Need Expert Advice on Calculating Pressure Buildup in Fixed Volume Vessel 4

[_METCON]

"Hello community,

I'm currently facing a challenging situation and would greatly appreciate some expert advice. Here's the scenario:

I have a fixed volume vessel with a total volume of 15458 ft³. The volumetric flow rate through this volume is 90 MMSCFD. The initial temperature of the gas is 80°F, with a Compressibility Factor of approximately 0.91 and a Molecular Weight of 22.9. The source pressure is 300 PSIG, and the normal operating pressure is 100 PSIG.

My concern is understanding the time it would take for the pressure to reach 250 PSIG if the exit of the vessel were to be suddenly shut. Assume no heat transfer and no additional information on piping dimensions.

I'm reaching out to this knowledgeable community for assistance in calculating this pressure buildup time. Any insights, formulas, or guidance on how to approach this problem would be immensely helpful.

Your expertise is highly valued, and I'm eager to hear your thoughts on this matter. Thank you in advance for your time and assistance!"

Feel free to make any adjustments or let me know if you have specific preferences!

 

[1503-44]

P1 V1 = P2 V2
265 x 15458 = 4096370 ft3 x psia
115 x 15458 = 1777670 ft3 x paia

Increase in volume factor
4096370 / 17776701 = 2.304

New gas volume
V2 = 2.304 x 15458 = 35621 ft3

Volume added to reach 250 psig
35651 -15458 = 20163 ft3

90 MMSCFD x 15 / (250+15) = 6.36 MMCFD @ 250 psig

20163/6360000 = 0.00317 days
0.0761 h
4.56 minutes


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[LittleInch]

A lot of variables.

How is your 300 psi source controlled down to 100 psi?

Does it control on pressure or flow?

If fixed flow it's about 35 minutes. However if flow is impacted by the pressure in this very large vessel then it can change.
[Edit] I was working in different units....
Mr 44, you've mixed actual volume and std volume.... Or maybe not

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

Where? The 15458 ft3 is actual volume
I think I remember you thought the same with that last LNG calc I did,, but no, it's not mixed here.
Actual volume / actual flow rate = actual time.

If it takes 30m you've added 90E6 x 15 / 265 x 0.5/24 = 432,629 ft3 @ 265 psia

90,000,000 x 15/265 = 5.1 MMACFD
Or about 3500 Actual ft3/minute

However we can use the average flow rate to get a better time if you are filling from 100 to 250 psig.

90x15/115 = 11.7 MMACFD @ 100 psig
90x15/265 = 6.36 MMACFD @ 250 psig

9 MMACFD

20163/9E6 = 0.0022 Days = 3.2 minutes

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[pierreick]

Hi,
My answer is 2'28" with STP (20C and 14.7 PSI)
Mass increase 4570 kg,
Mass flowrate 30.73 kg/s
hypothesis: T=cte, Mass flowrate =cte,
Pierre

 

[LittleInch]

Yup,

I got that wrong. I thought 30 mins sounded a bit too high.

But does depend if this is flow controlled or not.

This is pretty basic stuff though....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

OK. Thought so, but I was worried until I checked. Thought I bungled the compression ratio or something. I wanted to do it with PV=ZnRT, but those are all relatively constant, so it reduces to just PV terms. I hate it when I get those "that looks high" moments and I just keep on with the running the calcs. Have to listen to those moments. It's experience talking.

Pierre seems to have a lower number.

Yes, FCV Apparently at 90 MMSCFD.



--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[_METCON]

@1503-44, @Littleinch, and @pierreick

I approached this a bit differently...I approached this from the work equation by integrating for change in pressure. w= ∫ VdP = P1*V1*ln( P2/P1 ). Finding the work to fill the volume from 100PSIG to 250PSIG, then finding the work done by compression , inlet flowrate 90MMSCFD and source pressure 300PSIG. Then dividing the work to fill the volume by the work done by compression to find the time. Flowrate provides the time unit. Is this the correct approach? Maybe more importantly, is my integration correct? It has been a while...

Find work to fill volume:
System Volume = V1 = 15458 ft3 = 12830.26 m3
Final Target Pressure = P2 = 264.7 PSIA = 1825106.5 Pa
Initial Operating Pressure = P1 = 114.7 PSIA = 790856.5 Pa

Work to fill volume = ∫VdP = P1V1 ln( P2/P1 ) = 8485613776.28 J

Find Compression work:
Volume Rate of Flow = V2 = 90 MMSCFD = 29.5 m3/s
Pressure of delivery = P3 = 314.7 PSIA = 2169857 Pa

Compression Work = ∫VdP = P1V2 ln( P3/P1 ) = 23544611.13 J/s

Time to fill volume: Work to fill volume/Compression work = 360.41s = 6.01 min

@1504-44
I tried to take an PV=nRT approach to this with the molar flow rate of the gas but I can't seem to make it work...I kept having the same thought as you...

@ Littleinch
How is your 300 psi source controlled down to 100 psi? DP is pressure drop in the line to the vessel, no control valves.
Does it control on pressure or flow? I would assume, in the real world, flowrate would decrease, and pressure would increase as the flow out has been blocked. In the above approach I have assumed flowrate to be constant.

Thank you all.

 

[LittleInch]

But you have no compressor so there is no "work". I really don't think that approach would work, but is fairly novel.

"no control valves" - Aaaaah.

