gas network analyses

CNG Alternative Network Analysis –Part III

CHAPTER 1

NETWORK ANALYSIS


1 Network analyses

1.1 Overview

The solution of a liquid/gas network means

estimating the flow rates in all the network

branches and calculating pressures at all node.

The hydraulic analysis of networks is

important for the hydraulic and mechanical

design of the different networks on the ship.

Generally we should have at least five

networks onboard the ship; the

loading/unloading network, the relief network,

the inert gas network, the CO2 fire fighting

network and the water fire fighting network.

For the loading/unloading network it is

necessary to find the maximum flow rate in the

network and to find the maximum pressure at

each branch, thus allowing the diameter and

thickness selection of each branch. Also we

need to find the loading/unloading time of the

ship. For CO2 fire fighting network, it is

important to find the pressure drop in the

network due to a certain flow rate and hence

select the network nozzles. The water fire

fighting network should be solved to determine

the required pump power. This chapter shows

the Hardy Cross method for network analyses

and the other bases of the analyses of gas

networks. It discusses two programs used to

solve the network, one is made using

MATLAB and the other is made using

Microsoft Office Excel. Then we go through

the transient analyses of the loading and

unloading process of the network.

1.2 Hardy Cross Method

The Hardy Cross method is the method used to

solve liquid and gas networks. We are now to

discuss this method and how to use it for gas

network analyses.

1.2.1 Network Elements

As shown in figure, the branch is a pipe

connecting two nodes. The loop is a group of

pipes connected in series thus the last point

lays on the first one. A single pipe is a pipe

connecting two separate loops. The tank forms

a node at which the pressure is known but the

flow entering the network or exiting it is

unknown. Only one known-pressure node is

needed in each problem to determine the

pressures at all nodes in the network. If there is

more than one tank in the network, they should

be connected by a virtual pipe. The virtual pipe

is an imaginary pipe that connects two tanks to

form a closed loop. The energy difference on

the virtual pipe should be given, thus there is

only one given pressure in each network.

Virtual pipes should be connected in series to

form one open loop.

Supply

Consumption

Node

Virtual

Pipe

Loop

Branch

Tank

Tank

Single pipe


1.2.2 Governing Equations

1.2.2.1 Continuity Equation

The continuity equation should be satisfied at

all nodes in the network. That is:

ΣQin = ΣQout (Or ΣQ = 0

The algebraic summation of all flow rates

entering and exiting the whole network should

equal zero. It should be noted that in case of

gas networks the flowrate used is the standard

flowrate which is equivalent to the actual

flowrate in liquid networks.

1.2.2.2 Energy equation

Across any branch 12; the change in the

energy of the fluid due to flow through branch

should be equal to the energy loss due to

friction in this branch. That is:

LD

2Cf2

21

lossh

2

aveZ

2

aveT

2R

2H

1H

2

aveP

2M

2

2P

2

1P

aveT

aveZR2

M1000

2u

1u

ln2

C

Where;

u1 & u2 = flow velocities at sections 1 and 2

respectively, m/s

P1 & P2 = average pressure between points

1&2, kpa

Pave = average pressure between sections 1 and

2

2P1P

2P

1P

2P1P3

2aveP

T1&T2 = temperatures at sections 1 and 2

respectively, k

Tave = average temperature between

sections 1 and 2, k

2

2T1T

aveT

R = universal gas constant = 8.314

kj/kg-k

M = gas molecular weight = 17.4

kg/kmol (assuming 90% methane and 10%

ethane)

Zave = compressibility factor at Pave and Tave

From CNGA equation;

3.925T

G1.785105272.466P

1

1Z

2D

bZ

bTRπ

bPM

bQ4

A

.

mC

Qb = standard flow rate (m3/s)

Pb & Tb= standard pressure and temperature;

100 kpa and 298 k respectively

Zb = compressibility factor at Pb and Tb

4 CLOSED LOOPS

Tank 1

Tank 2

Tank 3

CNG PROJECT Network Analyses

3

Zb = 1

D = pipe diameter, m

To calculate the friction factor, f, we select to

use Panhandle B equation.

