Comparison of Isothermal and Non-Isothermal Pipeline Gas Flow Models

Home , Isothermal flow

NOTICE: This is the author's version of a work that was accepted for publication in Chemi- cal Engineering Journal. Changes resulting from the publishing process, such as peer re- view, editing, corrections, structural formatting, and other quality control mechanisms, may not be reflected in this document. Changes may have been made to this work since it was submitted for publication. A definitive version was subsequently publication in Chemical Engineering Journal, Volume 81, Issue 1-3, Pages 41-51 (January 2001), DOI: 10.1016/S1385-8947(00)00194-7

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 Comparison of isothermal and non-isothermal pipeline gas flow models

Andrzej J. Osiadacz, Maciej Chaczykowski, Warsaw University of Technology, Institute of Heat- ing and Ventilation, Nowowiejska 20, 00-653 Warsaw, Poland tel (48-22) 660-53-11, fax (48-22) 825-29-92, E-mail: [email protected], [email protected]

Summary The transient flow of gas in pipes can be adequately described by a one-dimensional approach. Basic equations describing the transient flow of gas in pipes are derived from an equation of motion (or momentum), an equation of continuity, equation of energy and state equation. In much of the literature, either an isothermal or an adiabatic approach is adopted. For the case of slow transients caused by fluctuations in demand, it is assumed that the gas in the pipe has sufficient time to reach thermal equilibrium with its constant - temperature surroundings. Similarly, when rapid transients were under consideration, it was assumed that the pressure changes occurred instanta- neously, allowing no time for heat transfer to take place between the gas in the pipe and the surroundings. For many dynamic gas applications this assumption that a process has a constant temperature or is adiabatic is not valid. In this case, the temperature of the gas is a function of distance and is calculated using a mathematical model, which includes the energy equation. In the paper comparison of different (isothermal and non-isothermal) models is presented. Practical ex- amples have been used to emphasize differences between models.

Keywords: gas flow, mathematical modelling, simulation, gas networks.

Notation:

A - cross-section area of the pipe, m 2 cp - specific heat at constant pressure, J/(kgK), cV - specific heat at constant volume, J/(kgK), D - pipe diameter , mm, f - Fanning friction coefficient, -, g - the net body force per unit mass (the acceleration of gravity) m/s 2 k - pipe roughness, mm kL - heat transfer coefficient, W/(mK), L - pipeline length, m, M -mass flow, kg/s, p(x) - pressure at x, Pa q - the heat addition per unit mass per unit time, W/kg 3 Qn - flow (under standard conditions) m /h, R - specific gas constant, J/(kgK), t - time, s T - gas temperature, K Tsoil - soil temperature, K, w - flow velocity, m/s x- spatial coordinate, m, Z - compressibility factor, -.

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Greek symbols α - the angle between the horizon and the direction x, λ - thermal conductivity coefficient of gas, W/(mK), µ - viscosity of natural gas, N s/m 2, ρ - density of gas, kg/m 3

Note: Flow rate Qn is shown in the standard conditions of 273.15 K, 0.1 MPa

1. Introduction It is a well established fact that flow in gas pipelines is unsteady. Conditions are always changing with time, no matter how small some of the changes may be. When modelling systems, however, it is sometimes convenient to make the simplifying assumption that flow is steady. Under many conditions, this assumption produces adequate engineering results. On the other hand, there are many situations where an assumption of steady flow and its attendant ramifications produce un- acceptable results. Dynamic models are just a particular class of a differential equation model in which time derivatives are present. During transport of gas in pipelines the gas stream loses a part of its initial energy due to friction- al resistance which results in a loss of pressure. This is compensated for by compressors installed in compressor stations. Compression of the gas has the undesired side effect of heating the gas. The gas may have to be cooled to prevent damage to the main transmission pipeline. If the cooler is installed heat from the gas is passed to the air in a force draught heat exchanger in which one or more fans operate, depending on the number of compressors in service. Cooling of the gas is desirable because it improves the efficiency of the overall compression process. As always it is a matter of balancing capital and maintenance costs against operating costs.

