gas transportation

8/12/2019 Gas Transportation

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Prepared by Group ( A )

Group Leader: Mohammed Al Yami



Introduction.o Reynolds Number ( NRE )

o Friction Factor ( )

o Relative Roughness ( e / D )

Pipeline Flow Calculations.

Types of Gas Flow through Pipelines.o Gas Flow in Series

o Gas Flow in Parallel

o Gas Flow in Looped2



The transmission of gas to the consumer may be divided into fourdistinct units: the gathering system ,the compression station, the main trunk

line, and the distribution lines . Pipelines, which comprise the gathering system,

main trunk line, and distribution lines, provide an economical method of

transporting fluids over great distances.

Many factors must be considered in the design of long-distance gas pipelines.

These include the nature and volume of the gas to be transmitted , the length of

the line, the type of terrain to be crossed, and the maximum elevation of the

route.

Studies of the flow conditions of natural gases in pipelines have led to the

development of complex equations such as ( the Weymouth equation, the

Panhandle equation, and the Modified-Panhandle equation) for relating the

volume transmitted through a gas pipeline to the various factors involved, thus

deciding the optimum pressures and pipe dimensions to be used. From

equations of this type, various combinations of pipe diameter and wall thickness

for a desired rate of gas throughout the system.3



The Reynolds number can be used as a parameter to distinguish between laminar

and turbulent fluid flow. The change from laminar to turbulent flow is usually

assumed to occur at a Reynolds number of 2,100 for flow in a circular pipe.

If U.S. field units of ft for diameter, ft/sec for velocity, lbm/ft for density andcentipoises for viscosity are used, the Reynolds number equation becomes:

If a gas of specific gravity g and viscosity (cp) is flowing in a pipe with an inner

diameter D ( in ) at flow rate q ( Mcfd) measured at base conditions of Tb ( oR )

and Pb ( psia ), the Reynolds number can be expressed as:

5



If a gas of specific gravity g and viscosity (cp) is flowing in a pipe with an inner

diameter D ( in ) at flow rate q ( ft3 / hr ) measured at base conditions of Tb ( oR )

and Pb ( psia ), the Reynolds number can be expressed as:

6



In addition to the size of a pipeline, the pressure and capacity of gas

transmission of a pipeline are primarily limited by the resistance to flow from the

pipe wall. The lost work is usually calculated using a friction factor by

dimensional analysis. It can be shown that the friction factor is a function of the

Reynolds number and of the relative roughness of pipe.

The equation for the friction factor in terms of the Reynolds number and relativeroughness varies based on type of fluid flow in pipeline.

The friction factor for laminar flow can be determined by following formula:

7



Studies of turbulent flow have shown that the velocity profile and pressure

gradient are very sensitive to the characteristics of the pipe wall, that is, the

roughness of the wall.

Therefore, the following equation is recommended for all calculations requiring

friction factor determination of turbulent flow was presented by Jain (1976):

This correlation is comparable to the Colebrook correlation. For relative

roughness between 10-6 and 10-2 and the Reynolds number between 5 x 103 and

10 108, the errors were within ±1.0% when compared with the Colebrook

correlation.

8



The frictional losses of fluid energy and pressure depend on the roughness of the

inside wall of a pipe. Wall roughness is a function of pipe material, method of

manufacture, and the environment to which it has been exposed.

Relative roughness ( e D ) is defined as the ratio of the absolute roughness to the

pipe internal diameter:

where and D have the same unit.

If no information is available on roughness, a value of = 0.0006 inches is

recommended for tubing and line pipes.

9



Engineering of long-distance transportation of natural gas by

pipeline requires a knowledge of flow formulas for calculating capacity

and pressure requirements. There are several equations in the petroleum

industry for calculating the flow of gases in pipelines.

In the early development of the natural gas transmission industry,pressures were low and the equations used for design purposes were

simple and adequate. However, as pressure increased to meet higher

capacity demands, equations were developed to meet the new

requirements. Probably the most common pipeline flow equation is the

Weymouth equation, which is generally preferred for smaller-diameter

lines (D ≤ 15 in. ±). The Panhandle equation and the Modified Panhandleequation are usually better for larger-sized transmission lines.

10



Qh = gas flow rate, ft3/hr @ P b & T b

T b

= base temperature, °R

P b= base pressure, psia

P1= inlet pressure, psia

p2= outlet pressure, psia

D= inside diameter of pipe, in.

