|

Co2 Pipeline
Started by Boualem, Dec 29 2011 11:48 PM
8 replies to this topic
Share this topic:
#1
Posted 29 December 2011 - 11:48 PM
Hello every one.
I calculated the flow rate of a liquid CO2 (Supercritical) through a pipeline, the result I found was in MMACFD (Actual cubic feet), I am wondering if there is a formula to convert this volume (flow rate) into standard conditions of pressure and temperature.
What are the conditions to take into account while sizing pipelines and equipment dealing with Liquid CO2.
Thank you
I calculated the flow rate of a liquid CO2 (Supercritical) through a pipeline, the result I found was in MMACFD (Actual cubic feet), I am wondering if there is a formula to convert this volume (flow rate) into standard conditions of pressure and temperature.
What are the conditions to take into account while sizing pipelines and equipment dealing with Liquid CO2.
Thank you
#2
Posted 30 December 2011 - 03:01 AM
Use General gas equation for convert MMACFD to MMSCFD,
#3
Posted 30 December 2011 - 06:20 AM
Boualem,
Paul
- Find the density of the liquid CO2 at the operating temperature.
- Convert your volume flow to mass flow
- Convert the mass flow to molar flow by dividing by 44
- From molar flow you can work out the equivalent gas volume at standard gas conditions in MMSCFD. One lbmol = 379.3 SCF. One kgmol = 23.63 Sm3
Paul
#4
Posted 30 December 2011 - 11:23 AM
Thank you for your reply,
I tried the method mentioned by Paul, i think I didn't work for the folloowing conditions: P = 2200 Psig and T = 100 F (Fluid is: Liquid CO2 at supercritical conditions).
I am wondering if we can use this relations: Q(St) = Q(Act)*(Pact/Pst)*(Tst/Tact)
Thank you
I tried the method mentioned by Paul, i think I didn't work for the folloowing conditions: P = 2200 Psig and T = 100 F (Fluid is: Liquid CO2 at supercritical conditions).
I am wondering if we can use this relations: Q(St) = Q(Act)*(Pact/Pst)*(Tst/Tact)
Thank you
#5
Posted 30 December 2011 - 12:30 PM
Boualem,
The equation you give is p1.V1/T1 =p2.V2/T2 which is the ideal gas law.
You do not have an ideal gas. You do not have even a non-ideal gas. You have a LIQUID. Therefore this relation is not applicable.
To calculate the equivalent standard gas volume flowrate from your liquid volume flowrate, you have to calculate mass flow and molar flow, like I said.
Now, what is the density of the liquid CO2?
Paul
The equation you give is p1.V1/T1 =p2.V2/T2 which is the ideal gas law.
You do not have an ideal gas. You do not have even a non-ideal gas. You have a LIQUID. Therefore this relation is not applicable.
To calculate the equivalent standard gas volume flowrate from your liquid volume flowrate, you have to calculate mass flow and molar flow, like I said.
Now, what is the density of the liquid CO2?
Paul
#6
Posted 30 December 2011 - 12:55 PM
Paul
At these conditions, P = 2200 Psig and T = 80 F is (based on an online program) 778 Kg/m3.
The product is a supercritical (P over or equal to 72 bar and T over or equal to 31 C).
Thank you
At these conditions, P = 2200 Psig and T = 80 F is (based on an online program) 778 Kg/m3.
The product is a supercritical (P over or equal to 72 bar and T over or equal to 31 C).
Thank you
#7
Posted 30 December 2011 - 03:48 PM
Referred CO2 seems to have pressure 2200 psig and temperature 80 - 100 oF. As the diagram of http://en.wikipedia...._carbon_dioxide indicates, at 80 oC it is liquid, at 100 oC supercritical fluid. Critical temperature of CO2 is 31.1 oC = 88 oF. Difference may not be significant, apart from the fact that supercritical fluid will not develop two phases, as long as it remains in the supercritical region.
Density at 2200 psig is calculated through the compressibility factor method, offered by breizh (critical pressure=73.9 bara, see http://www.cheresour...n-control-valve), post No 5. Result is ρ=706 kg/m3 at 100 oF and 767 kg/m3 at 80 oF, according to this method.
