Energy saving in the process of gas pipeline overhaul739068/FULLTEXT01.pdf · Energy saving in the...

Master of Science Thesis

KTH School of Industrial Engineering and Management

Energy Technology EGI-2014-071MSC EKV1044

Division of Heat and Power Technology

SE-100 44 STOCKHOLM

Energy saving in the process of gas

pipeline overhaul

Alexey Mitrokhin

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Master of Science Thesis EGI-2014-071MSC

EKV1044

Energy saving in the process of gas pipeline

overhaul

Alexey Mitrokhin

Approved

2014-06-16

Examiner

Supervisor

Vladimir Kutcherov

Commissioner

Contact person

Abstract

The problem of energy saving during overhaul of a linear part of gas trunkline is regarded in this paper.

This issue has been analyzed from different perspectives.

Thermodynamic analysis of gas evacuation from a string that is off operation for the overhaul to a parallel

or adjacent string with the use of mobile compressor systems was made. Economical attractiveness of

mobile compressor systems applications was proved.

Various methods of gas trunklines linear parts overhaul have been considered. For each of the methods

problems of minimization of necessary for the overhaul amount of energy consuming machines have been

solved.

The results can be used in the development of overhaul projects of the gas trunklines linear parts.

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Table of Contents

Abstract ........................................................................................................................................................................... 2

1 Introduction ............................................................................................................................................................ 4

2 Gas evacuation to a parallel or adjacent string before the start of the overhaul ......................................... 6

2.1 General description of the works ............................................................................................................... 6

2.2 Economical efficiency calculation framework ........................................................................................11

2.3 Calculation example ....................................................................................................................................18

2.4 Conclusion ....................................................................................................................................................20

3 Possibilities of minimization of necessary for overhaul amount of energy consuming machines .........20

3.1 Separate method of overhaul .....................................................................................................................22

3.1.1 Process modeling .................................................................................................................................22

3.1.2 Preset parameters .................................................................................................................................33

3.1.3 Initial solution .......................................................................................................................................35

3.1.4 Numerical method construction .......................................................................................................37

3.1.5 Iterative process ...................................................................................................................................39

3.2 Combined method of overhaul .................................................................................................................51




3.2.4 Iteration process...................................................................................................................................55

3.3 Combined method with all works performed inside trench.................................................................61




3.3.4 Numerical algorithm construction ....................................................................................................64

3.3.5 Iteration process...................................................................................................................................65

3.4 Conclusion ....................................................................................................................................................70

4 Conclusion ............................................................................................................................................................70

Bibliography .................................................................................................................................................................70

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1 Introduction

Energy saving is one of the most important trends of the world’s development. With the energy sources reserves running out that field becomes especially important.

Among international standards that regulate the relations in that field the group of standards ISO 50001 [1] needs to be mentioned. ISO 50001:2011 includes a list of requirements to the organizations. These requirements are aimed on the following:

Development of effective energy use strategy;

Choosing the means to complete this strategy;

Use of information for more effective energy utilization;

Evaluation of strategy application efficiency;

Energy conversion improving.

In the Russian Federation the main legislation that governs the relations in that sphere is the Federal Law of the Russian Federation of November 23, 2009 N 261-FZ "On energy saving and energy efficiency improvements and on Amendments to Certain Legislative Acts of the Russian Federation" which has the main purpose to "create the legal, economic and institutional framework to stimulate energy conservation and energy efficiency" [2].

In order to maximize the efficiency of natural fuel and energy resources use, to improve the performance of energy sector for the economy growth and to improve the standard of living in the country in the year 2000 the Energy Strategy of Russia till the year 2020 was approved [3].

Considering gas industry, the following JSC “Gazprom” [4] standards should be noted:

Gazprom Standard 4-2005 Regulation on the procedure of JSC "Gazprom" monitor for the effective use of gas [5] The document provides a framework for:

Monitoring the use of gas efficiency on JSC “Gazprom” subsidiary companies’ objects during construction, installation, reconstruction and operation of gas-consuming equipment that use the natural gas for its’ own technological needs;

Controlling the organization and execution of energy inspection in the JSC “Gazprom” subsidiary companies.

Gazprom Standard 2-1.20-601-2011 Methodology of calculating the effect of saving energy resources spent on own technological needs of gas trunkline transport [6] The standard sets main rules and also the calculation framework for evaluating the energy-saving effect on the objects of gas transport system.

Gazprom Standard 2-1.20-114-2007 Methodology of gas transport objects energy audit [7] The standard provides:

The calculation framework for evaluating the energy efficiency indicators on the objects of gas transport system;

The algorithm for analyzing the efficiency of gas consumption for technological needs of gas trunkline transport systems;

The algorithm for calculating the gas transport system energy balances;

The framework for the input data to energy passport of a gas transport system object;

The requirements for energy saving measures development. One of the last Gazprom standards, that regulates the relations in the field of energy saving is R Gazprom 2-1.20-673-2012 Energy saving regulation system of JSC “Gazprom” [8].

According to energy saving potential analysis, in the gas industry of Russia it is possible to achieve the total reduction of energy losses by 60 ÷ 70 % [9].

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At the moment workers of Russian gas field are searching the ways to increase energy efficiency [10], [11]. Particularly, LLC “Gazprom VNIIGAZ” has proposed an approach for JSC “Gazprom” objects energy saving potential assessment. This assessment provides the possibility of determining the maximum reserve of energy efficiency of these objects in the process of purposeful activity in the implementation of energy saving programs.

Methodology to assess the energy saving potential of technological facilities of JSC "Gazprom" is based on a comparison of their energy efficiency to the best national and world examples. Corporate methodology of JSC "Gazprom" in this field is formed according to comprehensive experience in the implementation of research in order to develop the concept of energy saving and energy efficiency of JSC "Gazprom" for the period 2011 - 2020 years (approved by JSC "Gazprom" order on December 8,

2010 № 364 [12] and is accumulated in the regulatory document R Gazprom 2-1.20-742-2013 Methodology for determining the energy saving potential of technological objects [13]).

According to [14] energy saving potential of technological objects of JSC "Gazprom" is to be calculated separately for each kind of energy resources used and in total in standard fuel dimension: gas saving potential, potential electricity energy savings, potential saving of thermal energy, total potential savings of all energy resources.

In this thesis paper the problem of energy saving during a gas pipeline overhaul is regarded. The Unified Gas Supply System of Russia includes, among others, approximately 170 thousands kilometers of gas trunklines and laterals and most of these pipelines have been in operation for more than thirty years. Hence in order to improve the reliability of gas supply, each year a large scope of overhaul work needs to be performed. Thus the problem of energy losses minimization is especially relevant in the case of gas pipelines overhaul.

Above stated problem is regarded from 2 sides:

The possibilities of energy source losses minimization before the start of the overhaul process. Today approximately 2 billiards cubic meters of gas are released to the atmosphere each year in Russia. Assuming the gas price at the level of 3000 rubles per cubic meter (that is the level of gas price inside the Russian Federation [15]) these two billiards correspond to:

In addition, taking into consideration the fact that 1/3 of total produced gas is exported for 400$ per cubic meter, the total amount of economy losses appears to be even more impressing:

The possibilities of minimization of the necessary for overhaul works performance amount of energy consuming machines.

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2 Gas evacuation to a parallel or adjacent string before

the start of the overhaul

In order to reduce the amount of gas released to the atmosphere the following technics may be used:

An increased amount of gas transferred to the consumer;

Gas evacuation to a parallel or adjacent string using mobile compressor stations;

With the use of stationary gas compressor units gas evacuation to the pressure levels, below which surge mode may occur;

Overhauls with the use of hot tapping technic. The technology of using mobile compressor stations appears to be very attractive. Applying such a system has the following economical advantages:

Economy of a sufficient amount of gas (that is especially relevant as the gas price is constantly rising);

Reduction of payments for the negative impact on the environment (according to Russian

legislation, 1.2 appendixes to the RF Government Resolution № 344 of 12.06.2003 [16]). Besides this, positive ecological effect of this technology in terms of global warming potential should be noted as methane is one of the so-called greenhouse gases.

2.1 General description of the works

As it has already been mentioned, it is reasonable to use a mobile compressor system to evacuate gas to a parallel or adjacent string of gas pipeline. In order to improve its mobility such systems are usually installed inside a truck. One of mobile compressor systems is presented on the picture (figure 1.1) [17]:

figure 1.1 – mobile compressor system layout

The general scheme of a mobile compressor system in operation is presented on the following picture (figure 1.2):

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figure 1.2 – Scheme of a mobile compressor system in operation

One of the most important requirements to such compressor system is their power as the total period of the overhaul works is usually limited. Besides this, mobile compressor systems obviously must provide the necessary pressure at the end of the process (at the inlet to a parallel or adjacent string).

In order to increase the system power and hence the speed of gas evacuation several compressor units may be used in a single system. The general dependence of gas evacuation speed on the amount of gas compressor units is presented on the picture (figure 1.3) [17]:

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figure 1.3 - Dependence of gas evacuation speed on the amount of gas compressor units.

For the optimization of gas evacuation process the compressors in the system may be operated in two ways:

1. One step regime – in the beginning of gas evacuation process. The regime is characterized by:

High efficiency in terns of the evacuated gas volume;

Maximum amount of cylinders in operation.

The scheme of a mobile compressor system operated in the one step regime is presented on the picture (figure 1.4):

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figure 1.4 - one step regime of a mobile compressor system

2. Two step regime – in the end of the gas evacuation process. The regime is characterized by:

Two steps of gas compression: one part of cylinders are working on the first step, the other – on the second;

High efficiency in terms of compressor ratio.

