Please see attachment 5
Attached are documents that can help. Also the work needs to be word document with calculation being shown.
Using the Beggs and Brill model, find the length of pipe between the points at 1000 psia and 500 psia with the following data in both vertical and horizontal cases. Compare the results for the flow pattern, liquid hold up and the pipe length in both cases. The flow and PVT data at average T and P conditions are given as:
QO=400
GLR=500 scf/STB Qw=600 bbl/D (STB/D) API=22
Gas gravity= 0.65
D= 1.995 in Bo=1.063 Rs= 92 scf/STB Z=0.91
μL= 17 cp
μw= 0.63 cp μg= 0.013 cp σL= 30 dyns/cm σw= 70 dyns/cm
>2 Phase Fluid
> eggs & Brill Method
.co 0001
0- ug-1 CheGuide 95
0
mm 71
inch 0.00
m 8 oot
9
19.7
3/min
tandard Conditions
.7 PSI, 60°F
m3/h .8
Kg/m3 lb/ft3 cP lb/ft.s 28.0 sults
0.462 1.055 0.487 Kg/m3 lb/ft3 bar psi 1.445 7713
157713 0.0042 Fanning 0
Fanning 0.0060 Fanning bar psi bar psi 0.989 bar psi 3.95 2
ft3/s ft/s ft/s lb/ft2.s ft/s .99
lb/ft3 8
Intermittent FALSE FALSE flow regime
Intermittent A B 0.4619 uphill 0.1844 1.055 EL(0) β B(θ) EL(θ) 0.4926 1.0669 0.4619 0.1844 1.055 0.487 23.20 lb/ft3 52.85 psi 0.3680 157713 1 2 3 4 5 6 0.0166 f 0.0166 0.0166 0.0166 0.0166 0.0166 0.0166 0.0042 3.77 psi http://cheguide.com/ Beggsand Brill method The Beggs and Brill method works for horizontal or vertical flow and
gD N mFR = (2-38)
m
l
l u
u λ
(2-39, 40) 302.01 316 lL λ=
4684.2 −= lL λ (2-41, 42) 4516.13 10.
−= lL λ 4 5.
Determining flow regimes For segregated, intermittent and distributed flow regimes use the following: ψ0ll yy = c
FR
b l N y =0 (2-43, 44)
with the constraint of that yl0 >= λl. ( ) ( )[ ]θθψ 8.1sin333.8.1sin1 3−+= C ( ) ( )gFRfvlell NNdC λλ ln1 −= (2-45,46) Where a, b, c, d, e, f and g depend on flow regimes and are given in the For transition flow, the liquid holdup is calculated using both the segregated ( ) ( )ntIntermitteBySegregatedAyy lll += (2-47) 2 3
3 LL A FR − ggll yy ρρρ += _
144 _ cPE g dl =⎟ ⎠ ⎜
The frictional pressure gradient is calculated using:
Dg dl mmtp
F
22 ρ ⎠ ⎜ ggllm λρλρρ += tp ff = (2-53,54)
The no slip friction factor fn is based on smooth pipe (ε/D =0) and the m mm Du μ Re = where ggllm λμλμμ += (2-55,56)
ftp the two phase friction factor is (2-57) Sntp eff = [ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 xxx S = (2-58)
and l y λ
Since S is unbounded in the interval 1 < x < 1.2, for this interval
)2.12.2ln( −= xS (2-60)
Using Beggs & Brill oF 3 ρg = 2.6 lb/ft
First find the flow regime, calculate NFR, λl, L1, L2, L3, and L4. NFR = 18.4, λl = .35, L1=230, L2=.0124, L3= .456, L4= 590. So .01 < λl < .4 and L3 < NFR < L1 so flow is intermittent.
