Gas Volume Calculation
82
Pipe dia 8" 6" 3" Total L (m) 6.4950 112.8300 9.1310 128.4560 D (m) 0.2191 0.1683 0.0889 10.7047 R (m) 0.1096 0.0842 0.0445 0.010705 A (m2) 0.0377 0.0222 0.0062 0.0661 V (m3) 0.2448 2.5088 0.0566 102.455 V total 2.810 m3 Keterangan: L Panjang pipa D Diameter pipa R Jari-jari pipa A Luas area penampang pipa V Volume Vtotal Asumsi volume gas yang diperlukan Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter Kalkulasi: Kalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dik Faktor pengali/konstanta 0.372 konstanta untuk s 1000000 MM Pipa 8" press barg 900 psig panjang 26 km 85301.8374 ft 85.30184 dia 28 inc volume gas 262483.2 SCF/K ft 22390299.247 SCF 22.3902992 MMSCF pipa 6" press barg 500 psig panjang 112.83 m 370.177 ft 0.370177 dia 6 in volume 6696 SCF/Kft 2478.705192 SCF 0.002478705 MMSCF
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Transcript of Gas Volume Calculation
Pipe dia8" 6" 3" Total
L (m) 6.4950 112.8300 9.1310 128.4560D (m) 0.2191 0.1683 0.0889 10.7047R (m) 0.1096 0.0842 0.0445 0.010705A (m2) 0.0377 0.0222 0.0062 0.0661V (m3) 0.2448 2.5088 0.0566 102.455
V total 2.810 m3
Keterangan:L Panjang pipaD Diameter pipaR Jari-jari pipaA Luas area penampang pipaV VolumeVtotal Asumsi volume gas yang diperlukan
Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter
Kalkulasi:Kalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
Faktor pengali/konstanta 0.372 konstanta untuk setiap /1000 ft1000000 MM
Pipa 8"press barg 900 psigpanjang 26 km 85301.8374 ft 85.30184dia 28 inc
volume gas 262483.2 SCF/K ft22390299.247 SCF
22.3902992 MMSCF
pipa 6"press barg 500 psigpanjang 112.83 m 370.177 ft 0.370177dia 6 in
volume 6696 SCF/Kft2478.705192 SCF
0.002478705 MMSCF
pipa 3"Press 5 barg 72.5189 psigpanjang 9.131 m 29.9573 ft 0.029957dia 3 in
Volume 242.7932772 SCF/Kft7.2734310431 SCF0.0000072734 MMSCF
Total Pemakaian
source :
How do you calculate the volume of natural gas in a pipeline?Answer:You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
8 km = 26246 ft = 26.246 kft 26246.7192
65 Bar = 942.7 psi
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)
65x (Pi x 0.1778 x 0.1778 x 8000/4) = 0.9 x n x 8.314x10-5)x (273+6)
n = 618523 moles
1 kmol of a gas occupies 22.441 Nm3 at standard conditions
t.f 618.523 kmol should ocupy 618.523 x 22.441 = 13880.274643 Nm3.
