15 replies to this topic

[Indu]

Hi all,

This is my first time posting so please excuse any errors I might make here.

I have a tank with 4500 L that needs to be inerted and purged 6 times in 20minutes.

The inert gas source is 35 psi N2 supply from a header which is 3" and then the supply pipe itself is 1" sch 40.

The tank needs to be held under pressure at 18psi and then vented out (through a vent header) down to 4 psi. This process repeats 6 times and the requirement is that this be done in 20 minutes.
The vent pipe is 2", sch 40.

I need to find out the flow rate, check if the pipe and other gauges and filters in the line can handle it and also find the total nitrogen required.

Also, how do I calculate the time for each purge cycle?

Thanks,
Indu

[Padmakar Katre]

Dear Indu,
I wish if you would have searched in the che forums regarding your query probably now you would have an answer for it.Anyways the same thread is discussed in che forums and the links are as follow
http://www.cheresour...amp;hl=nitrogen
http://www.cheresour...amp;hl=nitrogen

Now after going through these posts still you have any doubt you are welcome.

Regards,
Padmakar Katre

[Indu]

Yes, i did look at the previous examples. But since my pressurizing is from 4psi to 18 psi and the de-pressurizing is back to 4 psi from 18 psi, wanted to know if I should be doing in increments since the flow will change as the pressure will change.

The way I did it was (with someone's help on this forum offline), get the lb/s flow and calculated the time to purge and de-purge the system with the total volume of gas and check to see if my pipe size can handle it.

But my question still remains if I shoudl just consider the average flow or if I should be doing the calculations in increments (the flow will be largest when the fill starts at 4 psi and reduces as the pressure in the tank increases).

The things I have not included which makes this more complex are thtat there is a PRV on the nitrogen line, a filter entering and exiting the tank and some other valves and nozzles. teh Ks for all need to considered while accomodating the 6 purge cycles in 20 minutes.

Any suggestions or recommendations?

Thanks,
INdu

[JEBradley]

I spent about 5 minutes trying to put together an answer to your question the other day. I dont think the other threads really addressed what you were asking.

Im sorry but I didnt have time to work it out - my maths is really crap. But can I suggest a way to solve it.

Ok make a sketch plot of how the pressure will vary. Plotting time on the x-axis and pressure on the y-axis. Draw horizontal lines at 35 psi (header pressure), 18 psi (pressurised state) and 4psi. Now starting from 0 psi (g) imagine the N₂ valve is opened. The vessel will fill with nitrogen and will pressurise to 35 psi. Of course it will stop at 18 psi either through a PSV or a PT driving the nitrogen valve. Anyway the rate of pressurisation can be solved as a 1st order ODE. If, like me, your maths is crap - solve it incrementally like you suggest - start at maybe 5s and then adjust til you have acceptable accuracy. There are many other posts about gas flow rate calculations on this board - i'd use the colebrook-white equation, work out the volumetric flow given the pressure difference and then from this volume work out the new pressure in the vessel - ideal gas equation. This creates a new pressure difference which you use for the next increment.

The de-pressurisation is the same but in reverse. I think you should set up a spreadsheet to speed things up, unless you choose to do it using an analytical method.

I hope this provides some ideas

[Indu]

OKay...Now I am troubled.

I have been calculting this in various forms and to make it simple just assumed pressurizing takes 75% of the time and de-pressurizing takes 25% and hence I will have to pressurize this in 2.5 minutes per cycle.
I used several formulae and was pretty happy untill I hit this.
I used to the eq. 1.7 in Crane 410 which gives the lb/s for P drop between 35 and with f of 0.028 at 60 deg F and a d of 1.37". I get a w of 0.918 lb/s, which is 220 cfm with a rho of 0.2495.

I wanted to check again and so used the formula 1.7a qhich is the same but gives the answer in cf/h and I get a cfm of 746.
Can someone tell me where I am doing wrong. I have spent 2 hours just checking this!!!!

Also, does any one know how to find the K value for a filter? I have not specified the filter yet, but need to be able to get a filter such that I can still do these cycles in 20 minutes.

Any information is helpful/.


Thank you all in advance!

-Indu

[JEBradley]

Oh sorry

I think you mentioned yourself earlier in the thread that you didnt think you could use something like the equation in crane because the pressure difference will be constantly changing. I agree with you on that point.

As for the K value on filters - phone up your local filter supplier and ask them. They probably won't have a K value but theyll tell you a fairly accurate pressure drop. I had to do this recently and was really surprised by the extremely low pressure drop you get through a filter these days.

Btw if it helps you - Im trying to solve this problem myself now (how sad am I??)

[marthin_was]

Hi Indu,

like the other Jedi replied, there is no exact equation for purging system. Hopefully following article will be usefull for you and others...

regards,
Marthin

Attached Files

•  Properly_Purge_and_Inert_Storage_Vessel.pdf   180.8KB   177 downloads

[Indu]

MArtin, Thank you for the article and I did read it earlier.

Jedi, thank you for trying it out. Hope you can solve it and I owe one for you!

But I still dont understand why I am getting such different answers from teh same equation in Crane!!!! it is driving me nuts!!!!!!!!!!

[JoeWong]

MArtin, Thank you for the article and I did read it earlier.

Jedi, thank you for trying it out. Hope you can solve it and I owe one for you!

But I still dont understand why I am getting such different answers from teh same equation in Crane!!!! it is driving me nuts!!!!!!!!!!

Indu,
You have mentioned several times that you got different answer from same equation in Crane. I still don't see how you apply those equations...if you can submit your calculation, i guess some of us may be able to help.