Using the table to get a, b and c:
454.0 0173.0
5351.0
0 === c b y Find C and ψ, d, e, f and g from table: ( ) ( ) ( ) ( ) 0351.06.29*28.10*35.*96.2ln35.1ln1 0978.04473.0305.0 =−=−= −gFRfvlell NNdC λλ [ ] [ ] 01.1)8.1(sin333.)8.1sin(0351.1)8.1(sin333.)8.1sin(1 33 =−+=−+= θθθθψ C Find yl
The in-situ average density is _ Potential gradient is ftpsi dl cPE
/169. 1*29.24 _ ===⎟ ⎜
For friction gradient 3/1.1965.*6.235.*9.49 ftlbggllm =+=+= λρλρρ The Reynolds Number 109184
709. Re === mm μ
From Moody plot fn is .0045, solve for S 66.1
459. 2 === l [ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 =
[ ] [ ]( ) 379.)66.1ln(01853.0)66.1ln(8725.0)66.1ln(182.30523.0 42 =+−+−
Solve for ftp 0066.0045. 379. === eeff Sntp Find the friction gradient ftpsiftlb dl /032./62.4 94.10*1.19*0066.*22 3 ====⎟ ⎜ 1)Using the Beggs and Brill method find the length of pipe between the d = 1.995” γg = .65 oil 22
o API qo = 400 stb/day @ average conditions
Pipe Fittings in Horizontal flow To find the pressure drop through pipe fitting such as elbows, tees and These equivalent lengths have been determined experimentally for the most So to find the equivalent length for a 45o elbow in 2 inch pipe, find the Beggsand Brill method The Beggs and Brill method works for horizontal or vertical flow and
gD N mFR = (2-38)
m
l
l u
u λ
(2-39, 40) 302.01 316 lL λ=
4684.2 −= lL λ (2-41, 42) 4516.13 10.
−= lL λ 4 5.
Determining flow regimes For segregated, intermittent and distributed flow regimes use the following: ψ0ll yy = c
FR
b l N y =0 (2-43, 44)
with the constraint of that yl0 >= λl. ( ) ( )[ ]θθψ 8.1sin333.8.1sin1 3−+= C ( ) ( )gFRfvlell NNdC λλ ln1 −= (2-45,46) Where a, b, c, d, e, f and g depend on flow regimes and are given in the For transition flow, the liquid holdup is calculated using both the segregated ( ) ( )ntIntermitteBySegregatedAyy lll += (2-47) 2 3
3 LL A FR − ggll yy ρρρ += _
144 _ cPE g dl =⎟ ⎠ ⎜
The frictional pressure gradient is calculated using:
Dg dl mmtp
F
22 ρ ⎠ ⎜ ggllm λρλρρ += tp ff = (2-53,54)
The no slip friction factor fn is based on smooth pipe (ε/D =0) and the m mm Du μ Re = where ggllm λμλμμ += (2-55,56)
ftp the two phase friction factor is (2-57) Sntp eff = [ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 xxx S = (2-58)
and l y λ
Since S is unbounded in the interval 1 < x < 1.2, for this interval
)2.12.2ln( −= xS (2-60)
Using Beggs & Brill oF 3 ρg = 2.6 lb/ft
First find the flow regime, calculate NFR, λl, L1, L2, L3, and L4. NFR = 18.4, λl = .35, L1=230, L2=.0124, L3= .456, L4= 590. So .01 < λl < .4 and L3 < NFR < L1 so flow is intermittent.
Using the table to get a, b and c:
454.0 0173.0
5351.0
0 === c b y Find C and ψ, d, e, f and g from table: ( ) ( ) ( ) ( ) 0351.06.29*28.10*35.*96.2ln35.1ln1 0978.04473.0305.0 =−=−= −gFRfvlell NNdC λλ [ ] [ ] 01.1)8.1(sin333.)8.1sin(0351.1)8.1(sin333.)8.1sin(1 33 =−+=−+= θθθθψ C Find yl
The in-situ average density is _ Potential gradient is ftpsi dl cPE
/169. 1*29.24 _ ===⎟ ⎜
For friction gradient 3/1.1965.*6.235.*9.49 ftlbggllm =+=+= λρλρρ The Reynolds Number 109184
709. Re === mm μ
From Moody plot fn is .0045, solve for S 66.1
459. 2 === l [ ] [ ]( )42 )ln(01853.0)ln(8725.0)ln(182.30523.0 =
[ ] [ ]( ) 379.)66.1ln(01853.0)66.1ln(8725.0)66.1ln(182.30523.0 42 =+−+−
Solve for ftp 0066.0045. 379. === eeff Sntp Find the friction gradient ftpsiftlb dl /032./62.4 94.10*1.19*0066.*22 3 ====⎟ ⎜ 1)Using the Beggs and Brill method find the length of pipe between the d = 1.995” γg = .65 oil 22
o API qo = 400 stb/day @ average conditions
Pipe Fittings in Horizontal flow To find the pressure drop through pipe fitting such as elbows, tees and These equivalent lengths have been determined experimentally for the most So to find the equivalent length for a 45o elbow in 2 inch pipe, find the Turnitin Report Formatting Title Page Citation Outline
Two Phase Flow –
B
CheGuide
m
Line Number
P-
1
Chemical Engineer’s Guide
Description
Feed Pipe
Date
3
A
5
User Input
By
Pipe Data
Metric
English
Inner Diameter
50.
6
7
mm
1.
