Pipeline Engineering Fluid Flow

7/30/2019 Pipeline Engineering Fluid Flow

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FLUID FLOW

Mechanical Energy Balance

' '

2

2(1-1)

potential energy expansion work Kinetic energy Work added/ Sum of frictionchange change subtracted by losses

compressorsor pumps/expanders

Note that the balance is per unit mass. In differential form

GG (1-2)

Rewrite as follows

! " " # $ %

U G G (1-3)

Divide by& '

('

is the length of pipe)

( )

( 0 1

( 2

( 0 3

(

3

( 0

4

0

5

0

6 7 8

7

U U UG

GU

G

G(1-4)

or:

dp

dL

dp

dL

dp

dL

dp

dLTot elev accel frict

(1-5)

(

G

G

W

L

o

is usually ignored, as the equation applies to a section of pipe)

The above equation is an alternative way of writing the mechanical energy balance. It is not adifferent equation.

The differential form of the potential energy change is


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Natural Gas Basic Engineering Copyright: Miguel Bagajewicz.No reproduction allowed without consent

2

dZ

I

dL

gdZ

dLg sin I (1-6)

Friction losses: We use the Fanning or Darcy-Weisbach equation (Often called Darcy equation)

G4

3 9

@

( 0

2 2

(1-7)

an equation that applies for single phase fluids, only (two phase fluids are treated separately).The friction factor, in turn, is obtained from the Moody Diagram below.

A B C D E F G H G I

P Q Q

( R

@ S T

1 U

T V

Friction factor equations. (Much needed in the era of computers and excel)

Laminar FlowW

16

Re(1-8)


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3

Turbulent Flow W X0046.

Re(1-9)

smooth pipes: a=0.Iron or steel pipes a=0.16

Turbulent Flow1

23 7

2 5110Y ` Y

log

.

.

Re

H(Colebrook eqn) (1-10)

Equivalent length of valves and fittings: Pressure drop for valves and fittings is accounted foras equivalent length of pipe. Typical values can be obtained from the following Table.

a

b c d

F G H G I

e f g h i p q r s t q r s u t v w

Y x y

i p

y

h

x

g w

Y

h t t h s u w

h t t h s u

45O elbows 1590O elbows, std radius 3290

Oelbows, medium radius 26

90O elbows, long sweep 2090O square elbows 60

180O

close return bends 75180O medium radius return bends 50Tee (used as elbow, entering run) 60Tee (used as elbow, entering branch) 90Gate Valve (open ) 7Globe Valve (open ) 300Angle Valve (open) 170

Pressure Drop Calculations

B B C B F F E F D E F E D E E F E B E F F

j k l l m n o k o

a) Isothermal (Uis constant)

z { | }

}

~ ~ ~ ~

U

(1-11)

for a fixed I constant = 0


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4

22

GG

(1-12)

' '

U 2 2 (1-13)

b) Nonisothermal

It will not have a big error if you use U(Taverage), v(Taverage)

Consider the flow of liquid water (@ 20oC) through a 200 m, 3 pipe, with an elevation change

of 5 m. What is the pressure drop?

Can the Bernoulli equation assuming incompressible flow be used for gases? The next figureillustrates it.


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5

A B C D E F G H I

Error in Bernoulli equation

In conclusion, if

d 0 2 0 3. . using the assumption of incompressibility is OK.

a) Relatively small change in T (known)


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6

For small pressure drop (something you can check after you are done) can use Bernoulli and

fanning equation as flows

(1-14)

Then

(1-15)

but

, where

= Velocity (m/sec) = Specific volume (m3/Kg)

= Mass flow (Kg/sec)

= Cross sectional area (m2)

Then,

-

(1-16)

Now put in integral form

2

2 2

21

(1-17)

Assume

(1-18)

2

3

which comes from

(1-19)

(1-20)

The integral form will now be


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U

-

2

2 2

2

' ln (1-21)

Now use

, where

: Molecular weight. Then

U , which leads to:

2 2 2 22 2

U

(1-22)

Therefore;

2 2

2 2 2 ln 22

-

UU

'

(1-23)

but,

(1-24)

Then

2 2

2 2 2 ln 22

-

UU

' (1-25)

To calculate

Kays rule is used. This rule states that the reduced pressure and temperatureof the gas is obtained using the average pressure and temperature (as above calculated) and a

pseudo critical pressure and temperature.

,

(1-26)

,

(1-27)

In turn the critical pressure and temperatures are obtained as molar averages of the respectivecomponents critical values.

,

,

(1-28),

,

(1-29)With these values the

factor comes from the following chart:


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8

!

