Compressible Fluid Flow Through Orifice
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^ -^ J
COMPRESSIBLE FLUID FLOWTHROUGH AN ORIFICE
byHERSCHEL NATHANIEL WALLER, JR., B.S.
A THESISIN
MATHEMATICS
Submitted to the Graduate Facultyof Texas Tech University inPartial Fulfillment ofthe Requirements forthe Degree of
MASTER OF SCIENCE
Chairmarr of the Committee
Accepted
Dean/of theIGraduate/School
May, 1973
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^ ACKNOWLEDGEMENTS
I would like to thank Dr. Wayne Ford for allottingtime to direct the writing of my thesis and for the interest he has shown in my work. I am also indebted to Dr. L.R. Hunt for consenting to serve as a member of my committee
1 1
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TABLE OF CONTENTS
ACKN0V7LEDGMENTS iiLIST OF ILLUSTRATIONS iv
I. INTRODUCTION
II. EQUATIONS OF CONTINUITYIII. EULER'S EQUATIONS
IV. THE THREE TYPES OF FLUID MOTION 12
V. ROTATIONAL MOTION AND EULER'S EQUATIONS . . 16
VI. NAVIER-STOKES EQUATIONS 20VII. BERNOULLI'S EQUATIONS 35
VIII. FLOW EQUATIONS FOR THE ORIFICE METER . . . . 41
IX. SUMMARY AND CONCLUSIONS 51
LIST OF REFERENCES 53
1 1 1
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LIST OF ILLUSTRATIONS
Figure page1. An incompressible fluid element 3
2. Forces on a fluid element 8
3. A fluid element in two-dimensional flow . . . . 124. Viscous fluid elements, (a) at rest,
and (b) in motion 215. A diagrammatic comparison of one-dimensional
(a) nonviscous flow and (b) viscous flow ina pipe 22
6. Stresses on an infinitesimal volume of aviscous fluid 23
7. An orifice type differential meter withU-tube manometer 41
IV
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CHAPTER I
INTRODUCTION
The purpose of this thesis is to show the development of the fluid flow equation used almost everywhere inthe United States to calculate the rate of flow of naturalgas through an orifice.
This purpose is accomplished in, essentially, twosteps:
(1) Starting with the most fundamental relationships, the Navier-Stokes equations for compressible fluidsare derived. These equations allow for not only the usualhydrostatic forces but also the forces due to frictionbetween adjacent fluid elements and between the fluid andits container. The various types of fluid flow are discussed, and the Euler equations are developed.
(2) Using a multitude of assumptions the Navier-Stokes equations are reduced to the fluid flow equationused in the natural gas industry to calculate flow ratethrough an orifice. Several of the assumptions are listedand discussed.
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CHAPTER II
EQUATIONS OF CONTINUITYIncompressible Fluids
Consider an element of incompressible fluid volume.Let the element be a rectangular parallelepiped with sidesdx,dy, and dz parallel to, respectively, the mutuallyperpendicular axes x, y, and z. Let the instantaneousvelocity of the fluid be V with magnitude |v|, and letthe scalar components of the velocity vector parallel to theX, y, and z axes be, respectively, V , V and V .
As seen in Figure 1, the volume of fluid enteringthe left yz face of the element is
V^dydz,
and the volume leaving the right yz face of the element is
^^x(V^ + ^-^ dx)dydz.X 9x
Therefore the net change in volume for the x-direction is
9Vir-^ dxdydz.dx
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For the y-direction, the volume change is
9V9y dxdydz
and for the z-direction is
9Vj dxdydz
9V(V^+-y|dz)dxdy
V^dydz
9V y^(V + -Z dy)dxdzy ^Y V dxdy
V dxdzy
9VX(V^+^3^x) dydz
Fig.1.An incompressible fluid element
Therefore,the total change in volume is9V 9V 9V(^ + ^rr^+ y^) dxdydz9x 9y
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This,in other words, is the net outflow volume. Thevolume of fluid leaving the element must equal the volumeentering the element because the fluid being considered isincompressible. Therefore,
9V 3V 3V
Equation (1) is called the equation of continuity for incompressible fluids.
The sum of the partial derivatives of the scalarcomponents of velocity (the left side of Equation (1)) iscalled the divergence of V, abbreviated div V. Hence, foran incompressible fluid,
div V = 0. (2)
Compressible FluidsNow suppose the fluid is compressible; that is,
volume is a function of pressure. The equation of continuity for a compressible fluid must be based not upon theconstancy of volume but upon the constancy ofmass.