2. Basic equations The transient flow of gas in pipes can be adequately described by a one dimensional approach. Basic equations describing the transient flow of gas in pipes are derived from an equation of motion (or momentum), an equation of continuity, equation of energy and state equation [7]. In prac- tice the form of the mathematical models varies with the assumptions made as regards the conditions of operation of the networks. The simplified models are obtained by neglecting some terms in the basic model as a result of a quantitative estimation of the particular elements of the equation for some given conditions of operation of the network [3], [8]. 2.1 Conservation of mass: continuity equation Generally, the continuity equation is expressed in the form: ∂b ρwg ∂ρ − = (1) ∂x ∂t where: w - flow velocity, ρ - density of gas Substituting M = ρ w A , we have: 1 ∂M ∂ρ − = (2) A ∂x ∂t

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 where: A - cross-section area of the pipe, M - mass flow 2.2 Newton's second law of motion: momentum equation According to [2] the basic form of momentum equation can be expressed in the form

2 ∂p 2 fρ w2 ∂b ρwg ∂c ρw h − − −gρsin α = + (3) ∂x D ∂t ∂x where: f - Fanning friction coefficient, g - the net body force per unit mass (the acceleration of gravity) α - the angle between the horizon and the direction x. ∂b ρwg 2 fρ w2 ∂ The constituent factors , ,ρg sin α andcρw2 h define the gas inertia, hydraulic ∂t D ∂x friction force, force of gravity and the flowing gas dynamic pressure respectively. 2.3 State equation An equation of state for a gas relates the variables p, ρ, and T. The type of equation which is commonly used in the natural gas industry is [6], [11]: p = ZRT (4) ρ

where the deviation from the ideal gas law is absorbed in the compressibility factor Z, which is a function of p and T. 2.4 Conservation of energy The basic form of energy equation, according to [10] is the following ∂ 2 ∂ 2 ρ =L ρ F +w + IO +L ρ F +p + w + IO q Adx MbAdx gG cv T gz JP M bwAdx gG cv T gz JP (5) ∂t N H 2KQ ∂x N H ρ 2 KQ where: q - the heat addition per unit mass per unit time, T - gas temperature, cV - specific heat at constant volume. Before going over to analysis of transient conditions, the simple case of steady state conditions will be considered.

3. Steady state non-isothermal model The temperature of gas, as a function of length of the pipe is calculated using the heat balance equation, assuming that heat transfer process is quasi steady state, expressed by the following equation = − − cMdTp k L( T T soil ) dx (6)

where: cp - specific heat at constant pressure, J/(kgK), M -mass flow, kg/s, kL - heat transfer coefficient, W/(mK), T - gas temperature, K, Tsoil - soil temperature, K. The form of equation (6) results from the following transformations of the equation (11). Since under steady state conditions

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 ∂ 2 L ρ F +w + IO = MbAdx gG cv T gz JP 0 ∂t N H 2 KQ we can write the energy equation describing the flow of gas throw the horizontal pipe in the form ∂ 2 ρ =L ρ F +p + w IO q Adx MbwAdx gG cv T JP ∂x N H ρ 2 KQ The quantitative analysis of energy equation (paragraph 4.1) has shown that under steady state conditions term (III) (spatial derivative of the kinetic energy) can be neglected in comparison with other convective terms. Substituting the enthalpy and neglecting term (V) (heat conduction through the gas along the pipeline) we finally obtain the energy equation in the form of the heat balance equation. Equation (6) can be written in the form: dT k = − L dx − (T Tsoil ) cp M

By integrating between T(0) (T x=0 ) and T(x), x ∈(0,L] we get: T() x dT k x = − L dx z − z T()0()TTsoil cp M 0 Finally: = + − −βx T()() x Tsoilb T0 T soil g e (7)

where: β=k L/(c pM)