Equation (1-25) can be further simplified. First neglect the acceleration term because it isusually small compared to the others, to obtain:


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9

2

2 2 2 2 0

2

" #

" # $ % &

'

" #

" #

( )

0 1 2

3 4

UU

'

(1-30)

Form this equation we can get

, as follows:

2 2

2 22 52

32 2

5 6

7 8 9 @ A

5 6 5 6

5 6 5 6 5 6 5 6 5 6

B

C D

B

E

F

G H

E

E

S

'

(1-31)

But the volumetric flow at standard conditions is given by I I I

P

E

B

where the

subscript

stands for standard conditions. Therefore:

2

2 2

2 22 52

2

2

32 2

Q R

S T U V W

X X Q R Q R

X Q R Q R Q R

Y

` ` a b

c d c e d e f

g

Y ` c d h i

S

'

(1-32)

Now, ifp' =0, we get

2 22 22 52232 2

q r s t u

v v

v w x w x w x

y y

y

S

(1-33)

which can be rearranged as follows:

2 2 2

(1-34)

where2

2 2 2 5

64

j k

S and is known to be l u m n a product of a resistant factor l

times the lengthm o

With this, we have2

2 2 2 5

64 I 5 6 5 6 5 6

I I

l

S .

To calculate pressure drop we recognize that average pressures are a function of , which is

unknown. Then we propose the following algorithm:

a) Assume

9 @ A

and calculate

5 6

b) Calculate2 ( ) ( ) ( )

( )

2 2 2 5

64

z { | { | { |

z z

} ~

S

b) Use formula to get a new value 1 2 ( ) 2


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d) Continue until

~ ~

~

Hd

Depending on the choice of friction term expression, several formulas have been reported forequation (1-34). They are summarized in the following table.

~


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-

~

Natural gas (84,000 std m3/hr at 49 atm and 38oC) is sent from a gas refinery to a city, througha 16 pipeline. The distance is 170 Km. The gas reaches the other end at ground temperature,(5 oC). The gas to have the following molar fractions: Methane: 98%, ethane: 1.2%, propane:0.75%, and water: 0.05%. We also assume Re~5 106 and /D =0.01.

As a first approximation, we recommend using the Panhandle A equation: 2 2 1.855 ~ ~

W= 2.552 10-4

0.855/D4.586 (Wilson G.G., R. T. Ellington and J. Farwalther, 1991,

Gas distribution Home Study Course) , where s is thegas gravity (=Mgas/Mair)=0.65 for natural gas),

is in oR and D in inches)

What is the pressure drop?


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Heat Transfer Effects

To account for temperature changes due to heat transfer, we use total energy balance

( )

G G (1-35)

where the following is identified: Potential energy change: Rate of work done on the fluid element by pressure forces: ( ) Kinetic energy change: Internal energy changes: Heat transfer:

G . This is given per unit mass flowing (Kcal/h)/(m3/h)

Work added: G . This term is due to pumps and compressors. Since we will treat

these separately, this term is usually set to zero for pipes.

But the heatG is given by interactions with the ambient surroundings:

( )

SG (1-36)

where is the heat transfer coefficient, is the outside pipe temperature,

S (seenext figure) and

is the flowrate.

Then, (ignoring G because there are no pumps) to get:

(1-37)

Integrate and solve for (use

in the heat transfer equation)

2 2

2 12

S

(1-38)

But

(1-39)


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Finally, to obtain the outlet temperature, one would need to obtain it form the enthalpy and

pressure in the outlet

(1-40)

The procedure suggested is then:a) Assume

b) Use mechanical energy balance to obtain

c) Use total energy balance to obtain

d) get temperature

e) Go to b) and continue until convergence

SCENARIO IIOne has a turbine or Compressor/pump and needs

We use total energy with G = 0 and =0

G (1-41)

Integrating, one obtains:

!

' ' (1-42)

In this expression, we have "

given in Joules/Kg,

in Joules/sec and

in Joules/Kg. Thus,the work of the compressor/pump is given by:

#

$

% & ' (

' '

(1-43)

For compressors,%

is positive, while for turbines, it is negative. However, ''

is known for

liquids because enthalpy does not vary much with pressure. In addition, there isnt muchtemperature change in pumps). However, for gases, '

'

is much harder to obtain. Therefore wego back to the Mechanical Energy equation for pumps/compressors. Indeed, the Bernoulliequation gives

)

$

% & 0 1 2 & 0 1 2 (

' |

(1-44)

where the acceleration term has been neglected. For pumps, the density is constant, so oneobtains:


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WG

pU

' (1-45)

