Sable"Gas"Produc6on"
600"
2,400"
Cumula6ve"produc6on"
500"
2,000"
400"
1,600"
300"
1,200"
Max. flow rate was 590 MMCFD
200"
800"
Cumula6ve"produc6on"(BCF)"
MMCFD"
Daily"produc6on"
500 million cubic feet per day
100"
0"
Jan"00"
400"
Jan"02"
Jan"04"
Jan"06"
Jan"08"
Jan"10"
Jan"12"
soep_monthly_Jan_2014.xlsx
23-­‐02-­‐13 Jan"14"
0"
Jan"16" Dec"17"
23-02-10
Sable gas problem.pptx 1 USD/MMBTU
1
USD
MMBTU 1.35 CAD
CAD
×
×
= 1.3
MMBTU 1.05 GJ
USD
GJ
13 CAD/GJ
6.4 CAD/GJ
4.22 CAD/GJ
23-­‐02-­‐13 Sable gas problem.pptx 2 eastward energy rates for February 2023 1
USD
MMBTU 1.35 CAD
CAD
×
×
= 1.3
MMBTU 1.05 GJ
USD
GJ
13.45 CAD/GJ
eastward energy is the distributor
of natural gas in Nova Scotia
23-­‐02-­‐13 Sable gas problem.pptx 3 Venturi Tube Flow Meter • Due to the constricted area at section 2 with respect to section 1, the fluid
velocity is increased from V1 to V2.
• This causes a reduction in the static pressure from location 1 to location 2.
• The Bernoulli equation for incompressible flow in a horizontal plane
ρV12
ρV22
p1 +
= p2 +
2
2
P1
P2
23-­‐02-­‐13 Sable gas problem.pptx 4 Venturi Tube Flow Meter ρV12
ρV22
p1 +
= p2 +
2
2
V22 − V12 =
⎛ A2 ⎞ 2 2 ( p1 − p2 )
2
V2 − ⎜ ⎟ V2 =
ρ
⎝ A1 ⎠
2
2 ( p1 − p2 )
ρ
V2 =
A2V2 = A1V1
2 ( p1 − p2 )
⎡ ⎛ A ⎞2⎤
ρ ⎢1 − ⎜ 2 ⎟ ⎥
⎢⎣ ⎝ A1 ⎠ ⎥⎦
Q = A2
23-­‐02-­‐13 Sable gas problem.pptx V1 =
A2
V2
A1
Q = A2V2
2 ( p1 − p2 )
⎡ ⎛ A ⎞2⎤
ρ ⎢1− ⎜ 2 ⎟ ⎥
⎢⎣ ⎝ A1 ⎠ ⎥⎦
5 Venturi Tube Flow Meter Q = A2
m! =
2 ( p1 − p2 )
⎡ ⎛ A ⎞2⎤
ρ ⎢1− ⎜ 2 ⎟ ⎥
⎢⎣ ⎝ A1 ⎠ ⎥⎦
A2
⎛ D2 ⎞
1− ⎜ ⎟
⎝ D1 ⎠
4
=
2 ( P1 − P2 ) ρ
Qideal = EA2 2 ( p1 − p2 ) ρ
A2
⎛A ⎞
1− ⎜ 2 ⎟
⎝ A1 ⎠
E=
2
2 ( p1 − p2 ) ρ
E is the velocity of
approach factor
1
⎛D ⎞
1− ⎜ 2 ⎟
⎝ D1 ⎠
4
This is the flow based on
the assumption of no losses.