Panhandle B equation

0.01961

D

Gb

Q20.8255

f

1

0.03922b

Q2

0.01961

D

G20.8255

1f

Where G = gas gravity = M/Mair

From the energy equation and Panhandle B

equation and by neglecting the elevation

difference and the velocity change; we can

conclude n

bQr

lossh

2

2D

bZ

bTRπ

0.01961

D

G20.8255

bPM4

D

L2r

n = 1.96078

Where r is the resistance of the pipe and n is

the head loss index.

1.2.2.3 Solution algorithm:

The network problem is a complicated

problem that can only be solved by successive

iterations which end with reaching a

predefined tolerance. Here are the general

steps for solving gas/liquid networks.

First of all, we have to check the

continuity equation over the whole

network. The algebraic summation of all

the flow rates entering or exiting the

network should be zero. This condition

may not be satisfied if exists one or more

tanks in the network. Since the tanks are

nodes which absorb or release unknown

amounts of flow.

Define the flow resistances of all pipe in

the network

Distribute the external flow rates entering

or exiting the network among the branches

of the network. Try to distribute flow rates

in inverse proportionality with the

resistances of the pipes. This will reduce

the total number of iterations needed to

reach required tolerance.

The problem will be solved when the

summation of the energy difference

around each loop equals zero. So an

iteration scheme is held by changing the

flow rates in the branches till reaching the

required tolerance. Note that in each

iterate, the continuity equation should be

satisfied at all nodes.

The Hardy Cross method offers an iteration

algorithm for solving networks in a directive

manner that reduces the number of iterations.

Hardy Cross algorithm

1. Find the algebraic summation of the

energy differences of the branches

forming each loop, that is:

1n

0Q0ΣrQloss

Σh

2. If this summation is not equal to zero, the

usual case, the new flow rates are

calculated as following:

ΔQ1n

0QΣrn

1n

0Q

0ΣrQ

lossΣh

sorder term...higher ΔQ1n

0nQ

n

0Qr

nΔQ

0Qr

nrQ

lossh

ΔQ0

QQ

To reach the solution of the network, this value

should equal zero then:

1n

0QΣrn

1n

0Q

0ΣrQ-

ΔQ

All flow rates in the loop should be corrected

by this value ΔQ. This is then applied to all the

remainder loops. Note that the common

branches between two loops should be

corrected twice once in each loop. Note also

that the continuity equation will be

automatically satisfied in the next iteration as

the correction ΔQ is added to all branches in

the loop.

The equations listed above states that the

problem should now be solved (Σhloss = 0).


4

However, due to the double correction of the

common branches, this does not occur.

3. At the end of each iterate, the error should

be compared with the predefined

tolerance.

TOLERANCEΔQΣ

Where, Σ|ΔQ| is the summation of the absolute

values of corrections of all loops.

4. For virtual pipes, the energy difference is

defined as the difference in fluid energy,

kinetic, potential and pressure energy,

between the two terminals. The slow in

the virtual pipe is set to zero and is never

corrected.

After getting the flow rates in all branches,

now we can get the pressures in all nodes.

Once we know the pressure in only one

node in the network, we can solve the

whole network by using the energy

equation in the consequent branches.

nb

Qrloss

h2

2P

2

1P

aveT

aveZR2

M1000

Then,

M1000

aveT

aveZR2

n

bQr

2

1P

2P

1.3 Network Solution using MATLAB

Our MATLAB program is a general program

that can be used to solve any gas/liquid

network. The algorithm of the program is

listed below:


5

INPUT LOOPS AND NODES

START

INPUT PIPE LENGTHS, DIAMETERS

INPUT GAS/LIQUID PROPERTIES

GAS or LIQUID

ESTIMATING A SUITABLE VALUE FOR THE

INITIAL GUESS

FINDING FLOW RATES IN ALL BRANCHES ACCORDING TO THE INITIAL GUESS SUCH

THAT THE CONTINUITY EQUATION IS SATISFIED AT ALL NODES THERE IS NO

BALANCE AT THE EXTRA NODE

APPLYING THE ITERATION SCHEME TO FIND THE

FLOW RATES IN ALL BRANCHES

WITH TOLERANCE = 0.00000000001 M3/S

FIND THE ASSISTANT PARAMETERS AND BUILD THE

ASSISTANT ARRAYS

DETERMINING PRESSURES IN ALL NODES BY APPLYING THE ENERGY EQUATION

TO ALL BRANCHES THERE IS NO BALANCE AT THE EXTRA BRANCH IN EACH

LOOP OR AT THE VIRTUAL NODES

END

APPLYING THE ITERATION SCHEME TO FIND THE

FLOW RATES IN ALL BRANCHES

WITH TOLERANCE = 0.00000000001 M3/S


6

1.4 Solution using Microsoft Office

Excel

What is well known about Excel is that it is an

intelligent calculator that can maintain several

mathematical operations in a very short time,

but Excel is also a very powerful tool in

solving trial and error problems. The solution

using Excel gives a complete control of the

solution. For example, we can change the pipe

diameter or resistance to get a certain flowrate

distribution. Besides that the complete solution

can be chassed which allows validating

MATLAB results. The problem with Excel is

that it can not solve the transient problem so it

will be used for steady state or for a snap shot

during the transient problem. The transient

problem can be then solved using MATLAB.

The following lines explain the procedures

followed to solve networks using Excel:

Excel Sheet Set-up Procedure

1. Open a new Excel sheet and name it "MyNetwork".

2. Build a table including all loops with their branches organized in order.

3. Beside each branch; set the gas properties and other constants; n, G, M, R, Tb, Pb and Zb.


7

4. Set branch length and diameter

5. Set a column for "r" and set its equation.


8

Note that virtual pipes take zero "r" values.

6. Set a column for Q which contains the initial guess; it may be kept empty

Note that in case of repeated branches, e.g. 7-5 and 5-7; Q of one branch should be referred to negative

Q of the other branch.

7. Build a column for head loss across all branches.


01

NOTE that in case of virtual pipes the head loss is set to the energy difference between the two

terminals which is equal to the L.H.S of the energy equation.

8. Now build two columns named "LOOP BALANCE" and "NODE BALANCE" each is

divided into two columns

LOOP BALANCE:

o The first column contains names of all loops

o The second column contains summation of head loss around each loop

NODE BALANCE:

o The first column contains names of all nodes that we should satisfy continuity at (one

arbitrary, usually virtual, node is excepted from the balance)

o The second contains summation of flows entering each node


00

9. set-up Excel solver function as following;

"Tools" menu → Add-Ins... → check "Solver Add-in"


01

10. Use "Solver" to get the flow rates:

"Tools" menu → Solver

Parameters

Select one of the "LOOP BALANCE" or "NODE BALANCE" cells to be the target cell and

set it to zero.


02

Select changing cells of flow rates Qb, except the cells of the repeated branches and virtual

pipes.

Add two constraints ; LOOP BALANCE cells and NODE BALANCE cells equal to zero

11. Click "Solve" button. The program will change the flow rates to satisfy the preset conditions.

When the iteration ends we get the required flow rates in place.

12. Now, the pressure in all branches can be calculated from the following equation

loss

h

aveTaveZR2

M22

P2

1P

NOTE that while P2 is function of Zave, Zave itself is function of P1 and P2 Excel usually sends error

message when a cell is referred directly or indirectly to itself. So Excel iteration function should first be

checked;

"Tools" menu → "Options" → "Calculations" tab → check "Iteration"

13. Finally we can check for the erosion velocity.


03


04

1.5 Ship network solution

1.5.1 First trial

Now we are going to solve our

loading/unloading network. We will start with

an initial schematic of the network, we will

solve it using both Matlab and Excel and

compare results of both solutions.

i) Drawing Network Schematic

according to Hardy Cross method

principles

Assumptions:

Solving for half of the network for

symmetry.

Qb = 500 MMSCFD = 164 m3/s

Then Qb for half of the network = 82 m3/s.

Considering the shell of bottles as one

large tank.

The energy difference on all virtual tanks

is set to zero.

