unit-4 gas flow fundamentals
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UNIT-5
GASFLOWFUNDAMENTALS1
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FLOWEQUATION
Flow of natural gas and accompanying liquidsthrough gathering systems, process equipment andtransmission pipelines requires pressure drop as thedriving force.
All fluid flow equations are derived from a basicenergy balance which for a steady state system canbe given as:
Change in internal energy + change in kinetic energychange in potential energy+workdone on the fluid
+ heat energy added to the fluidshaft work done byfluid on surroundings = 0. ----------- (1)
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dU + dv2/2gc+ g/gcdz + d(pv)+ dQ-dws=0 --- (2)
U= internal energy ft-lb f/lbm
v= fluid velocity, ft/sec
z= elevation above a given datum plane , ft
p= pressure , lbf/ft2
V= volume of a unit mass of the fluid,ft
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/lbmQ= heat energy added to the fluid ft-lbf/lbm
ws= shaft work done by the fluid on the surroundings.
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Equation (2) can be converted to a mechanical energybalance using the following thermodynamic relations:
du + d(pv) = dh = Tds + Vdp ---------(3)
h= specific fluid enthalpy, ft-lbf/lbmT= temperature ,oR
s= specific fluid entropy, ft-lbf/lbmnow equation (2) becomes
Tds + Vdp + dv2/2gc+ g/gcdz + dQ -dws=0 ---------(4)For an ideal process ds=-dQ/T ----------(5)
Since no process is ideal
ds -dQ/T -----------(6)
Tds = -dQ + dlw ------ --------(7)lw= lost work due to irreversibilities due to friction.
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On further substitution to equation (4)
-dQ + dlw+ Vdp + dv2/2gc+ g/gcdz + dQ -dws=0(8)
Neglecting the shaft work wsand multiplying throughout
by density
dp + dv2/2gc+ g/gc dz + dlw=0 ---------- (9 )
And it can also be written as
p + v2/2gc+ g/g
c z + p
f=0 --------(10)
pfpressure drop due to friction= fv2l /2gcd
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FRICTIONINPIPE
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The term lwrepresents all energy losses resulting from
Irreversibilities of the flowing stream.
Friction lossesInternal losses due to viscosity effects
Losses due to roughness of the wall of the pipe
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FRICTIONFACTOR
Friction factor f,defined as the ratio of the wallshear stress and the kinetic energy per unitvolume and is used in computing the magnitudeof the pressure drop due to friction.
f=w/(v2
/2gc)For steady state flow in a uniform circular conduit
such as pipe this results in well known Fanningequation: pf= 2fLv
2/gc d
d is the inside pipe diameter.Friction factor f is called the Fanning frictionfactor.
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Usually , the moody friction factor equal to 4f is used.
In terms of the Moody friction factor f, the Fanning
equation becomes: pf= fL v2/2gc d
Moody friction factor is therefore a function of Reynolds
number and relative roughness.
f=f(NRe,//d)
hfs=4fLv2/2gcd
Head loss in terms of height of liquid flowing
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REYNOLDSNUMBER
It is named after Osborne Reynolds
(1842-1912), who proposed it in
1883.
= Inertia forces /viscous forces
NRe= dv/
d= inside diameter of the conduit through which the fluid is
moving
v= velocity of the fluid =density of the fluid
=viscosity of the fluid
http://en.wikipedia.org/wiki/Image:Oreynolds.jpg -
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Thus, it is used to identify different flow regimes, such as
laminar or turbulent flow.
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Laminar flow
Re < 2000
'low' velocity
Dye does not mix with water
Fluid particles move in straight lines
Simple mathematical analysis possibleRare in practice in water systems.
Transitional flow
2000 > Re < 4000
'medium' velocity
Dye stream wavers in water - mixes slightly.
Turbulent flowRe > 4000
'high' velocity
Dye mixes rapidly and completely
Particle paths completely irregular
Average motion is in the direction of the flow
Cannot be seen by the naked eyeChanges/fluctuations are very difficult to detect. Must use laser.
Mathematical analysis very difficult - so experimental measures are used
Most common type of flow.
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For cross sections other than circular, an equivalent diameter ,de,
defined as four times the hydraulic radius Rhis used instead of d.
de= 4 Rh= 4( area of flow /wetted perimeter)
E.g. 1
E.g. 2
deof annulus of inside dia diand
outside dia do=
dePipe of circular cross-section of
dia d=
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The units for parameters in the Reynolds number should
be consistent, so that a dimensionless number is
obtained.
d - ft
v - ft/sec
- lb m/ft3
- lb m /secFor all practical purposes the Reynolds number for
natural gas flow problems may be expressed as
NRe=20q g/ d
q = MMscfd
= cp
g = gas gravity
d = inches
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PIPEROUGHNESS
Absolute roughness: the absoluteroughness e of a pipe wall is defined as themean protruding height of relativelyuniformly distributed and sized, tightlypacked sand grains that would give thesame pressure gradient behavior as theactual pipe wall.
Relative roughness: it is the ratio of theabsolute roughness to the diameter of thepipe.
Relative roughness= /d15
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LAMINARSINGLEPHASEFLOW
The pressure drop for laminar flow is given by HagenPoiseuille relationship as follows:
p=32vl/gcd2
fv2 l /2gcd = 32 v l /gcd2
f= 32 *2/d v
= 64 /d v
f =64/NRe
Thus the friction factor is independent of pipe roughness
in the laminar flow regime.
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PARTIALLYTURBULENT& FULLYTURBULENT
SINGLEPHASEFLOW
For partially turbulent flow, friction factor is a
function of both Reynolds number and pipe
roughness. For fully turbulent flow , however the
friction factor is only very slightly dependentupon Reynolds number.
Several correlations have been reported for the
dependence of friction factors on Reynolds
number and pipe wall roughness.
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For smooth pipes:
f=0.5676NRe-0.3192for intermediate flow
f=16 log (NRef0.5/0.7063) for partially turbulentflow
f-0.5=2 log (NRef0.5/0.628) for fully turbulent flow
f=0.3614NRe
-0.25for NRe
up to 105
For rough pipes:
f-0.5=-2log [/(3.7d)+0.628/(NRef0.5)]
For very rough pipes:
f-0.5=-2log /(3.7d)
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ALLOWABLEWORKINGPRESSUREFORPIPES
To achieve higher throughputs a pipe can operate at high
pressure.
It is limited by the maximum stress the pipe can handle.
The maximum allowable internal working pressure can
be determined using ANSI specification:
p max=2(t-c)SE/do-2(t-c)Y
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Where
pmax=maximum allowable internal pressure, psig
t= pipe thicknessc=sum of mechanical allowances (thread and groove
depth), corrosion , erosion etc.,in.
S= allowable stress for the pipe material, psi
E=longitudinal weld joint factordo= outside diameter of the pipe, in.
Y= temperature derating factor
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ALLOWABLEFLOWVELOCITYINPIPES
High flow velocities in pipes can cause pipe erosion
problems, especially for gases that may have a flow
velocity exceeding 70 ft/sec.
The velocity at which erosion begins to occur is
dependent upon the presence of solid particles ,
their shape etc.
ve=C/0.5
In most cases C is taken to be 100.
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If c and are substituted then ve= -------------.
ve= 100(ZRT)0.5/(28.97pg)
0.5
The gas flow rate at standard conditions for erosion tooccur ,(qe)sc,can be obtained as follows:
(qe)sc= 1,012.435 d2(p/gZT)
0.5
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