Qideal = CEA2 2 ( p1 − p2 ) ρ
C is the discharge coefficient for a
venturi meter; typically C = 0.98 – 0.99
23-­‐02-­‐13 Sable gas problem.pptx 6 Cast Iron Venturi d
D
inches
D = inlet diameter
p1
d = throat diameter
p2
125# FLANGE
D
4
d
Do
A
B
A-1.800
2.9
3.3
B-2.400
3.4
3.6
C-2.900
3.7
3.9
A-2.700
4.2
4.9
C
T
9
15/16
D =B-3.600
20 inches;
d = 14.5
inches
6
4.9
5.4
11
8
C-4.350
5.4
5.8
A-3.600
5.6
6.5
B-4.800
6.6
7.2
C-5.800
7.2
7.7
23-­‐02-­‐13 250# FLAN
1
E
F
8
G
C
J
T
7 1/2
10
6 15/16
1 1/4
9 1/2
12 1/2
9 11/16
1
7/16
11 3/4
15
11
15/16
1 5/8
7/8
13 1/2
1 1/8
Sable gas problem.pptx 7 RF
Cast Iron Venturi 23-­‐02-­‐13 ∆ p = 20 inches of
water (0.72 psi,
4.9 kPa)
Sable gas problem.pptx 90°F = 32.2°C
ρair = 1.74 kg/m3
8 Venturi Tube Flow Meter 7.5 PSIG = 22.2 PSIA = 152 kPa
ρ at 101 kPa
15°C
1.22 kg/m3
32.2°C, 152 kPa
ρ = 1.74 kg/m3
∆ p = 20” water
(0.72 psi, 4.98 kPa)
27,358 ft3/s
D1 = D = 20" = 508 mm
D2 = d = 14.5" = 368 mm
π D22 π × 3682
=
= 0.106 m 2
E = 1 1− ( D2 D1 ) = 1 1− ( 368 508 ) = 1.17 A2 =
4
4
Q = CEA2 2 ( p1 − p2 ) ρ = 0.98 × 1.17 × 0.106 2(4,980) /1.74
Q = 9.20 m 3 /s = 552 m 3 / min = 19, 200 ft 3 /min @ 192 kPa (absolute)
1.74 3
Q = 19,200 ×
ft /min = 27, 400 SCFM @ 101 kPa (absolute)
1.22
23-­‐02-­‐13 Sable gas problem.pptx 9 4
4
Natural Gas Flow Meter •
•
•
•
Natural gas is flowing at the rate of 500 MMCFD (based on STP conditions).
500 MMCFD = 14.16 million m3/day = 164 m3/s
We need the mass flow rate. m! = ρQ
347,000 SCFM
We need the density at STP.
p
101 kPa
=
= 0.677 kg/m 3
RT 0.518 × 288
m! = ρQ = 0.677 × 164 = 111 kg/s
• We need the density at pipeline condition (11,700 kPa, 20°C).
p 11, 700 kPa
ρ=
=
= 77.1 kg/m 3
RT 0.518 × 293
m! 111
A reasonable ∆p;
Q= =
= 1.44 m 3 /s Q = CEA2 2 ( p1 − p2 ) ρ
not too high and
ρ 77.1
ρ=
• Calculate the ∆ p across the venturi.
2
not too low.
2
⎛
⎞
⎛
⎞
Q
1.44
∆ p=⎜
=⎜
= 5.42 kPa
⎟
⎟
⎝ CEA2 2 ρ ⎠
⎝ 0.98 × 1.17 × 0.106 2 77.1 ⎠
23-­‐02-­‐13 Sable gas problem.pptx 10 Natural Gas Flow Meter • What are the velocities within the venturi meter?
• We calculated the mass flow rate and the density at pipeline conditions.
m! = 111 kg/s
ρ = 77.1 kg/m 3
• Calculate the velocity where D1 = 20 inches = 506 mm.
m! = ρ AV = ρ A1V1 = ρ A2V2
π 2 π
A1 = D1 = 0.506 2 = 0.201 m 2
4
4
m!
111
V1 =
=
= 7.2 m/s
ρ A1 77.1× 0.201
• Calculate the velocity where D2 = 14.5 inches = 368 mm.
π 2 π
D2 = 0.368 2 = 0.106 m 2
4
4
m!
111
V2 =
=
= 13.6 m/s
ρ A2 77.1× 0.106
A2 =
23-­‐02-­‐13 Sable gas problem.pptx Both velocities are
acceptable for a gas
11 Sable Island Natural Gas Pipeline • The pressure of the natural gas at Sable Island was initially 11.7 MPa.
• The temperature was 20°C.
p
• Previously we calculated the density using the formula ρ =
RT
• This equation is derived from the “equation of state for an ideal gas”: p = ρ RT
• At normal pressures and temperatures most gases behave according to the
ideal gas law.