Connecting these tanks with one open

virtual loop

Diameters are assumed as follows:

Branch Diameter (in)

Main 36

Sub 30

Terminal 20

The solution parameters are listed in the

following table

n G M (kg/kmol) R (kj/kmol-k) Tb(k) Pb (kPa) Zb L(m) D(m)

1.961 0.6 17.4 8.314 298 101.3 1 38 0.911

LOOP 1

LOOP 2

LOOP 3

LOOP 4

LOOP 5

LOOP 6

LOOP 7

LOOP 8

LOOP 9

LOOP 10

LOOP 11

4

3 5 9 13 17 21 25

7 11 15 19 23 27

8 12 16 20 24 28

6 10 14 18 22 26

Qb = 82 m3/s

38 m

38 m

10 m

5 m


05

MATLAB solution

Branch Qb Branch Qb

3 4 34.775 19 17 -3.2002

4 6 34.775 17 13 -6.4568

6 8 16.839 13 15 6.1007

8 7 0 14 18 4.7546

7 5 -22.868 18 20 2.3566

5 3 -47.225 20 16 0

7 11 0 16 14 -4.4924

11 9 -11.799 19 23 0

9 5 -24.357 23 21 -1.8332

5 7 22.868 21 17 -3.2566

6 10 17.936 17 19 3.2002

10 12 8.6887 18 22 2.3981

12 8 0 22 24 1.3499

8 6 -16.839 24 20 0

11 15 0 20 18 -2.3566

15 13 -6.1007 23 27 0

13 9 -12.558 27 25 -1.4234

9 11 11.799 25 21 -1.4234

10 14 9.2471 21 23 1.8332

14 16 4.4924 22 26 1.0482

16 12 0 26 28 1.0482

12 10 -8.6887 28 24 0

15 19 0 24 22 -1.3499

Excel Solution:

Branch Qb Branch Qb

3 4 34.78 19 17 -3.2

4 6 34.78 17 13 -6.457

6 8 16.84 13 15 6.101

8 7 0 14 18 4.755

7 5 -22.87 18 20 2.357

5 3 -47.22 20 16 0

7 11 0 16 14 -4.492

11 9 -11.8 19 23 0

9 5 -24.36 23 21 -1.833

5 7 22.87 21 17 -3.257

6 10 17.94 17 19 3.2

10 12 8.689 18 22 2.398

12 8 0 22 24 1.35

8 6 -16.84 24 20 0

11 15 0 20 18 -2.357

15 13 -6.101 23 27 0

13 9 -12.56 27 25 -1.423

9 11 11.8 25 21 -1.423

10 14 9.247 21 23 1.833

14 16 4.492 22 26 1.048

16 12 0 26 28 1.048

12 10 -8.689 28 24 0

15 19 0 24 22 -1.35

Both Results are similar


06

The Excel is a very powerful solution tool, we

can assume any parameter to be constant and

solve for the other parameters. One solution

was to fix all the flow rates supplied to the

shells to the same value and get the

corresponding extra resistances, valve

openings, on the shell inlet branches.

1.6 Loading/Unloading network

mechanical design

For the loading/unloading network, it is

required to use the largest possible standard

diameter. This is because the network is used

as gas storage during transportation. For the

mechanical design we would use the following

design formula:

factorlocation L

factordesign F

FLJTD

t2SP

From the CSA Z662-96, assuming F.L = 0.5

J = joint factor, = 1.00

T = temperature factor, = 1.00 (T < 120 C)

S = specified minimum yield strength, SYMS,

= 100,000 psi, (X-100 steel)

P = internal pressure, 3,552.5 psi

For the first trial take D=36" then t = 1.3"

which is larger than the maximum standard

thickness corresponding to the selected pipe

diameter. Try a smaller diameter, 34". Then t =

1.208" which is less than 1.25", the maximum

standard thickness corresponding to diameter.

So now all the main pipes are made of 34"

pipes with 1.25" thickness.