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Amiel
http://wiki.answers.com/Q/How_do_you_calculate_the_volume_of_natural_gas_in_a_pipeline
incfoot1000 footm2inc2
Kalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
konstanta untuk setiap /1000 ft
kft 85301.84
tinggal masukkan panjang dan pressure
22.3902992 MMSCF 8"
Kft
0.002478705192 MMSCF 6"
Kft
0.0000072734 MMSCF 3"
22.3927852253 MMSCF
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
http://wiki.answers.com/Q/How_do_you_calculate_the_volume_of_natural_gas_in_a_pipeline
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Pipe dia8" 6" 3"
L (m) 6.4950 112.8300 9.1310D (m) 0.2191 0.1683 0.0889R (m) 0.1096 0.0842 0.0445A (m2) 0.0377 0.0222 0.0062V (m3) 0.2448 2.5088 0.0566
V total 2.810 m3
Keterangan:L Panjang pipaD Diameter pipaR Jari-jari pipaA Luas area penampang pipaV VolumeVtotal Asumsi volume gas yang diperlukan
Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter
Kalkulasi:Kalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
Faktor pengali/konstanta 0.372 konstanta untuk setiap /1000 ft1000000 MM
Pipa 8"press barg 700 psigpanjang 300 m 984.2520 ftdia 8 inc
volume gas 16665.6 SCF/K ft16403.149631232 SCF
0.0164031 MMSCF
pipa 6"press barg 500 psigpanjang 112.83 m 370.177 ftdia 6 in
volume 6696 SCF/Kft2478.705192 SCF
0.002478705 MMSCF
pipa 3"Press 5 barg 72.5189 psigpanjang 9.131 m 29.9573 ftdia 3 in
Volume 242.7932772 SCF/Kft7.27343104306356 SCF
0.0000072734 MMSCF
Total Pemakaian
128.4560 inc10.7047 foot
0.0107047 1000 foot0.0661 m2
102.455 inc2
konstanta untuk setiap /1000 ft
0.98425197 kft
0.0164031 MMSCF 8"
0.370177 Kft
0.002478705192 MMSCF 6"
0.0299573 Kft
0.0000072734 MMSCF 3"
Total Pemakaian 0.0188891283 MMSCF
Volume of Gas in Cylinder
To find the volume of gas available from a compressed gas cylinder, we apply the Ideal Gas Law (PV = nRT). In a high-pressure cylinder, the volume will be affected by the content's compressibility factor Z (PV = ZnRT). For example, an AL cylinder of pure helium may contain 134 cu. ft. of gas while the same cylinder of pure air may contain 144 cu. ft. under the same conditions. For these practical calculations, however, we assume ideal gas behavior for simplicity.
The Ideal Gas Law PV = nRT
Where:
P is pressure
V is volume
n is the number of moles
R is the gas constant
T is the absolute temperature
When the temperature is kept constant, we can derive the equation:
P (1) x V (1) = P (2) x V (2)
Where:
P (1) is the pressure of the compressed gas in the cylinder (psi)
V (1) is the internal volume of the cylinder, often referred to as water volume (liter)*
P (2) is the atmospheric pressure (1 atm - 14.7 psi)
V (2) is the volume of gas at pressure P (2) (liter).
For example, an AL sized cylinder is filled with nitrogen at 2000 psi. What is the gas volume of nitrogen from the cylinder?
P (1) is 2000 psi
V (1) is the internal volume of AL cylinder 29.5 liter*
P (2) is 14.7 psi
V (2) is the unknown volume of gas
Solving the equation above for V (2) gives:
V (2) = [p (1) x V (1)]/P (2) = (2000 psi x 29.5 liters)/14.7 psi = 4013 liters (approximately 140 cu. ft.)
Calucation :untuk suhu 100 psi
PV1=nPV2 590 liters20.835 ft3 untuk 1 tabung
To find the volume of gas available from a compressed gas cylinder, we apply the Ideal Gas Law (PV = nRT). In a high-pressure cylinder, the volume will be affected by the content's compressibility factor Z (PV = ZnRT). For example, an AL cylinder of pure helium may contain 134 cu. ft. of gas while the same cylinder of pure air may contain 144 cu. ft. under the same conditions. For these practical calculations, however, we assume ideal gas behavior for simplicity.
For example, an AL sized cylinder is filled with nitrogen at 2000 psi. What is the gas volume of nitrogen from the cylinder?
To find the volume of gas available from a compressed gas cylinder, we apply the Ideal Gas Law (PV = nRT). In a high-pressure cylinder, the volume will be affected by the content's compressibility factor Z (PV = ZnRT). For example, an AL cylinder of pure helium may contain 134 cu. ft. of gas while the same cylinder of pure air may contain 144 cu. ft. under the same conditions. For these practical calculations, however, we assume ideal gas behavior for simplicity.
To find the volume of gas available from a compressed gas cylinder, we apply the Ideal Gas Law (PV = nRT). In a high-pressure cylinder, the volume will be affected by the content's compressibility factor Z (PV = ZnRT). For example, an AL cylinder of pure helium may contain 134 cu. ft. of gas while the same cylinder of pure air may contain 144 cu. ft. under the same conditions. For these practical calculations, however, we assume ideal gas behavior for simplicity.
To find the volume of gas available from a compressed gas cylinder, we apply the Ideal Gas Law (PV = nRT). In a high-pressure cylinder, the volume will be affected by the content's compressibility factor Z (PV = ZnRT). For example, an AL cylinder of pure helium may contain 134 cu. ft. of gas while the same cylinder of pure air may contain 144 cu. ft. under the same conditions. For these practical calculations, however, we assume ideal gas behavior for simplicity.