For non-isothermal steady state flow, pressure at specific point of the pipe can be expressed by the following equation [1]: px() =b p(0) g2 − KM 2 (8)

where: p(0) - pressure at x=0, Pa, M - mass flow, kg/s, K - coefficient defined by the equation: L T(0)− T T(0)− T O =ZR 4f F + soil− soil −βx I − − K 2 M GTsoil x eJ 2b T(0) T(x) gP (9) A N D H β β K Q where: x - spatial coordinate, m, f - Fanning friction coefficient, Z - compressibility factor, R - specific gas constant, J/(kgK), A - cross section area of the pipe, m 2, Specific heat at constant pressure can vary strongly with temperature of the gas. In the gas transportation systems the temperature ranges are modest and the value of specific heats may be assumed constant [4]. Such assumption, however, cannot be applied with regard to the compressibility factor. Its value varies significantly with temperature and pressure of the gas and separate calculations are carried out for every discretization section of the pipeline.

3.1 Steady state simulation Differences between two models (isothermal and non-isothermal) were analyzed using a part of existing real gas system, Yamal – West Europe. This gas transportation system (Fig. 1) consists of five compressor stations, installed on the Polish terrain. At each compressor station there are

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 installed between 2 and 3 centrifugal compressors, driven by gas turbines. For the purpose of our investigation, one pipe between two compressor stations was taken. German Belarus border compressor border station 5 4 3 2 1

139.0 km 177.0 122.0 117.0 126.5 2.5

Fig. 1. Structure of gas transportation system.

Calculations were carried out for the following parameters: pipe diameter D = 1422 mm, pipe wall thickness 19.2 mm, pipeline length L=122 km, pressure at x=0 (discharge pressure) p1 = 8.4 MPa, temperature at x=0 (discharge temperature) Tx=0 =42.5 °C, 3 density of natural gas (under standard conditions) ρn = 0.7156 kg/m , viscosity of natural gas µ=0.135 ×10 -4 N s/m 2 3 flow (under standard conditions) Qn=2019950 m /h, soil temperature Tsoil =12 °C, heat transfer coefficient kL=25 W/(mK), pipe roughness k=0.03 mm. Fanning friction coefficient and compressibility factor were calculated for every discretization section of the pipeline using Nikuradse [1] and SGERG 88 [5] equations respectively. Results of the investigations are shown in Fig. 2 and Fig. 3.

= o =T CONST, T o 12 C 40 ≠ = o T CONST, T o 42.5 C ≠ = o T CONST, T o 30 C 35

25 C) o T ( T 20

0 0 12.2 24.4 36.6 48.8 61 73.2 85.4 97.6 109.8 122 x (km)

o o Fig. 2 Temperature profile along the pipeline for a) T=const, T o=12 C, b) T ≠const, T o=42.5 C, o c)T ≠const, T o=30 C.

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 8.5

= o =T CONST, T o 12 C ≠ = o 8.4 T CONST, T o 42.5 C ≠ = o T CONST, T o 30 C

8.3

8.2 p(Mpa) 8.1

7.9

7.8 0 12.2 24.4 36.6 48.8 61 73.2 85.4 97.6 109.8 122 x (km)

o o Fig. 3 Change of pressure along the pipeline for a) T=const, T o=12 C,b) T ≠const, T o=42.5 C, c) o T≠const, T o=30 C. Maximum pressure difference between isothermal and non-isothermal flow:

p− p − − δ = isotherm x= J n isotherm x= J ⋅ = 7.. 894 7 879 ⋅ = max 100% 100 019. % pisotherm x= J 7. 894 Maximum pressure difference between non-isothermal flow (without cooling system) and non- isothermal flow (with cooling system):

p− − p − − − δ = n isotherm x= J n iso c x= J ⋅ = 7.. 894 7 885 ⋅ = max 100% 100 011. % pn− isotherm x= J 7. 894 Discharge temperature Capacity Capacity difference

oC m3/h m3/h %

42.5 (without cooling system) 2019949.7 - -

30.0 (with cooling system) 2032010.2 12 060.5 + 0.6

12.0 (soil temperature) 2049998.4 30 048.7 + 1.5

Table 1 Comparison of capacity of the pipeline for above analyzed cases.