Consider Figure 1 again. If p is the density ofthe fluid, then the mass of the fluid entering the left yzface of the element in time dt is
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(pV^)dydzdt,
and the mass leaving the right yz face of the element is
9(pV )(pV + _ J i _ dx)dydzdt.X dX
Therefore, the net outflow of mass in the x-direction is
9(pV^) dxdydzdt.dX
Similarly, the net outflow for the y-direction is9(pV ) ^ ^ dxdydzdt
and for the z-direction is9(PV^)^r dxdydzdt9z
The total net outflow is, then.9(pV ) 9(pV ) 9(pV^)[ ^ ^ ^ + - ^ + - ^ ] dxdydzdt
Because of this outflow, however, the mass inside theelement is reduced by
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- (||-) dxdydzdt
For the total mass to remain unchanged, the following equation must hold:
9(pV ) 9(pV ) 3(pV )9x 9y 9z ^9t' '
or30 ^(PV,,) 9(PV^) ^ 9^ JHL + L_ + 1_ + :9t 9x 9y 9zz = 0. (3)
Equation (3) is called the equation of continuity for compressible fluids. If density is constant, as for incompressible fluids. Equation (3) reduces to Equation (1).
Another way to express Equation (3) is
1^ + div (pV) = 0 . (4)d t
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CHAPTER IIIEULER'S EQUATIONS
Again consider an element of fluid volume as inFigure 1. In this case, hov/ever, consider the forces actingupon the element. If the pressure is denoted by p, asseen in Figure 2, the differential force caused by thepressure across the yz faces of the element is
- (1 | dx)dydz.
Similarly, for the xz faces, the force is
- (l 1^ dy) dxdz
and for the xy faces,
- (Jc | dz) dxdy.d Z
The vectors i, 3, and k are the unit vectors parallel tothe X-, y-, and z-axis, respectively. The total differential force F is, therefore,
^= - ( ^i ? i 5^lf )^ ^y^^ ^
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If the operator de l, V, is defined to be
8
V = ^ 9x ^ = 3y ^ ^ 3z 'then
F = - Vpdxdydz. (6)
pdxdy + (^ dz)dxdy
pdydz
pdxdz + (- dy)dxdz
pdxdz
pdydz+(-^dx)dydz
pdxdy
Fig. 2.Forces on a fluid element
Also, if the mass of the element is dm, the density, p,of the fluid is defined to be
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dmp =dxdydz
Therefor e, Equation (6) becomes
^ n d m , ,F = - Vp - ^ . (7)
Even though the fluid element may change in shape as itmoves, its mass remains constant. Hence, if external forces(such as gravity) are ignored, Newton's second law ofmotion gives
^ dV ,F = ^ dm , (8)
where t is time. Substituting Equation (8) into Equation(7),
dV . _ ^ dm_ dm - - Vp
dV ^P ^ + Vp = 0. (9)
Vel oci ty, in general, depends not only upon position (x,y,z)and time (t) but also upon initial position (x^,y^,z ) at areference time (t). Assuming a fixed initial position andtime.
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10
^ = ^ + 9 V ^ + 9 V d z . 9V ,, vdt 9x dt ^ 9y dt 9z dt ^ It ^^^^
Letting
V = ^ , V = ^ V = ^x dt' V dt' ^z dt'
^X33^ ^ \ ^ ^ ^ Z 3 7 = ^-^- (1^)
Using Equation (11), Equation (10) becomes
i =(^V)V. Il. (12)Substituting Equation (12) into Equation (9),
9^p[(V.V)V+ 1^]+ Vp = S . (13)
This is the vector form of Euler's equations of motion.Equation (13) can be separated into three scalar equationsThe first of these is
9V 9V 9V 9V .o(v ^ + V - + V - + ^r-^ ) + -^ = 0;P^ X 3x y 9y z 3z 9t ' 9x
or9V 9V 9V 9V , J.X 3x y 9y z 9z 9t p 9x
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11
If body forces (external forces, such as gravity)are considered, their vector sum, B, can be denoted asfollows:
S = Xi + Y^ + zS ,
where x, Y, and Z are the scalar components of the bodyforces per unit mass in the x-, y-, and z-direction, respectively. Equation (14) then becomes
9V 9V 9V 3VX . ,, X , X + _^Ji + 1 9p^-^ + V ^-^ + V ^-^ + ^x^ + ^ - X = 0. (15)x 3 x y 9 y z 9z 9t p9 x
The vector form of Euler's equations with body forces is
(V.V)V + | ^ + V p - g = J . (16)dt P
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CHAPTER IV
THE THREE TYPES OF FLUID MOTION
Fluid motion is of three basic types:(1) Translation(2) Rotation(3) Deformation
To see the relationships among these types, considertwo-dimensional fluid flow. Figure 3 shows a point 0(x,y)in the fluid and a point 0'(x+dx, y+dy) a distance= i 0 odr = V(dx) + (dy) from 0. The velocity at 0 is V
dy
I x# Xdx
Fig. 3 . A fluid element in two-dimensional flow
12
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13
and at O' is ^ + dV. Now, V has components V , in thex-direction, and V , in the y-direction. Therefore thecomponents of V' = V + dV are
9V 9VV' = V + ^ dx + ^ dyX X 9x 9y -and
Let
Then
9V 9VV' = V + ^ dx + ^ dy .y y 9x 9y -
a = 9V^ 9V T 3V 3V_ J i b = ^ c = i( ^ + ^ )9x ' ^ 9y ' ^ 2^ 9x ^ 3y ^'and (17)
9V 9Ve = i(2^9x 9y
= 1( ^ _ ^ ).