The scheme of a mobile compressor system operated in the two steps regime (for the case, when the compressor has 4 cylinders and 3 of them are operated on the first step) is presented on the figure 1.5:

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figure 1.5 - two steps regime of a mobile compressor system

The process of gas evacuation is usually performed in two periods: the preparation and the main period.

The preparation period includes:

Connection junction installment on the by-pass pipelines;

Preparing the platform for the mobile compressor system;

Temporary trains installment and testing;

Flare valves distant control system installment and adjustment;

Means of communication testing;

Setting the mobile compressor system to work automatically. Choice of the place for the mobile compressor system is defined by the mobile compressor system connection junction conditions. The availability of the path for the mobile compressor system to reach the pipeline is also necessary. Mobile compressor system is connected to the flare valves through the temporary trains. The temporary trains have a smaller diameter in comparison to the diameter of the pipeline.

It is especially important to be able to calculate the amount of gas that may be saved in consequence of applying a mobile compressor system. With this data it is possible to evaluate whether gas evacuation to a parallel or adjacent string is economically reasonable or not.

From the perspective of calculation methodology the Gazprom standard 3.3-2-024-2011 Valuation methodology of natural gas spent on own technology needs and on gas trunkline transport losses [18] should be pointed out. In addition in this paper a special approach based on known thermodynamics laws is proposed.

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2.2 Economical efficiency calculation framework

According to Gazprom standard 3.3-2-024-2011 the total amount of gas, which may be evacuated from a string that is off operation for the overhaul, is calculated as follows:

(

)

(1.1)

Here:

– volume of the pipeline string off operation (m3);

– initial pressure in the pipeline string off operation (Pa);

– final pressure in the pipeline string off operation (Pa);

- initial temperature in the pipeline string off operation (K);

– final temperature in the pipeline string off operation (K);

- standard conditions temperature (K);

- standard conditions pressure (Pa);

- compressibility factor before the start of the process of gas evacuation;

- compressibility factor after the end of the process of gas evacuation.

The gas compressibility rates before and after the process of gas evacuation respectively may be calculated through (the same Gazprom standard):

Here:

– natural gas density under standard conditions (kg/m3).

The geometrical volume of the part of the pipeline is calculated as follows:

(1.2)

Here:

- internal diameter of the pipeline (m);

– length of the pipeline that is off operation for the overhaul (m). The gas temperature at the end of the process of gas evacuation needs to be specially analyzed. In the already mentioned Gazprom Standard this temperature is assumed to be equal to the initial gas temperature:

However, at least at first this statement doesn’t seem to be obviously correct. In order to check it’s correctness let us consider the process of gas evacuation in details.

On the following picture (figure 1.6) the scheme of gas pipeline at the start of gas evacuation process is presented:

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figure 1.6 – Pipeline scheme before the start of the evacuation process

After the start of the process of gas evacuation the gas begins to flow. For further analysis we should assess, whether the gas flow isothermal or not, in other words, whether the temperatures at different sides

of the pipeline are equal to each other or not (is equal to ?) (figure 1.7):

figure 1.7 – Pipeline scheme during the evacuation process

According to [19], in order to define, if the gas flow isothermal or not, the so-called Shukhov parameter may be used:

(1.3)

Here:

– heat transfer coefficient from the ground to the gas (

);

– gas mass flow rate in the pipeline (kg/s);

– specific mass heat capacity (

).

With the Shukhov parameter lower than 4 the gas flow cannot be regarded as isothermal. That means, that for the gas flow not to be isothermal the following inequality must be fulfilled:

Solving this inequality subject to the gas mass flow rate yields:

(1.4)

Thus, the gas flow rate is not isothermal, if the inequality (1.4) is fulfilled.

According to [19] the heat transfer coefficient may be assumed equal to

. Let us consider several

possible combinations of the parameters in the expression above (1.4) (table 1.1):

𝑇𝑔𝑟

𝑇 𝑝0

𝐿

𝑇 𝑇

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Table 1.1 – Combination of parameters for the assessment

№

1 1.380 50000 2790

2 1.190 40000 2675

3 1.000 30000 2560

According to the criteria (1.4), the flow is not isothermal if the mass flow rate through the pipeline is more than:

{

With the mobile compressor systems used, these inequalities are not fulfilled, because such systems

provide the mass flow rate not more than

. And usually the mass flow rate values are even

smaller.

In our case, because of relatively low mobile compressor systems power or, in other words because of small mass flow rate value G the Shukhov-parameter value is always larger than 4.

While gas is being evacuated from the pipeline the gas pressure is dropping. At the same time the gas is being cooled down according to the throttling effect:

(1.5)

Here:

– pressure drop in the pipeline because of the gas evacuation process (Pa);

- Joule-Thomson coefficient ( ). However, the Joule-Thomson effect is not the only effect that should be taken into account. As the gas is cooled down due to the throttling effect, the process of the heat transfer from the ground to the gas takes place. It should be noted, that the ground temperature may be assumed constant. The equation that defines the heat transferred from the ground to the gas on a meter of the pipeline may be written as follows:

( )

(1.6)

Here:

– ground temperature (K);

– gas temperature at a given moment of time (К).

From the other side, the amount of heat gas received may be calculated from the following formula (the influence of pipeline wall is not taken into account as the pipeline is considered to be thinwalled):

(1.7)

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Here:

– mass of the meter of the gas in the pipeline (kg/m);

– difference between ground and gas temperatures (K).

After the string had been switched off and all the preparation works have been completed, the gas has the

pressure 0 and the temperature . Let us consider the very first moment of the process of gas

evacuation. Let the pressure has already dropped for a small value . According to the throttling effect (1.5) a certain temperature drop corresponds to this pressure drop:

Here:

– Joule-Thomson coefficient, that corresponds to the initial gas parameters ( 0 and );

- temperature, that the gas would have in the absence of heat transfer from the ground (K).

The beginning of the process of gas evacuation may be represented with the following curve (figure 1.8):

figure 1.8 – Temperature dependence on pressure according to throttling effect

While the pressure and the temperature are changing by the values and respectively the heat transfer from the ground to the gas takes place. In order to calculate the amount of heat transferred for a given period of time, taking into account the linear dependence, it is reasonable to take the gas

temperature constant and equal to the mean value between и (figure 1.9):

𝑇

𝑝

𝑝0 𝑝

𝑇

𝑇

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figure 1.9 – Choosing the temperature

Hence, the amount of heat transferred from the ground to a meter of the gas within the period of time is calculated as follows:

( ) (

) (

)

Then, knowing it is possible to find the temperature - the real temperature after the gas

pressure dropped by the value :

(

)

(

)

Here:

– pressure drop time (s).

Solving the equation above subject to yields the following expression:

(

)

(1.8)

Here:

𝑇

𝑝

𝑝0 𝑝

𝑇

𝑇𝑔𝑟

𝑇𝑚

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- pressure drop speed (

).

Taking into consideration the formula above (1.8), in order to evaluate the possible values of , it is enough to compare two components from the numerator:

and

Or

and

If than in the process of gas evacuation the temperature drop takes place, otherwise – the

appears to be more than , that is theoretically impossible and that is why .

Under small values of (in the beginning of the process of gas evacuation) the throttling effect is, roughly said, stronger than the heat transfer from the ground. That means there is also a temperature drop.

However, with the temperature difference between the gas and the ground increased, the amount of heat transferred is obviously increased too. Let us assume, that the temperature already dropped for several

grades. Let be the gas temperature at this time. Herewith the Joule-Thomson effect occurs (1.5):

Here the variables have the same meaning as they had before.

In order to evaluate the amount of heat transferred the mean value between the initial gas temperature and the one, which the gas would have in the absence of heat transfer, is taken:

The amount of heat transferred from the ground to a meter of the pipeline for a given time is calculated as follows:

( ) (

) (

)

The real gas temperature after a period of time ( ) (with both throttling effect and heat transfer taken into consideration) is calculated similar to the case of the beginning of the process of gas evacuation:

(

)

(

)

Solving the equation subject to yields:

(

)

is larger than if is larger than . Or, in other words, if

( ) is over . Thus, the gas in the pipeline is heated up, if:

( )

( )

(1.9)

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As in the case of research, whether the gas flow isothermal or not, different combinations of parameters

in the last formula (1.9) will be regarded (again assuming

) (table 1.2).

Table 1.2 – Parameters combinations for the research

№

1a 1.38 56.19 2733 20 19

1b 1.38 56.76 2745 20 17

1c 1.38 57.34 2757 20 15

2a 1.19 47.21 2629 20 19

2b 1.19 47.69 2639 20 17

2c 1.19 48.17 2648 20 15

3a 1.00 39.5 2550 15 14

3b 1.00 39.7 2554 15 13

3c 1.00 39.9 2558 15 12

In order for gas heating to occur the following inequalities should be fulfilled:

( )

{

The pressure drop temperature is extremely rare higher than 10

(with 4 or compressors working

simultaneously). Besides this, even with the speed slightly over 10

, gas heating takes place after the

temperature drops by only 1 – 1.5 grades Celsius. Hence, for the next considerations it is reasonable to

assume , that means the temperature within the process stays the same. As in the case of

isothermal / non-isothermal study this could be explained be relatively low mobile compressor systems’ power.

Thus, it was proved, that . Hence, all of the parameters in the formula for (1.1) are known, the possible amount of gas saved due to the process of gas evacuation may be found.