Fig.4 shows that for non-isothermal flow, the pressure drop along the pipeline is greater than in the isothermal case. It is resulted by the decrease of the density of the gas, allowing the smaller mass of gas to be transported at the specific velocity.

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 7.95 7.916 7.903 7.894 7.9 7.875 7.879 7.859 7.85

7.8 T ≠ CONST, = o T o 42.5 C 7.75 p (MPa) 7.7

7.656 =T CONST, = o 7.65 7.639 T o 12 C

7.6

7.55 2 3 4 5 COMPRESSOR STATION #

Fig. 4. Comparison of suction pressures for the whole transportation system. On the Fig. 5 difference in HP used by compressor stations is presented. We can see significant difference in energy consumption by the drivers of the compressors of each case. It means that the cost of running of the transportation system is a function of discharge temperature. The low- est costs correspond to the isothermal flow.

1.740

1.729

1.720

1.700 1.698 HP (MW) 1.680

1.660 1.657

1.640 12.0 42.5 30.0

T ( oC) 0 Fig. 5. HP as a function of discharge temperature for the whole transportation system.

It is clear that decreasing the discharge temperature of the gas, the effectiveness of the transportation process can be significantly increased. Page 8

Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 4. Transient non-isothermal model With respect to the conveyance of gas in pipelines, there are two technically relevant extreme cases of pipe flow: – flow without heat exchange with the ground outside: adiabatic and more especially, isentropic flow, – flow with heat exchange with the ground outside, which is regarded as being a heat storage unit of infinite capacity with constant temperature T 0: isothermal flow. The flow processes of greatest concern here are those in which temperature equalization with the ground outside cannot take place. The temperature profile is a function of pipeline distance. In this case the transient, non-isothermal flow of gas in a horizontal pipe ( ρg sinα = 0 , ∂ bρwAgzdx g = 0 ) is described by the system of equations 2 ÷5. ∂x Two contradictory constraints are imposed on the above equations. The requirement is that on the one hand the description of the phenomenon is accurate, and on the other there is sufficient sim- plification to allow solving of this model by reasonable computation means. As a rule simplified models are sought which present a reasonable compromise between the accuracy of the description and the costs. The simplified models are obtained by neglecting some terms in the basic (accurate) equation. This results from the quantitative estimation of particular elements of the equation, carried for some given conditions of operation of the pipeline. This means that the model of transient flow used for simulation should be fitted to the given conditions of operation of the pipe. A necessary condition for proper selection of the model is therefore the earlier analysis of these conditions. Estimation of the particular terms of the equation (5) for given operating conditions and a given geometry of the pipe is given below.

4.1 Energy equation simplifications Assuming that heat transfer is limited only to conduction through a walled tube and the gas along a pipeline the following equation can be written ∂ ∂ ρ = Fλ T I − − q Adx A G Jdx kLb T T soil g dx (10) ∂x H ∂x K where: λ - thermal conductivity coefficient of gas, W/(mK), kL - heat transfer coefficient, W/(mK). Combining equations (5) and (10), the final version of the equation can be put into the form ∂ ∂ ρ ∂ ρ 3 ∂ ∂ ∂ ρ + F wAp I + F Aw I + ρ Fλ T I + bwAcv Tdx gG dx J G dx J bwAgzdx g - G A dx J ∂x ∂x H ρ K ∂x H 2 K ∂x ∂x H ∂x K 144 244 3 144 244 3 144 244 3 144 244 3 144 244 3 I IV V II III (11) ∂ ∂ ρ2 ∂ + − +ρ + F Aw I +ρ = kLb T T soil g dx b Acv Tdx g G dx J b Agzdx g 0 144 244 3 ∂t ∂t H 2 K ∂t VI 144 244 3 144 244 3 14 24 3 VII IX VIII By integrating the above equation between x = 0 and x = L (where L is the length of the pipe) for the following parameters of the system: pipe diameter D = 1422 mm, pipe wall thickness 19.2 mm, pipeline length L=122 km, pressure at x=0 (discharge pressure) p1 = 8.4 MPa, suction pressure p2 = 7.88 MPa,