γg= gas specific gravity (air = 1)

T= average flowing temperature, °R f= Moody friction factor

L= length of pipe, miles

= gas deviation factor at average

flowing temperature & average

pressure

5.052

2

2

123.3

L f T Z

D p p

p

T Q

g b

b

h

Z

11



Where:

e = base of natural logarithm = 2.718

s = 0.0375 γg ∆H/T

∆H = outlet elevation minus inlet elevation

(note that ∆H is positive when outlet is higherthan inlet).

5.0

g

52

2

s2

1

b

b

hZLf T

D) pe p(

p

T

23.3Q

5.0

g

52

2

s2

1

b

b

hZLf T

D) pe/ p(

p

T23.3Q

5.0

eg

52

2

s2

1

5.0

b

b

h ZLf T

D) pe p(

f

1

p

T

23.3Q

Le is the effective length of the pipeline.

15.0

f

= transmission factor

Where:

12



o To eliminate the trial-and-error procedure, Weymouth proposed that f vary as a function

of diameter in inches as follows:

D f

032.0

5.0

3/162

2

2

1062.18

LT Z D p p

pT Q

g b

bh

5.052

2

2

1

5.0

123.3

LT Z

D p p

f p

T Q

g b

b

h

3/1

13



Given:

Tb= 520 °R

Pb= 14.7 psia

P1= 400 psia

P2= 200 psia

D = 12.09 in.

γ g= 0.60

T= 520 °R

L= 100 mi

e= 0.0006 in.

Determine the line capacity.

SOLUTION:

For γg = 0.60ppc = 672 psia & Tpc = 358 °R

0.4464 1.4521

Z = 0.95 (From Fig 2-5)

672

)2/)200400((

pc

pr P

P P

358

520

pc

pr T

T T

16



Third trial:Qh = 1,000,000 ft3/hrRe = 2.2 × 106, f = 0.012Qh = 1,066,633 ft3/hr

5.0

3/16225.0

3/162

2

2

1

95.01005206.0

)09.12()200400(

7.14

520062.18062.18

LT Z

D p p

p

T Q

g b

bh

= 989,859 ft3/hr

Second trial:Qh = 500,000 ft3/hrRe = 1.1 × 106, f = 0.125Qh = 1,045,083 ft3/hr

18



= 28454875 ft3

/day

= 30972605 ft3/day

6182.2

4604.05394.02

2

2

1

07881.1

187.435 D

LT Z

p p

p

T Q

g b

b

6182.2

4604.05394.02207881.1

09.126.0

1

95.0100520

200400

7.14

52087.435

530.2

49011.0510.02

2

2

1

02.1

1737 D

LT Z

p p

p

T Q

g b

b

530.2

49011.0510.02202.1

09.126.0

1

95.0100520

200400

7.14

520737

19



20



21



22



23



Example: Natural gas with 24 MMcfd measured at standard conditions will

be delivered to a Riyadh city 300 miles from gas field. The gas is delivered to

the pipeline at 1500 psia pressure, and it is transmitted through a pipeline of

sufficient size, so that the pressure at the Riyadh is 50 psia. Assume specific

gravity of natural gas is 0.6 and the temperature of the flow is 100 oF.

What is the pipeline size required ?

Solution

I apply in Modified Panhandle equation to determine the required size:

D

g z LT

P P P b

T bq 53.2

51.0

961.0

22

21

02.1

737

24



According to average pressure & temperature, we can determine the deviation factor

( z ):

Pavg. = ( P1 + P2 ) / 2 = ( 1500 + 50 ) / 2 = 775 psia.

Tavg. = 100 oF = 560 oR

By computer program and based on Pavg. = 775 psia, Tavg. = 100 oF & g = 0.6, I get:

Z = 0.906

D = 7.597 7.6 inches

D MM 53.2

51.0

6.0 961.0906.0300)460100(

5021500202.1

7.14

4606073724

The minimum size required to transmitted 24 MMcfd is 7.597 7.6 inches

25



It is often desirable to increase the throughput of an existing pipeline by

gathering gas from new gas wells. A common economical solution to these problems is

to place one or more lines in parallel, either partially or throughout the whole length,

or to replace a portion of the line with a larger one. This requires calculationsinvolving flow in series, parallel, and series-parallel (looped) lines.

The philosophy involved in deriving the special relationships used in the solution of

complex transmission systems is to express the various lengths and diameters of the

pipe in the systems as equivalent lengths of common diameter or equivalent diameter

of a common length, there equivalent means that both lines will have the samecapacity with the same totally pressure drop.

