Natural Gas Part 13 ppt

Natural Gas472 NP R Then = = p dg W4  21.385242 0.02341 7- E 177086.42.32 6 - E 289431.54 = ××× ×  Alternatively . b T b z gp Rgd b Q b P g G88575.36 NP R  = = 0.02341453017 - E 177086.42.32 5 - E 052934.71447.141885750.36 ×××× ×××× = 21.385221 The equation of continuity for gas flow in a pipe is: === 2 v 2 A 21 v 1 A 1 W  Constant (10) Then, A v. W = In a cylindrical homogeneous porous medium the equation of the weight flow rate can be written as: W A v. p   (11) Equation (11) can be differentiated and solved simultaneously with the lost head formulas (equation 2, 3 and 4), and the energy equation (equation 1) to arrive at the general differential equation for fluid flow in a homogeneous porous media. Regarding the cross sectional area of the porous medium (A p ) as a constant, equation (11) can be differentiated and solve simultaneously with equations (2) and (1) to obtain. _ c v sin k d p d p d W d p A g p / 2 1 2 2                         (12) Equation (12) is a differential equation that is valid for the laminar flow of any fluid in a homogeneous porous medium. The fluid can be a liquid of constant compressibility or a gas. The negative sign that proceeds the numerator of equation (12) shows that pressure decreases with increasing length of porous media. The compressibility of a fluid (C f ) is defined as: d 1 C f d p    (13) Combination of equations (12) and (13 ) leads to: _ c v sin k d p d p W A g p / 2 1 2                        (14) Differentiation of equation (11) and simultaneous solution with equations (2), (1) and (13) after some simplifications, produces: _ c v sin 2 d p d p d p W C f A g p 32 2 1 2                          (15) Differentiation of equation (6) and simultaneous solution with equations (4), (1) and ( 1 3) after some simplifications produces: 2 f W p sin 2 2 A d p p d p d p W C f A g p 2 1 2                            - (16) Equation (16) can be simplified further for gas flow through homogeneous porous media. The cross sectional area of a cylindrical cross medium is: p A  d p 2 4  (17) Steady State Compressible Fluid Flow in Porous Media 473 NP R Then = = p dg W4  21.385242 0.02341 7- E 177086.42.32 6 - E 289431.54 = ××× ×  Alternatively . b T b z gp Rgd b Q b P g G88575.36 NP R  = = 0.02341453017 - E 177086.42.32 5 - E 052934.71447.141885750.36 ×××× ×××× = 21.385221 The equation of continuity for gas flow in a pipe is: === 2 v 2 A 21 v 1 A 1 W  Constant (10) Then, A v. W  = In a cylindrical homogeneous porous medium the equation of the weight flow rate can be written as: W A v. p   (11) Equation (11) can be differentiated and solved simultaneously with the lost head formulas (equation 2, 3 and 4), and the energy equation (equation 1) to arrive at the general differential equation for fluid flow in a homogeneous porous media. Regarding the cross sectional area of the porous medium (A p ) as a constant, equation (11) can be differentiated and solve simultaneously with equations (2) and (1) to obtain. _ c v sin k d p d p d W d p A g p / 2 1 2 2                         (12) Equation (12) is a differential equation that is valid for the laminar flow of any fluid in a homogeneous porous medium. The fluid can be a liquid of constant compressibility or a gas. The negative sign that proceeds the numerator of equation (12) shows that pressure decreases with increasing length of porous media. The compressibility of a fluid (C f ) is defined as: d 1 C f d p    (13) Combination of equations (12) and (13 ) leads to: _ c v sin k d p d p W A g p / 2 1 2                        (14) Differentiation of equation (11) and simultaneous solution with equations (2), (1) and (13) after some simplifications, produces: _ c v sin 2 d p d p d p W C f A g p 32 2 1 2                          (15) Differentiation of equation (6) and simultaneous solution with equations (4), (1) and ( 1 3) after some simplifications produces: 2 f W p sin 2 2 A d p p d p d p W C f A g p 2 1 2                            - (16) Equation (16) can be simplified further for gas flow through homogeneous porous media. The cross sectional area of a cylindrical cross medium is: p A  d p 2 4  (17) Natural Gas474 The equation of state for a non ideal gas is: p M z T R   (18) Where =p Absolute pressure =T Absolute temperature Multiply equation (11) with  and substitute A p in equation (17) and use the fact that: p d 2 pd 2 1 p d pdp  = Then _ f W zR p zR d g d d W zR C p f g d 2 2 2 sin 1.621139 5 2 2 1.621139 1 4                              (19) The compressibility of ideal gas   g C is defined as z C . g p z p 1 1 _  (20) For an ideal gas such as air, C . g p 1  (21) (Matter et al, 1975) and ( Ohirhian, 2008) have proposed equations for the calculation of the compressibility of hydrocarbon gases. For a sweet natural gas (natural gas that contains CO 2 as major contaminant), (Ohirhian, 2008) has expressed the compressibility of the real gas (C g ) as: p f C Κ = (22) For Nigerian (sweet) natural gas K = 1.0328 when p is in psia. Then equation (19) can then be written compactly as: _ AAp B p d p p C d p p p 2 2 ( ) (1 ) 2    (23) Where 4 p gMd zRT 2 KW p C , zRT sinM2 p B , M 5 p gd zRT 2 W p f621139.1 p AA = ==  The denominator of the differential equation (23) is the contribution of kinetic effect to the pressure drop across a given length of a cylindrical isotropic porous medium. In a pipe the kinetic contribution to the pressure drop is very small and can be neglected. What of a homogeneous porous medium? Kinetic Effect in Pipe and Porous Media An evaluation of the kinetic effect can be made if values are substituted into the variables that occurs in the denominator of the differential equation (23) Example 2 Calculate the kinetic energy correction factor, given that 0.75 pounds per second of air flow isothermally through a 4 inch pipe at a pressure of 49.5 psia and temperature of 90 0 F. Solution The kinetic effect correction factor is 2 p C _ 1 Where C for a pipe is given by, 4 gMd zRT 2 KW C = Here sec/lb75.0W = , ft 0.333333 ft 12/4inch4d === , psf 7128 psf 44 49.5 psia 45.5 p =×== , R550R)46090( F o 90 T °=°+== , fluid) theis(air 1.0 z = 2 secft / 32.2 g ,1545R == , 28.97 M = . Then, 58628.41504 4 333333.097.282.32 55015451 2 75.01 C = ×× ×××× = , gas idealan for 1 K = Steady State Compressible Fluid Flow in Porous Media 475 The equation of state for a non ideal gas is: p M z T R   (18) Where =p Absolute pressure =T Absolute temperature Multiply equation (11) with  and substitute A p in equation (17) and use the fact that: p d 2 pd 2 1 p d pdp  = Then _ f W zR p zR d g d d W zR C p f g d 2 2 2 sin 1.621139 5 2 2 1.621139 1 4                              (19) The compressibility of ideal gas   g C is defined as z C . g p z p 1 1 _  (20) For an ideal gas such as air, C . g p 1  (21) (Matter et al, 1975) and ( Ohirhian, 2008) have proposed equations for the calculation of the compressibility of hydrocarbon gases. For a sweet natural gas (natural gas that contains CO 2 as major contaminant), (Ohirhian, 2008) has expressed the compressibility of the real gas (C g ) as: p f C Κ = (22) For Nigerian (sweet) natural gas K = 1.0328 when p is in psia. Then equation (19) can then be written compactly as: _ AAp B p d p p C d p p p 2 2 ( ) (1 ) 2    (23) Where 4 p gMd zRT 2 KW p C , zRT sinM2 p B , M 5 p gd zRT 2 W p f621139.1 p AA = ==  The denominator of the differential equation (23) is the contribution of kinetic effect to the pressure drop across a given length of a cylindrical isotropic porous medium. In a pipe the kinetic contribution