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Knowing the total amount of gas saved it is possible to evaluate the economical effect of applying a mobile compressor system. The effect may be divided into two components: the economical benefit due to the gas not released to the atmosphere and reduction of payments for the negative impact on the environment (according to Russian legislation):

1. Money for the saved gas:

(1.10)

Here:

C –price on gas spent for own technological needs. 2. Reduction of payments for the negative impact on the environment:

(1.11)

Here:

– payments for a kilogram of methan released to the atmosphere;

– ecology factor coefficient in a certain region of Russian Federation;

– interest rate (from the year 2003).

Mobile compressor systems are usually gas driven and they consume a part of gas they save. The total gas consumption rate is usually from 3 to 5 %. For the purposes of this paper it is reasonable to take 4%:

(1.12)

Costs for the mobile compressor system operation may be calculated as follows:

(1.13)

Here:

– transportation cost of thousands of cubic meters of gas.

Thus, the total economical effect of applying a mobile compressor system before the start of the overhaul process is calculated in accordance with the following expression (1.10 – 1.13):

(1.14)

2.3 Calculation example

Let us consider an example of gas evacuation process. Let a pipeline part (that is off operation) be 30-kilometer long and have 1020mm diameter. Pipeline wall thickness is 10mm. The ground temperature is 15 grades Celsius. After the pipeline had been off operation for the overhaul and after the preparation period has been finished the gas in the pipeline has a pressure of 4.217 MPa. For the gas evacuation process the mobile compressor system Ariel JGA/6 with Caterpillar G3412C LE engine is involved. The mobile compressor system has the following characteristics:

Rated power: 626 kW;

Suction range: 0.5 – 7.5 MPa;

Maximum discharge pressure: 7.5 MPa;

The mobile compressor system may be in operation under the ambient pressures from 7 MPa to 1 MPa and under the ambient temperature from -40 to +30 grades Celsius.

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Own technological needs gas price is assumed to be 4150 rubles. Mobile compressor system operation costs 1600 rubles per thousand cubic meters. Payment for methane released to the atmosphere – 50 rubles per ton. Let the works be performed in the Central Region of Russian Federation; in this case ecology factor coefficient is 1.9. The interest rate (from the year 2003) – 1.79. Let us calculate the economical reasonability of applying the mobile compressor Ariel JGA/6.

As it was proved before, the gas temperature during the whole process of gas evacuation stays the same and is equal to the temperature of the ground.

Knowing the temperatures and pressures at the start and the end of the process of gas evacuation with the

help of known relations the relevant compressibility rates and are calculated:

Under and

Under and Volume of the part of the pipeline (1.2):

Evacuated gas volume (1.1):

(

)

(

)

Saved gas price (1.10):

Payment reduction benefit (1.11):

The total fuel consumption (1.12):

Mobile compressor system operation costs (1.13):

And, finally, the resulting economical effect (1.14):

As it may be observed from the calculations above the economical benefit is principally achieved through the energy source save. At the same time the largest losses come from the mobile compressor system operation costs.

The total costs and benefits of using the mobile compressor system are presented on the following graph (figure 1.10):

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figure 1.10 – Economical benefits and losses

It should be noted that though this diagram was constructed for a single case, the total proportion of economical benefits and costs stays almost the same for all other cases.

Thereby due to mobile compressor system application an economical benefit of 2 million rubles may be achieved for the case of particular pipeline part.

2.4 Conclusion

In this part of the paper the possibilities for mobile compressor systems use during gas trunklines overhaul were analyzed. Problem’s topicality was stated: a huge gas volume is each year released to the atmosphere. The steps of gas evacuation process were discussed; the framework for a mobile compressor system application economical benefit calculation was presented, analyzed and proved.

As a short conclusion it should be said, that generally mobile compressor systems applications lead to significant economical benefits during overhaul works execution. However, at the same time the use of mobile compressor systems is sometimes impossible due to the time pressure. The total period of gas evacuation works (including preparation and main periods) may exceed ten days, and for some cases such a significant increase of total overhaul works time is inappropriate.

3 Possibilities of minimization of necessary for overhaul amount of energy consuming machines

The overhaul is the main method of providing reliability for the gas transport system of Russian Federation. In the field of gas transport 2 – 2.5 thousands kilometers of gas trunklines are repaired each year. The main ways of executing the overhaul works are pipeline re-insulating and changing of defective pipeline parts. With these works performed a large number of corrosion, stress-corrosion and other defects is removed. Due to these means the number of accidents on the gas transport objects of JSC “Gazprom” is lowered and the gas transport to Russian and foreign consumers remains reliable.

Saved gas price Payment reduction Gas consumption System operation

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t (t

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usa

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In order to improve productivity, energy saving rate and works quality during the performance of gas trunklines overhaul it is necessary to apply new methods, as well as new technologies and new technological means, which include:

Repair of gas trunklines linear parts under pressure, including not stopping the gas transport;

Underwater lines repair with the use of special connectors, caissons and steel and composite coupling installment;

Underwater crossings surfaced and slack parts repair with the use of underwater constructions to temporarily change the river bed;

Automatic and semiautomatic welding of roundabout joints of large diameter pipelines in field and in a plant;

Automatic ultrasonic control of weld joints;

Use of pipeline laying optimized technology;

Concrete coated pipes applications;

Automatic external detectors applications. The main document that provides legislation for the overhaul works execution is Gazprom Standard 2-2.3-231-2008 Gas truknlines linear parts overhaul rules [20].

Local pipeline section repair is performed according to R Gazprom document 2-2.3-595-2011 Rules of method selection for defective pipeline sections of gas trunklines linear parts of Unified Gas Supply system repair [21].

Works planning is accomplished according to Gazprom Standard 2-3.5-302-2009 Planning of gas trunklines linear parts overhauls [22].

In order to re-insulate a pipeline the following should be done:

Defective pipeline section dismantling;

Primary pipeline scraping of old insulation coating;

Pipeline inspection and rejection;

Replacement of defective pipe segments with new pipes;

Final pipeline cleaning of old insulation coating;

Pipeline insulating;

Pipeline laying;

Pipeline burial;

High-pressure water, air or gas testing of the pipeline. Technical equipment and materials for the overhaul process, as well as overhaul technologies prior to their application must undergo the examination according to Gazprom Standard 2-3.5-046-2006 Examination framework for technical specification of equipment and materials, for technology certification and for organizations readiness assessment to perform the diagnostics and overhaul works of JSC “Gazprom” gas transport objects [23].

Pipeline laying process combined with the process of insulating is the most difficult and energy-consuming work among all gas trunklines overhaul steps.

The following main methods are used to perform the pipeline insulating and laying:

1) The separate method. Two columns of pipelayers are used to perform the work: the first one insulates the pipeline on the edge of the trench and the second one lays the pipeline into the trench. For a 1420 mm pipeline 4 and 5 90 tons bearing capacity pipelayers respectively are used.

2) The combined method. One column performs insulating and laying in a single process. For a 1420 mm pipeline 7 pipelayers are usually used.

3) The combined method with all the work performed inside the trench.

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In this case 4-5 90 tons capacity pipelayers are used for a 1420 mm pipeline.

At first sight, considering only the energy saving the third method should always be used. But sometimes it is not possible: for example for the case of a flooded trench. And sometimes the most energy-consuming method - the separated method is the only method that can be used, for example, if the insulation coating needs to be checked before the pipeline is layed into the trench. Hence, all of the methods should be considered and in all cases an energy-saving solution needs to be proposed.

Prior to the mathematical modeling it should be noted that all of the future energy saving solutions must also provide the reliability of the process.

3.1 Separate method of overhaul

Let us consider the separate method of overhaul of gas trunklines linear parts. The amount and the types of vehicles and equipment involved in the process of gas pipeline overhauls for the case of 1420 mm diameter pipeline are listed in the following table.

Table 2.1 – Technical equipment for the overhaul performed by separate method

Name Quantity

Pipelayer 9

Trolley pipe holder 7

Final cleaning machine 1

Heating unit 1

Priming machine 1

Insulating machine 1

Bitum melting boiler 1

Ladder 1

It should be noted that pipelayers are the most energy consuming elements in the process of gas pipelines overhaul. In the case of 1420 mm pipeline 9 pipelayers with 90 tones bearing capacity are used. Let us consider the possibilities for the minimization of their amount or for their replacement with less energy consuming pipelayers (with lower bearing capacity).

From the very name of the method it may be concluded that this method consists of two separate processes: pipeline insulating and pipeline laying. In this section only the second process is considered as from the modeling and calculating point of view the first one is absolutely similar to the combined method with all the works performed inside of the trench that is regarded in the third section of this chapter.

3.1.1 Process modeling

Let us consider the process of pipeline laying from the edge of the trench into the trench. Conventionally 5 90 tones bearing capacity pipeliayers are involved into the process. The scheme of this process is presented on the picture (figure 2.1):

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figure 2.1 – Pipeline laying scheme

Here:

– pipelayers’ lifting forces,

and – pipeline ground support reaction forces;

– distances between force application points;

– pipeline lifting heights;

– trench depth. It is necessary to calculate the scheme parameters so that the pipelayers lifting forces reach its minimum. Pipeline lifting and laying procedures can be described by the differential equation of the deflection curve:

(2.1)

Here:

– bending moment in any cross-section of the lifted pipeline;

– tensile modulus;

- moment of inertia.

Boundary conditions:

Where: = – length of the elevated section of pipeline, – depth of trench (can become zero in the event of symmetrical lift), y – displacement, i.e. in this context, lift height, y’ – angular deflection of pipeline cross-section area.

The process of pipeline laying can be fully modeled with the equation above and the problem of mathematical minimization can be stated. However, the parameters of the technological scheme, besides the minimum possible value of pipeline lifting forces, must also provide enough low value of pipeline stress in order to avoid its damage and to ensure the reliability of the process of pipeline laying. Such requirement may be taken into account with the help of additional constraints added to the system.