JoeWong

[Indu]

I used the equation
w = sqrt(144gA2(p1'2 - P2'2)/(vfL/D)

f = 0.0288
P1' = 49.7;P2' = 18.7
L = 25'
D = 1.37/12
v = 1/0.2945 lb/ft3

if I calcultaed this, my w is 0.905 lb/s == 217 cfm

But the same formula is expressed in Crane as
q'h = 114.2 sqrt ((P1'2 - P2'2)/f*Lm*T*Sg)d5) this formula gives me 700+ cfm..

that is my confusion. Not sure which number I should use....

[JoeWong]

I used the equation
w = sqrt(144gA2(p1'2 - P2'2)/(vfL/D)

f = 0.0288
P1' = 49.7;P2' = 18.7
L = 25'
D = 1.37/12
v = 1/0.2945 lb/ft3

if I calcultaed this, my w is 0.905 lb/s == 217 cfm

But the same formula is expressed in Crane as
q'h = 114.2 sqrt ((P1'2 - P2'2)/f*Lm*T*Sg)d5) this formula gives me 700+ cfm..

that is my confusion. Not sure which number I should use....

Indu,

Some simple comments without going in details...

i) You missed p1' for first equation.
ii) You have not advise the T & Sg used.
iii) Large DP --> Please read section 3-3, there have some limitation of equation to be used for compressible gas flow.
iv) pressure drop from source (35 psig) and receiver (4 psig) may be taken by throttling valve & line frictional lose. Majority of pressure drop may be taken by the throttling valve. Those line pressure drop will be much lower.
v) The flow across the throttling valve ma be under critical flow by looking at P2'/P1' (less than 0.5-0.55)
vi) You system pressure is rather low. Those the LN(P1/P2) section in equation 1-6 or 3-7 may seriously affect your accuracy.
vii) As you feed is at fix pressure (35psig) and your receiving system pressure will increase according to time, those you may need to you time-discrete method for better accuracy
viii) On the depressuring cycle, your system pressure (18 psig) will decrease to 4 (psig). Again, the DP will change with time.

Lastly, i doubt the question can be solved by just applying a simple equation.
I am sorry for raising quite a number of confusing comments. Please ignore if you think they do not help you.


JoeWong

[Indu]

Joe,
Thank you.
I missed the P1 in teh formula but I have used in my calculations so I still am amzed at the different answers.
the same thing seems to happen with the formulas 3-20 which are for more than 40% pressure change.
the q that I calculate from w, dividing by rho is much lower than if I used the formula directly for q.

I still am confused as to how to approach this problem.


any help is appreciated.

Thanks,
Indu

[JoeWong]

Joe,
Thank you.
I missed the P1 in teh formula but I have used in my calculations so I still am amzed at the different answers.
the same thing seems to happen with the formulas 3-20 which are for more than 40% pressure change.
the q that I calculate from w, dividing by rho is much lower than if I used the formula directly for q.

I still am confused as to how to approach this problem.


any help is appreciated.

Thanks,
Indu

Indu,
Just for clearing the equation issue, not the core subject...Can you pls advise the T & Sg used ?

JoeWong

[JEBradley]

Ive finished my calc for the initial pressurisation - just wanted to post it to see what people thought.

There are a few issues

1.) Initial velocity is over 900 m/s !!! impossible - but as it only last 1 s ive decided to overlook this point.
2.) Ive ignored gas compressibility. Ma<0.3 for all but the first few seconds so I hope this is a reasonable assumption.
3.) Final pressure (of each increment) is worked out by taking a volumetric flow from source (header) to destination (vessel). Assume this is transferred without expansion. Ive then considered the expansion separately and worked it out by normalising the volumes - e.g. 1m^3 at 2 bar + 2m^3 at 4 bar is normalised to 1x2 + 2x4 = 10 m^ at 1 bar. And then I took this combined volume and the volume of the vessel, along with the initial pressure to work out a new pressure (hope your following).

Lots of iffy points to consider but the end result seems promising. The graph is a nice curve and the pressurisation time is about 15 seconds. As you can see it shouldnt make a discernible difference if you're pressurising from 0-18 psig or 4-psig as the first bit of pressure flies in.

Hope you guys have a chance to look at it and please feel free to criticise - this is just a working calc and I hope we can refine it and get the correct answers eventually.

Attached Files

•  Vessel_Purging.xls   259.5KB   116 downloads

[Indu]

Joe,
Thank you.
I missed the P1 in teh formula but I have used in my calculations so I still am amzed at the different answers.
the same thing seems to happen with the formulas 3-20 which are for more than 40% pressure change.
the q that I calculate from w, dividing by rho is much lower than if I used the formula directly for q.

I still am confused as to how to approach this problem.


any help is appreciated.

Thanks,
Indu

Indu,
Just for clearing the equation issue, not the core subject...Can you pls advise the T & Sg used ?


Sg = 0.9666
and T is 60 deg F

JoeWong

[JoeWong]

Indu,

You just driving yourself crazy...


Eq. 1.7 will give you mass flow (w) in lb/s. You use the density (0.2945 lb/ft3) to get your Actual volumetric flow (~217 CFM or to be exact 217 ACFM).

On the other hand, eq. 1.7a will give you volumetric flowrate (q'h) at Standard condition. As you claimed you got 700+ CFM (to be exact is 700+ SCFM)

As temperature is 60degF, then no correction on the temperature but correction on pressure from Actual to Standard...

217 ACFM
=> 217 x (49.7 / 14.7) SCFM
=> 733 SCFM which is quite inline with your result, 700+ SCFM.


Hope this clear your doubt on the equations.

JoeWong