V' = V + adx + cdy - edyX X -^and (18)
V' = V + bdy + cdx + edxy yThe components V and V in Equations (18) are,X ytherefore, the components of strictly translational velocity.
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14
linear velocity in the x- and y-direction, respectively. IfV and V were the only components in Equations (18),^ y( a = b = c = e = 0 ) , then the rectangular fluid element inFigure 3 would remain rectangular at every point in thefield of flow; that is, the flow would be ideal parallelflow. The terms adx and bdy are expressions of thechange of velocity in the x- and y-direction, respectively;they represent the stretching rate of the edge of theelement in each direction. From Figure 3,
3V 9V^1 = 93E^ ^^^ ^2 = 97^ '
or3V 9V
Equation (19) represents the change of the angle betweenthe two edges of the rectangle at point 0. The terms a, b,and c, then, represent the deformation of the fluid elementbetween point O and point 0'.
Now assume a = b = c = 0. Then
3V 9V__J1 = ^9x 9y
so that
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15
Yi = - Y2 'or
- Yi = Y2
Therefore,
e= I (Yi - (-Yi))
e = Y-L .
Thus,
^ 9V 9V
Equation (20) is an expression of the angular velocity withwhich the rectangular element moves about an axis throughpoint O and normal to the plane of flow; that is, erepresents the rotational velocity of the element.
The terms V , V , a, b, c, and e, then,describe the three types of motion in a fluid: translation,rotation, and deformation. Equations (18) completelyexpress the relationship among these types of flow for thetwo-dimensional case.
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CHAPTER VROTATIONAL MOTION AND EULER'S EQUATIONS
As shown in Chapter IV, angular (rotational) velocity about an axis normal to the plane of flow can beexpressed as
T 9V 3V1( _jz: X2^ 3x 3y
for two-dimensional flow. If the axis normal to the planeof flow in Figure 3 is thought of as the z-axis, then theabove expression represents rotational flow about thez-axis.
For three-dimensional flow rotational velocity isrepresented by three terms; one is the above expression.The other two are
T 3V 3V1( _Ji 2 .2^9z 9x ^
for rotational velocity about the y-axis, and
9V 9V1( _ ^ y )2^ 9y 9z ^
16
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17
for rotational velocity about the x-axis.Tne vector,V ,formed from the three expressions.
1-J.9V 9V 3V 9V T 9V 9V
is called the vorticity vector.Now,since
^ 9V 3V 3V 3V _ 9V 3VVX V = i( )+ tf 1\+t( Z \ f o i \
^ 9 y 9z' ^9z 93r^+ ^alT 9^^'^ 1)the following relationshipisestablished:
V*= (VXV). (22)
Equation 21) is an expression of the curl of the velocityvector; that is.
c u r l V = V X V . 23)
From the theory developedinChapter III,ifthe flowisirrotational,
V*=|(curl V)= ;
then
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18
V X V = ? .
Euler's equations of motion, as developed in ChapterIII,consist of three scalar equations or one vector equation. The scalar form can be written as follows:
9V 9V 9V 9V T .^ v ^ + V^, + V ^ ^ + ^ = X - i ^ (24a)X dx y oy z 3z 9t p 9x
9V 9V 9V 9V T .V ^ ^ + V ^ + V ^ + ^ = Y - 1 | P (24b)x 9x y 9y z 9z 9t P 9y
^^z ^^z ^ \ ^^z 1 9 PV TT-^ + V -r^ + V ,7-^+ -^ = z - - 4^ (24c)X 3x y 3y z 9z 3t p 3z ^
Consider the left side of Equation (2 4a) ,the Euler equation for the x-direction:
3V 3V 9V 9VV ^ + V ^ + V ^ + ^X 9x y 9y ^ ^z 9z ^ 9t (25)9V 9V 9V 9V 9V
Since1 9V 9V^1(_ J L ^2^9x 9y
represents rotational velocity about the z-axis, and
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19
T 9V 9V1 / X z X2^9z ~ 9x ^
represents rotational velocity about the y-axis, the middletwo terms on the right side of Equation (25) have coefficients that are merely twice the rotational velocitiesabout the axes perpendicular to the direction of flow. Theleft sides of Equations (24b) and (24c) can be writtensimilarly.