𝑅𝐵 𝐾

𝐾

𝐾𝑛

𝐾

𝑛

𝑙

𝑙0

𝑙

𝑙𝑛

𝑙𝑛

𝑡

𝑅𝐴

-24-

Forces acting on the pipeline are as follows: ground support reaction forces on the left / right ( and

), pipelayers’ lifting forces ( ), and pipeline weight. In order to determine the correlation between the loads (bending moments, lift force, pipeline ground support reaction force) and linear parameters (lift height in points of load application and distances between lift points) we can write the equation for bending moments acting on the pipeline as well as pipeline lift height and angular deflection equations (produced by integration). Now we can present expressions for bending moments through the left part of the pipeline lift scheme (q = pipeline specific weight):

under 0

0

under 0

∑

, under

Thus, bending moments in the pipeline lifting points are:

0 0 0

0 0

∑

, m = 3,4,…,n,n+1

Let us consider a specific case: m = n+1:

∑

However, it is known from boundary conditions that . Therefore, the equation can be written as follows:

∑

Now we can introduce the following definitions:

0

(2.2)

Common practices with insulating and laying operations during the overhauls of pipelines demonstrate that the bending radii in areas of pipeline lifting by pipelayers have an equal sign. The change in curvature sign occurs near pipeline ground suspension points. Then, since the bending moment dependence of

pipeline curvature follows

, ( – curvature), bending moment at end points has one sign, and at lift

areas – the other. If we assume that the lifted pipeline section between pipelayers experiences the change in curvature sign, in the first place it will be due to a marked increase in span length between pipelayers, which in turn will result in gain in weight of a liftable pipeline section, stronger forces in lift points and, respectively, the necessity to use more powerful and costly equipment. Further, increase in size of the actual scheme hinders synchronising of pipelayer’s movement during lifting and laying operations. In a diagram, the bending moment dependence on coordinate position in general looks like this (figure 2.2):

-25-

figure 2.2 – Bending moment dependence on the coordinate

Hence, the pipeline lifting points (expressions for the respective moments, , were written

above) and also the point between points 0, where function M(x) reaches a maximum, appear the most critical in terms of possible bent fractures. We can write the expression for the moment in this point

( ). In the beginning we can find this point as:

0 0

0

0

(2.3)

In order to limit the maximum absolute values of the relevant constraints must be added to the model.

It is known that EIβ(x) = EIy’(x) with regards to boundary conditions, where β(x) – bending curve

angular deflection in point x. By analogy with source [24] we can find β(x) for various values of x. Under

0 :

∫(

)

From the boundary conditions :

Under 0 :

∫( 0

)

0

From the above expression for under 0 we have:

l0 l1 l2 ln ln-1

M

x

-26-

0

0

0

0

0 0

0

Lemma: under , the expression for becomes:

∑

(2.4)

or .

Proof:

We prove it by induction. For m = 2 it has been proved above. We assume that the statement is true for m

(i. e. under ), we prove it for m + 1 (under ):

∫( ∑

)

∑

The induction hypothesis produces:

∑

∑

∑

∑

Hence, the expressions for angular deflection in pipeline lifting points:

0 0

0

∑

Let us consider separately the case when m = n + 1:

∑

But it is known from the boundary conditions that . In the same manner we can write the equation:

∑

(2.5)

Finally, from bending curve equation we have:

∬

By integrating similar to derivation of angular deflection equations we get equations of lift height. For:

-27-

0 0

0

For 0 :

0

0

0

0

0

0 0

0

Lemma: under the expression for is set by:

∑

Proof:

Under m = 2 it has been proved above. Further, once again by induction, let us assume that it is valid

under :

∑

Then under:

∑

∑

∑

Hence, the expressions for heights of lift in pipeline lifting points can be written as:

0 0

0

0

∑

(2.6)

-28-

Let us consider separately the case when m = n + 1:

∑

But it is known from boundary conditions that . . In the same manner we can write the equation:

∑

(2.7)

Next, we can put:

0

Further, as the pipeline is at equilibrium, the resultant of all forces acting on it is equal to zero, or:

∑

(2.8)

Where qln = the overall weight of the lifted pipeline section.

Heights of lift in end lift points (h1 и hn) are the preset values because the heights must be big enough in order to enable the proper work of the pipelayers. At the same time, these values must not be too large, as it will result in the maximum pipeline stress increase.

All of the equations that define the pipeline behavior are derived above. Besides these equations, as it has already been discussed, in order to provide the reliability of the pipeline laying process additional constraints that limit the maximum absolute value of the bending moment must be added into the model:

{

(2.9)

What is the exact value of that is reasonable to take will be considered later.

Any vector of parameters that satisfies all of the equations and inequalities above is feasible. Particularly very large distances can theoretically also be feasible for the model. From the scientific point of view there is no any contradiction in it, however, from the practical point of view it is extremely difficult to synchronize the work of a very long pipelayers group. Thus, it is necessary to add other additional constraints to the model:

(2.10)

What is the exact value of a that is reasonable to take will be considered later.

Hence we have the mathematical model (2.2 – 2.10). Within this model the pipelayers’ forces must be

minimized in dependence on other technological scheme parameters :

-29-

{

∑

0 0

∑

0

0

(

∑

)

∑

∑

∑

∑

0

Here:

– preset parameters;

- varying parameters;

0 – accessory parameters.

Restrictions on problem solution area: 3n inequalities, n+3 equations with 3n + 1 unknowns.

Note: In the model detailed above the distances can be defined either from end-point of pipeline lifting

or between the lifting points, i. e. ∑ 0 . The form of equations describing the model behaviour

should stay unchanged in this connection.

Note 2: Because of physical meaning, all variables describing the model are nonnegative.

This model is essentially a multi-purpose one as it is suitable to describe both lifting and laying operations during construction and repair of all pipeline types: gas, oil, water pipelines, etc.

Let us consider the equation describing the model. By analogy literature [24] we can make the following change to variables:

√

√

√

-30-

√

√

Further, we consider a set of equations describing the model with respect to new dimensionless variables,

. When presenting pipeline lifting scheme parameters in dimensionless form with a

dimensional component, we produce interdependences with reducing dimensional components. In that case equations and inequations, which describe the model take this form:

{

0

0

∑

(

)

0

0

(

∑

(

)

)

∑

∑ (

)

∑

( )

∑

( )

0

When passing to such dimensionless parameters we can produce the universal solutions for any diameter pipelines of any wall thickness, and produced of any material. In addition, in passing to dimensionless

parameters only ratio values,

and

, are used in the computations instead of values of elastic modulus

(E), moment of inertia (I), pipe specific weight (q), and height values, .

In order to construct a numerical algorithm that solves the problem, the number of lifting point n needs

to be chosen. As with the forces at each point may be too large and, from the other side, as larger the number of such points is, that harder it is to synchronize the work of the pipelayers, it is decided to

take .

So, let us write the problem of mathematical optimization under :

-31-

{

∑

0

0

∑

(

)

0

0

(

∑

(

)

)

∑

∑ (

)

∑

( )

∑

( )

0

We need to note here that the equation ∑

can be excluded from the system as it

has variable which is never used. Now we exclude this equation and rewrite the problem in standard

form:

-32-

{

∑

0

0

∑

(

)

0

0

(

∑

(

)

)

∑ (

)

∑

( )

∑

( )

We have a conditional minimisation problem where the opportunity set is defined by 12 inequalities and 7

equalities. Next, we convert the problem as follows: introduce nonnegative variables, , i = 1,..,12, with

the help of which we exclude the inequalities by rewriting each inequality, , in the form of equation,

. Further, we can convert the objective function using a logarithmic boundary approach subject to non-negativity constraint for all controllable variables. Finally, we produce the following optimisation equation set:

-33-

{

∑

∑

0

∑

0

0

(

( 0)

)

(

∑

( )

)

(

∑

( )

)

∑ (

)

∑

( )

∑

( )

0

0

( 0)

( 0) ( )

∑

( )

0

0 0

There are 24 control variables, and the opportunity set is defined by 19 equalities. Note here that the objective function is a convex function while constraints are nonlinear functions.

3.1.2 Preset parameters

The algorithm will be constructed for the case of a pipeline with 1420 mm external diameter and 16.5 mm wall thickness, which is to be layed into a 2.4-meters depth trench. Let us choose the maximum absolute bending moment value. With the dynamic loads taken into consideration the maximum pipeline stress value must be sufficiently lower than pipeline steel yield strength. The stress is defined through the bending moment as follows:

-34-

(2.11)

Here:

is the moment of resistance:

(2.12)

Here:

- pipeline external diameter;

- pipeline wall thickness. Taking into account the fact that the pipeline steel yield strength is usually not lower than 350 MPa, let us set the maximum acceptable stress at the level of 235 MPs. Hence, the maximum bending load for the “dimensional” case is calculated as follows:

In the “non-dimensional” case it corresponds to:

√ (2.13)

The pipeline tensile modulus is equal to:

(2.14)

And the moment of inertia may be calculated:

(2.15)

The first lifting height must provide smooth work of the pipelayers. From the common practices of these works it is known that the first height should not be lower than 0.4 meters above the ground level. As in

our cases the heights are measured from the level of the trench bottom is equal to 2.8 meters.

Pipeline specific weight is calculated as follows:

(2.16)

Here:

– steel density.