Pipe dia1/2" 1" 3" 4"
L (m) 1.0000 15.1600 0.4000 51.0500D (m) 0.0127 0.0254 0.0762 0.1016
V (m3) 0.00012668 0.00760063 0.001824150 0.413878
P (1) x V (1) = P (2) x V (2)
Where:
P (1) is the pressure of the compressed gas in the cylinder (psi)
V (1) is the internal volume of the cylinder, often referred to as water volume (liter)*
P (2) is the atmospheric pressure (1 atm - 14.7 psi)
V (2) is the volume of gas at pressure P (2) (liter).
I. Perhitungan N2 untuk Purging
Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter
Kalkulasi: Berdasarkan Pendekatan Pamjang PipaKalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
Faktor pengali/konstanta 0.372 konstanta untuk setiap /1000 ft1000000 MM
Pipa 8"press barg 700 psigpanjang 6.495 m 21.3091 ftdia 8 inc
volume gas 16665.6 SCF/K ft355.12893696 SCF
0.0003551 MMSCF
pipa 6"press barg 680 psigpanjang 112.83 m 370.177 ftdia 6 in
volume 9106.56 SCF/Kft3371.03906112 SCF
0.003371039 MMSCF
pipa 3"Press 5 barg 72.5189 psigpanjang 9.131 m 29.9573 ftdia 3 in
Volume 242.7932772 SCF/Kft7.273431043064 SCF
0.0000072734 MMSCF
Total Pemakaian
http://www.webqc.org/ideal_gas_law.htmlsource :http://wiki.answers.com/Q/How_do_you_calculate_the_volume_of_natural_gas_in_a_pipeline
How do you calculate the volume of natural gas in a pipeline?Answer:You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
8 km = 26246 ft = 26.246 kft
65 Bar = 942.7 psi
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)
65x (Pi x 0.1778 x 0.1778 x 8000/4) = 0.9 x n x 8.314x10-5)x (273+6)
n = 618523 moles
1 kmol of a gas occupies 22.441 Nm3 at standard conditions
t.f 618.523 kmol should ocupy 618.523 x 22.441 = 13880.274643 Nm3.
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Amiel
Kalkulasi 2 : Berdasarkan Rumus Gas Ideal
Masukkan dan Cari Data Berikut ini :65 barg 6.5x10^6 pa 6500000 papanjang pipa 8 km 8000 meterdiameter pipa 7 inc 0.1778 meter 0.0889Temperature 6 deg celcius 279 kelvinVolume pipa 198.629Z 0.9R reinol numberPV=ZnRT
65x10^5 x (22/7)x 0.0889^2 x 8000 = 0.9 x n x 8.314 x (6 + 273)
1291610320 = 2087.6454 nn = 618692.389042698 mol
1 kmol = 22.41 Nm3 at standard condition Note1 Nm3 = 37.326 SCF Nm3
n 618.6923890427 Kmol Sm313864.89643845 Nm3 SCF517521.1244615 SCF517.5211244615 MSCF0.517521124461 MMSCF
measure at 0 C at 1 atm or 14.73 psiameasure at ………measure at 14.73 psia at 60 F
6" Total94.7000
0.1524
1.7274 2.15082946 m3
konstanta untuk setiap /1000 ft
0.0213091 kft
tinggal masukkan panjang dan pressure
0.0003551 MMSCF 8"
0.370177 Kft
0.00337103906112 MMSCF 6"
0.0299573 Kft
0.0000072734 MMSCF 3"
0.0037334414 MMSCF
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)0.0889
1 12916.1032 12916.103202 0.020876454 n 198.70928
n 618692.38904 198.70928
2790.00008314
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Input Processpsig 900 97397781.6708 paKM 26 26000 meterInch (dia) 28 0.7112 meterR 0.3556 meterVol 10332.88256
Z 0.9R 0.000083148F 80 353ZNRT 0.0264161196 n
Vol (1) 1006399839608.9n (2) 38097943787659.7 mol
38097943787.6597 Kmol853774920281.455 Nm3 1 kmol = 22.41 Nm3 at standard condition31868002674425.6 SCF 1 Nm3 = 37.326 SCF
Volume Gas 31868002.6744256 MMSCF n
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
914.696
1 kmol = 22.41 Nm3 at standard condition1 Nm3 = 37.326 SCF
3.81E+10 Kmol8.54E+11 Nm33.19E+13 SCF3.19E+10 MSCF
31868003 MMSCF
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Pipe dia1/2" 1" 3" 4" 6"
L (m) 1.0000 15.1600 0.4000 51.0500 94.7000D (m) 0.0127 0.0254 0.0762 0.1016 0.1524
V (m3) 0.00012668 0.00760063 0.001824150 0.413878 1.7274
P (1) x V (1) = P (2) x V (2)
Where:
P (1) is the pressure of the compressed gas in the cylinder (psi)
V (1) is the internal volume of the cylinder, often referred to as water volume (liter)*
P (2) is the atmospheric pressure (1 atm - 14.7 psi)
V (2) is the volume of gas at pressure P (2) (liter).