1.7 Solving the network at the start of

loading and start of unloading

1.7.1 Start of loading:

When the loading process starts, the gas

pressure in tanks is about 30 bar, whereas the

pressure in the loading line is kept at 245 bars

from the last loading time. Unlike our first

trial, in this solution we know the pressures at

all terminals, while the flowrates are

completely unknown. The network will have

the following schematic:

Solving the network gives the following results:

BRANCH Qb (m3/s) Pave(kpa) Zave us / ue (%)

LOOP1

3 4 49288.77591 11484.04696 0.803189019 2070.899721

4 6 49288.77591 8665.586262 0.843953791 2272.966825

6 8 33219.56682 7347.266455 0.864476237 1594.930042

8 7 0 7000 0.870049331 0

7 5 -33220.37072 7347.28233 0.864475984 1531.964815

5 2 -49288.67792 8665.609755 0.843953434 2070.894047

2 3 -49288.67792 11484.05464 0.803188914 1819.849753

LOOP2 7 11 0 7000 0.870049331 0

LOOP 1

LOOP 2

LOOP 3

LOOP 4

LOOP 5

LOOP 6

LOOP 7

LOOP 8

LOOP 9

LOOP 10

LOOP 11

4

3

5 9 13 17 21 25

7 11 15 19 23 27

8 12 16 20 24 28

6 10 14 18 22 26

26.5 m

10 m

6.5 m

LOOP 12

2

1

26.5 m 26.5 m

4 m

12 m

24 m


07

11 9 -10837.55223 7039.963174 0.869404327 517.7731223

9 5 -16068.30721 7385.981426 0.863859341 740.9935754

5 7 33220.37072 7347.28233 0.864475984 1594.968639

LOOP3

6 10 16069.2091 7385.915274 0.863860395 767.7247826

10 12 10830.3299 7039.911206 0.869405165 519.983256

12 8 0 7000 0.870049331 0

8 6 -33219.56682 7347.266455 0.864476237 1531.930382

LOOP4

11 15 0 7000 0.870049331 0

15 13 -3595.611082 7004.61188 0.869974846 172.5328021

13 9 -5230.75498 7044.557594 0.869330235 249.9036943

9 11 10837.55223 7039.963174 0.869404327 520.3300128

LOOP5

10 14 5238.879201 7044.345694 0.869333652 251.3888439

14 16 3531.182157 7004.451319 0.869977439 169.5382886

16 12 0 7000 0.870049331 0

12 10 -10830.3299 7039.911206 0.869405165 517.4313567

LOOP6

15 19 0 7000 0.870049331 0

19 17 -1631.765394 7000.980132 0.8700335 78.3343954

17 13 -1635.143898 7005.591582 0.869959025 78.46119954

13 15 3595.611082 7004.61188 0.869974846 172.631635

LOOP7

14 18 1707.697044 7004.94769 0.869969423 81.98448967

18 20 1153.572595 7000.496581 0.87004131 55.38505656

20 16 0 7000 0.870049331 0

16 14 -3531.182157 7004.451319 0.869977439 169.4446018

LOOP8

19 23 0 7000 0.870049331 0

23 21 -2.322696243 7000.000003 0.870049331 0.111516747

21 17 -3.378503294 7000.000025 0.87004933 0.162207906

17 19 7 7000.000022 0.87004933 0.336082356

LOOP9

18 22 554.124449 7000.557748 0.870040322 26.60429067

22 24 396.4738952 7000.061171 0.870048343 19.03541157

24 20 0 7000 0.870049331 0

20 18 -1153.572595 7000.496581 0.87004131 55.38163857

LOOP10

23 27 0 7000 0.870049331 0

27 25 -1.055807051 7000.000001 0.870049331 0.05069116

25 21 -1.055807051 7000.000003 0.870049331 0.05069116

21 23 2.322696243 7000.000003 0.870049331 0.111516747

LOOP11

22 26 157.6505538 7000.085315 0.870047953 7.569058657

26 28 157.6505538 7000.038261 0.870048713 7.569068091

28 24 0 7000 0.870049331 0

24 22 -396.4738952 7000.047054 0.870048571 19.03530024

LOOP12

1 3 98577.45384 19491.52545 0.706267666 3618.345562

3 2 49288.67792 11723.90881 0.799900869 2045.286079

2 5 49288.67792 8994.215749 0.838988806 2227.27384

5 9 16068.30721 7775.758788 0.85769722 749.3020366

9 13 5230.75498 7454.377469 0.862771648 244.8579828

13 17 1635.143898 7417.840074 0.863352356 76.57339082

17 21 3.378503294 7414.436542 0.86340649 0.158214487

21 25 1.055807051 7414.436522 0.863406491 0.049443187

25 27 1.055807051 7414.436519 0.863406491 0.049443187

27 1 0 17370.37037 0.732986119 0

As we can see, the velocity of gas at certain

branches exceeds the erosional velocity limit.