Therefore, for irrotational flow. Equation (24a)reduces to
3V1 9 , 2 2 2, X 1 9p._ ^ (V + V + V ) + TTT^ = X - - ^ .2 9x ^ X y z' 9t p 9x
Similarly, Equations (24b) and (24c) become, respectively.3V1 3 f2 , v2 2 y - Y - ^ -P2 9? ^^x*^y ^ ^z^ ^ 9t ^ P 9y
and3V
2 9z X y z 9t P 9z
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CHAPTER VI
NAVIER-STOKES EQUATIONSIn the preceding derivations friction forces have
been ignored. Friction between one fluid element andanother and between the fluid and its container must beconsidered if a truly general fluid flow equation is to bedeveloped.
That property of a real fluid which causes shearing(friction) forces is called viscosity. A fluid whose flowis affected by viscosity is called a viscous fluid.
Incompressible FluidsConsider, first, simple parallel flow of a viscous
incompressible fluid, illustrated in Figure 4.In Figure 4(a) the fluid is at rest, fluid element
E^ lies atop fluid element E^, and viscosity has noeffect. In Figure 4(b ), however, the fluid is in motion.As shown, E, has scalar velocity v, and E2 has scalarvelocity v + dv. The friction force (or shear stress) perunit area, T , is defined as follows:
dv I ^c\T = y ^ , (26)
20
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(a)
(b)
Fig. 5. A diagrammatic comparisonof one-dimensional (a) nonviscous flowand (b) viscous flow in a pipe.
22
Consider Figure 6, which illustrates the three
dimensional case of viscous incompressible fluid flow. Thefigure shows both normal, or direct, stresses (i. e.,stresses due to pressure) and shear stresses (i. e.,stresses due to friction) that affect a parallelepiped ofinfinitesimal volume dxdydz. Note that, in viscous fluids,even the normal stresses are dependent upon the orientationof the axes, as shown by the subscripts x, y, and z on p.
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24
7 P T T/ x xy xzT P Tyx ^y yz
W_.. T Pzx zy ^z
However, to avoid rotation of the infinitesimal element.
^xy = V x ' ^xz = ^zx' ^ ' ^yz = ^zy
The differential force in the positive x-direction is formulated as follows:
9p^ x = [P x- (P, - 33^dx)]dydz
9T+ [T - (T ^ dy)]dxdzyx yx 9y -9T+ [T - (T ^ dz)]dxdyzx zx 9z
9p 9T 9TF = (_ii + - ^ +_2X)(jxdydz (27a)^x 9x 9y 9z ^Similarly, for the y- and z-direction,
9p 9T 9TF = (-1Z + - ^ + - ^ ) dxdydz (27b)y 9y 9z 9xand
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25
9p 9T 9T^z = ^JT* - ^ - - l i Ay^z. (27c)
The relationships between shear and normal stresses willnow be developed.
As already mentioned, in a viscous fluid the normaland shear stresses depend upon the orientation of the coordinateaxes. The stress system can be divided into thehydrostatic pressure p and any additional normal andtangential stresses that cause only deformation of the fluidby the action of viscosity.
For plane flow, three terms with coefficients a, b,and c, as defined in Equations (17), characterize the rateof deformation of a fluid element. The coefficient crepresents half the angular rate of deformation betv/een thetwo edges of the plane rectangular fluid element. Theserates of deformation must be proportional to the extranormal and tangential stresses; the constant of proportionality is 2y, where the 2 is required for agreementbetween Equation (26) and Equations (18) . Therefore, forthe three-dimensional case, the additional normal stressescaused by the action of viscosity are
9V 9V , 9V^Px = ^^-^ Py = 2 ^ - ^ ' ^^^ P- ^ ' ^ '
so that the total normal stresses are
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and
26
9VPx = -P ^ Px = -P - 2 y - ^ , (28a)
9VPy = -P + Py = -P + 2 y - ^ , (28b)9VP^ = -P + p; = -P + 2y-3f . (28c)
The hydrostatic pressure term p has a negative signbecause p , p , and p were assumed to be positiveX y z ^outward.
The shear stresses are related to the velocity ofangular deformation (cf. Equations (17)) as follows:
3V 3Vxy yx ^ 3x 9y
^^z ^^xT = T = y ( ^ + - ^ ) ; (29b)xz zx ^ 3x 9z3V 3V
T = T = y (-^ + -5^) . (29c)yz zy 9z dySubstituting Equations (28a), (29a),and (29b) into Equation (27a),
9V . 9V 9V
3 ^^z ^^X
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27
rearranging.
