line Sizing

1. LINE SIZING OF THE MAIN PRODUCTION HEADER
(A GAS / LIQUID TWO PHASE LINE)
Parameters to calculate includes:
1. Internal diameter of pipe
2. Thickness of pipe
3. Pressure drop across pipe length
4. Length of pipe
CALCULATING INTERNAL DIAMETER OF PIPE
Important formulas and equations
1. Specific Gravity SG = 141.5
API+131.5
For light crude API = 35° SG L = 0.84985
For medium crude API =27° SG M = 0.892744
For heavy crude API = 19° SG H = 0.940199
SG for water SG W = 1
S1 = Average specific gravity for hydrocarbon water mixtures at standard conditions
S1 = SGL+SGM+SGH+SGW
= 0.92069825
4
2. Density of gas/liquid mixture ‘ρm’ (Ibs/ft 3 )
ρm =
12409S1P + 2.7RSgP
198.7P + RTZ

Where
P = operating pressure, psia.
S1 = Average specific gravity for hydrocarbon water mixtures at standard conditions
R = gas/liquid ratio, ft3/barrel at standard
T = operating temperature, °R.
Sg = gas specific gravity at standard
Z = Gas compressibility factor
From project P = 550psia
S1 = 0.92069825
R= 0.0825
T = 645.57 °R
Sg = Mgas
= 25.39
= 0.8731
Mair 28.96443
Z = 0.9314 (using DAK-EOS)
Therefore
ρm =
12409S1P + 2.7RSgP
198.7P + RTZ
= 57.4734 Ibs/ft 3
3. Fluid erosion velocity ‘V e ’ (ft/s)
Ve =
c
√pm
Where c ‘empirical constant’ = 100 (Corrosion anticipated API 14 E)
Ve =
c
√pm = 100
√57.4734
= 13.1907 ft/s

4. Minimum pipe internal cross-sectional area required ‘A’ (in 2 /1000barrels)
A =
9.35 + (
ZRT
21.25P )
Ve
= 9.35 + (0.9314 ∗ 0.0825 ∗ 645.57
21.25 ∗ 550 )
13.19067
= 0.7092
Minimum internal area = 0.7092* raw crude flow rate (in 1000 barrels/day)
= 0.7092*272.76 (a surge factor of 20 % added API 14E)
= 193.4414 in 2
Internal diameter = Di = √ 4A π
= 15.6939 in
CALCULATING THICKNESS OF PIPE
Important formulas and equations
t =
Pi Do
2(SE + PiY)
Where:
t = pressure design thickness, inches;
Pi = internal design pressure, psig. = 535.304psig (550psia)
D o = pipe outside diameter, inches = Di + 2t
E = longitudinal weld joint factor = 1.00 (from ANSI B31.3 assuming seamless)
Y = Temperature factor = 0.4 (for ferrous materials at < 900 F when t <D/6).
S = maximum allowable stress = 20,000 psig (ASME B31.3 – Process piping)
Assumption: pipe is an ASTM A106, Grade B, seamless pipe,

Pi Do
t =
2(SE + PiY)
=
Pi (Di + 2t)
2(SE + PiY)
t = Di [
Pi
2(SE+PiY)−2Pi
] = 0.2206 in
Thickness = 0.2206 in
Outer Diameter D o = Di +2t = 16.1352 in
NPS = 18 inch (available manufactured size)
CALCULATING PRESSURE DROP ACROSS PIPE LENGTH
Important formulas and equations
1. Weymouth approximation of Moody friction: f = 0.032(Di) 0.333
2. W = total liquid plus vapour rate, (lbs/hr.) = 2,830,000
= 0.032(15.6939) 0.333
= 0.08

3. The pressure drop in a two-phase steel piping system may be estimated using a
simplified Darcy Equation (from the GPSA Engineering Data Book, 1981)
ΔP = 0.000336f (W2 )
di 5 pm
Where:
ΔP = Pressure drop, psi/100feet.
di = Pipe inside diameter, inches= 15.6939
f = Moody friction factor, dimensionless 0.08
ρm = gas/liquid density at flowing pressure = 57.4734 lbs/ft3
W = tota1 liquid plus vapour rate, lbs/hr.
ΔP = 0.000336f (W2 )
di 5 pm
= 0.000336 ∗ 0.08 ∗ (2, 830,0002 )
(15.6939) 5 ∗ 57.4734
= 4.0105 psi/100ft
Pressure drop = 4. 0105 psi/100ft
CALCULATING TOTAL LENTH OF PIPE
Method Used: Flow Head Loss Method
So the pressure at the exit of the pipe is assumed to 400 psia (80% of separator operating
pressure)
Pressure drop across total length of pipe = inlet pressure – outlet pressure
ΔP(total) = 500 – 400 = 100 psia

Recall:
Pressure drop across 100ft of pipe length = 4.0105 psi
Total actual length of pipe = 100∗100
4.0105
= 2500 ft (approximately)
2. EQUIPMENT SIZING OF A 3 PHASE SEPARATOR
Parameters to calculate includes:
1. Seem to Seem length
2. Demister sizing
SEEM TO SEEM LENGTH
Step 1. Calculation of drag coefficient CD
Archimedes number “A r ” calculation
From project
Ar = dv3 (ρd−pc)ρc∗g
µc 2
d v = dispersed bubble size = 0.05 cm
ρ d = dispersed phase density = 0.9545g/cm 3
ρ c = continuous phase density = 0.8272g/cm 3
g = gravitational constant = 981cm/sec 2
µ c = continuous phase viscosity = 0.02734poise