This may result in premature failure of the

piping system. This can be solved by using


08

extra pipe resistance, partially closed valves,

on the branches subjected to severe erosion.

These valves are controlled to keep the gas

velocity always under the erosional velocity

limits. The problem with valves is that they

reduce the loading/ unloading rate and hence

causes extra cost due to ship delay in Sidi-Krir

dock. So another suitable solution is to use

internal pipe coating for pipes subjected to

excessive erosion.

1.7.2 Start of unloading: This time, the ship network and tanks are at

245 bar, relieves to the unloading line at 30

bars. The results are listed in the table below:

BRANCH Qb (m3/s) Pave(kpa) Zave us/ue (%)

LOOP1

3 4 50752.55493 9987.78499 0.824326902 2361.383628

4 6 50752.55493 6146.058023 0.884064318 2963.4826

6 8 34206.12215 3807.986428 0.924854111 2413.709522

8 7 0 3000 0.939839485 0

7 5 -34206.95009 3808.019487 0.924853508 1997.35949

5 2 -50752.454 6146.095887 0.884063687 2361.375653

2 3 -50752.454 9987.795021 0.824326756 1939.471341

LOOP2

7 11 0 3000 0.939839485 0

11 9 -11159.40759 3104.522919 0.937873654 763.2183617

9 5 -16545.50391 3897.438853 0.923224405 966.0995548

5 7 34206.95009 3808.019487 0.924853508 2413.767944

LOOP3

6 10 16546.43278 3897.291836 0.923227079 1131.692885

10 12 11151.96924 3104.389622 0.937876156 786.9238794

12 8 0 3000 0.939839485 0

8 6 -34206.12215 3807.986428 0.924854111 1997.32305

LOOP4

11 15 0 3000 0.939839485 0

15 13 -3702.407106 3012.280813 0.939608085 260.257896

13 9 -5386.096324 3116.527383 0.937648405 368.3679067

9 11 11159.40759 3104.522919 0.937873654 787.4487564

LOOP5

10 14 5394.463536 3115.977355 0.937658723 379.2498416

14 16 3636.051273 3011.854181 0.939616122 256.5731228

16 12 0 3000 0.939839485 0

12 10 -11151.96924 3104.389622 0.937876156 762.7386771

LOOP6

15 19 0 3000 0.939839485 0

19 17 -1680.310715 3002.615224 0.939790198 118.4719055

17 13 -1683.689218 3014.888929 0.939558956 118.353655

13 15 3702.407106 3012.280813 0.939608085 261.2554339

LOOP7

14 18 1758.412263 3013.175907 0.939591223 124.0285523

18 20 1187.831416 3001.325204 0.93981451 83.81774425

20 16 0 3000 0.939839485 0

16 14 -3636.051273 3011.854181 0.939616122 255.6272482

LOOP8

19 23 0 3000 0.939839485 0

23 21 -2.322696243 3000.000006 0.939839485 0.163897971

21 17 -3.378503294 3000.000063 0.939839484 0.238399588

17 19 7 3000.000056 0.939839484 0.493945691

LOOP9

18 22 570.5808475 3001.488437 0.939811433 40.26021942

22 24 408.2483846 3000.163281 0.939836408 28.80750436

24 20 0 3000 0.939839485 0

20 18 -1187.831416 3001.325204 0.93981451 83.78297511

LOOP10

23 27 0 3000 0.939839485 0

27 25 -1.055807051 3000.000001 0.939839485 0.07450162

25 21 -1.055807051 3000.000008 0.939839485 0.07450162

21 23 2.322696243 3000.000006 0.939839485 0.163897972


11

LOOP11

22 26 162.3324629 3000.227728 0.939835193 11.45454311

26 28 162.3324629 3000.10213 0.93983756 11.45463916

28 24 0 3000 0.939839485 0

24 22 -408.2483846 3000.125602 0.939837118 28.80637092

LOOP12

1 3 101505.0089 19083.86016 0.71063338 3849.841815

3 2 50752.454 10284.32637 0.820049682 2309.899209

2 5 50752.454 6647.874354 0.875774263 2784.046508

5 9 16545.50391 4666.713461 0.909442773 1008.394862

9 13 5386.096324 4042.48739 0.920593984 333.4613644

13 17 1683.689218 3965.950467 0.921980092 104.4143309

17 21 3.378503294 3958.74763 0.922110753 0.209518574

21 25 1.055807051 3958.747589 0.922110754 0.065476091

25 27 1.055807051 3958.747584 0.922110754 0.065476091

27 1 0 16551.51515 0.742330035 0

As we can see, only the flow passing through

the main pipe exceeds the erosional velocity