I. Perhitungan N2 untuk Purging
Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter
Kalkulasi: Berdasarkan Pendekatan Pamjang PipaKalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
Faktor pengali/konstanta 0.372 konstanta untuk setiap /1000 ft1000000 MM
Pipa 8"press barg 700 psigpanjang 6.495 m 21.3091 ftdia 8 inc
volume gas 16665.6 SCF/K ft355.12893696 SCF
0.0003551 MMSCF
pipa 6"press barg 680 psigpanjang 112.83 m 370.177 ftdia 6 in
volume 9106.56 SCF/Kft3371.03906112 SCF
0.003371039 MMSCF
pipa 3"Press 5 barg 72.5189 psigpanjang 9.131 m 29.9573 ftdia 3 in
Volume 242.7932772 SCF/Kft7.273431043064 SCF
0.0000072734 MMSCF
Total Pemakaian
http://www.webqc.org/ideal_gas_law.htmlsource :http://wiki.answers.com/Q/How_do_you_calculate_the_volume_of_natural_gas_in_a_pipeline
How do you calculate the volume of natural gas in a pipeline?Answer:You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
8 km = 26246 ft = 26.246 kft
65 Bar = 942.7 psi
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)
65x (Pi x 0.1778 x 0.1778 x 8000/4) = 0.9 x n x 8.314x10-5)x (273+6) 12
n = 618523 moles n
1 kmol of a gas occupies 22.441 Nm3 at standard conditions
t.f 618.523 kmol should ocupy 618.523 x 22.441 = 13880.274643 Nm3.
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Amiel
Kalkulasi 2 : Berdasarkan Rumus Gas Ideal
Masukkan dan Cari Data Berikut ini :65 barg 6.5x10^6 pa 6500000 papanjang pipa 8 km 8000 meterdiameter pipa 7 inc 0.1778 meter 0.0889 radiusTemperature 6 deg celcius 279 kelvinVolume pipa 198.629Z 0.9R reinol numberPV=ZnRT
65x10^5 x (22/7)x 0.0889^2 x 8000 = 0.9 x n x 8.314 x (6 + 273)
1291610320 = 2087.6454 nn = 618692.38904 mol
1 kmol = 22.41 Nm3 at standard condition Note1 Nm3 = 37.326 SCF Nm3 measure at 0 C at 1 atm or 14.73 psia
n 618.6923890427 Kmol Sm3 measure at ………13864.89643845 Nm3 SCF measure at 14.73 psia at 60 F517521.1244615 SCF517.5211244615 MSCF0.517521124461 MMSCF
Total
2.15082946 m3
0.0213091 kft
tinggal masukkan panjang dan pressure
0.0003551 MMSCF 8"
0.370177 Kft
0.00337103906112 MMSCF 6"
0.0299573 Kft
0.0000072734 MMSCF 3"
0.0037334414 MMSCF
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)0.0889
12916.1032 12916.103200.020876454 n 198.70928
618692.389 198.70928
2790.00008314
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
measure at 0 C at 1 atm or 14.73 psia
measure at 14.73 psia at 60 F
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Pipe dia1/2" 1" 3" 4"
L (m) 1.0000 15.1600 0.4000 51.0500D (m) 0.0127 0.0254 0.0762 0.1016
V (m3) 0.00012668 0.00760063 0.001824150 0.413878
P (1) x V (1) = P (2) x V (2)
Where:
P (1) is the pressure of the compressed gas in the cylinder (psi)
V (1) is the internal volume of the cylinder, often referred to as water volume (liter)*
P (2) is the atmospheric pressure (1 atm - 14.7 psi)
V (2) is the volume of gas at pressure P (2) (liter).
