pipe. 𝜌𝑜= 0.8 g/cm3, 𝜇𝑜= 2 cp, 𝜎𝐿= 30

dynes/cm

• 1 g/cc = 62.428 lbm/ft3, Tpc= 374 oR, Ppc= 717 psia, 𝛾𝑔=

0.709, 𝜇𝑜= 0.0131 cp

Dr. Almarri

Eng. Aldousari


Engineering 164

• 2.3.3 Multiphase flow in pipe lines

• Multiphase flow – the simultaneous flow of two or more

phases of fluid – will occure in almost all oil pipelines and

gas pipelines.

• In an oil pipeline, whenever the pressure drops below the

bubble point, a gas will evolve, and from that point, gas

liquid flow will occure.

• Sometimes water also flows wicj will results in 3 phase

flow.

Dr. Almarri

Eng. Aldousari


Engineering 165

• Holdup behavior

• In two phase flow, the amount of the pipe occupied by a

phase is often different from its proportion of the total

volumetric flow rate.

•

• If we have this vertical pipe

• Liquid is denser than gas

• Gas will be moving faster because of this.

Dr. Almarri

Eng. Aldousari


Engineering 166

• Holdup phenomena occurs

• The volume fraction of the denser phase will be greater

than the input volume fraction of the denser phase, that is,

the denser phase is "held up" in the pipe relative to the

lighter phase.

• 𝐻𝐿 =𝑉𝐿

𝑉

• VL: Volume of liquid

• V: Volume of pipe segment.

Dr. Almarri

Eng. Aldousari


Engineering 167

• Another parameter used in describing two phase flow is

the input fraction of each phase, 𝜆, defined as

• λ𝐿 =𝑞𝐿

𝑞𝐿+𝑞𝑔

• λ𝑔 = 1 − λ𝐿

• q is the volumetric flow rates of the two phases.

Dr. Almarri

Eng. Aldousari


Engineering 168

• Slip velocity is defines as the difference between the

average velocities of the two phases. Thus

• v𝑠 = v𝑔 − v𝐿

• 𝑣𝑠 =1

𝐴

𝑞𝑔

1−𝐻𝐿−

𝑞𝐿

𝐻𝐿

Dr. Almarri

Eng. Aldousari


Engineering 169

• Example:

• If the slip velocity for a gas-liquid flow is 60 ft/min and the

superficial velocity of each phase is also 60 ft/min, what is

the holdup for each phase.

Dr. Almarri

Eng. Aldousari


Engineering 170

• Begges and Brill Method

• Begges and Brill method is applicable to any inclination

and flow direction.

• This method is based on the flow regime that would occur

if the pipe were horizontally; corrections are then made to

account for the change in holdup behavior with inclination.

Dr. Almarri

Eng. Aldousari


Engineering 171

• It is based on the following parameters:

• 𝑁𝐹𝑅 =𝑣𝑚2

𝑔𝐷

• λ𝐿 =𝑣𝑠𝐿

𝑣𝑚

• 𝐿1 = 316 λ𝐿0.302

• 𝐿2 = 0.0009252 λ𝐿−2.468

• 𝐿3 = 0.1 λ𝐿−1.4516

𝐿4 = 0.5 λ𝐿−0.738

Flow regimes used in Begges-Brill are segregated, transition, intermittent, and distributed

Dr. Almarri

Eng. Aldousari


Engineering 172

Flow through chokes

• The flow rate from almost all flowing wells is controlled

with a wellhead choke

• The chock is a device that places a restriction in the flow

line.

• Chock advantages:

• Prevention of water conning

• Prevention of sand production

• Satisfying production rate or pressure imposed by surface

equipment.

Dr. Almarri

Eng. Aldousari


Engineering 173

single phase liquid flow

• The flow rate is related to the pressure drop across the

chock by

• 𝑞 = 𝐶𝐴2𝑔𝑐∆𝑝

𝜌

• Where C is the flow coefficient of the choke and A is the

cross sectional area of the chock.

• C is given by Figure 10-11

Dr. Almarri

Eng. Aldousari


Engineering 174

Oil field units

• 𝑞 = 22,800𝐶𝐷22 ∆𝑝

𝜌

• Where

• Q is bbl/d

• D2: chock diameter in in.