to the pressure drop is very small and can be neglected. What of a homogeneous porous medium? Kinetic Effect in Pipe and Porous Media An evaluation of the kinetic effect can be made if values are substituted into the variables that occurs in the denominator of the differential equation (23) Example 2 Calculate the kinetic energy correction factor, given that 0.75 pounds per second of air flow isothermally through a 4 inch pipe at a pressure of 49.5 psia and temperature of 90 0 F. Solution The kinetic effect correction factor is 2 p C _ 1 Where C for a pipe is given by, 4 gMd zRT 2 KW C = Here sec/lb75.0W = , ft 0.333333 ft 12/4inch4d === , psf 7128 psf 44 49.5 psia 45.5 p =×== , R550R)46090( F o 90 T °=°+== , fluid) theis(air 1.0 z = 2 secft / 32.2 g ,1545R == , 28.97 M = . Then, 58628.41504 4 333333.097.282.32 55015451 2 75.01 C = ×× ×××× = , gas idealan for 1 K = Natural Gas476 The kinetic effect correction factor is 999183.0 2 7128 58628.41504 _ 1 2 p C _ 1 == Example 3 If the pipe in example 1 were to be a cylindrical homogeneous porous medium of 25 % porosity, what would be the kinetic energy correction factor? Solution Here, d p = d = 333333.0 25.0 = ft1666667.0 0212.344046 4 166667.097.282.32 55015451 2 75.01 p C = ×× ×××× = Then, 993221.0 2 7128 0212.3441046 1 2 p p C _ 1 == The kinetic effect is also small, though not as small as that of a pipe. The higher the pressure, the more negligible the kinetic energy correction factor. For example, at 100 psia, the kinetic energy correction factor in example 2 is: 998341.0 2 )144100( 0212.3441046 _ 1 = × Simplification of the Differential Equations for Porous Media When the kinetic effect is ignored, the differential equations for porous media can be simplified. Equation (14) derived with the Darcy form of the lost head becomes: d c v p sin d k p /               (24) Equation (15) derived with the (Ohirhian, 2008) form of the lost head becomes: d p c v sin 2 d d p p 32               (25) Equation (16) derived with the (Ohirhian, 2008) modification of the Darcy- Weisbach lost head becomes: 2 f W d p p sin 2 d p 2 A d p p                 (26) In terms of velocity (v) equation (26) can be written as: 2 f v d p p sin d 2 d p p                 (27) In certain derivations (for example, reservoir simulation models) it is required to make v or W subject of equations (24) to (27) Making velocity (v) or weight (W) subject of the simplified differential equations When v is made subject of equation (24), we obtain: d p - k v sin / d c p               (28) When v is made subject of equation (25), we obtain: - d p d p v sin 32c d p 2               (29) When v 2 is made subject of equation (27), we obtain: - 2 g d p d p v sin f d p p 2               (30) When W 2 is made subject of equation (26), we obtain: - 2 g d A p p d p W sin f d p p 2 2               (31) Let S be the direction of flow which is always positive, then equation (28) can be written as: d p d z - k v s ds ds _ 6 10 1.01325           (32) Where: Steady State Compressible Fluid Flow in Porous Media 477 The kinetic effect correction factor is 999183.0 2 7128 58628.41504 _ 1 2 p C _ 1 == Example 3 If the pipe in example 1 were to be a cylindrical homogeneous porous medium of 25 % porosity, what would be the kinetic energy correction factor? Solution Here, d p = d = 333333.0 25.0 = ft1666667.0 0212.344046 4 166667.097.282.32 55015451 2 75.01 p C = ×× ×××× = Then, 993221.0 2 7128 0212.3441046 1 2 p p C _ 1 == The kinetic effect is also small, though not as small as that of a pipe. The higher the pressure, the more negligible the kinetic energy correction factor. For example, at 100 psia, the kinetic energy correction factor in example 2 is: 998341.0 2 )144100( 0212.3441046 _ 1 = × Simplification of the Differential