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 temperature at x= 0 (discharge temperature) Tx= 0=42.5 °C, temperature at x=L (suction temperature) Tx=L =13 °C, 3 density (under standard conditions) ρn = 0.7156 kg/m , 3 flow (under standard conditions) Q n=2019950 m /h, soil temperature Tsoil =12 °C, heat transfer coefficient kL=25 W/(mK), -2 thermal conductivity coefficient of gas λave =3.4 ×10 W/(mK), specific gas constant R = 518.8 J/(kgK), 3 specific heat at constant pressure cp = 2.278 ×10 J/(kgK), 3 specific heat at constant volume cv = 1.759 ×10 J/(kgK), we get the following values for each term of the equation

L ∂ ρ≈ ρ − = bwAcv T g dxn Q n c v c T= T = h z ∂x x L x 0 (I) 0 → O(10 7) 2019950 =0. 7156 ⋅ ⋅1759 . ⋅ 103 ⋅b 286 − 3155 .. g = − 20810 ⋅ 7 W 3600

L ∂ ρ F wAp I ≈ρ − z G Jdxn Q n ZRc Tx= L T x =0 h ∂x H ρ K 6 (II) 0 → O(10 ) 2019950 =0. 7156 ⋅ ⋅097 .. ⋅ 5188 ⋅b 286 − 3155 .. g = − 60 ⋅ 10 6 W 3600

L ∂ ρ 3 F Aw I ≈1 ρ 2 − 2 = z dxn Q n w w ∂x G 2 J 2 ex= L x =0 j (III) 0 H K → O(10 2) 1 2019950 = ⋅0. 7156 ⋅ ⋅c496 .2 − 513 . 2 h = − 3410 . ⋅ 2 W 2 3600

L ∂ bρwAgz g dx≈ ρ Q g z − z (IV) z ∂ n n c x= L x =0 h 0 x

Assuming that the difference of levels is e.g. 100 m, we get

ρ − = ⋅2019950 ⋅ ⋅ = ⋅ 5 → 5 nQ n gc z= z = h 0. 7156 981 . 100 39 . 10 W O(10 ) x L x 0 3600 L ∆− ∆ L ∂ ∂T O TT= = (V) − FλA I dx≈ −λ A x L x 0 zM∂ HG ∂ KJP ave ∆ 0 N x x Q x from the steady state analysis we have ∆T 4. 8 = K/m ∆ x x=0 6100

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 ∆T 0. 2 = K/m ∆ x x= L 6100

∆− ∆ 2 TT − π ⋅13836...4 8− 0 2 − −λ A x= L x =0 = −34. ⋅ 10 2 ⋅ ⋅ =39. ⋅ 10 5 W → O(10 -5) ave ∆x 4 6100

L ()− ≈ ∆ (VI) ∫ kL T T soil dx kLt L s 0

Substituting the mean temperature difference between the gas and the soil ∆Ts=15.75 K we have ∆ = ⋅ ⋅5 ⋅ = ⋅ 7 → 7 kL L T s 25 122. 10 1575 . 48 . 10 W O(10 ) L ∂ ∆T (VII) ρAc T dx≈ ρ Ac L z ∂ bv g v ∆ 0 t t Let the increment of the temperature be 5 K and be achieved within ∆t = 1h. Substituting the mean value of gas density along the pipeline ρ =64./ 27 ⋅ kg m3 we get ∆T π ⋅13836. 2 5 ρAc L =64. 27 ⋅ ⋅1759. ⋅ 103 ⋅ 122 . ⋅ 10 5 ⋅ =2 . 88 ⋅ 10 7 W → O(10 7) v ∆t 4 3600 L ∂ F ρAw 2 I Lw ∂b ρAw g Lwρ ∆ Q (VIII) dx ≈ ≈ n n z ∂ G J ∂ ∆ 0 t H 2K 2t 2 t