Probably it would be better to express rate of CO2 as mass flow, multiplying volumetric flow rate at known temperature and pressure by the resulting density. Standard conditions refer to 1 Atm a (=1.01325 Bara) and 15 oC (*). Gas molar volume at standard conditions is 22.414* (273+15)/273=23.65 m3/mol, and corresponding density of CO2 44/23.65=1.86 kg/m3. This assumes ideal gas at standard conditions, method as above would give ρ=1.87 kg/m3, which is more precise. So flow at standard conditions can be found from mass flow divided by density at standard conditions.
Example: 1000 MMACFD at 2200 psig and 80 oF is equivalent to 1000*767/1.87 = 410160 MMSCFD of CO2 gas.
* see http://www.cheresour...and-actual-m3hr.
Density at 2200 psig is calculated through the compressibility factor method, offered by breizh (critical pressure=73.9 bara, see http://www.cheresour...n-control-valve), post No 5. Result is ρ=706 kg/m3 at 100 oF and 767 kg/m3 at 80 oF, according to this method.
Probably it would be better to express rate of CO2 as mass flow, multiplying volumetric flow rate at known temperature and pressure by the resulting density. Standard conditions refer to 1 Atm a (=1.01325 Bara) and 15 oC (*). Gas molar volume at standard conditions is 22.414* (273+15)/273=23.65 m3/mol, and corresponding density of CO2 44/23.65=1.86 kg/m3. This assumes ideal gas at standard conditions, method as above would give ρ=1.87 kg/m3, which is more precise. So flow at standard conditions can be found from mass flow divided by density at standard conditions.
Example: 1000 MMACFD at 2200 psig and 80 oF is equivalent to 1000*767/1.87 = 410160 MMSCFD of CO2 gas.
* see http://www.cheresour...and-actual-m3hr.
Edited by kkala, 30 December 2011 - 04:11 PM.
#8
Posted 30 December 2011 - 05:00 PM
Boualem,
OK, so the liquid density is 778 kg/m3 at your operating conditions. This equals 778 x 0.0624 = 48.5 lb/cu.ft.
You said
Suppose your volume flowrate is V expressed in million actual cubic feet per day of this liquid.
We convert to mass flow: m = V x 48.5 million lb/day
to molar flow; M = m/44 = Vx 48.5/44 million lbmol/day
This quantity of CO2 when converted to gas at standard conditions has a volume of
Q (gas) = M x 379.3 mmSCFD (this conversion is valid for any gas)
Q (gas) = V x 48.5/44 x 379.3 mmSCFD
There is your answer, I leave the arithmetic to you.
Paul
OK, so the liquid density is 778 kg/m3 at your operating conditions. This equals 778 x 0.0624 = 48.5 lb/cu.ft.
You said
I calculated the flow rate of a liquid CO2 (Supercritical) through a pipeline, the result I found was in MMACFD (Actual cubic feet)
Suppose your volume flowrate is V expressed in million actual cubic feet per day of this liquid.
We convert to mass flow: m = V x 48.5 million lb/day
to molar flow; M = m/44 = Vx 48.5/44 million lbmol/day
This quantity of CO2 when converted to gas at standard conditions has a volume of
Q (gas) = M x 379.3 mmSCFD (this conversion is valid for any gas)
Q (gas) = V x 48.5/44 x 379.3 mmSCFD
There is your answer, I leave the arithmetic to you.
Paul
#9
Posted 02 January 2012 - 09:19 PM
Thank you so much for your reply Paul, KKala and Zak, I really appreciated it.
Similar Topics
Determining Pipeline Design PressureStarted by Guest_applepe_* , 05 Dec 2024 |
|
![]() |
||
![]() ![]() Energy Saving In Pump By Reducing Pressure Drop In Downstream PipelineStarted by Guest_Srinivas Agastheeswaran_* , 19 Nov 2024 |
|
![]() |
||
![]() ![]() Steam Condensate Pipeline Water HammerStarted by Guest_kaidlut_* , 06 Sep 2024 |
|
![]() |
||
![]() Achieving Ultra-Low Water Content (<0.1 Ppmv) For Pipeline PiggingStarted by Guest_Finchh_* , 21 Aug 2024 |
|
![]() |
||
Pipeline Material SelctionStarted by Guest_BabRafiq1_* , 06 Sep 2022 |
|
![]() |