limits.

1.8 Transient flow analysis

The flow in the network during loading and

unloading is not steady. It is transient flow as

the flow rate in each branch and pressure at

each node is continuously varying. This is due

to the fact that the pressures in tanks, bottles

are continuously varying due to the gas

accumulation in the bottles.

To solve the loading problem we can imagine

the network to consist only of one pipe

connecting two tanks; the first is a high

pressure tank, while the second is a low

pressure tank. Now consider the natural flow

from the high pressure tank to the low pressure

tank. From the defined pressures across the

pipe, tank pressures, we can find the flow rate

in the connecting pipe by applying the energy

equation.

n2

2

2

1 rQPP

The flow from the high pressure tank between

moments: t1&t2 results in a pressure reduction

in the tank. The new pressure can be found

from the state equation of the gas.

constantetank volumV

2RT

2Z

2mV

2P

step timepredefinedt

2t&

1 tat times tank pressurehigh in masses gas1m&2m

tρQΔVmm 21

In the same manner we can find the new

pressure in the low pressure tank. The process

is repeated with a time step, t, till the

difference in pressure between the two tanks

becomes within a predefined tolerance.

Now consider the forced flow, through a

compressor, from the low pressure tank to the

high pressure tank; this time the flow rate can

be evaluated by solving both the system curve

and compressor curve. The new tank pressures

can be found from the state equation as

previously mentioned. Note that the discharge

pressure from the compressor is calculated

based on the tank pressure and not vice versa.

The first case, natural flow, can be solved

simply while the solution of the second case

requires the curves of the compressors selected

for the loading and unloading processes. Then,

we can use the MATLAB to solve the transient

problem. The solution parameters are listed in

the table below:

Pressure

Flow rate

Operating

point Compressor

curve

System

curve


10

G 0.6

Pb (kpa) 101.3

Tb (k) 288

Tave (k) 298

Zb 1

Zave 0.63

R (kj/kmol-k) 8.314

Mair (kg/kmol) 28.9

Pressure in bottles (kpa) 24500

Pressure in loading line (kpa) 7000

Tolerance (kpa) 1000

External flow entering all branches = 0

Tolerance (m3/sec) 0.00001

Volume per shell (m3) 4750

Time Step (sec) 3.6

Required Pressure (kpa) 7000

The solution of the problem is based on a

constant pressure at the end of the loading line,

2.5 km long. Also we considered the solution

of half the network for simplicity.

The natural-flow unloading process takes

98,071 seconds, that is 27.24 hours.

Several phenomena can be noticed in the

transient analyses of the loading and unloading

process; some of these are the following:

From the snap-shot solution for the

start of loading moment mentioned

above we can expect the tanks close

to the supply line to be filled very

soon, while the tanks remote from the

supply line to be filled after a very

long period of time. This is not true as

the fast flow into the close tanks, at

the first, moment results in a pressure

rise in these tanks which in turn forms

a resistance to more inlet flow. The

flow then goes with relatively larger

quantities to the remote tanks, till the

pressure in these tanks rises

somewhat. So the whole network will

be filled almost at the same time. The

same thing happens during the

unloading process, the fast flow out of

the close tanks, at the first moment,

releases their pressures and hence

more flow comes out from far tanks,

the whole network is unloaded almost

in the same time.

