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    NATURAL GAS MEASURMENT (CONT’D)

     The method of measuring liquid or gas by

     volume is referred to as volumetricmeasurement.

     Producers account for gas in units of mcf(1000 cu.ft)

     The total mass of a substance in a cu ft of gasdepends partly on its absolute pressure (psia)and partly on temperature (oR).

     The American Petroleum Institute (API) and

     American Gas Association (AGA), since 1967,have used 14.73 psia and 60of (520oR) as theirstandard (base) conditions.

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    ORIFICE METER

     This is a type of differential pressure method of meteringgas

     The orifice meters compound of two elements

    •  Primary Element

    • Secondary Element

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    PRIMARY ELEMENT

     The primary element is the differential pressure-

    producing device. It composed of the followingparts:

     The meter tube – a length of special pipe through which the gas flows.

     The orifice plate holding and positioning device-an orifice flange or an orifice filting installed as anintegral part of the meter tube to hold the orificeplate in a position perpendicular and concentric to

    the flow of gas. The orifice plate- a flat circular plate with acentrally bored, sharp-edged orifice machined toan exact.

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    Predetermined dimension that forms acalibrated restriction to the flow of gas throughthe meter tube and is the source of thedifferential.

     Pressure taps – Precisely located holesthrough the pipe walls or orifice plate may bemeasured.

     Straightening vanes – a device that may beinserted in the upstream section of the metertube swirling in the gas stream.

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    SECONDARY ELEMENT

     The secondary element is called the differential

    gauge and is connected to the upstream anddownstream pressure taps of the primaryelement.

     It measures the differential pressure and static

    pressure which are recorded on a circular chart.

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    BASIC ORIFICE METER EQUATION

    From the general energy conservation equation

    for any points in the meter tube

    ∫ ∫ ∫    −−=++2

    1

    2

    1

    2

    1lwwdz 

     gc

     g 

     gc

    udvVdp

     Where,

     v = sepcific volume = cu ft/ibmp = pressure, ibf/sq ftu = average linear – flow velocity ft/secgc = conversion factor: 32.17 (ibm/ibf) (ft/sec

    2)

    g = acceleration due to gravity ft/sec2z = vertical distance above datum, ft

     w = work done by flowing fluid, ft ibf/ibmLw = work energy lost due to frictional effect, ft-

    ibf/ibm.

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    BASIC ORIFICE METER EQUATION (CONT’D) From the assumption, dx = o, w = o

    O gcduU Vdp   =+∫ ∫ 

    2

    1

    2

    1

     From the above equation, it can be shown that,

    ∆   

      

     =

     z T 

    hp DC 

     P 

    T q

     g b

    b

    γ  β 41

    59.2182

    2

     Therefore,

     f  w g    phC q  1=

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    GENERALIZED ORIFICE FLOW EQUITATION

     The generalized orifice flow equation is given as:

      f  wh   P hC q

      1=

      f  wT h

     Where,qg is the gas flow rate, scf/hr

    h w is the differential pressure (in inches of water @ 60o

    f)pf is the static pressure (in psia)

    CI is the orifice flow constant

    Is the pressure extension.

     This constant is a function of several other orificeconstants given by:

    CI = (F b) (Fr) (Y) (Fpb) (Ftb) (Fg) (Fpv) (Fm) (Ft) (Fa)

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    GENERALIZED ORIFICE FLOW EQUATION (CONT’D)

    Where

    F b = Basic orifice factor

    Fr = Reynolds number factor (viscosity)

    Y = Expansion factor

    Fpb = Pressure base factor (14.73) contracted pressure base

    Ftb = Temperature bases factor (T b/520)

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    Ftf= Flowing temperature factor

    Fg= Specitic gravity facator,

    Fpv = Super compressibility factor

    Fm = Manometer factor for me-cury meter.

    Ft= Guage location factor

    Fa = Orifice thermal expansion factor

    = 1 + [0.000,018,5 (of – 68)] 304 and 316 stainless steel

     =1 + [0.000, 015, 9 (of – 68)] monel

    etemperatur  flowactual +400/520

        

      

     g γ   1

        

      z 

     Zb

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    QUESTION ON METERING Given an orifice meter equipped with flange taps withstatic pressure from the down stream taps and the

    following:D1= line size = 8.071 inches of actual 1D

    D2 = orifice size = 1.000 inch

    Flowing temperature = 65of

    Ambient temperature = 70

    o

    fP b = contract pressure base = 14.65 psia

    Temperature base = 50of = 510.oRγg = Specific gravity = 0.570

    H w

     = total heating value = 999.1Btu cu.ft

    Xn = mole fraction nitrogen content = 0.011

    Xc = mole fraction of carbon (iv) oxide content = 0.000

    H w average differential head = 50 inches water

    Pf= average downstream gauge pressure = 370 psia Calculate the orifice flow constant and the uantit of

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    0008.1

    1239.01301.0/int

    ),(25.