Thus, the maximum bending moment absolute value is calculated in the following way:

-35-

√

√

As a next step the maximum distance between two nearby pipelayers for the non-dimensional case needs to be calculated. From already mentioned common practices such distance should not exceed 30 meters. Thus, in the non-dimensional case:

√

√

The only thing left is to calculate the non-dimensional parameters and

. As it has already been stated

before, dimensional values of and must provide reliable and smooth pipelayers' operation. For this

purpose it is reasonable to make them 0.4 higher than the trench depth. Then, ,

and yield:

3.1.3 Initial solution

The initial solution in the stated problem is a vector of 24 components, from which an iterative process of numerical optimization begins. The relevant numerical algorithm will be constructed later. From the numerical methods of optimization point of view sometimes the initial solution must be feasible, i.e. it must fulfill the equalities and inequalities that define the model and for same cases it doesn’t have to be feasible. On our case the method (that will be described later) doesn’t require the feasibility of the initial

solution. However, it is crucial for all of the variables to be non-negative, including all . It means obligatory fulfillment of the inequalities in the initial model, i.e. fulfillment of the equalities (8’) – (19’) in the modified model.

On the basis on these concerns the following initial solution is chosen:

-36-

0

[

0

00

0

0

0

0

0

0

0

0

0

0 0 0 0 0 0 0 0 0 0 00 0 0]

[

]

Let us now assume that – residual vector of constraints when solving , where i-th coordinate

corresponds to residual in i-th equation within the system of optimisation problem. 0 becomes:

0

[

]

In the first eight components relatively large residuals are observed, but, as it has already been mentioned before, the most important is non-negativity of the variables from 8th to 19th.

-37-

3.1.4 Numerical method construction

In order to find the solution of this problem we will construct a method of numerical optimization that is based on Newton method for constrained minimisation as described in one source [25]. The method was described there for the following optimisation problem:

{

where is convex.

We substitute the original problem with the following one in order to define the step value:

{

with variable , i. e. we expand the objective function in Taylor series up to second-order at point. Step

represents a vector that should be added to to solve the optimisation problem in case that the objective function expansion in Taylor series is used instead of the objective function. Earlier it was

concluded that we need to solve the following system of equations in order to calculate step :

[

] [

] [

]

where is dual variable vector.

We cannot apply this approach directly to our problem, as its constraints are nonlinear equalities. We need

to linearize them. Assume that the current solution is . The linearised system will be:

Hence, instead of , the constraints to define step will be as follows:

and the resulting system to define the step value :

[

] [

] [

] (2.17)

We need the objective function gradient and Hessian in order to define the step value. They can be found through:

-38-

[

0

0

]

Hessian is a diagonal matrix of 24х24 order with the following items on the principal diagonal:

,

, 0, 0, 0 ,0,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

.

-39-

We still have the constraint gradient matrix to produce. Because of the large order of the matrix, its standard form is not presented in the body text of this graduation paper. The standard form of the constraint gradient matrix is given in Appendix 1.

We now have all necessary data to determine the step . We still have the value of logarithmic

boundary, , to determine. In accordance with one source [25], we need first to solve the optimisation

problem under . Here we can produce some optimal . After that we solve the equation under

, taking as the initial approximation. Having solved the equation under , we produce

and take it as the initial value to solve the problem under . And so on until we get the desired accuracy for the optimum value. The accuracy will be as follows:

Here:

– the highest value of ;

- number of inequalities within the original system.

3.1.5 Iterative process

To begin, we solve the problem under . The following system needs to be solved to get the step value:

[ 0 0

0 ] [

] [

0

0 ]

We get the following value for the step of the solution:

[

]

The solution has converged, and corresponds to the optimal solution, if [25]:

-40-

(2.18)

where

– is the norm of vector, , defined by the objective function’s

Hessian in the current solution point; – accuracy required for optimal solution.

Let us calculate the expression

in the point 0 considering previously obtained value of the step

:

, i.e. the adjusted original solution is not optimal; we need to take Newton

method step.

From literature we have , where should be adjusted by bisecting starting with 1, until the objective function at a new approximation becomes less than it is at the current approximation [25].

However, applying not the classical, but the “modified” Newton method of numerical optimization the algorithm must take into account not only the object function change: the change of the residual vector of constraints must also be considered.

The object function value and the residual vector of constraints norm in the initial point look the following:

0

‖ 0 ‖

Let us take and calculate 0 and corresponding object function value and the residual vector of constraints norm:

0

[

]

[

]

[

]

-41-

‖ ‖

As is can be observed from the values above, with the vector is positive and provides decrease of the object function as well as the residual vector of constraints norm. Hence, the first step of the method is made.

Considering the fact that the problem is non-linear and there is a relatively large amount of non-equalities

in the initial problem, the universal use of the above mentioned algorithm to find the value leads to negative variables among the components of the vector of solution, which is inacceptable from the very essence of the problem point of view.

In this paper the following method for -finding is proposed: starting from 1 -value will be decreased until the residual vector of constraints norm and the object function value for the next solution point are

not lower, than for the previous one; at the same time will not be made lower than a certain value

( defined for each internal cycle empirically). Then if the requirement for object function and

residual vector of constraints norm decrease is achieved and if under received all of the components of

the next point solution appear to be non-negative, than the step is made with this , otherwise – it is

decreased until the next solution point is non-negative disregarding .

Besides this, the following must be added to classical Newton method of optimization: the internal cycle

completion criterion is not only a certain value of

, but also a sufficiently low value of the residual

vector of constraints norm (‖ ‖).

The algorithm scheme is presented on the following flowchart (figure 2.3):

-42-

𝜆 𝑥𝑘

𝜀

��

𝑥𝑘 𝑥𝑘 �� 𝑥𝑛𝑡

𝑥0 𝑡

𝑡 𝑡

𝑡 𝑡𝑚𝑎𝑥

‖𝐺 𝑥𝑘 ‖ 𝜀

𝑥𝑘 𝑥𝑘

‖𝐺 𝑥𝑘 ‖ ‖𝐺 𝑥𝑘 ‖

��

�� 𝑡 𝑚𝑖𝑛 𝑥𝑖𝑘

𝑓 𝑥𝑘 𝑓 𝑥𝑘

𝑥 𝑥𝑘

figure 2.3 – Method flowchart

-43-

Hence, the first step of the method is completed. Let us continue the procedure until the necessary value

of

is received. The further process is illustrated in the next table.

Table 2.2 – Iteration process

Iteration

number k ‖ ‖

0

1 1.25755812587859

2 0.874812547506552 0.0860933759954594507 16.61489408389

94

3 0.377514835714825 0.0541835814985909686 16.28375927547

27

4 0.131265659908108 0.0394780791784377638 15.97075259506

12

5 0.00170228154209354 0.0202365207815823449 15.96595293713

23 1

6 0.0000335252529422287 0.0000652955729088800630 15.96690141602

05

As it can be observed from the table, the process is converged in 6 steps: vector is feasible (‖ ‖

) and optimal (

). The optimal vector for the case of the first logarithmic barrier

value looks as follows:

[ ]

-44-

Then let us apply the same method to the problem with barrier , taking as the initial solution. The solution path is again illustrated with a table (table 2.3):


Iteration number k

‖ ‖

0 0.233437183755816 0.0000652955729088800630 5.24637437787

105

1 0.191484301622780 0.000549229749093236696 5.21810437070

624

2 0.182262960290801 0.000572627244153597979 5.21217341577

094

3 0.173375161099434 0.000594342563807389321 5.20652871692

006

4 0.164823889679771 0.000614421804880991586 5.20115991242

443

5 0.156609962151581 0.000632913977976057589 5.19605654692

549

15 0.0920663354050498 0.000742521410564762419 5.15452588180

371

25 0.0534179951267075 0.000753658231121238291 5.13319997066

500

35 0.0312080878464407 0.000712065900275558319 5.12083462509

398

45 0.0000669545708328142 0.000634877463811888798 5.10915042296

673

55 0.00000940714879362183 0.000502491127958326752 5.10499594625

900

174 0.0000989363486892666857 5.10254251302

449

So, the optimal problem solution for the case of logarithmic is received. Let us take it as the initial solution for the next step of external cycle. As it has been done before, the path is illustrated with a table:


Iteration number k

‖ ‖

0 0.907047367667444 0.0000989363486892666857 3.91988617924

990

1 0.758128477067502 0.000744624901772509250 3.90183945974

799

2 0.697223058991001 0.000871923387260424608 3.89427918174

443

-45-

Iteration number k

‖ ‖

3 0.642435479373060 0.000986358659533918261 3.88732668154

827

4 0.593102655317617 0.00108928585283004481 3.88092090249

312

5 0.548622321196679 0.00118191590584660841 3.87500740610

858

15 0.160837253509211 0.00191574036994971190 3.81125617755

674

25 0.102227068590039 0.00195958180852912360 3.79876398390

258

35 0.0692264914542298 0.00193954902044947679 3.79055760088

550

45 0.0368694539507497 0.00189635156543304685 3.78066481229

209

55 0.0212565026397900 0.00177474176271020341 3.77528644695

921

155 0.000698785071053849 0.000673865024622414947 3.76578073978

082

345 0.00000930376986705176 0.0000997658800596067910 3.76542781391

617

Let us take the optimal solution of the previously solved problem for the case of . The algorithm of the internal cycle stays the same:


Iteration number k

‖ ‖

0 0.197021666483706 0.0000997658800596067910 3.59104234820

807

1 0.145695501585263 0.00108699538057780852 3.58712922006

312

2 0.112979795553050 0.00163656753319690873 3.58423207119

186

3 0.0905285203805620 0.00196772496481714384 3.58198634590

553

4 0.0743124871308499 0.00217785782572545378 3.58018760586

049

5 0.0621507879312205 0.00231548748938941677 3.57871167865

852

15 0.0196055496148430 0.00253844878309996075 3.57163479943

948

25 0.00845160529949861 0.00236419354230522731 3.56919377780

152

-46-

35 0.00272296778700467 0.00197989041639020681 3.56735673586

642

45 0.00115844250370885 0.00161095338613785415 3.56671542383

497

55 0.00115844250370885 0.00130673554541319253 3.56642764508

452

155 0.000324048682126321 0.0000890083855520210402 3.56566606507

509

397 0.00000208451821986833 0.0000988631754215766028 3.55957938614

147

Let us repeat the procedure for the case of :


Iteration number k

‖ ‖

0 0.0202229780396081 0.0000988631754215766028 3.53053371715

831

1 0.0124865794136163 0.0000594866757073910269 3.52973570899

958

2 0.00847026221643867 0.000181865211676468963 3.52924183907

194

3 0.00613659324370013 0.000144616928249382984 3.52890686003

223

4 0.00464009365105799 0.000178738645350907434 3.52866421845

311

5 0.00363026640567796 0.000159424145978677710 3.52848085218

186

15 0.000804063559156254 0.000147287259875356899 3.52777480753

338

25 0.000276407488823655 0.000120923019305931536 3.52760019648

511

35 0.000118325634305180 0.0000980460150106598458 3.52753011125

051

39 0.0000952859554010995 0.0000922862684664201867 3.52751914681

144

And, finally, for the case of :


Iteration number k

‖ ‖

0 0.0202229780396081 0.0000988631754215766028 3.53053371715

831

-47-

1 0.00508853255636798 0.0000983393735755306393 3.52392013602

844

2 0.00403487318427090 0.0000933138128916935798 3.52381913400

177

78 0.0000982054995831079002 3.52337600530

565

As a result the vector is obtained:

[

]

[

0

0

]

This vector is optimal with the following accuracy:

Let us analyze the obtained solution from the point of view of physics. The zero value of the variable

should be pointed out. That means that from the energy-saving point of view pipeline lifting must be

performed in three points. Proximity of to zero can also be highlighted. In the sense of

primary system’s values it means equality of the bending moments in all lifting points to each other.

Let us return to the dimensional values with the obtained non-dimensional solution. For a 1420 mm external diameter pipeline with 16.5 mm wall thickness the following values were calculated earlier:

-48-

Earlier it was also assumed that the trench depth is equal to 2.4 meters and lifting height at the first lifting point is equal to 2.8 meters above the level of the bottom of the trench. With these preset data the optimal parameters of the technological scheme for the dimensional case are calculated as follows:

√

0 0

√

√

√

√

√

√

√

√

√

According to the restrictions in the model the maximum pipeline stress that corresponds to the calculated scheme is not higher than 235 MPa.

The obtained solution provides a sufficient benefit in terms of necessary for the work performance amount of energy-consuming machines. In order to evaluate this benefit let us consider the number of pipelayers needed for the overhaul work on the linear part of gas trunklines. On the following chart the dependence of lifting forces of typical pipelayers on their arms [26] is presented (figures 2.4 – 2.6):

-49-

figure 2.4 – Pipelayer lifting power dependence on its arm (90 ton bearing capacity)


Lif

ting f

orc

e

kN

Arm

Arm

Lif

ting f

orc

e

Arm

Arm

kN

-50-


As it may be observed from the graphs above the lifting power highly depends on the pipelayer’s arm, which depends on the trench width. According to Gazprom standard 2-2.3-231-2008 [19] dependence of slope steepness on the type of ground looks as follows (table 2.8):

Table 2.8 – Slope steepnesses depending on trench depth and kind of ground

Types of ground Slope steepness (depending on the depth of the trench, not more than)

1.5 3 5

Made ground 1:0.67 1:1 1:1.25

Sand and gravel 1:0.5 1:1 1:1

Sandy loam 1:0.25 1:0.67 1:0.85

Clay loam 1:0 1:0.5 1:0.75

Clay 1:0 1:0.25 1:0.5

Loesslike 1:0 1:0.5 1:0.5

Arm

Arm

Lif

tin

g f

orc

e

kN

-51-

As the 2.4 meters trench depth was assumed for our case only the second column of the table must be regarded. For example, for clay ground the pipelayer’s arm is approximately 2.5 meters; that means that applying optimal from the energy-saving point of view scheme only 3 pipelayers are enough for the works to be performed: two of them with 90 tones bearing capacity and one with 40 tones bearing capacity. For cases of grounds with lower density, in order to provide reliability to the process and to avoid pipelayers’ fall into the trench one reserve pipelayer must be added to the third and to the first lifting points. However, even in that case sufficient energy-saving may be achieved because additional pipelayers with lower (compared to the classical scheme) maximum lifting power may be used (table 2.1).

The calculated scheme in details is presented in the Appendix 2.

3.2 Combined method of overhaul

Let us consider the combined method of gas trunkline linear part overhaul. The amount and the types of machines and equipment used in that case are usually similar to the previous one (table 2.1). The only difference lies in the amount of pipelayers involved in the overhaul. For the considered case there are usually 7 pipelayers. And the pipelayers remain the most energy-consuming element among other machines used. Let us consider the problem of pipelayers’ amount minimization and/or applying less powerful types of them.


In terms of mathematical modeling the problem is similar to the one previously considered. With the help of the procedure described above a problem of mathematical optimization is set and transformed in a certain way.

-52-

{

∑

∑

0

∑

0

0

(

( 0)

)

(

∑

( )

)

(

∑

( )

)

∑ (

)

∑

( )

∑

( )

0

0

( 0)

( 0) ( )

∑

( )

0

0 0


As in the previous case the solution will be found for the case of 1420 mm external diameter pipeline with 16.5 mm wall thickness, which is to be layed into 2.4 meters depth trench. Hence, all pipeline parameters remain the same (2.14, 2.15, 2.16):

-53-

Due to overhaul’s method change there are other dimensional (and non-dimensional) values of and

( and

). At this time the solution has not only to provide smooth and reliable operation of the pipelayers. Unlike the first method swabbing and cleaning machines operate in the first and in the forth point respectively. That’s why the heights should be sufficient to enable unhindered machinery operation as well as the accessibility for servicing (for example, for swab replacement). Pipe swabbing machine is located in the lead of the convoy for visual surveillance and instrumental examination of the pipeline for defects prior to applying insulation. Coating machine is located within the last pipelayer zone in order to prevent damage to insulating coating by trolley pipe holders (suspension devices).

Based on these concerns it is reasonable to take the values and equal to each other and 0.7 higher

than the trench depth. Then, knowing that , and we have:

The maximum bending moment value and maximum acceptable distance between the pipelayers remain the same in comparison to the previous case:

However, due to the change of the value , the relevant non-dimensional values should be re-calculated:

√

√

√

√


The problem differs from the previous one only in terms of the preset parameters; hence it makes sense to use the same method that was used before. Thus, the choice of the initial solution is based on the same

concerns: main requirement is non-negative and their proximity to zero. It is reasonable to assign the initial solution to the following vector:

-54-

0

[

0

00

0

0

0

0

0

0

0

0

0

0 0 0 0 0 0 0 0 0 0 00 0 0]

[

]

As it has been before, let be the residual vector of constraints norm in the point x, where i-th

coordinate corresponds to residual in i-th equation within the system of optimisation problem. 0 becomes:

0

[

]

In the first seven components relatively large residuals are observed, but, as it has already been mentioned before, the most important is non-negativity of the variables from 8th to 19th.

Previously constructed numerical method will be used to solve the problem.

-55-

3.2.4 Iteration process

As the whole algorithm was described in the previous chapter, now only the path of the solution is presented.

So, under (table 2.9):


Iteration number k

‖ ‖

0

1 2.63193280656348

2 1.45976517102185 0.111535770588206948 19.49621105227

62

3 1.05301448320891 0.0883353604610064291 18.87184278450

19

4 0.647347568103385 0.0660549812115270307 18.05936608108

36

5 0.304368717664325 0.0425999649102095646 17.30757522948

45 1

6 0.116073924513472 0.00636853332350989683 16.95180498487

49

7 0.0471670095442469 0.00518362820186726424 16.86126684209

38 1

8 0.000905593656912840 0.00392541843106027945 16.80961298466

94

Under :


Iteration number k

‖ ‖

0 0.00392541843106027945

1 0.162610361037930 0.000444624887254109447 5.429802784236

72

2 0.132611084016454 0.000814325682637962183 5.410136082414

39

3 0.126108788135541 0.000822122396882211883 5.406034911075

03

4 0.119874695945976 0.000828780028612427630 5.402135520164

69

5 0.113905829619986 0.000834349271734592147 5.398375002873

89

15 0.0675092906919452 0.000841371074430799786 5.370304758566

62

-56-

Iteration number k

‖ ‖

25 0.0398574709208196 0.000791309290027137420 5.353717166982

02

35 0.0237747828862972 0.000716523826587452182 5.343927211055

82

45 0.0143919915037685 0.000635008780910378844 5.338084693776

12

55 0.00884333826743524 0.000555659289303819196 5.334551962120

15

155 0.000112153944083447 0.000121493251198405999 5.328908098233

87

168 0.0000685978989276207 0.0000990985434923947307 5.328891817601

89

Under :


