Lyons W.C. (ed.). Standard handbook of petroleum and natural gas engineering.2001- Volume 1
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Fluid Mechanics
175
0.051
0.041
0.03(
a
.-
P
c
0
.-
g
p
0.02(
>
0.01!
0.01(
(C)
Figure
2-20.
Viscosity
of
natural gases: (a)
0.6
gravity;
(b)
0.7
gravity; (c)
0.8
gravity:
(d)
1
.O
gravity.
pipe the pressure
is
2,000 psi and the elevation is
150
ft.
At the outlet the elevation is
100
ft. What will be the pressure at the outlet when the temperature is 32"F? When
the temperature is
104"F?
y
=
(0.907)(62.4 Ib/ftg)
=
56.60 Ib/ft'
A
=
[(2
in.)/(12 in./ft)]'
=
0.0278
ft'
C
=
4(2 in.)/(12 in./ft)
=
0.6667 ft
176
General Engineering and Science
Values
of
(VD)
for
Water at
60°F
(velocity
in
fps
X
diameter
in
inches)
0.1
02
0.4
08
0.81
2
4
6
8
10
20
40
6080100
200 400
600
lp00
2,WO
4000
Spa0
l0.W
0.1
0.09
0.08
0.07
0.06
0.05
-?
0.04
2
B
s?
>
0.03
I1
0.025
-
-
3
s
'E
0.02
c
L
0.015
0.01
0.009
0.008
ReynoldsnumkrR=
(Vlnrps
dmn.
vmtzpersec)
Figure
2-21.
Friction factor
for
flow
in pipes
[l].
D
=
2A/C
=
0.0833 ft
v
=
Q/A
=
[(1,000 gal/min)(0.1337 fts/ga1)/(60 s/min)]
=
.0278 ft'
=
80.21 ft/s
=
7,380
x
g
=
32.2 ft/s2
lb-s/ft2
@
32°F
=
647
x
Ib-s/ft'
@
104°F
At
32°F
(56.60)( 0.0833)( 80.21)
=
1,
59
R=
(32.2)(7,380x
lod)
Because
R
<
2,000 the flow
is
laminar
and
f
=
64/R
=
0.04022. Assuming that the fluid is
incompressible implies that
yl
=
y,
and
v1
=
v2,
and Equation 2-54 can be rewritten as
1
(0.04022)(
100)(80.21)'
2( 32.2)( 0.0833)
/(
144 in.'/ft2
=
124psi
Fluid Mechanics
177
At 104°F
Because
R
>
4,000
the flow is turbulent. From Table 2-9,
E
=
150
x
Interpolating from Figure 2-21 yields
f
=
0.03. Now apply the Colebrook equation:
and
E/D
=
0.0018.
3.7 (18,150)(0.03)"
0.0018
+
I
1
-
=
-2lOg,"
-
fi
f
=
0.02991
(0.02991)(100)(80.21)2
=
610
psi
P,
=
2,000+(~)[(l;0-100)- 2( 32.2)(0.0833)
I
In
pzping system
fittings, valves, bends, etc., all cause additional pressure drops.
For such components the pressure drop can be estimated by modifying the frictional
component of Equation 2-54 with a
resistance coefficient,
K
=
fL/D,
or
an
equivalent
length,
L/D.
Typical resistance coefficients are given in Table 2-12 and typical
equivalent lengths are given in Table 2-13. To correctly apply either the resistance
coefficient or the equivalent length, the flow must be turbulent.
Table
2-12
Representative Values
of
Resistance Coefficient
K
[l]
Sudden contraction
-
Grodual reduction Sudden enlargement
Gradual enlargement
K
K'[1
-(d/Dlz]'
(D-dVSLI
0.05l0.10
lO.ZOlOWb.40
lO.5010.80
dB1
K'
lo.irlo.zolo.~rlo.~Io.9sIi
os[t.to
Exit
loss=(sharpeoged,pojcctinp.Rounded
1,
K.1.0
Compiled
from
datagivcn
in"Pip
Friction
Manunl.'idrd..
Hydriulic
instirutc.
1961.
178
General Engineering and Science
Table 2-13
Representative Equivalent Length in Pipe Diameters
(UD)
of
Various Valves and Fittings [l]
Globe valves, fully open
.....................................................
450
Angle valves, fully open
.....................................................
200
.........................................
13
V4
open
.....................
.........................................
35
Gate valves, fully open
'I2
open
............................................................................
160
'I4
open
.....
.............................................
900
In line, ball check valves, fully open
.................................. 150
Swing check valves, fully open
..........................................
135
Butterfly valves, 6 in. and larger, fully open
.........................
20
90"
standard elbow
..............................................................
30
45"
standard elbow
..............................................................
16
...........................
20
90"
street elbow
...................................................................
50
45"
street elbow
...................................................................