9
inch
Pipe Roughness
0.001
8
0.0000
Pipe Length
10
3
28.0
4
f
Flow Direction
uphill
Pipe Inclination angle
90
° with horizontal
Gas
Flowing Temperature
32.2
Deg C
90.00
Deg F
Inlet Pressure
11
bar
17
psi
Volumetric flowrate
9.0
m3/h
5.31
ft
@
S
944.9
Nm3/h @ 1 atm, 0°C
587.6
SCFM @
14
Density
141.3
Kg/m3
8.823
lb/ft3
Viscosity
0.020
cP
1.34E-05
lb/ft.s
Liquid
Volumetric flowrate
4.75
20.9
16
US gpm
Density
6
13
38.320
Viscosity
0.500
3.36E-04
gas/liquid Surface tension
dyne/cm
Re
Flow pattern map
Intermittent
Pressure drop due to head
Holdup volume fraction,
EL(0)
0.462
Inclination correction factor,
β
1.055
Inclined holdup v.f,
EL(θ)
0.487
Mixture density
371.61
23.20
DP due to Head
3.64
52.85
Pressure drop due to friction
Friction factor ratio
1.445
No Slip Reynold’s Number
15
No Slip friction factor, fNS
0.0042
Fanning
Two phase firction factor, fTP
0.006
DP due to friction
0.26
3.77
Total pressure drop
3.90
56.62
Correction due to acceleration
0.989
Corrected Total Pressure drop
3.95
57.23
Pressure drop / Length
bar / 100 mt
17.44
psi / 100 ft
Calculation (In English Units)
Area of Pipe
0.0217
ft2
Gravity constant, g
32.174
ft/s
Liquid Gas
Flowrate
0.0466000002
ft3/s
0.08855
Superficial Velocity
2.147
4.079
Weight Flux, G
82.26
lb/ft2.s
36.0
Mixture Velocity, Vm
6.23
Liquid Hold up, CL
0.345
Froude Number, Fr
7.25
Mixture Viscosity, µm
1.25E-04
Mixture density
18
Flow Pattern Map
Check Flow regimes
L1*
229.1043
Segregated
FALSE
L2*
0.0
12
TRUE
L3*
0.4691
Distributed
L4*
652.8663
Transition
Selected
Hydrostatic pressure difference
Liquid Holdup Volume Fraction Horizontal Flow EL(0)
Segregated
0.4926
-14.85
Intermittent
0.4619
15.85
Distributed
0.5077
Selected 0.4619
Check for if greater than CL
Liquid Holdup Correction Factor
Liquid velocity number, Nvl
4.50
Flow direction option menu
Correction factor, β
downhill
Uphill Flow
Segregated
1.0669
Intermittent
0.1844
Distributed 0.0000
Downhill Flow
All Flow regimes
0.7380
Selected 0.1844
Check for negative values
Inclination angle, θ
1.57
radians
B(θ)
Liquid Holdup EL(θ)
For Transition flow
-1.9257006375
Segreg.
1.319
0.650
Selected for transition flow
0.3448
Intermit.
Selected Liquid holdup
0.4874
Two phase density
DP due to hydrostatic head
Pressure drop due to friction loss
Friction factor ratio
y
1.4517
ln(y)
0.3727
S
0.3680
S final
Friction factor ratio, fTP/fNS
1.4448
No Slip Reynolds number
Friction factor estimate using Colebrook White equation
Epsilon / D
3.6E-05
Re 157713
Iteration
f
0.0004
0.0261
0.0159
0.0167
0.0166
Iteration 7 8 9 10 11 12
f 0.0166 0.0166 0.0166 0.0166 0.0166 0.0166
Iteration 13 14 15 16 17 18
Fanning friction factor
Two phase friction factor
ft 0.006
DP due to friction loss
Total pressure drop 56.62 psi
Correction factor due to acceleration
Ek
0.011
Total pressure drop 57.23 psi
everything in between. It also takes into account the different horizontal
flow regimes. This method uses the general mechanical energy balance and
the average in-situ density to calculate the pressure gradient. The following
parameters are used in the calculations.
u
2
=
2 0009252.
738.6
−= lL λ
Segregated if
λl < .01 and NFR < L1 or λl >= .01 and NFR < L2
Transition if
λl >= .01 and L2 < NFR <= L3
Intermittent if
.01 <= λl <.4 and L3 < NFR <= L1 or λl >= .4 and L3 < NFR <= L4
Distributed if
λl < .4 and NFR >= L1 or λl >= .4 and NFR > L4
l
a
λ
following table
& intermittent equations and interpolating using the following:
NL
−
= AB −= 1 (2-48,49)
sin
θρ
g
dp
⎞
⎝
⎛ (2-50,51)
uf
dp
c
=⎟
⎞
⎝
⎛ (2-52)
n
ntp f
f
Reynolds number,
m
N
ρ 1488
where
)ln(
x
+−+−
2
l
x
= (2-59)
qo = 2000 bpd qg = 1 mmcfpd Temp = 175
μo = 2 cp μg = .0131 Pipe = 2.5”
ρo = 49.9 lb/ft
3 Pressure = 800 psi
6.29
35.*845.