Iteration number k

‖ ‖

0 0.0000990985434923947307

1 0.718501085202477 0.000305424296191871914 4.08283282933

801

2 0.561566876369847 0.000604668794443130250 4.06179234224

689

3 0.499867940338580 0.000709766755742552002 4.04528052670

023

4 0.446917549912473 0.000792711156234169028 4.04273817463

827

5 0.401269813444435 0.000857780412315491039 4.03832949776

547

15 0.164573523224817 0.00102717568564567475 3.99618533621

346

25 0.0847282641179876 0.000939265700904121635 3.97716056125

355

35 0.0499520177667782 0.000833576048049094044 3.96679581801

824

45 0.0322183739670753 0.000744958846198365168 3.96045953120

373

55 0.0221608855489775 0.000673766139345116622 3.95627076476

704

-57-

Iteration number k

‖ ‖

155 0.00252679920206464 0.000300794130989010281 3.94438961312

908

455 0.0000991730383604558 0.0000293290446953209678 3.94148804063

500

Under :


Iteration number k

‖ ‖

0 0.0000293290446953209678

1 0.168439178313696 0.000526322960725646356 3.75407715903

896

2 0.142941891654406 0.000838925019332507405 3.75145644285

965

3 0.133085109890066 0.000880615661562282885 3.75034216176

892

4 0.124347970419009 0.000910703567068538622 3.74930473199

094

5 0.116542988678104 0.000931402533572088108 3.74833542222

273

15 0.0677400501396594 0.000898504929639602225 3.74121864323

503

25 0.0433769609421876 0.000782537162690038340 3.73688485586

407

35 0.0292673865322432 0.000701121304585072087 3.73404527792

605

45 0.0206000296420117 0.000648388105485505815 3.73209441649

135

55 0.0150350245694305 0.000611024327174249549 3.73069969186

609

155 0.00185724762288559 0.000403258514677051012 3.72625107497

718

512 0.0000372830928890085 0.0000996674803408824594 3.72515666296

632

Under :

-58-


Iteration number k

‖ ‖

0 0.0000996674803408824594

1 0.0164834686719429 0.0000981561363394718305 3.69782921037

331

2 0.0125494763710338 0.0000994220836569720132 3.69727273273

833

3 0.00933014101375491 0.0000962008765730420243 3.69688398114

550

4 0.00819190315313158 0.0000974977769730383083 3.69673883171

946

5 0.00725834605527624 0.0000965599431465731913 3.69661139941

940

15 0.00283784775936026 0.0000898383869608921785 3.69586725409

055

25 0.00144697512228944 0.0000831183898705082318 3.69554009889

248

35 0.000846257876289527 0.0000768686298612774569 3.69536300416

745

45 0.000538435850546247 0.0000710789929465373835 3.69525562457

068

55 0.000362679575187229 0.0000657216738219719909 3.69518566263

849

98 0.0000978688509565343 0.0000472813631265727260 3.69505728334

737

Under :


Iteration number k

‖ ‖

0 0.0000472813631265727260

1 0.00454833998315154 0.0000468087215337222045 3.69163012052

562

2 0.0101033038583491 0.0000535600444111234694 3.69159467159

860

3 0.291024327162408 0.000442016255616443259 3.69151592533

417

4 0.272775147057900 0.00161819697776090644 3.69093436161

381

-59-

Iteration number k

‖ ‖

5 0.267492463241250 0.00274605172124211747 3.69038984826

920

15 0.150735653392002 0.0112733005267551759 3.68621072674

930

25 0.0682012403437864 0.0159496273273172749 3.68389940773

572

35 0.0322044316598228 0.0177832391199201842 3.68297809834

425

45 0.0151676184681273 0.0185177610240296400 3.68259406573

454

55 0.00732251231480171 0.0187446551960411857 3.68245987276

354

155 0.000307498908912260 0.0172700486054822022 3.68304424028

478

7607 0.00000999423428728148973 3.69110254044

905

Hence, the method is converged; optimal solution is obtained. As in the previous case, under the last logarithmic barrier value a higher number of iterations was made compared to other logarithmic barrier values in order to achieve maximum preciseness of the resulting solution.

Vector of optimal scheme parameters in the non-dimensional case looks as follows:

-60-

[

]

[

0

0

]

This vector is optimal with the following proximity:

The obtained vector, of course, is not the same as in the previous problem. However, in some sense these two solutions are similar to each other: in both of the cases the forth lifting point is «unloaded», i.e., as it has been before, the optimal in terms of energy-saving pipeline lifting has to be performed in three points.

Let us analyze the solution for the dimensional case:

√

0 0

√

√

√

√

-61-

√

√

√

√

√

The obtained values of pipelines’ lifting powers are slightly higher than in the case of overhaul separate method. It can be easily explained: under the combined method the pipeline must be lifted higher. But a sufficient energy-saving benefit may be achieved all the same in this case.

Detailed illustration of the optimal from the energy-saving point of view scheme of this overhaul method is presented in the Appendix 2.

3.3 Combined method with all works performed inside trench

The scheme of the process is presented on the figure 2.7:

figure 2.7 – Technological scheme of the overhaul with all works performed inside trench


As pipeline position is obviously symmetrical, it’s modeling, compared to other cases, should be done in a different way. As it has been done before the modeling is based on the differential equation of the bending curve. However, on the symmetry basis it is enough to consider only the half of the lifted pipeline. The boundary condition is equality of the angular deflection of pipeline cross-section area in the mid-point to zero:

(2.19)

Bending moment expressions for the first and for the middle point of the elevated pipeline are generally the same as in the previous cases:

0

0

𝐾

𝐾

𝐾 𝑅0 𝑅0

𝑥

𝑦

𝑙 𝑙 𝑙 𝑙

-62-

According to the bending curve differential equation these expressions are integrated:

0

0

However, from the boundary condition it is known that . Hence, the second expression yields:

0

(2.20)

From the first expression the expression for the lifting height in the first point is obtained:

0

(2.21)

Further, as the pipeline is at equilibrium, the resultant of all forces acting on it is equal to zero, or:

0 (2.22)

According to the symmetry the received equations are enough to describe pipeline position in the considered case. The model looks as follows (2.20 – 2.22):

{

0

0

0

To simplify the calculation’s procedure and to make solution more universal, by analogy with the previous chapters the system is conversed to non-dimensional values:

{

0

( )

0

0

( )

( )

Besides this, the maximum bending moment absolute value restrictions must be added to the model. As a result, the final mathematical model for this case looks as follows:

{

0

( )

0

0

( )

( )

0

0

0 ( )

-63-

Within this model the problem of pipelayers’ lifting forces minimization is set. Let us write the problem in a standard way:

{

0

0

0 ( )

0

0

0

( )

( )

0

( )

Now, let us transform the problem according to the logarithmic barrier method:

{

0

0

0

0 ( )

( )

0

0

( )

( )

(2.23)


As it has been made before the solution will be constructed for the case of 1420 mm external diameter pipeline with 16.5 mm wall thickness that is to be layed into 2.4 meters depth trench. Such a pipeline characteristics have been already calculated (2.14, 2.15, 2.16):

As in the case of combined method with the works executed on the edge of the trench the height must provide smooth operation of swabbing and coating machine. Usually it is taken at the following level:

Again, the maximum acceptable pipeline stress is assumed to be 235 MPa. Than the maximum bending moment for the dimensional case is equal to:

In non-dimensional values it corresponds to:

-64-

√

√


The initial solution in the stated problem is a vector of 8 components, from which an iterative process of numerical optimization begins (the numerical algorithm has been constructed earlier). The initial solution

can be not feasible. But it is crucial for it to have all, including non-negative components.

Based on these concerns the following vector is taken as the initial solution:

0

[ 0

0

0

0

0

0 0 0 0 ]

[

]

Let be the residual vector of constraints when solving , where i-th coordinate corresponds to

residual in i-th equation within the system of optimisation problem. 0 becomes:

0

[

]

3.3.4 Numerical algorithm construction

The problem will be solved with the same method that was used before. The only difference lies in the dimension of the equation system to obtain the step value of the algorithm.

The object function gradient and hessian respectively look as follows:

-65-

[

0

]

Hessian is a diagonal matrix of 8х8 order with the following items on the principal diagonal:

,

,

, ,

,

,

,

,

.

3.3.5 Iteration process

The numerical method path looks as follows (tables 2.15 – 2.19):

Under :


Iteration number k

‖ ‖

0 1.76825793116024554

1 1.49689575437186 1.32874621642303192 2.63406389153

693

2 0.987022153118459 0.550170551272630703 3.64883896744

013

3 1.01618143635999 0.107595807564438067 4.33374176069

398

4 0.546714750032141 0.0340246116094954365 3.46292561264

230

5 0.309344373778541 0.0226360820992658439 3.22513329747

593

6 0.218181131076011 0.0203622152660151955 3.15723562374

300

7 0.146764840745074 0.0186665425724495855 3.11068892147

682

8 0.0948674148450070 0.0172436120623901366 3.08070422365

816

-66-

Iteration number k

‖ ‖

9 0.0105093636539118 0.0169803851574453421 3.04723252382

956 1

10 0.00843165009443124 0.0137392037730173274 3.05329397369

234 1

11 0.0159343494608055 0.0105280610808404351 3.05135744882

829

12 0.0000712776663436064 0.00931924561380164350 3.04568434451

324 1

13 0.000212435942050038 0.0000858112451959491113 3.05614071225

104

Under :


Iteration number k

‖ ‖

0 0.0000858112451959491113

1 11.8519269940195 0.00121078793599176614 4.91714194283

167

2 9.52557156282447 0.00499360190036487583 4.79893438838

251

3 7.77152171930664 0.00768450558694030571 4.70394431154

486

4 6.42831884558860 0.00963672432444012683 4.62646144008

327

5 5.38371757535443 0.0110788525614468978 4.56238599795

132

15 2.31775035907204 0.0102635546000870003 4.26025130094

836

25 1.44283419392098 0.0153700054982753142 4.16313627217

536

35 0.308009639065886 0.0152405794150547065 4.11937125663

332

45 0.177157002693660 0.0146386973100074112 4.09621094087

072

55 0.109792780675670 0.0139946467733927538 4.08279210489

751

155 0.00434882745403299 0.00854277109583189212 4.06092319127

409

-67-

Iteration number k

‖ ‖

581 0.000999891306671146091 4.06856831546

383

Under :