26
Standard tee:
90"
long-radius elbow
.....
Flow through run
...............................................................
20
Flow through branch
.........................................................
60
Compiled from dta given
in
"Flow
of
Fluids," Crane Company Technical Paper
41
0,
ASMF,
1971.
Example 2-14
Water flows from a horizontal 4-in.
ID
pipe into a horizontal 1-in.
ID
pipe at a rate
of 1,000 gal/min (see Figure 2-22). If the transition is abrupt, what
will
be the pressure
change across the connection?
Q
-
[(1,000 gal/min)(0.1337 ftJ/gal)(60 s/min)]
v
=--
=
25.53 ft/s
I
A,
[E
4
(
')'ft.]
12
[(l,000)(0. 1337)/(60)]
vp
= =
408.6
ft/s
[a(&)']
y1
=
y,
=
62.4 Ib/ft3
From Equation 2-54,
From Table 2-8,
DiJdoUt
=
4;
k
=
0.45.
PI
-
P,
=
AP
=
L[v;-v:+kv']- --[v~(l+k)-v~]
Y
2g
-2g
1
62'4
[408.6'(1
+
0.45)
-
25.53'1-
2(32.2) 144
1,625 psi
Fluid Mechanics
179
4
Figure
2-22.
Diagram for Example
2-14.
The 1,120 psi of this pressure drop is a dynamic loss due to the change in velocity,
and 505 psi is a frictional loss due to the fitting.
Components of
a
piping system that are connected in
series
produce additive
pressure drops, while components that are connected in
parallel
must produce the
same pressure drop.
While the modified energy equation provides for calculation of the flowrates and
pressure drops in piping systems, the
impulse-momentum
equation is required in order to
calculate the reaction forces on curved pipe sections. The impulse-momentum equation
relates the force acting on the solid boundary to the change in fluid momentum. Because
force and momentum are bothvector quantities, it is most convenient to write the equations
in terms of the scalar components in the three orthogonal directions.
(2-62)
where
M
is the fluid mass flow rate,
ZFx
is the sum
of
forces in the
x
direction,
v,
is
the initial fluid velocity in the
x
direction, etc.
Example
2-1
5
Water flows through a 120" reducing bend at
a
rate of
100
gpm. The inlet diameter
of the bend is 2 in. and the outlet diameter is
1
in. (see Figure
2-23).
What
is
the
reaction force on the bend?
Assuming that the flow is incompressible,
h;l=
(100 gal/min)(0.1337 ft3/ga1)(62.4 lb/ft')
[(32.2
ft/s2)(60 s/min)]
Ib
-
s4
ft
M
=
0.4318-
=
0.4318 slugs/s
A
=
"(
2)'
=
0.02182
ft2
'
4 12
A
=
E(L)'
4
12
=
0.005454ft2
180
General Engineering and Science
Figure
2-23.
Diagram
for
Example
2-15.
vyp
=
-v2
sin
30"
=
-20.43 ft/s
vxp
=
v2
cos3Oo= 35.38 ft/s
lb
s
ft
Fx
=
M(v,,
-vXJ
=
0.4318-((0-35.38 ft/s)
=
-15.28
Ib
FY
=
M(v,,
-
vY,)
=
0.4318(-10.21+20.43)
=
4.413 Ib
Flow through
chokes
and
nozzles
is
a
special case
of
fluid dynamics. For
incompressible
flu&
the problem can be handled by mass conservation and Bernoulli's equation.
Bernoulli's equation is solved for the pressure drop across the choke, assuming
that the velocity
of
approach and the vertical displacement are negligible. The
velocity term is replaced by the volumetric flow rate times the area at the choke
throat to yield
Q'r
AP
=
___
2gC2A2
(2-63)
C is a constant introduced
to
account for frictional effects. In general, 0.94
5
C
I
0.98.
Fluid Mechanics
181
Example
2-1
6
Assume
100
ft3/min of water is to be pumped through a nozzle with a throat
diameter of
3/4
in. What pressure drop should be expected?
100
60
Q
=
-
=
1.67
ftg/s
y
=
62.4
lb/ft3
Assume
C
=
0.95;
then
=
2,210
psia
(1.67*)( 62.4)
2( 32.2)(0.95*)( 3.068 x
lo-')*(
144)
AP
=
To
analyze
compressible flow through chokes
it is assumed that the entropy of the fluid
remains constant. The equation of
isentropic
flow is
p,v;
=
P,Vi
(2-64)
where
PI
and VI are the pressure and specific volume of the fluid at point
1,
immediately upstream of the choke, and
P,
and V, are the pressure and specific
volume immediately downstream of the choke. Equation
2-64
can be combined with
the ideal gas law to provide an estimate for the temperature drop across the choke
T,
=T,
2
(
rlii
(2-65)
where
T,
and T, are temperatures in
OR.