3 s \ d \ d \9x 9y 9z
. 9V 3V 3V
If F^ is defined to be the force per unit volume in thex-direction,
2 2 2r 9 V^ 9 V 9 V
F' = - ^ + u( - + - + - )X dx ^^ ^ 2 ^ 2 ^ ^ 2 ^dx 3y 3z
(30a)^ 9 ,^^x ^ ^ ^ ^ ^^z ,
Corresponding substitutions give the following expressionsfor force per unit volume in the y- and z-direction:
2 2 23 V 9 V 3 Vy ^y 3x2 9y2 3z2
^ ^ 9?^~93F ^ - 3 ? ^ 3 ^ '(30b)
and2 2 23 V 3 V 3 V^
= ^ ^ 3x 2 3^2 3 ^ ^( 3 0 c )
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28
However, for incompressible fluids, the equation of continuity states that
9V3V 3Vii +- J :4- ^ = 0.3x3y 3zTherefore, Equations (30) become
and
22:?cs 3 V 3 V 3 V^x 33F ^ ^ 2 ^ T + ^ ^'(31a)9x3y 3z2 2 Pa^3 V3 V 3 Vy=-|f ^ ( i ^ ~ ^ - i ) '(31b)^ 3x 3y dz22 ?:r. 9 V^ 9 V3V
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299V 3V 3V 3V T ^^V - ^ + V - ^ + V - ^ + - ^ = Z - ^ | ^ . (32c)X 3x y 9y z 3z 3t p 3z
Substituting F', F', and F' into Euler's equations for- -^f - ^ / and - ^ , respectively, gives
3V 3V 3V 3VV -ii + V -rrii + V -;rii + ^X 3x y 3y z 3z 3t2 2 2
n c. 9 V 9 V^ 9 V^P ^^ P 3x2 3^2 3^2
3V 3V 3V 3VV - + V r ^ + V T^ + r^X dx y 3y z 3z 3t2 2 2, , 3 V 3 V 3 VP 9y p .^2 . 2 2
(33a)
(33b)
3x 3y 3zand
3V 3V 3V 3V^ x ^ ^ ^ ^ z ^ -Tt
(33c)9 2 2n . 9' V 9' V 9''VP 9z p 3^2 3^2 g zi
Equations (33) are the scalar Navier-Stokes equations forincompressible fluids. In vector notation. Equations (33)become
(^.V)^+ ll = 6 -ivp+ ^V . (34)
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30
Since
i l = ^ - v ) ^ . I l,Equation (34) can also be stated as follows
|V = g _ 1 y^2^dt p ^ p (35)
Compressible FluidsTo obtain the Navier-Stokes equations for compres
sible fluids. Equations (33) must be modified slightly. Aterm proportional to
3V 3V 3V3x 3y 3z
must be added to Equations (2 8).Let e be the constant of proportionality. Then
Equations (2 8)become3V 3V 3V 3Vp.= - P + 2y + e ( - ^ + ^ + ^ ) , (36a)X 3x ^ 3x 3y 3z3V 3V 3V 3VPy = - P ^ 2 y ^ . e ( - ^ + - 3 ^+ ^ , (36b)
and 9V 3V^ 3V 3VPz = - P - 2 y - ^ . e ( - 3 | . ^ . ^ ) . (36c)
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31
Summing Equations (36),
3V 3V 3VPx ^ Py ^ Pz = -^P ^ (2y f 30) ( ^ + ^ + ^ ) . (37)
Now,if the fluid were incompressible. Equation (37) wouldbe
Px + Py + Pz = -3P (38)
by Equation (1) . Assuming Equation (3 8)holds for compressible fluids. Equation (37) becomes
3V 3V 3V3p = -3p + (2y + 39)( ^ '3x 3y 3zSolving for G,
e = - I y . (39)
Substituting this result into Equations (36),3V ^ ...3V 3V 3V^
3V o 9V 9V 9V^Py = -P + 2p 33^ - I y ( ^ + a / + g /), (40b)and
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329V 3V 3V^ 3V
Since Equations (29) are unaffected by compressibility,substituting Equations (40a), (29a),and (29b) into Equation(27a) gives
a 9V 3V^ 3V 3V
g 8V 3V 3V 8V^87(^(3^+5 / ^3^(v''33r+ a/))]dxdydz;or, the force per unit volume, F', is
; 3V^ -^ 3V 3V 3V
. 9V, 3V^ . 3V 3V^^('^(a^^-a^r" ^ 3 '^'
Simplifying,2 2 2
Sir. 3 V 3 V 3''vF'= - lE + u( - + ^ + ^ )^x 3 x ^ ^ . 2 ^ , 2 ^ ^ 2 ^dx 3y 3z, ^ 9V 9V 9V3 ^ 9x ^9x 9y 9z ^*
(41a)
Corresponding substitutions give the following expressionsfor force per unit volume in the y- and z-direction:
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33
a n d
2 2 29 V 9 V 9' VF ' = - | P + u ( ^ + Z + Z )y ^y dx^ 9 y 2 3 z 2T . 3V 3V 3V3 * 3 y ^ 3 x 3 y 3 z ^
2 2 2. 3 V 3 V 9 VT n ' o p , , Z , Z , Z F _ = - ^ + y ( 2 ^ 2 * 5 ^= ^^ 3x 3y 3 z ^T ' 9V 9V 9V^+ 1 u - ^ (i^ + Z + ^ )^ 3 ^ 9 z ^ 9 x 9y ^ 9 z ^ '