Equations for Porous Media When the kinetic effect is ignored, the differential equations for porous media can be simplified. Equation (14) derived with the Darcy form of the lost head becomes: d c v p sin d k p /               (24) Equation (15) derived with the (Ohirhian, 2008) form of the lost head becomes: d p c v sin 2 d d p p 32               (25) Equation (16) derived with the (Ohirhian, 2008) modification of the Darcy- Weisbach lost head becomes: 2 f W d p p sin 2 d p 2 A d p p                 (26) In terms of velocity (v) equation (26) can be written as: 2 f v d p p sin d 2 d p p                 (27) In certain derivations (for example, reservoir simulation models) it is required to make v or W subject of equations (24) to (27) Making velocity (v) or weight (W) subject of the simplified differential equations When v is made subject of equation (24), we obtain: d p - k v sin / d c p               (28) When v is made subject of equation (25), we obtain: - d p d p v sin 32c d p 2               (29) When v 2 is made subject of equation (27), we obtain: - 2 g d p d p v sin f d p p 2               (30) When W 2 is made subject of equation (26), we obtain: - 2 g d A p p d p W sin f d p p 2 2               (31) Let S be the direction of flow which is always positive, then equation (28) can be written as: d p d z - k v s ds ds _ 6 10 1.01325           (32) Where: Natural Gas478  s v Volumetric flux across a unit area of porous medium in unit time along flow path, S cm / sec 2 sec / cm 980.605 gravity, todueon Accelerati g cc / mass gm , fluid ofDensity Mass cc weight / gm , fluid of weight Specificg = = ==   cm / atm refers, s which v point to at the S alonggradient Pressure sd pd = atm cm sq / dynes 6 10 1.01325 darcys. medium, theofty Permeabili k cm downwards, positive considered ,coordinate Vertical z scentipoise fluid, theofViscosity =× = = =  According to (Amyx et al., 1960), this is “the generalized form of Darcy law as presented in APT code 27 “. Horizontal and Uphill Gas Flow in Porous Media In uphill flow, the + sign in the numerator of equation (23) is used. Neglecting the kinetic effect, which is small, equation (23) becomes dp A A B p P P d p 2 2    (33) zTR 2Msinθ p B , M 5 p gd 2 zTRW p 1.621139f p AA = = An equation similar to equation (33) can also be derived if the Darcian lost head is used. The horizontal / uphill gas flow equation in porous media becomes. Where Mk 2 p d zTRWc546479.2 Mk 2 2 d zTRWc8 Mk p A zTRWc2 / p AA    ′ = ′ = ′ = Solution to the Horizontal/Uphill Flow Equation Differential equations (33) and (34) are of the first order and can be solved by the classical Runge - Kutta algorithm. The Runge - Kutta algorithm used in this work came from book of (Aires, 1962) called “Theory and problems of Differential equations”. The Runge - Kutta solution to the differential equation ( ) given that xat x y,xf dx dy n == is x x at 0 y y 0 ==   y y k k k k 1 2( ) 0 1 2 3 4 6      (35) where k Hf x y o o k H f x H y k o k H f x H y k o o k Hf x H y k o x x n o H n n sub ervals steps ( , ) 1 1 1 ( , ) 2 0 1 2 2 1 1 ( , ) 3 1 2 2 ( , ) 4 0 3 int ( )             Application of the Runge - Kutta algorithm to equation (33) leads to: dp AA B p p p d p 2 / 2    (34) Steady State Compressible Fluid Flow in Porous Media 479  s v Volumetric flux across a unit area of porous medium in unit time along flow path, S cm / sec 2 sec / cm 980.605 gravity, todueon Accelerati g cc / mass gm , fluid ofDensity Mass cc weight / gm , fluid of weight Specificg = = ==   cm / atm refers, s which v point to at the S alonggradient Pressure sd pd = atm cm sq / dynes 6 10 1.01325 darcys. medium, theofty Permeabili k cm downwards, positive considered ,coordinate Vertical z scentipoise fluid, theofViscosity =× = = =  According to (Amyx et al., 1960), this is “the generalized form of Darcy law as presented in APT code 27 “. Horizontal and Uphill Gas Flow in Porous Media In uphill flow, the + sign in the numerator of equation (23) is used. Neglecting the kinetic effect, which is small, equation (23) becomes dp A A B p P P d p 2 2    (33) zTR 2Msinθ p B , M 5 p gd 2 zTRW p 1.621139f p AA = = An equation similar to equation (33) can also be derived if the Darcian lost head is used. The horizontal / uphill gas flow equation in porous media becomes. Where Mk 2 p d zTRWc546479.2 Mk 2 2 d zTRWc8 Mk p A zTRWc2 / p AA    ′ = ′ = ′ = Solution to the Horizontal/Uphill Flow Equation Differential equations (33) and (34) are of the first order and can be solved by the classical Runge - Kutta algorithm. The Runge - Kutta algorithm used in this work came from book of (Aires, 1962) called “Theory and problems of Differential equations”. The Runge - Kutta solution to the differential equation ( ) given that xat x y,xf dx dy n == is x x at 0 y y 0 ==   y y k k k k 1 2( ) 0 1 2 3 4 6      (35) where k Hf x y o o k H f x H y k o k H f x H y k o o k Hf x H y k o x x n o H n n sub ervals steps ( , ) 1 1 1 ( , ) 2 0 1 2 2 1 1 ( , ) 3 1 2 2 ( , ) 4 0 3 int ( )             Application of the Runge - Kutta algorithm to equation (33) leads to: dp AA B p p p d p 2 / 2    (34) Natural Gas480 p p y a 2 2 2 1   (36) Where ( ) 3 a x72.0 2 x48.1x96.4 6 2 2 p 3 a x36.0 2 a x5.0 a x1 a p aa a y aa )( +++ +++= + ( ) 2 a x72.0 a x96.196.4 6 a p u ++ )L 2 S p2 (AA a p aa += R 2 T 2 z 2 2 psinM2 2 S , M 5 p gd 2 RW 2 T 2 z p f621139.1 2p AA  = = R av T a av z LsinM2 a x , M 5 p gd 2 RW av T av z p f621139.1 a p u  = = Where: p 1 = Pressure at inlet end of porous medium p 2 = Pressure at exit end of porous medium f p = Friction factor of porous medium. θ = Angle of inclination of porous medium with horizontal in degrees. z 2 = Gas deviation factor at exit end of porous medium. T 2 = Temperature at exit end of porous medium T 1 = Temperature at inlet end of porous medium z a v = Average gas deviation factor evaluated with T a v and p a v T a v = Arithmetic average temperature of the porous medium given by 0.5(T 1 + T 2 ) and p a v a = a p aa 2 2 p + In equation (36), the component k 4 in the Runge - Kutta algorithm was given some weighting to compensate for the variation of temperature (T) and gas deviation factor (z) between the mid section and the inlet end of the porous medium. In isothermal flow where there is little variation of the gas deviation factor between the mid section and the inlet end of the porous medium, the coefficients of x a change slightly, then, ( ) ) 2 a x5.0 a x25( 6 2 p u ) 3 a x5.0 2 a x2 a x5( 6 2 2 p 3 a x25.0 2 a x5.0 a x1 a p aa a y +++ +++ +++= Application of the Runge-Kutta algorithm to equation (34) produces. p p y b 2 2 1 2   (37) ( ) 3 b x36.0 2 b x5.0 b x1 b p aa b y +++= ( ) 2 b x72.0 2 b x48.1 b x96.4 6 2 2 p +++ ( ) 2 b x72.0 b x96.196.4 6 2 p u +++ [...]... Ohirhian, P.U (2009) Equations for the z-factor and compressibility of Nigerian Natural gas, Advanced Materials Research, Vols 62-64, pp 484-492, Trans Tech Publications, Switzerland Vibert T (1939) Les gisements de bauxite de I’Índochine, Gẻnie Civil 115, 84 Natural gas properties and flow computation 501 21 X Natural gas properties and flow computation Ivan Marić and Ivan Ivek Ruđer Bošković Institute... capacity of ideal gas at constant pressure, R is the universal gas constant, Z is the compression factor and  mI and  m are the corresponding molar densities of ideal and real gas at temperature T After substituting the first and the second derivative of the AGA-8 compressibility equation (Starling & Savidge, 1992, ISO-12 213- 2, 2006) Z  1  B  m  r  C   C b 18 n 13 58 * n * n n 13 n  kn  