    0005.149000.01

    64.138

    0680.011

    0680.0

    1071.81,23.

    38.200

    1071.81,22.

    64.13865.1437050

    1239.0071.8

    000.1

    2

    1

    2

    =

    ==

    =+=

    +=+=

    =

    −−

    =

    −−

    =+=

    ===

    γ  

    β 

    γ  

    β 

    and  ph for erpolating by

     pressure staticdownstream for  Table !rom

     ph

    b ! 

    b

    line Dinanin plateina for  Table !rom

     ! 

    line Dmanin plateina for  table !rom

     " ph

     D

     D

     f  w

     f  w

    b

     f  w

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    )()()()()()()()()()()(

    tan

    9993.0

    )36.(int

    ,570.070,370

    054.1),(35.

    1,35.

    3245.1

    570.0,34.

    9952.0

    65,33.

    9808.0

    50,32.

    0055.1

    65.14,31.

    1

    at m pv g tf  tb pbr b

    m

     g 

    o

     f  

     pr 

     pv

     g 

    tb

    o

    tb

    o

     pb

    b

     !  !  !  !  !  !  !  ! #  !  ! C 

    ist cons floworificetheThen

     ! 

     Tableerpolationby

    and  f  etemperatur ambient  psig  P  !or 

     ! a Table !rom

     !  Table !rom

     ! 

     gravity specific for  Table !rom

     ! 

     f  etemperatur  flowing  for  Table !rom

     ! 

     f  baseetemperatur  for  Table !rom

     f  

     P  for  Table !rom

    =

    =

    ===

    =

    ==

    ==

    ==

    =

    ==

    γ  

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    hr  $scf  

    hr  ft 

     "

     phC qh

    hour one for  flowof  ratetheThrn "

     f  w

    /05.37

    /052,37

    4.3845025.267

    ,25.267)9993.0()0254.1()3245.1()9952.0(

    )9808.0)0055.1()0008.1()0005.1()38.200(

    3

    ===

    =

    =

    =

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    CHART

     Two principal types of chart are:

     Uniform scale direct – reading chart for differential

    pressure – line are spaced on equal distance

     Square – root chart – reads the square root of the

    percentage of the full scale range of the gauge.

    DIRECT – READING CHART

     Circular charts for recordity the differential andstatic pressure guages are usually 12 in. in

    diameter

     Scale ranges for fuel gas metering are given below.

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    Common differential pressure ranges, in.

    of water

    Common static pressure ranges,

    psig

    0 to 10

    0 to 20

    0 to 50

    0 to 100

    0 to 200

    0 to 100

    0 to 250

    0 to 500

    0 to 1000

    0 to 2500

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    SQUARE – ROOT CHART For this type of chart, a chart factor (CF) may bedefined as: 

    100

     $%C!  =

     Actual pressure

    rangemeter  "C%

    C!  "C%

    2

    2

    10

    )(

         =

    =

     Where, MR = Meter Range  CR = Chart Reading

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    EXAMPLE ON CHART Given that for a 50in by 100 Ib gauge, the differential pressuregauge Rh = 50 in and static pressure; Rp = 100 psi. assume square

    root chart readings; differential = 7.2 in and static = 9.4 psi.Calculate the actual differential and static pressure.

    SOLUTION 

     psia "

     % "reading chart 

     pf  

     pressure&tatic

    water in "

     % "reading chart 

    h

     pressureal  Differenti

     p

    hw

    36.8810010

    4.9

    10

    .92.255019

    2.7

    10

    ,

    2

    2

    2

    2

    =   

      =

       

      =

    =   

      =

      

     

     

     =

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     VENTURI METER

     The venturi tube operates on exactly the same

    principle as the orifice meter. Discharge coefficients of venturi are larger than

    those for orifices and vary from about 0.94 to

    0.99. It gives a definite improvement in power losses

    over an orifice and is often indicated for

    measuring very large flow rates where power

    losses can become economically significant.

     The initial higher cost of a venturi over an orifice

    may thus be offset by reduced operating costs.

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    INSTRUMENT CALIBRATION Calibration entails the adjustment of a measurementdevice so that the value from the measurement device

    agrees with the value from a standard. The international science organization (ISO) hasdeveloped a number of standards specially directed tocalibration of measurement devices. Within most companies, the responsibility for

    calibration measurement device is delegated to a sspecific department.  The frequency of calibration is normallypredetermined but earlier action may be dictated if

    the values from the measurement device becomesuspected. Calibration of some measurement devices involvescomparing the measured value with the value fromthe working standard.

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    THANK YOU