• 20°C is a normal temperature; 11.7 MPa is a very high (not normal!) pressure.
• At abnormal temperatures and pressure the ideal gas law is not valid for
engineering purposes.
• However, we continue to use the ideal gas law by introducing a factor, Z,
which we label the compressibility factor: p = Z ρ RT
• The calculation of Z is complex, therefore, we use a chart – the
Compressibility Chart.
• Z is a function of the reduced temperature, Tr and reduced pressure, Pr.
T
Tr =
Tcr
23-­‐02-­‐13 p
Pr =
pcr
Tcr and pcr are the critical T and p
Sable gas problem.pptx 12 Sable Island Natural Gas Pipeline • For natural gas, which is 96 – 97% methane, the critical temperature is -83°C,
or 190 K – we have to employ the absolute temperatures.
• The critical pressure for natural gas is 4.87 MPa.
• Evaluate the reduced temperature and pressure for Sable gas:
Tr =
T 293 K
=
= 1.54
Tcr 190 K
p 11.7 MPa
=
= 2.40
pcr 4.87 MPa
Pr =
• Evaluate the reduced temperature and pressure for natural gas at STP
(standard temperature and pressure), 288 K and 0.101 MPa:
Tr =
23-­‐02-­‐13 T
288 K
=
= 1.52
Tcr 190 K
Pr =
p 0.101 MPa
=
= 0.02
pcr
4.87 MPa
Sable gas problem.pptx 13 Compressibility Chart for Gases Tr = T/Tcr
Tr = 1.54
Z = 0.82 for Sable
gas at 11.7 MPa
Pr = p/pcr
Pr = 2.40
23-­‐02-­‐13 Sable gas problem.pptx 14 Compressibility Chart for Gases Tr = 1.52
Z = 1.00 for all
gases for Pr ≈ 0
Pr = 0.02
23-­‐02-­‐13 Sable gas problem.pptx 15 Sable Island Natural Gas Pipeline • The compressibility factor for Sable gas is Z = 0.82
• Recalculate the density.
p = Z ρ RT
p
ρ=
ZRT
Previously we
calculated 77.1 kg/m3
p
11, 700 kPa
=
= 94.0 kg/m 3
ZRT 0.82 × 0.518 × 293
• Recalculate the pressure drop across the venturi meter.
• Recall that ∆ pventuri ∝ ρ
2
2
⎛
⎞
Q
⇒ ρ ∝ρ
∆ p=⎜
⎟
⎝ CEA2 2 ρ ⎠
Previously we calculated that ∆ p was 5.4 kPa across the venturi.
The revised ∆ p is
ρ=
( )
⎛ 94.0 ⎞
∆ p = 5.4 kPa ⎜
= 6.6 kPa
⎝ 77.1 ⎟⎠
23-­‐02-­‐13 Sable gas problem.pptx 16 The Sable Gas Pipeline 220-km long natural gas
pipeline from Sable Island
to Goldboro, Guysborough
County
23-­‐02-­‐13 Sable Island
Sable gas problem.pptx 17 The Sable Gas Pipeline Sable Island
220-km Sable
gas pipeline
Sable Island
23-­‐02-­‐13 Sable gas problem.pptx 18 Sable Island Natural Gas Pipeline •
•
•
•
•
•
•
Determine the pressure drop over the Sable gas pipeline.
The Sable gas pipeline was 220 km long (220,000 m).
The outer diameter of the pipeline was 26.0 inches (660 mm).
The inside diameter was 24.6 inches (625 mm). Ai = 0.307 m2.
The density of the gas at the production platform was 94.0 kg/m3.
The mass flow rate was 111 kg/s.
The head loss is given by
⎛ L ⎞ ⎛V2 ⎞
hl = f ⎜ ⎟ ⎜ ⎟ expressed in meters head of gas
⎝ Di ⎠ ⎝ 2g ⎠
• The pressure loss is given by
⎛ L ⎞ ⎛ ρV 2 ⎞
∆ p = ρ ghl = f ⎜ ⎟ ⎜
expressed in Pa (pascals)
⎟
⎝ Di ⎠ ⎝ 2 ⎠
• Calculate the velocity at the entrance to the pipeline.