Let the increment of the load of the pipeline at x=L be 0.5 ×Qn and be achieved within ∆t=1h. We then get Lwρ ∆ Q 122...⋅ 105 ⋅ 504 ⋅ 07156 1009975 n n = ⋅ =618. ⋅ 10 7 W → O(10 7) 2 ∆t 2 3600 L ∂ ∂z (IX) ρgzA dx≈ ρ gAL = 0 (for each gas pipeline) z ∂ b g ∂ 0 t t Based on above given analysis we can write the simplified form of the energy equation ∂ 2 ∂ − =L ρ F + w IO +L ρ F + pIO kLb T soil Tdx gM bAdx gG u JP M bwAdx gG u JP ∂t N H 2 KQ ∂x N H ρKQ The character of the results cannot be generalized. This can only be the starting point, which al- lows the forwarding of the hypothesis that in the case when the selected parameters do not change rapidly, transient non-isothermal flow through the horizontal pipe can be represented be the set of equations in the form

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 R ∂b ρwg ∂ρ |− = | ∂x∂ t ∂ ρ 2 ∂ ρw ∂ ρw2 |−p −2 f w =b g + c h | ∂ ∂ ∂ x D t x S ∂ 2 ∂ | − =L ρ F + w IO +L ρ F + pIO kLb T soil Tdx gM bAdx gG u JP M bwAdx gG u JP | ∂t N H 2 KQ ∂x N H ρKQ | p | = ZRT T|ρ 4.2 Transient simulation The investigations were carried out using the method of lines [9] for the following values of parameters compressor compressor station 1 station 2

D(1422 ×19.2 mm)

122.0 km

Fig. 6 Structure of gas transportation system.

pipe diameter D = 1422 mm, pipe wall thickness 19.2 mm, pipeline length L=122 km, temperature at x=0 (discharge temperature) Tx= 0=42.5 °C, 3 density (under standard conditions) ρn = 0.7156 kg/m , soil temperature T soil =12 °C, heat transfer coefficient kL=25 W/(mK), -2 thermal conductivity coefficient of gas λave =3.4 ×10 W/(mK), specific gas constant R = 518.8 J/(kgK), 3 specific heat at constant pressure cp = 2.278 ×10 J/(kgK), 3 specific heat at constant volume cv = 1.759 ×10 J/(kgK), p(,)0 t = const = 8.4 MPa U boundary conditions = V QN (,)() L t f t W = The function QN (,)() L t f t is shown in Figure 5. For given range of variation of the load flow behavior in the pipeline is approximated accurately by the rough-pipe flow law [1]. This law is represented by the horizontal lines in the Moody diagram of friction factor, with each line corre- sponding to a specific value of the relative roughness of the pipe.

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3500

3000

/h) 2500 3 (m 3 x10

n 2000 Q

1500

1000

500 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 t (h)

Fig. 7 Change of flow with time ( x=L) – (boundary condition). Results of calculation are presented in Figs. 8, 9, 10, 11, 12.

7.8

7.6

7.4 p(Mpa) 7.2

= o =T CONST, T o 12 C 6.8 ≠ = o T CONST, T o 42.5 C ≠ = o T CONST, T o 30 C

6.6 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 t (h)

Fig. 8 Change of pressure at x=L. Maximum pressure difference between non-isothermal flow (without cooling system) and isothermal flow:

p− p − − δ = isotherm x= J n isotherm x= J ⋅ = 7145.. 7 067 ⋅ = max 100% 100 109. % pisotherm x= J 7145.

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 Maximum pressure difference between non-isothermal flow (without cooling system) and non- isothermal flow (with cooling system):

p− p − − δ = isotherm x= J n isotherm x= J ⋅ = 7145.. 7117 ⋅ = max 100% 100 0. 39 % pisotherm x= J 7145.

15 C) o 13 T ( T

= o =T CONST, T o 12 C ≠ = o 7 T CONST, T o 42.5 C ≠ = o T CONST, T o 30 C

5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 t (h)

Fig. 9 Change of temperature at x=L. Profile of temperature at x=L is caused by load variations and by distributed velocity of the gas along the pipeline.