I. Perhitungan N2 untuk Purging
Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter
Kalkulasi: Berdasarkan Pendekatan Pamjang PipaKalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
Faktor pengali/konstanta 0.372 konstanta untuk setiap /1000 ft1000000 MM
Pipa 8"press barg 700 psigpanjang 6.495 m 21.3091 ftdia 8 inc
volume gas 16665.6 SCF/K ft355.12893696 SCF
0.0003551 MMSCF
pipa 6"press barg 680 psigpanjang 112.83 m 370.177 ftdia 6 in
volume 9106.56 SCF/Kft3371.03906112 SCF
0.003371039 MMSCF
pipa 3"Press 5 barg 72.5189 psigpanjang 9.131 m 29.9573 ftdia 3 in
Volume 242.7932772 SCF/Kft7.273431043064 SCF
0.0000072734 MMSCF
Total Pemakaian
http://www.webqc.org/ideal_gas_law.htmlsource :http://wiki.answers.com/Q/How_do_you_calculate_the_volume_of_natural_gas_in_a_pipeline
How do you calculate the volume of natural gas in a pipeline?Answer:You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
8 km = 26246 ft = 26.246 kft
65 Bar = 942.7 psi
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)
65x (Pi x 0.1778 x 0.1778 x 8000/4) = 0.9 x n x 8.314x10-5)x (273+6)
n = 618523 moles
1 kmol of a gas occupies 22.441 Nm3 at standard conditions
t.f 618.523 kmol should ocupy 618.523 x 22.441 = 13880.274643 Nm3.
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Amiel
Kalkulasi 2 : Berdasarkan Rumus Gas Ideal
Masukkan dan Cari Data Berikut ini :65 barg 6.5x10^6 pa 6500000 papanjang pipa 8 km 8000 meterdiameter pipa 7 inc 0.1778 meter 0.0889Temperature 6 deg celcius 279 kelvinVolume pipa 198.629Z 0.9R reinol numberPV=ZnRT
65x10^5 x (22/7)x 0.0889^2 x 8000 = 0.9 x n x 8.314 x (6 + 273)
1291610320 = 2087.6454 nn = 618692.389042698 mol
1 kmol = 22.41 Nm3 at standard condition Note1 Nm3 = 37.326 SCF Nm3
n 618.6923890427 Kmol Sm313864.89643845 Nm3 SCF517521.1244615 SCF517.5211244615 MSCF0.517521124461 MMSCF
measure at 0 C at 1 atm or 14.73 psiameasure at ………measure at 14.73 psia at 60 F
6" Total94.7000
0.1524
1.7274 2.15082946 m3
konstanta untuk setiap /1000 ft
0.0213091 kft
tinggal masukkan panjang dan pressure
0.0003551 MMSCF 8"
0.370177 Kft
0.00337103906112 MMSCF 6"
0.0299573 Kft
0.0000072734 MMSCF 3"
0.0037334414 MMSCF
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)0.0889
1 12916.1032 12916.103202 0.020876454 n 198.70928
n 618692.38904 198.70928
2790.00008314
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Input Processpsig 900 973.9782 barKM 26 26000 meterInch (dia) 28 0.7112 meterR 0.3556 feetVol 10332.88256 M3
Z 0.9R 0.00008314F 90 305.372222222222ZNRT 0.0228497819 n
Vol (1) 10064002.3566002n (2) 440441943.850685 mol
440441.943850685 Kmol9870303.96169384 Nm3 1 kmol = 22.41 Nm3 at standard condition368418965.674184 SCF 1 Nm3 = 37.326 SCF
Volume Gas 368.418965674184 MMSCF n
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
914.696
1 kmol = 22.41 Nm3 at standard condition1 Nm3 = 37.326 SCF
440441.9 Kmol9870304 Nm3
3.68E+08 SCF368419 MSCF
368.419 MMSCF
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Pipe dia1/2" 1" 3" 4"
L (m) 1.0000 15.1600 0.4000 51.0500D (m) 0.0127 0.0254 0.0762 0.1016
V (m3) 0.00012668 0.00760063 0.001824150 0.413878
P (1) x V (1) = P (2) x V (2)
Where:
P (1) is the pressure of the compressed gas in the cylinder (psi)
V (1) is the internal volume of the cylinder, often referred to as water volume (liter)*
P (2) is the atmospheric pressure (1 atm - 14.7 psi)
V (2) is the volume of gas at pressure P (2) (liter).