For compressors, one needs to obtain an expression of volume knowing that the evolution isisentropic (or nearly isentropic). Thus

3

2 0 4

= constant (5 6 7

2 8

7

0

for ideal gases5 9 7

2 8

7

0

for

real gases). Substituting v p vsn

s

n

1

1

1

Uintegrate to get

1

11

@

@

A B C

D

@

D

@

D

@

EF

G H

E I

F E

(1-46)

The above expression does not include the compressibility factor. A better expression, whichincludes the efficiency, is

1

11

1 2

@

@

D

@ A B C A B C

D

@

P

D

@

Q Q

EF

G H R S

F EK

(1-47)

The efficiency factor is usually between 60 to 80% and normally given by the manufacturer.One expression for such factor is:

1

1

@

@

A B C

D

@

D

@

P

A B C

D

@

E

S

E

S S

K

(1-48)

Finally, the outlet temperature is obtained fromT T

U

T

U

T

E I E I (1-49)

Using the gas law to obtain /D @ A B C I I in terms of temperatures and pressures, substituting and

rearranging, on e obtains:

> @1

1

V

V

V

W X Y W X Y

V

V

V

a b c d

a b

(1-50)

where

c d

is the compression ratio. Normally, manufacturers recommend not exceeding 300 oFat the outlet.


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The natural gas of exercise 1-2, is available originally at 2 atm. Calculate the compressionwork needed to reach delivery pressure (49 atm) using one compressor. Calculate the outlettemperature and determine the duty needed to cool the gas down to the corresponding inletconditions. Is it acceptable to use one compressor?

We now discuss the compression ratio. This is limited in compressors to the range 1.2 to 6.Extra compressors should be added if the CR >6, and after-coolers need to be added to controlthe temperature. If more than one compressor is to be used, the practice is to use the same CRfor all.

Consider two compressors.- Write the power expression for each one assuming the gas is cooled down to its inlet

temperature after compression.- Add both expressions to obtain the total work as a function of the intermediatepressure (the rest should not be a variable)

- Take first derivative and obtain the desired result that

c d

e f

c d

g

Obtain the set of compressors needed to compress the gas of exercise 1-2 properly, that is,limiting the temperature and using the right CR.


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Two Phase Flow

Two phase flow has several regimes, which are depicted in the next figure:

h i p q r s t u v w

a x y b y x

The two extreme cases are:

- Bubble: Vapor and Liquid in Equilibrium (Benzene 40%, Toluene 60%)- Dispersed: Liquid and gas (air and benzene).

The latter is common in gas pipelines; the former is common in crude pipelines, especially lightcrudes.

Dispersed

Annular

Stratified

Froth

Wavy

Slug

Plug

Bubble


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One important thing to recognize is that except for the extreme cases, the phases travel atdifferent velocities. Typical velocities are shown in the next table:

s t u w

a b y y x y b y x

REGIME LIQUID VEL(ft/sec) VAPOR VEL.(ft/sec)

Dispersed Close to vapor > 200Annular 20Stratified


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3)l

depends on the vapor/gas superficial velocity. It can be modified changing thediameter

4) Transition boundaries are not at all that sharp.

From this diagram, we notice that following change of regimes in a pipe. As the pressure dropis large, then the density of the vapor is lower.

1)n

O\ U n O\ U Abscissa decreases2)

1 1

O U 1

O U Ordinate increases

Thus trajectories are always "up" and "to the left". Thus a bubbly flow may become, plug, slug

or annular, an annular may become wavy or dispersed, depending on the starting position in theplot, and so on.

Lockart and Martinelli (1949) developed one of the first correlations. It is based on multiplyingthe pressure drop obtained by considering the vapor phase occupying the whole pipe, by afactor

2 z { | } ~ z {

I' ' (1-51)

In turn, the correction factor is given by I , where

''

. The following table

gives some typical values of the corresponding constants:

s t u

: Constants for Lockart and Martinellis correlation

Bubble 14.2 0.75Slug 1190 0.82Stratified 15400 1(horizontal)

Plug 27.3 0.86Annular 4.8 -0.3125 D(in) 0.343-0.021 D(in)

We notice that there are several more modern correlations, which will be explored later. Inturn, the pressure drop due to gravity, is given by

(1 ) sin

H U H U T

(1-52)

wheren

H is the (void) fraction of gas. We omit the pressure drop due to acceleration.