Iteration number k

‖ ‖

0 0.000999891306671146091

1 0.840266834740949 0.00121951078840909890 3.98581327044

341

2 0.673041581440838 0.00137912537456743472 3.97744023668

159

3 0.615987832136524 0.00149526783478361744 3.97073629549

436

4 0.457965223942436 0.00158164348592784070 3.96525549525

140

5 0.386379557188984 0.00164700535958588884 3.96069822760

192

15 0.122720830526601 0.00184109063847870077 3.93971027955

824

25 0.0515116250029967 0.00183937265983505978 3.93147184371

967

35 0.0263355898603806 0.00178123144701828836 3.92776770539

589

45 0.0150562807472507 0.00170855466892999228 3.92583119281

617

55 0.00924953381438453 0.00163242209100227117 3.92473196938

803

65 0.00597340617884814 0.00155670026182094562 3.92407864046

446

154 0.000327575828934667 0.000999987727711889587 3.92323961545

707

Under :


Iteration number k

‖ ‖

0 0.000999987727711889587

-68-

Iteration number k

‖ ‖

1 0.0432303442585118 0.00154190695931395684 3.89164001056

444

2 0.0238821046020961 0.00167312069661832005 3.89026284507

181

3 0.0146890677380645 0.00171672313192004528 3.88955232784

351

4 0.00986451329100003 0.00172780573536415760 3.88912695351

425

5 0.00855854658300433 0.00172375083113352270 3.88885101832

708

15 0.000834667664842733 0.00153907645095595130 3.88823540780

474

25 0.000224011203929135 0.00131590808198352584 3.88833056672

652

35 0.0000802226352049601 0.00112159785725076777 3.88849406012

446

43 0.0000431940016908913 0.000986617923580252604 3.88860798449

411

Under :


Iteration number k

‖ ‖

0 0.000986617923580252604

1 0.00349663199476340 0.00107488778858970012 3.88386168171

229

2 0.00122956203625823 0.00108866405220543164 3.88377020900

027

3 0.000147882005009593 0.00124016284762714507 3.88361491572

609

4 0.0000666214631664444 0.00116784332647250529 3.88366343178

370

5 0.0000361438773025627 0.00109539175628367656 3.88371774620

568

8 0.0000140007600829200 0.000961472148040637799 3.88382416684

354

22 0.0000965649620821320252 3.88455784520

117

-69-

As the solutions for the previous cases showed, further increase of the logarithmic barrier is not reasonable as the point of solution changes insignificantly.

So, the optimal solution is obtained. It looks as follows:

0

[ 0

]

[

]

Let us analyze the obtained solution. From the mathematical point of view it is absolutely correct. However, the distance from the middle point to the last point of the elevated pipeline appears to be

almost zero ( ). From the reliability point of view it is extremely undesirable to perform the pipeline lifting in one point. Hence, let us re-state the initial problem of mathematical optimization:

{

0

0

0

0 ( )

( )

0

0

( )

( )

(2.24)

The problem (2.24) differs from the problem (2.23) in the following: the pipeline lifting forces are “equally

important” from the mathematical optimization point of view (in the object function

instead of

). Applying the algorithm to this “modified” problem yields the following optimal solution:

0

[ 0

]

[

]

In this case the lifting force in the middle point is almost zero, but this is feasible from the point of view

of technology. Besides this, this solution has another advantage: all are significantly greater than zero.

It means that with the overhaul works performed according to this scheme the pipeline stress is much lower than even acceptable 235 MPa.

Let us calculate the dimensional parameters:

0 0

√

-70-

√

√

√

√

In addition, in order to understand the pipeline position better, let us calculate the height in the middle point:

0

The optimal scheme that was calculated in this chapter is presented in the Appendix 2 in details.

3.4 Conclusion

The technological schemes of pipeline coating and laying works execution were calculated in this chapter with the use of methods of mathematical optimization. The obtained solutions provide great energy saving for different methods of pipeline overhaul.

For the first time a numerical method of mathematical optimization was used to solve these types of problems. The proposed approach is universal and provides a framework for calculating optimal parameters of technological schemes for diverse cases of pipelines with different diameters and wall thicknesses.

4 Conclusion

The main problem of this paper was to analyze possible ways for energy consumption reduction during the execution of gas trunklines linear parts overhaul. This problem was regarded from two points of view: from the point of view of energy source safe before the start of the overhaul itself and from the point of view of reduction of energy-consuming machines number during the very process of the overhaul.

The main results of the paper are the following:

1. Thermodynamic analysis of gas parameters dynamics in the process of gas evacuation from a string that is off operation for the overhaul to a parallel or adjacent string with the use of mobile compressor systems.

2. Economical benefit evaluation for the case of mobile compressor system application while the string is being prepared for the overhaul.

3. Models of pipeline positions that correspond to diverse methods of the overhaul. 4. Overhaul technological scheme parameters optimization subject to maximum energy saving and

with the use of numerical methods of mathematical optimization. 5. Practically feasible technological schemes of gas trunkline linear part overhaul that suppose the

least amount of energy-consuming equipment used.

Bibliography

1. ISO 50001:2011

2. Federal Law of the Russian Federation of November 23, 2009 N 261-FZ "On energy saving and energy efficiency improvements and on Amendments to Certain Legislative Acts of the Russian

-71-

Federation" which has the main purpose to "create the legal, economic and institutional framework to stimulate energy conservation and energy efficiency"

3. Energy Strategy of Russia till the year 2020 4. http://www.gazprom.ru/about/ 5. Gazprom Standard 4-2005 Regulation on the procedure of JSC "Gazprom" monitor for the effective

use of gas 6. Gazprom Standard 2-1.20-601-2011 Methodology of calculating the effect of saving energy resources

spent on their own technological needs of gas trunkline transport 7. Gazprom Standard 2-1.20-114-2007 Methodology of gas transport objects energy audit 8. R Gazprom 2-1.20-673-2012 Energy saving regulation system of JSC “Gazprom” 9. A. Kalinin. Calculation, management and optimization of operation of gas compressor units.

Moscow. MPA-PRESS. 2011. 10. A. Karasevich, E. Kreinin. Prospects and reserves of energy conservation in Russia. 2010. 11. A. Ishkov, G. Khvorov, M. Yumashev, E. Yurov, L. Eshich. Current state and future development

of energy-saving trends in gas transportation. 2010.

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technological objects 14. G. Khvorov, M. Yumashev. Methodology for estimating the energy saving potential of technological

facilities of JSC "Gazprom" 15. Order of the Federal Tariff Service of Russia from 27.11.2012 N 273-e / 1

16. RF Government Resolution № 344 of 12.06.2003 17. Lebersdorfer Maschinenfabrik GmbH & Co.KG presentation "How new technologies and more

efficient pipeline operational management can improve gas transport efficiency" 18. Gazprom standard 3.3-2-024-2011 Valuation methodology of natural gas spent on own technology

needs and on gas trunkline transport losses 19. B. Porshakov, A. Kalinin, S. Kuptscov, A. Lopatin, K. Shotidi. Energy-saving technology in the gas

trunklines transport system. Moscow. MPA-PRESS. 2006. 20. Gazprom Standard 2-2.3-231-2008 Gas truknlines linear parts overhaul rules 21. R Gazprom document 2-2.3-595-2011 Rules of method selection for defective pipeline sections of

gas trunklines linear parts of Unified Gas Supply system 22. Gazprom Standard 2-3.5-302-2009 Planning of gas trunklines linear parts overhauls 23. Gazprom Standard 2-3.5-046-2006 Examination framework for technical specification of equipment

and materials, for technology certification and for organizations readiness assessment to perform the diagnostics and overhaul works of JSC “Gazprom” gas transport objects

24. V. Berezin, L. Telegin, E. Anikin. Methodological guidelines for the calculation of the strength of the pipeline during construction. Moscow. 1974.

25. V. Feodosiev. Strength of materials. Moscow. Publisher Bauman Moscow State Technical University. 2009.

26. S. Boyd, L. Vandenberghe «Convex Optimization». Cambridge. Cambridge University Press. 2004. 27. http://www.komatsu.ru/group.xgi?&category_id=131

Master of Science Thesis

KTH School of Industrial Engineering and Management

Energy Technology EGI-2014-071MSC EKV1044

Division of Heat and Power Technology

SE-100 44 STOCKHOLM

Appendix 1

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-73-

Appendix 2

Separate method schemes

22-23 m 69-70 m 69-70 m

0.7 m 0.7-0.8 m

Direction of movement

Cleaning machine

Pipelayer 90 t

Pipelayer 90 t

Coating machine

Direction of movement Pipelayer 90 t

Pipelayer 90 t

Pipelayer 40 t

0.4 m 0.6 m

0.7 m

8-9 m 9 m

-74-

Combined method scheme

Combined method scheme (with all work executed inside trench)

Direction of movement Pipelayer 90 t

Pipelayer 40 t

Cleaning machine Coating machine

22-23 m 69-70 m 69-70 m

0.7 m 0.7-0.8 m

Direction of movement

Cleaning machine

Pipelayer 90 t Pipelayer 90 t

Coating machine

0.7 m 1 m 1.1 m

12 m 13 m

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