Ar = 0.2253 (0.9545−0.8272)0.8272∗981
0.02734 2 = 349.8211
Reynolds number calculation:
Re = (√19.075 + 2.129√Ar − 4.3675) 2 = 10.9349
Drag coefficient
CD = (0.5423 + 4.737
Re 0.5)2 = 3.9
Step 2. Calculate vessel diameter and length for gas capacity.
Using:
Leff d = 421 TZQg
[ ( ρg
) CD
P pl−pg dm ]1/2
T= operating temperature o R
Leff = Effective length of vessel where separation occurs, ft.
Q g = Gas flow rate, MMscf/D
P = Operating Pressure, Psia
Z= Gas compressibility

CD= Drag Coefficient
d m = drop diameter,µm
ρl = liquid density lb/ft 3
ρ g =gas density lb/ft 3
d= vessel internal diameter, in.
From Project
P= 600 psia
T= 809.67 o R
Z= 0.9439 (using DAK-EOS)
Q g = 0.02433 MMSCF/D
ρ g = 1.468 lb/ft 3
ρl = 55.615 lb/ft3
d m = 500 microns
CD= 3.9
Vessel fill = 50%
Therefore
Leff ∗ d = 421 TZQg
[ ( ρg
) ∗ CD
P pl−pg dm ]1/2
= 421 ∗ 809.67∗0.9439∗0.02433
600
= 0.189727
1.468
[ (
) ∗ 3.9
55.615−1.468
500 ]1/2

Step 2. Calculate Leff, Lss = Leff + d/12 (gas) for different values of d
Lss = Seam to seam Length, ft.
Leff = (leff * d)/d
Assuming d = 0.5 Assuming d = 0.7 Assuming d = 1
Leff = 0.189727
0.5
Leff = 0.189727
0.7
Leff = 0.189727
1
= 0.379455 = 0.271039 = 0.189727
Lss = 0.421122 Lss = 0.329373 Lss = 0.273061
12Lss/d = 10.10692 12Lss/d = 5.646386 12Lss/d = 3.276729
d in Leff Lss 12Lss/d
0.5 0.379455 0.421122 10.10692
0.7 0.271039 0.329373 5.646386
1 0.189727 0.273061 3.276729
Table 1
Step 3 Calculate the vessel diameter and length for liquid retention time with below equation:
d 2 Leff = troQo + trwQw/1.4F1
t ro = oil retention time, minutes = t rw = Water retention time, minutes (API 12J)
Q O = Oil flowrate/D
Q w = Water flow rate, B/D
F l = Fraction of vessel crossectional area filled by liquid.

t ro = t rw = 10mins ( API 12J, recommended Oil retention Time)
Q 0 = Oil rate = 50000 bpd
Q w = 2500 bpd
d 2 Leff =
10( 50000 + 2500)
1.4 ∗ 0.5
d 2 Leff = 750000
Step 4. Calculate Leff, Lss = 4/3 Leff (liquid) for different values of d
Assume d = 16 Assume d= 24 Assume d= 30
L eff = 750000
L 16 2 eff = 750000
L 24 2 eff = 750000
30 2
= 2929.688 = 1302.083 = 3333.333
Lss = 3906.25 Lss = 1736.111 Lss = 1777.787
d in Leff Lss 12Lss/d
16 2929.688 3906.25 2929.688
24 1302.083 1736.111 868.0556
30 833.333 1111.111 444.444
135 41.15226 54.86968 4.8773
Table 1

Step 5. Selecting vessel that satisfies both gas and liquid capacity.
A comparison of Tables 1 and 2 shows that the liquid capacity is the dominant parameter.
Hence, a 135in. × 54.87-ft vessel is sufficient, as it has a slenderness ratio within the typical 3 to
5 range. This size should be rounded up to 11.25ft × 55ft
DEMISTER SIZING
Using the equation:
Vm = Kd[ ( ρ1−ρg)
] 1/2
pg
K d = Demister capacity Factor, ft. /sec
V m = Maximum velocity, ft/sec
ρ L = Liquid Density, lbm/ft 3
ρg = gas density, lbm/ft 3
ρ g = 1.468 b/ft 3
ρl = 55.615 lb/ft3
K d = 1.169 ft/sec (API 14E)
Vm = 1.169[ ( 55.615−1.468)
] 1/2
1.468
Vm = 7.0997 Ft/sec

DEMISTER AREA A d
Ad =
0.327TZQg
P
Kd [
ρ1 − ρg
ρg
]
1
2
T= 809.67 o R
Z= 0.9439 (using DAK-EOS)
Q g = 0.02433 MMSCF/D
P= 600 psia
K d = 1.169 ft/sec
ρl = 55.615 lb/ft3
ρ g = 1.468 lb/ft
Ad =
0.327∗809.67∗0.9439∗0.02433
600
1.169 [ 55.615−1.468
1
2
]
1.468
= 0.0014
Demister Area (A d ) = 0.0014ft 2
BY:
ADESANMI DAMILOLA
and
DADA TAIWO