• ∆𝑝: psi

• 𝜌: lmb/ft3

• The chock usually given in 64ths of an inch.

Dr. Almarri

Eng. Aldousari


Engineering 175

Example

• What will be the flow rate of a 0.8 specific gracity, 2 cp oil

through a 20/64-in chock if the pressure drop across the

chock is 20 psi and the line size is 1 in.

Dr. Almarri

Eng. Aldousari


Engineering 176

Pressure drop through pipe fittings

• When fluids pass through pipe fittings (tees, elbows, etc) or valves, secondary flows and additional turbulence create pressure drops that must be included to determine the overall pressure drop in a piping network.

• The effects of valves and fittings are included by adding the "equivalent length" of the valves and fittings to the actual lengths of straight pipe whrn calculating the pressure drop.

• The equivalent lengths of many standard valves and fittings have been determined experimentally (crane, 1957) and are given in Table 10-1

• The equivelant lengths are given in pipe diameters, this value is multiplied by the pipe diameter to find the actual length of pipe to be added.

Dr. Almarri

Eng. Aldousari


Engineering 177

Surface gathering systems

• In most oil and gas production installations, the flow from

several wells will be gathered at a central processing

station or combined into a common pipeline

• When individual flow lines all join at a common point, the

pressure at the common point is equal for all flow lines.

• The common point is typically a separator in an oil

production systems.

• The flowing tubing pressure of an individual well (i) is

related to the separator pressure by

Dr. Almarri

Eng. Aldousari


Engineering 178

• 𝑝𝑡𝑓𝑖 = 𝑝𝑠𝑒𝑝 + ∆𝑝𝐿𝑖 + ∆𝑝𝑐𝑖 + ∆𝑝𝑓𝑖

• Where ∆𝑝𝐿𝑖 is the pressure drop through the flow line

• ∆𝑝𝑐𝑖 is the pressure drop through the chock

• ∆𝑝𝑓𝑖 is the pressure drop through fittings

Dr. Almarri

Eng. Aldousari


Engineering 179

• In a gathering system where individual wells are tied into a

common pipeline, so that pipeline flow rate is the sum of

the upstream well flow rates.

Dr. Almarri

Eng. Aldousari


Engineering 180

Example

• The liquid production from three rod-pimped wells is

gathered in a common 2 in line, as shown.

• 1 in flow lines connect each well to the gathering line, and

each well line contains a ball valve and a conventional

swing check valve. Well 1 is tied into the gathering line

with a standard 90o elbow, while wells 2 and 3 are

connected with standard tees.

• The oil density is 53.04 lbm/ft3 and its viscosity is 5 cp. The

separator pressure is 100 psig. Assuming the relative

roughness of all lines to be 0.001, calculate the flowing

tubing pressures of the three wells.

Dr. Almarri

Eng. Aldousari


Engineering 181

Drag Reducing Agents

• Drag reducing agnets (DRA) are added to crude oil to

allow a higher flow rate in a pipeline for the same pressure

drop.

•

• They are effective with light crude oil.

• DRA are expensive and not recoverable; their use must be

economically justified.

• DRA increases the flow up to 90%

• The effect of the DRA is to reduce the frictional pressure

drop in the turbulent flow; they reduce turbulence by an

unkown mechanism

Dr. Almarri

Eng. Aldousari


Engineering 182

Heavy oil Transport

• The viscosity of heavy oils is very high and highly sensitive

to the oil temperature.

• Dilute oil wil a light hydrocarbon solvent lowers the

viscosity to an acceptable value and allows pumping at

reasonable flow rates.

•

• Leak Detectiong

• Visual Surveilance

• Monitoring of flow and pressure.

Dr. Almarri

Eng. Aldousari


Engineering 183

• Pipe desiging

• Petroleum pipelines should not

• Exceed the maximum allowable working pressure

(MAWP)

• Fall below the crude oil bubble point pressure.

Dr. Almarri

Eng. Aldousari


Engineering 184

Two Phase Flow

Example

• During the calculation of a pressure traverse in a gas

well producing liquid, the following conditions were

determine at the average pressure and temperature in

the pipe increment. Using the Hagedorn and Brown

method, determine the pressure gradient.