( 4 1 b )
( 4 1 c )
S u b s t i t u t i n g F ' , F , a n d F ' i n t o E u l e r ' s e q u a t i o n s^ X y z(Equations (32)) for - | ^ , - g ^ , and - ^, respectively,gives
3V 3V 3V 3V\ - ^ ^ \ ^ * z ^ - ^
2 2 2-. . 9' V 9' V 9^V^P ^^ P 3x^ 3y^ 3z^
^ 3V 3V 3V^1 y _3_ ,_jc + __Z + ^ ), 3 ^ 9x ^ 9x 9y 9z^'
3V 3V 3V 3VX dx y 9y z 9z 9t
9 2 2n . 9 V 9 X ^ vP ^y P 3x2 3 2 3 2
3V 3V 3V^_ 1 y 3 / X , _y_ . E )^ I ^ 3y ^~3^ 9y 9z''
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34
and
9V 3V 3V 3VV ^ + V - +V + -X dx y 3y z 3z 3t2 2 21 : r. n 3 V^ 3^V^ 3^V
^ 3x 3y 3z
3 p 3z ^ 3x 3y 3z^*Equations (42) are the scalar Navier-Stokes equations forcompressible fluids. In vector notation. Equations (42)become
(v-v)v. Il = g - i vp . v^^ . i v ( ^. % . )
But, using the definition of the divergence of V,
t ^ . V ) ^ + |V g _ 1 ^p Ev2^+ I Hv(divV). (43)
Again,sincei - v)v I l -
E q u a t i o n (4 3) c a n a l s o b e s t a t e d a s f o l l o w s :
dV ^ g _ 1 ^ + y v^V + i ^ V ( d iv V) . (44 )d t p ^ p 3 p
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CHAPTER VIIBERNOULLI'S EQUATION
The preceding chapters have shown the development ofincreasingly more general equations describing fluid flow.This and subsequent chapters will show how a multitude ofassumptions are used to reduce the general equations to anequation frequently used in industry to calculate the rateof flow of natural gas.
To obtain the first relationship, Bernoulli's equation,irrotational flow will be assumed; that is,
V x ^ = 5, (45)
as discussed in Chapter V. Moreover, the identity
V(v-V) = 2V.VV + 2V X (V X V) (46)
will be utilized (see [3], p. 313).Now,the relationship
/ V.d? = 0, (47)
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36
where the symbol }>denotes the line integral around anycclosed curve C and dr is the infinitesimal vector(dx)1 + (dy)] + (dz)k, can be shown to hold whenever Equation (45)does. This is a result of Stokes's theorem (see[3] ,p. 295). A trivial consequence of Equation (47) ,asproved in [6], pp. 264-265, is that
j V-dr
i s i n d e p e n d e n t o f t h e p a t h t a k e n fro m p o i n t P^. t o p o i n t
I n d e p e n d e n c e o f t h e p a t h o f i n t e g r a t i o n i m p l i e s t h a tV ' d r c a n b e w r i t t e n a s t h e d i f f e r e n t i a l o f a s c a l a rf u n c t i o n (^ ; t h a t i s ,
V - d r = d . (48 )
Another v/ay of expressing this is as followsP1 ^ .../ V-dr =
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V,.d-r = |i dx . |i dy . |i d.
or
V(j)-dr = d(J) . (50)
Subtracting Equation (50) from Equation (48 ),
V-dr - V(t).dr = 0 ;
therefore
(V -V(|)).dr = 0 .
This implies that the vector in parentheses is orthogonalto the vector dr. Bu t, since dr is arbitrary,
V - V ( t ) = 5 ,
or
V = V(j) . (51)
Equation (46) can now be greatly simplified using theassumptions and Equation (51 ). By Equation (45), Equation
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(46) becomes
V(^-V) = 2V-VV . (52)
But,the identity
V = 1^1 = . ) /2
further reduces Equation (52) to
^V^= V(^ V^) . (53)
Now E q u a t i o n (5 3) c a n b e s u b s t i t u t e d i n t o t h e v e c t o r fo rmo f E u l e r ' s e q u a t i o n s . E q u a t i o n ( 1 3 ) , t o o b t a i n
p V (i V^) + p | ^ + Vp = ^ ,
o rV (l v 2 ) + | i + ^ = t i . (54)2 d t p
Substituting Equation (51) into the above relationship gives
V(|v2) ^ ^ ( V * ) + ^ = i 5 ;
that is.