cn... the natural gas mixture J/(mol·K) j j cm , pi Ideal molar heat capacity of the gas component D d h K M p q R s T vm vmI yi Z β Δp Δ Upstream internal pipe diameter Diameter of orifice Specific enthalpy Size parameter Molar mass of the gas mixture Absolute pressure Mass flow rate Molar gas constant 8314.51 Specific entropy Absolute temperature Molar specific volume Molar specific volume of ideal gas. .. presentation of modern methods of estimating the physical properties of gases and liquids can be found in (Poling at al., 2000) Formulations explicit in the Helmholtz energy have been widely used to represent the properties of natural gas because of the ease of calculating all other thermodynamic properties by mathematical 502 Natural Gas differentiation (Lemmon & Starling, 2003, Span & Wagner, 1996, Span... 2003) for meta-modeling the effects of natural gas properties in flow rate measurements (Marić & Ivek, 2010) will be illustrated The practical examples of ANN and GMDH surrogate models for the compensation of natural gas flow rate measurement error caused by the thermodynamic effects, with the corresponding accuracies and execution times will be given The models are particularly suitable for implementation... thermodynamic properties of natural gas This section summarizes the procedure (Maric, 2007) for the calculation of specific heat capacity at constant pressure cp and at constant volume cv, JT coefficient μJT and isentropic exponent κ of a natural gas based on thermodynamic equations, AGA-8 extended virial type characterization equation (Starling & Savidge, 1992, ISO-12 213- 2, 2006) and DIPPR generic... can be rewritten in the following integral form: (5) 504 Natural Gas cm ,v  cm ,vI  T vm  2 p   2  dvm ,  T   vm v mI   ( T  const )   (6) where cm ,vI , vmI and  m denote the ideal molar heat capacity at constant volume and the corresponding molar volume of ideal and real gas at temperature T Real gases behave more like ideal gases as pressure approaches zero or vmI   After substituting... L.G., Brar, S.& Aziz, K (1975) ‘’Compressibility of Natural Gases, Journal of Canadian Petroleum Technology”, pp 77-80 500 Natural Gas Muskat, M (1949) “Physical Principles of Oil Production”, p 142, McGraw-Hill Book Company, Inc., New York Nutting , P.G (1930) Physical Analysis of Oil Sands, Bulletin of American Association of Petroleum Geologists, 14, 133 7 Ohirhian, P.U (2008) A New Dimensionless Friction... into the Eq (7) and after integration we obtain cm ,v  cm , pI  R  RT r 2C0  TC1  C2  , (9) with C0  C1  C2  58    2C  n 13 * n 18 C n 13 18 C n 13   * n  * n  B , K3  B  , K3 kn *  TC n   rbn 1e  cn  r ,   (10) (11) (12) Natural gas properties and flow computation 505   3 * where  r is the reduced density (  r  K  m ), B is the second virial coefficient, C... ( BB p S 1 = a  ) 1.62 1139 RL 0.270110 RL  5M 6 gd p gd p 5 M 2 M sin  p 5 2 1 6 gd p M ,x f = 2 M sin  L f z av Tav R (53) 492 Natural Gas z a v f = Gas deviation factor at the midsection of the porous medium calculated with p av T a v and f , where T av  0.5( T 1  T 2 ) f 2p1p2 and pav  p1  p2 During isothermal flow in which there is no significant variation of the gas deviation factors (z) . natural gas (natural gas that contains CO 2 as major contaminant), (Ohirhian, 2008) has expressed the compressibility of the real gas (C g ) as: p f C Κ = (22) For Nigerian (sweet) natural. natural gas (natural gas that contains CO 2 as major contaminant), (Ohirhian, 2008) has expressed the compressibility of the real gas (C g ) as: p f C Κ = (22) For Nigerian (sweet) natural. ideal gas such as air, C . g p 1  (21) (Matter et al, 1975) and ( Ohirhian, 2008) have proposed equations for the calculation of the compressibility of hydrocarbon gases. For a sweet natural