8.6

= o =T CONST, T o 12 C 8.4 ≠ = o T CONST, T o 42.5 C ≠ = o T CONST, T o 30 C 8.2

7.8 p(MPa)

7.6

7.4

7.2

7 0 12200 24400 36600 48800 61000 73200 85400 97600 109800 122000 x (m)

Fig. 10. Change of pressure along the pipeline for t=12 h.

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 1.3 = o =T CONST, T o 12 C ≠ = o T CONST, T o 42.5 C ≠ = o 1.25 T CONST, T o 30 C

1.2

ε 1.15

1.1

1.05

1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 t (h)

Fig. 11 Change of compressor ratio at compressor station No. 2.

Maximum compressor ratio difference between non-isothermal flow (without cooling system) and isothermal flow: ε− ε 118..− 126 ∆ε = isotherm n− isotherm ⋅100% = ⋅100 = 6. 78 % max ε isotherm 118. Maximum compressor ratio difference between non-isothermal flow (with cooling system): and isothermal flow: ε− ε 1074..− 1078 ∆ε = isotherm n− isotherm ⋅100% = ⋅100 = 0. 37 % max ε isotherm 1074.

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= o =T CONST, T o 12 C ≠ = o 1.2 T CONST, T o 42.5 C ≠ = o T CONST, T o 30 C

0.8 (MW) e N 0.6

0.4

0.2

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 t (h)

Fig. 12 Changes of HP at compressor station No. 2 for the isentropic compression.

Maximum HP difference necessary to keep constant discharge pressure at compressor station No. 2 for non-isothermal flow (without cooling system) and isothermal flow: NN− − ∆ = isotherm n− isotherm ⋅ = 0.. 666 1255 ⋅ = N max 100% 100 88. 43 % Nisotherm 0. 666 Maximum HP difference necessary to keep constant discharge pressure at compressor station No. 2. for non-isothermal flow (with cooling system) and isothermal flow: NN− − ∆ = isotherm n− isotherm ⋅ = 0.. 666 0807 ⋅ = N max 100% 100 2117. % Nisotherm 0. 666

5. Conclusions It is clear that cooling of the gas improves the efficiency of the overall compression process. There exists a significant difference in the pressure profile along the pipeline between isothermal and non isothermal process. This difference increases when the quantity of gas increases. This shows that in the case when gas temperature does not stabilize the use of an isothermal model leads to significant errors. The problem of choosing the correct model is a function of network structure and network complexity.

6. Bibliography [1] The American Gas Association 1990. Distribution – System Design . A.G.A., Arlington. [2] Daneshyar, H. 1976. One-dimensional Compressible Flow. Pergamon Press, New York. [3] Goldwater, M. H. and Fincham, A. E. 1981. Modelling of Dynamical Systems (ed. H. Ni- cholson). Peter Peregrinus, Stevenage. [4] Holman, J. P. 1980. Thermodynamics. McGraw-Hill Book Company, New York. [5] ISO 12213–3 Natural gas – Calculation of Compression factor – Part 3: Calculation using physical properties.

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Pobrano z http://repo.pw.edu.pl / Downloaded from Repository of Warsaw University of Technology 2021-09-25 [6] Kralik, J. et al. 1988. Dynamic Modelling of Large Scale Networks With Application to Gas Distribution . Elsevier, New York. [7] Osiadacz, A. J . 1989. Simulation and Analysis of Gas Networks . Gulf Publishing Company, Huston. [8] Osiadacz, A. J . 1996. Different Transient Models - Limitations, Advantages And Disad- vantage. PSIG The 28th Annual Meeting, San Francisco. [9] Schiesser, W. E. 1991. The Numerical Methods of Lines. Academic Press, London. [10] Shapiro, A. H. 1954. The Dynamics and Thermodynamics of Compressible Fluid Flow . The Ronald Press Company, New York. [11] Wylie, E. B., Streeter V. L. 1978. Fluid Transients . McGraw – Hill, New York.

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