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3 9
V ( | v2 + | 1 + E) = ^ . 55)
S i n c e the q u a n t i t y in p a r e n t h e s e s i s e v i d e n t l y i n d e p e n d e n to f p o s i t i o n , i t m u s t be a f u n c t i o n of t i m e o n l y . T h e r e f o r e ,
1 V^+ II- + / = f(t) . (56)^
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g = acceleration of gravity andz = elevation above some datum.
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CHAPTER VIIIFLOW EQUATIONS FOR THE ORIFICE METER
Further understanding is best served at this time bydescribing the setup for the standard orifice type differential meter used in the natural gas industry.
Figure 7 is a cut-away schematic drawing of anorifice type differential meter in which a manometric liquidis used to measure differential pressure. The fluid to be
Directionof flow
TD
Fig.7.Anorifice typedifferential meter with U-tube manometer
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42
measured, flowing from left to right, is partially obstructedby a metal plate. A, in which a concentrically-located holehas been bored. The purpose of this metal plate, called anorifice plate, is to produce a pressure drop. The greaterpressure is sensed at location 1, called the upstreampressure tap; the lower pressure is sensed at location 2,called the downstream pressure tap. Because of their locations,the particular pressure taps in Figure 7 are calledflangetaps. The upstream and downstream pressures arerelayed to a U-tube manometer, B, filled with mercury orsome other suitable manometric liquid.
The method of transformation of Equation (58) intoa form that utilizes data from the orifice meter to obtaina flow rate will now be outlined. For details, see [5],pp. 51-52,and [1], pp.78-79.
Assume that density, p, is constant. Then, forpressure tap locations 1 and 2,
2 2^1 Pi ^2 P2^ + z , + - i = ^ + z ^ + , (59)2g 1 Y 2g 2 y
where y = P^ ^^ specific weight. For horizontal pipe,rearrangement of Equation (59) gives
.2 2 . Pi P2v; - Vt =2g(--^ ^) . (60)2 1 ^ Y
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V ^ = ( ^ V , ) ^ ,
D = inside diameter of the pipe andd = diameter of the orifice.
V2 = ^ 1/2 (2g( ^ ^ ^ ) ) ^ ^ ^ (61) jCI - 4 ;
D
tal constant C, called the coefficient of discharge, is
V2 = 5-^ 72(2g(^L_^)) V2^ (^2)(1 - ^ )D
the resulting quotient j - ^ ^^ renamed K.(1 - 4Dation (62) then becomes
Pi - P2 1/2V2 =K(2g(-i^ -^))^/^
, since the quantity rate of flow, Q, through the
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44
is theproductof thevelocityof thefluidand the A, of theorifice.
Pi P2 1/2Q=KA(2g(-i ^ ) ) ^ / ^ . (63)
ation (63) is a form of the so-called hydraulic
Units will nowbeassignedto thequantitiesin
et '1Q=fluid flow rateat theaverage specific weight, 'Y,incubic feetpersecond;
A=orifice areainsquare feet;g=accelerationofgravityinfeetpersecondper
second;andK= =-y ,correspondingto theconditionof
(1- 4D
measurement.Pi~ P9 = is thedifferential head, h, of
the flowing fluidinfeetat theaverage specific weightttheorifice. Equation (63) therefore becomes
Q=KA(2gh) /2^ (64)
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45
re each quantity has the units designated above. Equa(64) is very unhandy to use practically; therefore, it
Qh = ^ (Vf'^^^' < >
\IQ, = hourly fluid flow rate at stated base conditions Iof temperature and pressure, I
h^ = differential pressure across the orifice in inches of water column, i
P_ = absolute static pressure in pounds per squareinch (psia) at a designated tap location, and
C' = orifice flow constant.To change Equation (6 4)into the practical form.
(65), several substitutions must be made:
g = 32.17 ft/sec^; (66a)
h Yh = - ^ ^ , (66b)12Y
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46
^w ~ 62.37 lb/ft = specific weight of water at60 F., and
Y = actual specific weight of the natural gas inpounds per cubic foot at flowing conditions;
A = - ^ , (66c)4(144)where d = diameter of the orifice in inches; and
PY = 0.08073 ^ i|^ G , (66d)14.7 - f
where0.08073 = specific weight of dry air at 14.7 psiaand 32F.,T^ = flowing temperature of the natural gas in
degrees Rankine (R.),andG = specific gravity of the flowing gas, where the
specific gravity of dry air is taken to be 1.000
Substituting Equations (66a) through (66d) into Equation(64) ,
,2 h (62.37)(14.7)T .Q = K( ^ )(2(32.17) { ^ ^))^^ . (67)
4(144) 12(0.08073)P^(49 2)G
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48
2 ^b 1 1/2 ^ 1/2C = 218.44 d' K p^ (^) (|) , (71)b f
re Z = compressibility factor at T^ and P-.To make computations easier. Equation (71) is, in
tice, subdivided into factors, as detailed in [1]. The
(1) Basic orifice factor, F, .