Furthermore, the first law of thermodynamics
can also be imposed, yielding the following equation for the volumetric flowrate:
(2-66)
where
Q
is
the volumetric flow rate in scfm,
C
is a
discharge coefficient
that accounts
for friction and velocity
of
approach (see Figure
2-24).
A
is the choke area in square
inches, PI is the inlet pressure in pounds per square inch absolute (psia), P, is the
outlet pressure in psia, TI is the inlet temperature in
OR,
and S is the specific gravity
of the gas.
Equations
2-65
and
2-66
apply only as long as the fluid velocity at the throat of the
choke is subsonic. Sonic velocity is the speed of a pressure wave in a fluid. Once
sonic
velocity
is achieved, the effects of the downstream pressure can no longer be transmitted
to the upstream side of the choke.
Flow
+
K?
-
Rqnulda
Niimkr
bared
OII
d,
Figure
2-24.
Flow
coefficient for nozzles
and
orifices (from Brown,
p.
226,
Fig.
3-19).
Thus there is a critical pressure ratio beyond which the
flow
at the throat
is
always
sonic. This
is
termed
critical
flow.
=
(
-L)k''k-')
(2-67)
Fluid Mechanics
183
If the pressure ratio is less than or equal to that specified by Equation
2-67,
the flow
will
be sonic at the choke throat and the temperature at the throat can be found from
T2
=
TI(
&)
The flow rate can be found from
(2-68)
(2-69)
For critical flow the discharge coefficient is dependent upon the geometry of the
choke and its diameter or the ratio
p
of its diameter to that of the upstream pipe (see
Figure
2-24).
Example
2-1
7
A
0.6
gravity hydrocarbon gas flows from a 2-in. ID pipe through a 1-in. ID orifice
plate. The upstream temperature and pressure are
75°F
and
800
psia, respectively.
The downstream pressure is
200
psia. Does heating need to be applied to assure that
frost does not clog the orifice? What will be the flow rate?
Check for Critical
Flow.
From Table
2-8
it is determined that k
=
1.299.
Checking
the pressure ratio, Equation
2-67,
gives
2
k'(k-l)
--
"
-
0.25
<
(-)
=
0.546
PI
k+l
Therefore critical flow conditions exist and the fluid velocity in the choke
is
sonic.
Temperature
in
the Choke.
From Equation
2-68
the temperature in the throat is
calculated as
T,
=
TI(-&-)
=
535(
)
=
465.4"R =5.4"F
1
+
1.299
Therefore, if the gas contains water, icing or hydrate formation may occur causing
the throat to clog.
A
heating system may be needed.
Flowrate.
Assume that the discharge coefficient is
C
=
1.0.
The choke area is
A
=
R(
1/2),
=
0.785
in?. Thus, from Equation
2-69,
(1.299t1)/(1.29%1)
0.5
[1.2gg(
1.299+1)
3
800
[(
0.
6)(535)]0'5
Q
=
610(1.0)(0.785)
=
14.021
scfm
184
General Engineering and Science
Now check on the discharge coefficient.
gas gravity as
The Reynolds number for gases can be calculated directly in terms of flowrate and
(2-70)
where
Q
is in scfm,
s
is the specific gravity of the gas, d is the pipe hydraulic diameter
in inches, and
c~
is in centipoise. From Figure 2-20 the viscosity of the gas is
p
=
0.0123 cp
and
(28.8)(
14021)( 0.6)
(0.0123)(2)
Re
=
=
9,850,000
From Figure 2-24, using
p
=
0.5, the value of the discharge coefficient is read as
C
=
0.62, and a new estimate of
Q
is:
Q
=
0.62( 14,021)
=
8,693 scfm
A
further iteration produces
no
change in the estimated flow rate for this case.
In subcritical flow the discharge coefficient is affected by the velocity of approach
as well as the type of choke and the ratio of choke diameter to pipe diameter. Discharge
coefficients for subcritical flow are given in Figure 2-24 as a function of the diameter
ratio and the upstream Reynolds number. Since the flow rate is not initially known,
it is expedient to assume C
=
1,
calculate
Q
use this
Q
to calculate the Reynold's
number, and then use the charts to find
a
better value of
C.
This cycle should be
repeated until the value of C
no
longer changes.
Example
2-1
8
A
0.65 gravity naturally gas
(K
=
1.25) flows from a two-in. line through a 1.5-in.
nozzle. The upstream temperature is 90°F. The upstream pressure is 100 psia while
the downstream pressure is
80
psia.
Is
icing a potential problem? What will be
the flowrate?
Check for critical flow using Equation 2-67.
The flow is clearly subcritical.
Check the outlet temperature using Equation 2-65.
T,
=
550(0.8°.25/'.25)
=
506.86'R
=
66'F
There will be no icing.
Calculate the flowrate.