It is obvious that the flow rate reduces

by time due to the continuous

reduction in the pressure difference

between tanks. That means that the

high flow rate during the start point

will stay only for a short time while

the rest of the loading/unloading

process will occur at much lower

rates.

0 1 2 3 4 5 6 7 8 9 10

x 104

-1.5

-1

-0.5

0

0.5

1

1.5x 10

5

Time (sec)

Flo

wra

te (

m3/s

ec)

0 1 2 3 4 5 6 7 8 9 10

x 104

0.5

1

1.5

2

2.5

3x 10

4

Time (sec)

Pre

ssure

(kpa)

Pressure

Flow rate Q1 Q2 Q3

CNG PROJECT CNG Technology

11

1.9 System Timing

We are now to calculate the time of the

different processes included in the CNG

transportation. This will help us determine the

number of ships required to satisfy the required

demand to the required destination. The CNG

comprises mainly three processes; loading,

shipping and unloading. The loading process is

accomplished using five high flow rate

compressors, each having average flow rate of

500,000 m3/hr and maximum discharge

pressure of 40 bar (5 × 40 = 200 bar + 70 bar

supply pressure = 270 bar > 245 bar). While

the offloading is maintained using two parallel

compressors of the same type (70 – 40 = 30

bar remaining in bottles)

So we can say that the average loading rate is

500, 000 m3/hr. the unloading process starts

naturally by the gas pressure. This continues

till the pressure in bottles reaches almost the

network pressure, 70 bar, then the compressor

is used to suck the remainder gas till the bottle

pressure reaches almost 30 bar. The average

unloading rate by compressor is 1, 000, 000

m3/hr. The ship speed is about 25.5 knots,

nautical miles per hour, that equal 25.5 ×

1.15077 = 29.34 miles/hr. The distance from

Egypt to Cyprus is 435.05 miles. The required

number of ships can be calculated from the

following formula:

Cargo

DemandTime ShippingTime UnloadingTime Loading

shipsN

speed Ship

Distance2Time Shipping

Rate UnloadingCompressor

Cargo Remaining

Time UnloadingNaturalTime Unloading

Rate Loading

CargoTime Loading

mile/hr 29.34Speed Ship

/hr3

M

1,000,000Rate UnloadingCompressor

/hr3

M500,000Rate Loading

SCM40,845,000Cargo

/hr3

M244,000,000/Demand

miles435.05Distance

hr 29.66Time Shipping

hr 30.7Time UnloadingCompressor

hr 27.24Time UnloadingNatural

hr 81.69Time Loading

Then

Nship = 0.7 ships

For a reliable system, we need at least two

ships. The waiting period of the ship when the

ship stays offshore till the next trip can be

calculated from the following formula:

Cargo

DemandPeriod WaitingTime ShippingTime UnloadingTime Loading

shipsN

In our case the waiting time will be 320.85

hours, 13.4 days.

The previous method is not an accurate method

for the determination of the system timing as

the loading/unloading rates of compressors are

widely alternating during loading/unloading

process.

CNG PROJECT CNG Technology

12

1.10 Loading and unloading

pipelines

These offshore pipelines were designed

according to the same principles mentioned

before the results are listed in the table below:

Loading pipeline

Pipe length (km) 4

Inlet pressure (bar) 247

Outlet pressure (bar) 245

Inlet temperature ( c ) 20

Surrounding temperature ( c ) 20

Flow rate (SCM/D) 5,184,000

Material X-80

Diameter(inch) 10.75

WT(inch) 0.356

Concrete thickness(cm) 15

Unloading pipeline

Pipe length(km) 2.5

Inlet pressure(bar) 245

Outlet pressure(bar) 244.8

Inlet temperature( c ) 20

Surrounding temperature( c ) 20

Flow rate (SCM/D) 10,368,000

Material X-80

Diameter(inch) 14

WT(inch) 0.406

Concrete thickness(cm) 15

The design is based on the compressor flow

rate rather than the natural flowrate. The

loading pipeline is solved backward while the

unloading pipeline is solved forward.

gas network analyses

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