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Hydrate Formation

Hydrates are crystalline structures between water and hydrocarbons. One typical example isgiven in the figure below:

h i p q r s t u w

The next figure shows the Pressure-Temperature diagram of water-hydrocarbon systems. Curve1-1 represents the curve for vapor pressure of the hydrocarbon.

h i p q r s t u w

Cage of water molecules

CH4 molecule in the center


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The next figure shows some specific cases of hydrocarbons:

h i p q r s t u w

-

-

-

-

Clearly, in high pressure pipelines, favorable thermodynamic conditions for hydrate formationcan be encountered. It is therefore important to keep in mind that these conditions need to beavoided. These hydrates can be prevented from forming through heating, pressure change (nota choice in pipelines) and the introduction of inhibitors. These inhibitors are salts, alcohols,glycols, ammonia and MEA. The most widely used is methanol. The next figure shows thedepression of hydrate formation temperature observed for various hydrates.

h i p q r s t u t w

-

-

-


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21

Pipeline Costs

Historical pipeline and compressors installed cost data were obtained from the Oil & GasJournal special report on Pipeline Economics, September 3, 2001. Pipeline per mile costdistribution for different pipe diameters and compressor installed cost for different horsepowerrequirement are plotted in the following figure. All cost figures are updated to 2005 dollarsusing Marshal & Swift cost indexes.

h i p q r s t u t t w

-

h i p q r s t u t w

-

-

-

-

-

Fixed Capital Investment were calculated by adding the installed cost of a pipe length (assumed5000 miles) and the cost of all required recompression stations. The Fixed Capital Investmentsobtained are then divided by the pipe length to obtain a

cost profile for different flowrates. The curve in the next figure shows that this cost profile takes a logarithmic shape.


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h i p q r s t u t w

-

A linear correlation gives the following form:001108.0)(*001659.0)/$(

Operating costs for pipelines were estimated as follows; an average of 5 operators is assumedto be the requirement for each compression station, with an hourly wage of $21. Directsupervisory and clerical labor is assumed to be 20% of operating labor. Compressor fuel

requirement is estimated at 8,000 Btu / BHP-HR, and fuel cost at $2.5 per million Btu.Maintenance cost is assumed to be 7% of the FCI for compressors and 3% for pipes whileinsurance is 1% for compressors and 0.4% for pipes. Operating cost per pipeline mile versuscapacity is plotted in the next figure:

h i p q r s t u t w

! " # " $ % $ &

This estimate should be reasonable with about 40% accuracy. Similar linear approximation tothat of the FCI is assumed. Linear regression was used to estimate the operating cost


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dependence of the capacity, ignoring capacities less than 100 MMSCFD. This gave a generalcorrelation of the following form:

00004.0)(*00007.0)//$(.

" $

'

Consider the pipeline of Exercise 1-8:- Vary the pipeline maximum pressure (1200 psia) to some lower and higher value.

Adjust the diameter accordingly and calculate the number of recompressionstations.

- Calculate the cost. Can you say that 1200 psia is the right pressure?

Pipeline Looping

Pipeline looping is the practice of designing pipelines with segments run in parallel. Thispractice increases the pipeline flow capacity without altering the final pressure. If temperatureis close to ambient temperature, the location of a loop does not change the final delivery

pressure. However, when temperature changes substantially, then the location of a loop has aninfluence. Thus, in these cases, for example, it is recommended to loop in the upstream region,where the gas is hotter. This allows the gas to cool down faster and therefore increase thedelivery pressure.

Consider the following example: A 100 Km length (20 OD) pipeline is used to send 289

MMSFD at an inlet pressure of 1,200 psia and a temperature of 45o

C. The pipe roughness is750 inches, and a soil temperature of 10

oC.

Three alternatives were studied for this pipeline. a) No looping, b) Looping the first 25 Km,and c) Looping the last 25 Km. The results of a simulation are shown in the next figure:

h i p q r s t u t v w

( $ ! $ ) "

0

" " #


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24

Verify the results of figure 1-15 using the simulator.

Retrograde condensation

One very common phenomenon in pipelines is retrograde condensation. Consider the P-T plotof the next figure. It corresponds to a gas with the following composition: Methane: 93.47 %,Ethane: 3.52%, Propane: 0.585%, n-butane: 0.16%, i-butane: 0.11%, pentane: 0.055%, i-

pentane 0.05%, hexane: 0.09%, heptane: 0.04%, octane: 0.03%, nonane: 0.01%, CO2: 0.0545,N2: 1.34%. Assume a 15, 200 Km pipeline starts at 60 atm and 15

oC. If the external

temperature is 5 oC (U=1 BTU/hr-ft2oF), then it is clear that there will be liquid formation inthis pipeline, even if the operation is isothermal.

h i p q r s t u t w

1 2 3

#

" ) 4

# $

3

" #

3

"

3

$ "

Interestingly, if the pressure at the other end is low enough, then the liquid might vaporizeagain. This means that the pressure drop regime inside the pipe might change and one has to becareful in performing the simulations.