For simplicity, illustrative examples will be based on the Weymouth equation.

26



Applying the Weymouth equation to each of the three segments gives:

28



29



where

qh: gas flow rate, ft3

/hrTb: base temperature, oR

Pb: base pressure, psia

D: inside diameter, inches

g: gas specific gravity ( air = 1 ), dimensionless

T: average following temperature, oRL: length of pipeline, miles

z: gas deviation factor at average flowing pressure & temperature

M: symbol = 1000

30



Example:

G =0 .6 z = 0.95 Total length = 10 miles

1. Determine the capacity of the pipe line ?2. Determine P3 at conjunction point ?

3. Determine LA’ if it is required to increase capacity 25%?

31



Solution

1.

Qh =18.062(Tb/Pb) (( P12- P2

2)D16/3/ Z T L ))0.5

=18.062*520/14.7*((4002-2002)*416/3/0.6*520*7*0.95))0.5

= 195905.96 ft3/hr

LA’=LB (DA/DB)16/3 =3*(4/6)16/3 =0.345 mile

Lequ. = LA’+ LA

=0.345+7=7.345 mile

∆Qh% = (1/LAeq)0.5-(1/L)0.5/(1/L)0.5 = ((1/7.345)0.5-(1/10)0.5)/(1/10)0.5 = 16.7%

Qh NEW = Qh +( 0 .167 * Qh ) = 228622.26 ft3/hr

32



or directly by the relation

Qt / Q1 = (( L / D116/3) / ( L1 / D1

16/3 + L2 / D216/3 ))0.5

= (( 10 / 616/3 ) / ( 3/616/3 ) + ( 7 / 416/3))0.5

=1.167

QhNEW = 228622.2615 ft3/hr

2.

(P12 - P3

2) LA ZA= ( P32 - P2

2) LA” ZB

ZA = ZB = 0

To eliminate trail and error

( 1064000 - P32 ) * 7 = 0.345 * ( P32 - 13120 )

P3 = 393 psia

33



3.

∆Qh%= (1/L Aeq) * 0.5 - ( 1 / L ) * 0.5 / ( 1 / L ) * 0.5

0.25 = ( 1 / L Aeq ) * 0.5 - ( 1 / 10 ) * 0.5 / ( 1 / 10) * 0.5

L Aeq= 6.39mile

L A’= X ( D A / DB )16/3 = X ( 4 / 6 )16/3

=0.115 X

6.39 = 0.115 X + ( 10 – X )

X = L A = 6.07 mile

L A’ = 0.329 mile

34



By adding new parallel pipelines , What would be the resulting increase in

capacity ?

What is the diameter that can obtain the given flow rate ( capacity ) ?

These questions are our target here

35



The new flow rate with both lines is

qt = q1 + q2

The length L is constant

L = L1 + L2

To calculate the increment in gas capacity, I apply in the following formulas:

or

36



Example Length of pipe line = 10 miles

Diameter of pipe line 1 = 4 inches

Diameter of pipe line 2 ( parallel to pipeline 1 ) = 6 inches

What would be the increase in capacity ?

37



Solution

The increase in the gas capacity = 211.33 %

38



It considered the most complex design of transmission system. The part ofpipeline is parallel. This is called looping or series-parallel system.

In the looped pipeline illustrated below, the original line consisted of segments

( A&B ) with the same diameter. In this system, a looping segment ( C ) has been

added to increase the capacity of the pipeline system.

AB

C

Looped System

P1 P3 P2

39



Consider a three-segment looped pipeline depicted in Figure below.

qt

L2L1

P1

L

q1 D1

q2

P2P3

D2

D3 qt

Series – Parallel Pipelines

To calculate the total flow rate in pipeline for parallel section, apply in the following

equ. :

40



To calculate the pressure drop in this section ( Parallel section ), apply in the

following equation:

where

qh: gas flow rate, ft3/hr

Tb: base temperature, oR

Pb: base pressure, psia

D: inside diameter, inches


T: average following temperature, oR

L: length of pipeline, miles


M: symbol = 1000

41



To calculate the flow rate in the second section ( Series ) with ( D3 ), apply in

the following formula:

To calculate the pressure drop ( P ) in the second section ( Series ) with ( D3 ),

apply in the following formula:

42



To calculate the flow rate throughout transmission system, apply in the

following formula:

To calculate the pressure drop P throughout transmission system, apply in

the following formula:

43



To calculate the increment in capacity or flow rate due to looped pipeline,


3/16

3

3

23/16

2

3/16

1

1

3/16

3

3

%

D

L

D D

L

D L

qq

qt

where

qh: gas flow rate, ft3/hr

Tb: base temperature, oR

Pb: base pressure, psia D: inside diameter, inches


T: average following temperature, oR

L: length of pipeline, miles


q ( % ): increment in gas capacity due to looped pipeline, percent 44



To calculate the increment in capacity or flow rate due to looped pipeline,


1

231.21

111

1

D R

Y

X

231.2

1

11

21

11

D R

X Y

45



where

X: increase in gas capacity, percent

Y: fraction of looped pipeline ( the ratio of looped pipeline length to the

original length ), fraction

R D: is ratio of the looping pipe diameter to the original pipe diameter,

dimensionless

46



The effects of looped line on the increase of gas flow rate for various pipe

diameter ratios are shown in Figure below:

47



48



Example: Consider a 4-in pipeline that is 10 miles long. Assuming that the

compression and delivery pressures will maintain unchanged, calculate gas

capacity increases by using the following measures of improvement:

(a) Replace three miles of the 4-in pipeline by a 6-in pipeline segment.( Series pipeline )

This problem can be solved by following formula:

L = 10 miles L1 = 7 miles L2 = 3miles

D1 = 4 in. D2 = 6 in.

3/163

33/16

2

23/16

1

1

3/16

11

D

L

D

L

D

L

D

L

qt q

49



The increase or improve in gas capacity = 16.68 %.

(b) Place a 6-in parallel pipeline to share gas transmission.


D1 = 4 in. D2 = 6 in.

3/16

1

3/163

3/162

3/161

1 D

D D D

q

t q

50



The increase or improve in gas capacity = 294.83 %.

(c) Loop three miles of the 4-in pipeline with a 6-in pipeline segment.


L = 10 miles L1 = 7 miles L2 = 3miles

D1 = 4 in. D2 = 6 in.

3/163

32

3/162

3/161

1

3/163

(%)3

D

L

D D

L

D

L

qqt q

51



The increase or improve in gas capacity = 17.91%.

52



53



Example: A portion of a large gas gathering system consists of 6 in. line 9.4

miles that handling 7.6 MMscfd with an average specific gravity of 0.64. The

upstream pressure is 375 psi and the average delivery pressure is 300 psi. The

average temperature is 73 oF. Due to new well completion, it is desired to

increase the capacity of this line 20% by looping with additional 6 in. line.

What length is required ?

Solution

I apply in the following formula to determine Y :

231.2

1

11

21

11

D R

X Y

54

1



231.2

)6/6(1

11

22.01

11

Y

25.01

694444.01

Y

75.0

305556.0Y 407408.0Y

Y length of looped pipeline

original length of pipeline

length of looped pipeline

9.4

407408.0length of looped pipeline

9.4

Length of looped pipeline = 9.4 0.407408 = 3.8296352 miles

It should be looped 3.83 miles with 6 in. to increase the capacity of this line 20%.

55



Example: A series pipeline consists of three lengths:

1. 30 miles of 6 inches pipe



Determine the equipment the equivalent lengths of 6, 8, & 10 inches ?

Solution

I apply in the following formula to determine the equivalent length at common

pipeline diameter:

6 in. 8 in. 10 in.

30 miles 40 miles 50 miles

D

L

D

L

D

L

D

Le3/16

3

33/16

2

23/16

1

13/16

56



I get the following:

57



58



59



Solution



I determine the equivalent length with common diameter to convert parallel

lines to series lines by following formula:

and I get for the section ( BC ):

L

D

L

D

L

D

Le

D

5.03

3/83

5.02

3/82

5.01

3/81

5.0

3/8

61

and for section ( AB ):



( )

To calculate pressure at ( B ), it need trail & error calculation. I assume the

pressure at ( B ), then I calculate the flow rate of gas and compare it with given

flow rate.

62

I apply in Weymouth equation for pipe line equation with size ( ID 15 inches ) to



calculate the flow rate of gas at pipe line conditions as follows:

I get:

5.0

_ _

3/162

2

2

1062.18

z LT g

D P P

P bT bqh

Then, f rom resul ts the pressure at ( B ) = 2289 psia 63

For section ( AB ):



For section ( AB ):

Finally, the press ure at ( A ) = 2685 ps ia

64



Thanks for Allah

Thank you for Attention

gas transportation

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