I. Perhitungan N2 untuk Purging
Pipa gas yang dihitung adalah mulai dari taping TGI sampai turbin meter
Kalkulasi: Berdasarkan Pendekatan Pamjang PipaKalikan luas penampang dalam pipa (inch) dengan pressure kerja (psig) dikalikan dengan 0.372
Faktor pengali/konstanta 0.372 konstanta untuk setiap /1000 ft1000000 MM
Pipa 8"press barg 700 psigpanjang 6.495 m 21.3091 ftdia 8 inc
volume gas 16665.6 SCF/K ft355.12893696 SCF
0.0003551 MMSCF
pipa 6"press barg 680 psigpanjang 112.83 m 370.177 ftdia 6 in
volume 9106.56 SCF/Kft3371.03906112 SCF
0.003371039 MMSCF
pipa 3"Press 5 barg 72.5189 psigpanjang 9.131 m 29.9573 ftdia 3 in
Volume 242.7932772 SCF/Kft7.273431043064 SCF
0.0000072734 MMSCF
Total Pemakaian
http://www.webqc.org/ideal_gas_law.htmlsource :http://wiki.answers.com/Q/How_do_you_calculate_the_volume_of_natural_gas_in_a_pipeline
How do you calculate the volume of natural gas in a pipeline?Answer:You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
8 km = 26246 ft = 26.246 kft
65 Bar = 942.7 psi
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)
65x (Pi x 0.1778 x 0.1778 x 8000/4) = 0.9 x n x 8.314x10-5)x (273+6)
n = 618523 moles
1 kmol of a gas occupies 22.441 Nm3 at standard conditions
t.f 618.523 kmol should ocupy 618.523 x 22.441 = 13880.274643 Nm3.
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Amiel
Kalkulasi 2 : Berdasarkan Rumus Gas Ideal
Masukkan dan Cari Data Berikut ini :65 barg 6.5x10^6 pa 6500000 papanjang pipa 8 km 8000 meterdiameter pipa 7 inc 0.1778 meter 0.0889Temperature 6 deg celcius 279 kelvinVolume pipa 198.629Z 0.9R reinol numberPV=ZnRT
65x10^5 x (22/7)x 0.0889^2 x 8000 = 0.9 x n x 8.314 x (6 + 273)
1291610320 = 2087.6454 nn = 618692.389042698 mol
1 kmol = 22.41 Nm3 at standard condition Note1 Nm3 = 37.326 SCF Nm3
n 618.6923890427 Kmol Sm313864.89643845 Nm3 SCF517521.1244615 SCF517.5211244615 MSCF0.517521124461 MMSCF
measure at 0 C at 1 atm or 14.73 psiameasure at ………measure at 14.73 psia at 60 F
6" Total94.7000
0.1524
1.7274 2.15082946 m3
konstanta untuk setiap /1000 ft
0.0213091 kft
tinggal masukkan panjang dan pressure
0.0003551 MMSCF 8"
0.370177 Kft
0.00337103906112 MMSCF 6"
0.0299573 Kft
0.0000072734 MMSCF 3"
0.0037334414 MMSCF
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.
Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the north sea)0.0889
1 12916.1032 12916.103202 0.020876454 n 198.70928
n 618692.38904 198.70928
2790.00008314
1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
Input Processbar 65 65 barKM 8 8000 meterInch (dia) 7 0.1778 meterR 0.0889 feetVol 198.70928 M3
Z 0.9R 0.00008314F 6 279ZNRT 0.020876454 n
Vol (1) 12916.1032n (2) 618692.389042698 mol
618.692389042698 Kmol13864.8964384469 Nm3 1 kmol = 22.41 Nm3 at standard condition517521.124461468 SCF 1 Nm3 = 37.326 SCF
Volume Gas 0.52 MMSCF n
You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
79.696
1 kmol = 22.41 Nm3 at standard condition1 Nm3 = 37.326 SCF
618.6924 Kmol13864.9 Nm3
517521.1 SCF517.5211 MSCF0.517521 MMSCF
Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.