It's a very conservative assumption for sure to just use 90 MMSCFD throughout. I assume there is a fairly long length of pipeline between your vessel and the source point. There will be a significant difference between flow with a DP of 200 psi and one with only 50.

However its still not going to be more than 10 minutes, but only a transient analysis will give you that to any level of accuracy. How accurate do you want it. Would somewhere between 5 to 10 minutes be enough or do you love the false accuracy of a computer analysis which would be 6 minutes, 32.7643 seconds??

Or do it in steps of say 50psi and integrate assuming flow stays the same until the pressure in the vessel increase by 50 psi. Then recalculate your incoming flow and do the same. a little bit conservative, but not as bad as a fixed 90 MMSCFD.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[_METCON]

@LittleInch can I not assume there is a "compressor" supplying 90MMSCFD at 300PSIG at the inlet of the system? Regardless if there is an actual compressor there or not?

I'm only looking for "rule of thumb" accuracy here and am not going to look into any transient analysis. Just need an approximate time required. I'm not sure which one to run with now.

 

[LittleInch]

Just use volume or mass flow.

Far easier.

You have your appro times already. Between 5 and 10 minutes.

Actually I forgot as the pressure increaes in the pipeline some of the mass flow will be stored in the pipeline so that's another factor increasing pressurisation time. [pre][/pre]

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Latexman]

I found a mass flow rate at a 200 psi differential (300 psi - 100 psi), and at a 50 psi differential (300 psi - 250 psi). Took the average of the two flow rates, and came up with 3-4 minutes. Same ball park as others found.

Good Luck,
Latexman

 

[1503-44]

Maybe. But the pipeline is upstream of the flow controllers set at 90MMSCFD. Given the initial information, I supposed that upstream segment is already at 300 psig and the flow control valves are properly sized to allow the passage of 90MMSCFD with an initial 200 psig pressure drop and 100 psig discharge into the vessel. As the FCVs are changed to discharge 250psig with a 50psi drop, that results in 90MMSCFD and 250 psig flow into the vessel, thus no significant additional time for pressure build up would be required, assuming the pipe from FCVs to vessel is relatively short and small D.

Note that I supposed that the flow rate was maintained at 90MM for the pressure range 100 to 250psig, as you did not mention any other flow rates.

If there is something different there, tell us now.

Should be the same answer. The mass cancels out.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Latexman]

But the pipeline is upstream of the flow controllers set at 90MMSCFD.

Maybe not.

DP is pressure drop in the line to the vessel, no control valves.

A sketch/PFD/P&ID sure would be nice.

Good Luck,
Latexman

 

[LittleInch]

Mr 44. My reading of what the OP has said is that there are no CVs upstream the vessel. -"
DP is pressure drop in the line to the vessel, no control valves"

There is a pressure source (well?, offtake? who knows), what must be quite a long pipeline (no data), then this mysterious vessel which is probably kept at 100psi which results in a flow of 90 MMSCFD.

The OP is now trying to work out what happens if the outflow stops. The flow rate will clearly fall as the pressure in the vessel rises assuming the start pressure of 300 psi stays the same.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

Yes, I know. As usual, not all info has arrived, It rarely does and solutions are subject to revision as it does.

"DP is a pressure drop in the line. No controllers."
Then LI's pipe fill scenario may be important, but only if it involves a lot of pipe volume.
What might affect it more is, without controllers, the pressure build will not be linear and the average flow rate could be a lot lower. Still, ìt may not change the fill time very much. Can't say until, or if we get pipe dimensions and a curve showing how the source pressure rises with flow rate. If there are no flow control valves, something has to do it. Is there a compressor, or what.

I only filled in those info gaps with a system description consistent with the solutions we have arrived at up to this point. It gives the OP an idea of what might need clarification. We should let him do that now.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[georgeverghese]

There are, I think, 2 possible scenarios here, all of which involve a short feed pipe:
a) Upstream flow is set on flow control at an upstream FCV - in this case, flow will increase when the outlet from the vessel is shut - the control valve will open more than it was initially.
b) Upstream flow is set at a fixed manual throttle valve - in this case, flow will decrease when the outlet from the vessel is shut. Flow is initially in choke flow mode, since p2 < 0.5x p1, but will later become subsonic as p2 > 0.5 x p1

Which one do you have ? Or is it some other configuration ? As you can infer, the result will differ considerably depending on configuration

 

[LittleInch]

Or, more likely IMO, the inlet to the pipe is pressure controlled to 300psi?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[pierreick]

Hi,
To add to the discussion :

https://community.ptc.com/t5/Mathca...l-finite-difference-approximation/td-p/748959

Pierre

 

[1503-44]

Only know 3 things.
Pipe pressure is initially 100psig.
It changes to 250 when the vessel is full.
Flow is 90MM.
Problem is that nobody knows the upstream boundary conditions, Q vs P.

We should probably assume the simplest config.
No valves at all, with a 100psig discharge when vessel is open through and 250 when finished.

That the vessel defines the downstream boundary conditions.
Pipe pressure rises according to the volume added to the vessel, initially at 100 psig and Q=90MM,
No valves at all, with a 100psig discharge when vessel is open through,
backing up to 250psig when vessel is closed.
BUT flow would decrease as the vessel pressure rises to 250, which violates the 90MM constant flow.
If that scenario is allowed, it can be solved iteratively, stepping through time.
And takes a bit longer than 4m, but probably not too much.

It also looked like too much work yesterday morning. It was a lot easier to assume a constant 90MM flow as stated.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."