V=
23-­‐02-­‐13 m!
111
=
= 3.85 m/s
ρ Ai 94.0 × 0.307
Sable gas problem.pptx 19 Sable Island Natural Gas Pipeline • We need the friction factor, f.
• To find f we need the Reynolds number.
ρ DV 94.0 kg/m 3 × 0.625 m × 3.85 m/s
6
ReD =
=
=
22
×
10
µ
1.05 × 10 −5 N·s/m 2
• We also need the relative roughness of the pipe.
• From Table 8.1 the roughness projections on commercial steel pipe are 0.045
mm.
relative rougness =
ε 0.045 mm
=
= 0.00007
Di
625 mm
• Use the Moody diagram, Figure 8.20, to find f, the friction factor
23-­‐02-­‐13 Sable gas problem.pptx 20 Moody diagram RelaEve roughness f =0.011 0.00007 ReD = 22,000,000 February 13, 2023 ENGI2103.1.W2023.pptx 21 Sable Island Natural Gas Pipeline ⎛ L ⎞ ⎛ ρV 2 ⎞
∆ p= f ⎜ ⎟⎜
expressed in Pa (pascals)
⎟
⎝ Di ⎠ ⎝ 2 ⎠
ρV 2 94.0 × 3.852
=
= 697 Pa = 0.70 kPa
2
2
L 220, 000 m
=
= 352, 000
f = 0.011
Di
0.625 m
•
•
•
•
•
•
∆ p = 0.011× 352, 000 × 0.70 kPa = 2, 700 kPa = 2.7 MPa
The pressure at the production platform was 11,700 kPa, or 11.7 MPa.
The pressure loss over the 220-km pipeline was 2.7 MPa.
The pressure at landfall is 11.7 – 4.7 MPa = 9.0 MPa.
However, as the pressure decreases, the density will decrease, and the
velocities will increase – causing higher pressure drops.
The above analysis assumes that the density and velocity remain constant over
the entire 220 km.
We should perform calculations over small sections of the pipeline, and
update densities and velocities at the start of each new section
23-­‐02-­‐13 Sable gas problem.pptx 22 Excel spreadsheet to calculate ∆p’s p = 11.7 - 0.30 = 11.40 MPa ⎡⎛ ε Di ⎞ 1.11 6.9 ⎤
1
= −1.8log ⎢⎜
+
equation to calculate f
⎥
⎟
ReDi ⎦
f
⎣⎝ 3.7 ⎠
23-­‐02-­‐13 Sable gas problem.pptx 23 Pressure"
V"
12"
p = 9.0 MPa with
a single step 11"
10"
11"
10"
p = 11.7 MPa 9"
8"
7"
8"
p = 8.0 MPa V = 3.8 m/s 7"
6"
6"
5"
5"
V = 6.3 m/s 4"
3"
Velocity"(m/s)"
9"
Pressure"(MPa)"
12"
4"
3"
2"
Calculating every
22 km (10 steps) 1"
Do we have the correct answer
with 10 steps? Try 44 steps? 0"
0"
February 13, 2023 22"
44"
66"
88"
110"
132"
154"
km"from"producBon"plaDorm"
ENGI2103.1.W2023.pptx 176"
198"
2"
1"
0"
220"
24 Pressure"
12"
12"
11"
11"
p = 7.9 MPa p = 11.7 MPa 10"
9"
9"
8"
8"
7"
V = 6.4 m/s V = 3.8 m/s 7"
6"
6"
5"
5"
4"
4"
3"
2"
Calculating every
5 km (44 steps) 1"
Very little change from
the previous iteration February 13, 2023 22"
44"
66"
88"
110"
132"
154"
km"from"producBon"plaDorm"
ENGI2103.1.W2023.pptx 3"
2"
1"
0"
0"
Velocity"(m/s)"
10"
Pressure"(MPa)"
V"
176"
198"
0"
220"
25 These are the
results from an
online
calculator of
gas pipeline
pressures. 23-­‐02-­‐13 Sable gas problem.pptx 26 I calculated
94.0 kg/m3 I used R = 0.518 23-­‐02-­‐13 Sable gas problem.pptx 27