3

3

1500 30

180 5

2.992 0.0006

0.012 50

0.45 8

25 /

sg

o

sL

g L

L g

ftP psia V

s

ftT F V

s

d in ft

lbmcp

ft

lbmcp

ft

dynes cm

m

m

Solution

• Before finding HL and f, some preliminary calculations are made:

3

2

22

3 3

5 30 35

50.143

30

(1 ) 5 0.143 8 (1 0.143) 14

2.992

123.1416 0.0487

4 4

50

2 2

25

m sL sg

sLL

sg

n L L g L

p

L

ftV V V

s

V

V

lbm

ft

dA ft

lbm lbm lbm

ft ft

dynes dynes

cm cm

3ft

dynes

cm

30.48 cm

1 ft

32.174lbm

. ft2

444820

s

dynes

2

3907.2

s

ft

Solution

• Before finding HL and f, some preliminary calculations are made:

1

141/ 4 4 42

3

2

1

141/ 4 4 42

3

2

1 15 907.2 5 28.2 11.52

32.18

1 130 907.2 30 28.2 69.13

32.18

LLv sL

Lgv sg

d

ft s ft sN V

ftg s ft s ft

s

ft s ft sN V

ftg s ft s ft

s

N d

1/ 22.992

32.1812

Lft

g ft

2s

2

907.2s

3ft

2

1

2

42.6

Solution • Before finding HL and f, some preliminary calculations are made:

1/ 42

3

3

32.181

0.45

50

25

L L

L

f

lbm

t

g sN cp

ft dynes

m

cm

.32.174

lbm ft

2

444820

s

dynes

30.48 cm

1 ft

1/ 43

1/ 4

322

3

62

3

32.18

0.45 18.14363402

50

32.18

0.45 5972.729287

50

L

L

ft

ssN cplbm

ft

ft

ssN cp

f

lbm

lb

t

m

4

1/ 4

1/ 44 4

3 4

.0.45 3844.048569

0.45L

ft scp

lbm lbm

N cp

.

7.874032746ft s

lbm

46.74 10lbm

.ft s

1cp0.002388194

Example

• Determine HL, from figure 3-6, CNL= 0.002

0.002

Example

0.1

0.575 0.1

0.1-5

0.575 0.1

0.38

2.14

1. 0.0024

2. .3 6, 0.002

( )3. , (14.7 )

11.52 (0.002) 15007.52 10

42.6 69.13 14.7

4. .3 7, 0.29

5.

L

L

L

L

Lv LH a

d gv a

H

L

gv L

d

N

From Fig CN

N CN PX where P base pressure psia

N N P

X

HFrom Fig

N NX

N

X

0.38

2.14

69.13 0.00240.002276199

42.6

Figure 3-7. Liquid holdup factor correlation, Pa=base pressure (14.7 psia).

0.29

Example

3

2 2

3

1 0.29 1 0.29

6. .3 8, 1

7. 1 0.29 0.29

(1 )

50 0.29 8 (1 0.29) 20.18

149.71

20.18

( ) ( ) 0.45 0.012 0.034

14

L L

LL

m L L g L

m

nf

m

H H

m L g

n mRem

m

From Fig

HH

H H

lbm

ft

lbm

ft

cp

lbm

V dN

m m m

m

3ft35

ft

s

2.992

12ft

0.034 cp

46.72 10lbm

ft . s

1cp

5347222.222

Figure 3-8. Correlation of second correction factor, ψ

Example

0.0006 ft

d

12 in

1 ft

2.992 in

0.9

Re

0.9

0.0024

.3 9 .3 15

0.025

1 21.251.14 2log

1 21.251.14 2 log 0.0024 6.373

5347222.2

0.157 0.024622659

m

from fig or Eq

f

d Nf

f

f f

Example

2

( cos )2

32.2

f m

m

c c

f VdP g

dh g g d

ft

dP

dh

2s

32.174lbm . ft

2.flb s

20.18lbm

3

0.025 9.71

cos(0)