2 ' b 1 ^/2^b =218.44 A \ ^ (^) , (72a)D fe K^ is found from a set of empirical equations. The
T, = 520*'R. ,b 'T^ = 520R.,P^ = 14.7 psia, andDG = 1.000
are assumed. Equation (72a) then becomesF^ = 338.17 d^KQ. (72b)
(2) Reynolds number factor, F^
F = 1 + B , (72c)r ,^ ^ ,1/2 '(Vf)
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49
ere b is calculated from a set of empirical equationspurpose of F is to allow for the difference between used to calculate F, in Equation (72b),and K,
in Equation (71).3) Expansion factor, Y. This factor allows for the change
fic weight of the gas across the orifice plate.
4) Pressure base factor, F , .
F ^ = M ^ , (72d)Pb P Ke P^ is the desired pressure base.
(5) Temperature base factor, F^j^.
F = -A- (72e)^tb 520 '
e T, is the desired temperature base.
(6) Flowing temperature factor, F^^
F = ( i 2 0 / / \ (72f)*tf T^'
(7) Specific gravity factor, FgF = ( ) . (72g)g G
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(8) Supercompressibility factor,FPV11/2
Usingthesymbolismoftheeight factors above. Equation(71)becomes
C-=F^F^YFp^F^j^F^^F^Fp^. (73)
Thentheflow rateincubic feetperhour, Q^, at T, andh b
Pj_ , is calculated using the follov/ing equation:
Qh=V r ^ V ^ t b^ t f V p v ' V f ) ' ^ - (74)
Three additional factors,not inuniversalusein thenaturalgasindustry,arealso developedin[1]. Thesefactorsare(1) themanometer factor,F ;(2)thelocationfactor,F.; and(3)theorifice thermal expansion factor,
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CHAPTER IXSUMMARY AND CONCLUSIONS
Starting with the most fundamental relationships, theNavier-S tokes equat ions, which allow for friction, weredevel oped; several assumptions were then made to reducethese very complicated partial differential equations to aform used in the calculation of gas flow across an orificeplate. The assumptions, several of which are cited anddiscussed in [5 ], pp.52-55, are listed below:
(1) The gas flow is irrotational. See Equation (45) .(2) Friction does not affect fluid flow. That is,thevelocity of the fluid is the same at all points across thediameter of the pi pe , and no energy is lost as the gaspasses through the orifice. This assumption results fromthe use of Euler's equations in the derivation of thehydraulic equation.(3) Fluid flow velocity is not time dependent. See Equation (57) .(4) Gravity is the only body force.(5) Compressible fluid flow across an orifice is incompressible; that i s, the specific weight of the fluid does not
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52
change as it passes through the orifice. This assumptionis required to obtain Bernoulli's equation.(6) The velocity at the upstream pressure tap is related tothe velocity at the downstream pressure tap as the orificearea is related to the cross-sectional area of the pipe.This assumption is made to derive the hydraulic equation.(7) Suction or impact effects at the pressure taps are nil.(8) The acceleration of gravity is 32.17 feet per secondper second.
All of these assumptions are at least partiallyincorrect. However, as discussed in[5] ,the effects ofthe assumptions are either negligible or are corrected bythe construction of the piping upstream and downstream ofthe orifice plate or by factors in Equation (74).
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LIST OF REFERENCES1. American Gas Association. Gas Measurement CommitteeReport No. 3. Orifice Metering of Natural Gas. NewYork: American Gas Association, 1969.2. Aris, Rutherford. Vectors, Tensors, and the BasicEquations of Fluid Mechanics. Englewood Cliffs,New Jersey: Prentice-Hall, Inc., 1962.3. Hildebrand, Francis B. Advanced Calculus for Applications. Englewood Cliffs, New Jersey: Prentice-
Hall,Inc., 1962.4. Kaufmann, Walther. Fluid Mechanics. New York: McGraw-Hill Book Company, Inc., 1963.5. Spink, L. K. Principles and Practice of Flow MeterEngineering, 8th Ed. Norwood, Massachusetts: TheFoxboro Company,1958.6. Wrede, Robert C. Vector and Tensor Analysis. NewYork: John Wiley & Sons,1963.
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