Generate the answers for the above example using the simulator. Change the pipe diameter andthe length to verify the statements.


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25

Pipeline Optimization Process

Conventional Pipeline design methods, which rely mostly on hand calculations or at best onsimple spreadsheet suggest that the compressor size and the pipe diameter be varied and thecost of service ($/(m3*Km) for the first year be plotted as a function of flowrate. Typicalassumptions are that there is no volume buildup in the pipe, the time value of money isneglected and that the facilities are designed to sustain the flows.

For example, Figure 1-17 shows one such exercise performed for three different diameters andparametric at different maximum operating pressures (MOP) and compression ratios.Efficient operating ranges that are flat are preferred

h i p q r s t u t w

5

1

! % $ ) " % " ! $

3

$

Explain why 5 -curves go through a minimum.


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J-Curves are a simplistic first approach but one that can provide a first approximation to theright diameter and compressors. Thus one needs to establish

- 6 7q 8 s w

In most cases this is defined by a variety of other factors and given to thedesigner.

-9

i @ s

i A s B A i 8 i

C

@

D

i 8 E w

Most pipelines are constructed taking into account thefact that demand at the receiving end(s) will increase through time. Thus, one isfaced with the decision of designing for future capacity and underutilize the pipelinefor some time or design for current or more short term capacity and use loops toexpand later.

-F G

@

A H i

7

A H w

If capacity expansions are considered, then they need to take placethrough looping. Not only the new loop has to be designed, but its timing andcapacity be selected.

-I

G

i P q P

7

@ s r

8 i A p @ r s H H q r s w

This choice has already been considered inconstructing the J-Curves. However, in more complex situations, one is faced withmultiple delivery points with different delivery pressures, etc.

-9

i @ s Q i R s w

This choice has already been considered in the J-Curve selection butneeds to be revisited anyway in view of the influence of the other factors.

-S 7

T

h

D

8

7

r w

This factor is the ratio between the average daily volume delivereddivided by the peak volume. If this ratio is too small, then storage facilities forinventory holding (salt caverns, underground caverns, abandoned reservoirs, etc ifavailable, or large LNG or high pressure, CNG, storage tanks) are more convenientthan more powerful compressors and larger diameters. The issue to resolve is when

these are substituted by inventory holding sites. In addition, the question remainswhere these holding sites should be located.

-C

7

P @ r s H H

7

r Q 8

8 i

7

A Q @

D

i A p w

While an earlier exercise suggests that whenmultiple compressors are used it is best to keep the compression ratio equal, thishypothesis needs verification. Reciprocating compressors are chosen when the

power requirement is smaller than 5,500 Kw. Compression ratios recommended forcentrifugal compressors are generally in the range of 1.25 to 1.35 and smaller than1.5 for reciprocating compressors.

Consider the pipeline of Exercise 1-8:

- Assume two compression stations will be used.- Determine (by inspection and using a simulator what is the best compression ratiofor each compressor. Aim at minimizing total work only.

-U

s

8 i A p

A

T C

7 7

i A p w

Clearly, cooling leads to significant savings becausepressure drop is reduced. However, money needs to be spent to install and run thecoolers. Thus the trade-off needs to be resolved.


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27

Consider the shown in the following figure

- The piping is in the ground and is not insulated. Assume a ground temperature of25

oC and a ground conductivity of 0.7 W/(m

oC). The gas elevation profiles are

provided in the following table:

Km Elevation (m)0 42115 7143 14.93323 60550 10609 120613 122630 235638 470

650 890

- The gas ((1.9% methane, 5% Ethane, 2% propane, 1% n-butane and 0.1% n-pentane) is supplied at the two points indicated in the diagram at 1,367 kPa and35oC in the first station (Km 0), and 1520 kPa and 30oC in the second (Km 143).

- Determine using simulations a) Piping diameter, b) Compressors at the supplystation, c) cooling required. Do not use a pressure above 5,600 Kpa. Use cost data

provided above.- Will new compressors be needed/beneficial?

Supply: 10,407,840 m3/d

2,148,200 m3/d

Supply: 5,722,000 m3/d

18,240 m3/d

6,595,200m3/d

134,400m3/d

64,600 m3/d

Km 0 115 143 323 550 609 630 Km 650

2,832,000 m3/d

613

638

3,617,400 m3/d

384,000 m3/d

336,,000m3/d


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Natural Gas Basic Engineering Copyright: Miguel Bagajewicz.32

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Pipeline Engineering Fluid Flow