lbm

ft

2

2

335

ft

ft

2s

2 32.174lbm

. ft

2.flb s0.249 ft

3 220.19630758 18.55927159 38.75 38.75

f flb lbdP

dh ft ft

211 ft

ft

2

2

144

10.269 0.269

f

in

lbdP psi

dh in ft ft

200

APPENDIX A

CONVERSION FACTORS

Thermodynamics

201

APPENDIX A: CONVERSION FACTORS Mass 1.0 Kg Mass 1.0 lbm

= 1000.0 g = 16.0 oz

= 0.0010 metric ton = 0.00050 ton

= 2.20462 lbm = 453.59300 g

= 35.27392 oz = 0.45359 Kg

Length 1.0 m Length 1.0 ft

= 100.0 cm = 12.0 in

= 1000.0 mm = 1/3 yd

= 1000000.0 microns meter = 0.3048 m

= 10000000000.0 angstroms = 30.4800 cm

= 39.370 in

= 3.280800 ft

= 1.093600 yd

= 0.000621 mile

Volume 1 m3 Volume 1 ft3

= 1000 liters = 1728 in3

= 1000000 cm3 = 7.4805 gal

= 1000000 mlt = 0.028317 m3

= 35.3145 ft3 = 28.317 liters

= 220.83 imperial gallons = 28317 cm3

= 264.17 gal = 0.1781076 bbl

= 6.2898106 bbl

= 1056.68 qt

Force 1 N Force 1 lbf

= 1 Kg.m/s2 = 32.174 lbm.ft/s2

= 100000 dynes = 4.4482 N

= 100000 g.cm/s2 = 444820 dynes

= 0.22481 lbf

202

APPENDIX A: CONVERSION FACTORS

Pressure 1 atm Energy 1 J

= 101325 N/m2 = 1 N.m

= 101325 Pa = 10000000 ergs

= 101.325 Kpa = 10000000 dyne.cm

= 1.01325 bars = 0.0000002778 Kw.h

= 1013250 dynes/cm2 = 0.23901 cal

= 760 mm Hg at 0 oC = 0.7376 ft-lbf

= 760 torr = 0.0009486 Btu

= 10.333 m H2O at 4 oC

= 14.696 lbf/in2

= 14.696 psi

= 33.9 ft H2O at 4 oC

= 29.921 in Hg at 0 oC

Power 1 W gc 1 Kg.m/N.s2

= 1 J/s = 1 g.cm/dyne.s3

= 0.23901 cal/s = 32.174 lbm.ft/lbf.s2

= 0.7376 ft.lbf/s

= 0.0009486 Btu/s g/gc 9.8 N/Kg

= 0.001341 hp 980 dyne/g

Temperature 1 oC T(oK) = T(oC) + 273.15

274.15 oK T(oR) = T(oF) + 459.67

33.8 oF T(oR) =1.8* T(oK)

493.47 oR T(oF) =1.8* T(oC) +32

203

APPENDIX A: CONVERSION FACTORS

Water density 1 g/cm3 H2O at 4 oC Air density 1.210 g/cm3 Air at 20 oC

1000 Kg/m3 H2O at 4 oC 1210 Kg/m3 Air at 20 oC

62.3664 lbm/ft3 H2O at 4 oC 75.5381 lbm/ft3 Air at 20 oC

Gas 1 lbmol

379 ft3

Gas Constant 8.314 m3.Pa/mol.oK 1 Kmol 1000.000 mol

8314 cm3.kPa/mol.oK 1000.000 gmol

83.14 cm3.bar/mol.oK 2.2046 lbmol

82.06 cm3.atm/mol.oK

62356 cm3.torr/mol.oK 1 lbm 0.45359 Kmol

0.08314 lt.bar/mol.oK 453.59 mol

0.08205 lt.atm/mol.oK 453.59 gmol

62.3601283 lt. mm Hg/mol.oK

0.7302 ft3.atm/lbmol.oR 1 mol 1 gmol

10.7310 ft3.Psia/lbmol.oR 0.001 Kmol

1545.0000 ft.lbf/lbmol.oR 0.0022046 lbmol

8.314 J/mol.oK

1.98712914 cal/mol.oK

1.98740775 Btu/lbmol.oR

transportation and storage handout 4

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