FR
l
l N
a
λ
459.01.1*454.0 === ψll yy
3
/29.246.2*)459.1(9.49*459. ftlbyy ggll =−+=+= ρρρ
g
g
dp
144
144
sin
⎠
⎞
⎝
⎛ θρ
First find the mixture density and viscosity
cpggllm 709.65.*0131.35.*2 =+=+= λμλμμ
1488*203.*39.13*1.191488
m
m
Du
N
ρ
35.
l
y
x
λ
)ln(
xxx
x
S
+−+−
)66.1ln(
=S
Dg
uf
dp
c
mmtp
F
203.*17.32
22
⎠
⎞
⎝
⎛ ρ
points at 1000psi and 500 psi with the following data. Both vertical and
horizontal cases.
qw = 600 bpd μg = .013 cp σo = 30 dynes/cm σw = 70 dynes/cm
GLR = 500 scf/stb
βο = 1.063 Rs = 92 scf/stb μo = 17 cp μw = .63
z = .91
valves an equivalent length is add to the flow line. This will account for the
additional turbulence and secondary flows which cause the additional
pressure drop.
of the fittings. These are found in the following tables. They are given in
pipe diameters, which are in feet.
equivalent length for the elbow in the table, 16, and multiply it by .166 feet,
which gives 2.66 feet. This is added to the length of the flow line, the
pressure drop for the system is then calculated using one of the methods for
horizontal flow.
everything in between. It also takes into account the different horizontal
flow regimes. This method uses the general mechanical energy balance and
the average in-situ density to calculate the pressure gradient. The following
parameters are used in the calculations.
u
2
=
2 0009252.
738.6
−= lL λ
Segregated if
λl < .01 and NFR < L1 or λl >= .01 and NFR < L2
Transition if
λl >= .01 and L2 < NFR <= L3
Intermittent if
.01 <= λl <.4 and L3 < NFR <= L1 or λl >= .4 and L3 < NFR <= L4
Distributed if
λl < .4 and NFR >= L1 or λl >= .4 and NFR > L4
l
a
λ
following table
& intermittent equations and interpolating using the following:
NL
−
= AB −= 1 (2-48,49)
sin
θρ
g
dp
⎞
⎝
⎛ (2-50,51)
uf
dp
c
=⎟
⎞
⎝
⎛ (2-52)
n
ntp f
f
Reynolds number,
m
N
ρ 1488
where
)ln(
x
+−+−
2
l
x
= (2-59)
qo = 2000 bpd qg = 1 mmcfpd Temp = 175
μo = 2 cp μg = .0131 Pipe = 2.5”
ρo = 49.9 lb/ft
3 Pressure = 800 psi
6.29
35.*845.
FR
l
l N
a
λ
459.01.1*454.0 === ψll yy
3
/29.246.2*)459.1(9.49*459. ftlbyy ggll =−+=+= ρρρ
g
g
dp
144
144
sin
⎠
⎞
⎝
⎛ θρ
First find the mixture density and viscosity
cpggllm 709.65.*0131.35.*2 =+=+= λμλμμ
1488*203.*39.13*1.191488
m
m
Du
N
ρ
35.
l
y
x
λ
)ln(
xxx
x
S
+−+−
)66.1ln(
=S
Dg
uf
dp
c
mmtp
F
203.*17.32
22
⎠
⎞
⎝
⎛ ρ
points at 1000psi and 500 psi with the following data. Both vertical and
horizontal cases.
qw = 600 bpd μg = .013 cp σo = 30 dynes/cm σw = 70 dynes/cm
GLR = 500 scf/stb
βο = 1.063 Rs = 92 scf/stb μo = 17 cp μw = .63
z = .91
valves an equivalent length is add to the flow line. This will account for the
additional turbulence and secondary flows which cause the additional
pressure drop.
of the fittings. These are found in the following tables. They are given in
pipe diameters, which are in feet.
equivalent length for the elbow in the table, 16, and multiply it by .166 feet,
which gives 2.66 feet. This is added to the length of the flow line, the
pressure drop for the system is then calculated using one of the methods for
horizontal flow. Order an Essay Now & Get These Features For Free:
