Equalization Pressure of Two Air Receivers

Have you re-read the answers you got a couple of years ago?

This is a different application. My last submittal did not include the formula to compute the answer.

Thanks for your input.

The test papers don't seem to have changed much in a few years.

OOps

Mr.JohnDG - wow! you do keep records.

Your problem interested me and as you did not get -on my opinion a satisfactory answer- I worked it over and here are the results.

The answers you got do not correspond to the question you set because of several aspects. The most important one is that the equations used are valid for a constant gas volume and in the containers the gas mass is variable although the volume remains constant ( hard shell which do not collapse when pressure decreases). The problem can be solved by computing the MASS FLOW between the 2 receivers and the gas masses in the 2 shells over the time. The comments were correct it means the pressure and temperature in #1 decrease and the same 2 parameters will increase in #2 but not in the way the answer was done.

The exact solution should be obtained integrating a system of 2 differential equations :

dp1/dt = - (p1o*v1o^k)/V1*m'

dp2/dt = +(p2o*v2o^k)/V2*m' where

p1 = pressure in shell #1

p1o = initial pressure in shell #1 at t=0

V1 =volume of shell #1

m' =mass flow between the 2 shells.

The other parameters have same meaning for shell "2.

The equation for m' ( dm/dt) which considers the friction in the tube and the ration L/d is nonlinear and leads to very complex equations in above system.

I tried to find a simpler solution and I considered the flow as constant for a short interval of time (from 1 to 3 s as you will see later this is short with respect to time to end value) which allowed a simple integration procedure using an EXCEL sheet.

The following graphs give some information about results.

Graph 1 shows that for the pressure value at end of transfer the time step has no influence, only the time to reach the end pressure is changed but due to the small precision request this has no influence.

Graph 1

The test was done for extreme p2 levels = -11 and 80 psig. The result shows that the procedure is robust.

I shall take the questions as you formulated them and give one by one answers via graphs:

1- How long will it take to equalize the pressure between these two tanks using the pipe fittings above for p1o=175 psig and p2o=80psig ?

Graph 2

As mentioned above the wall friction has to be considered. For the graph above a very high value was used to obtain information about the highest possible times in case of wall high friction. The graph shows some interesting aspects:

- The decrease/increase rate is the highest at the start as it was expected

- The shell 1 has a smaller volume so that the same mass flow has a stronger effect the volume being at the denominator (see above equations)

- Even with a very high friction the time to end pressure is less 6'.

Graph 2.1

The graph shows how temperatures changes during the process and how the mass flow will decrease progressively as the pressure difference decreases as well.

Graph 3

The influence of friction on time is almost linear for frictions above 0.003 which are normal for ducts of this diameter. The previous graph was computed with a value 0.01. Usual values are in the range of 0.005 or even less. This shows that time will not be a problem.

2- How long will it take to equalize the pressure between these two tanks using the pipe fittings listed above if p1o= 175 PSIG and p2o= 0 ?

It is difficult to make the computations with an absolute p2o=0. I was limited to -11 psig which is near to the value expected.

For same values for other parameters the pressure p2o was given different values and results are in following graph.

Graph 4

The blue curve shows how the end pressure changes as function of the p2o value which was between -11 and 150 psig. The red curve shows how the time to reach the end value evolutes as function of same p2o.

3- What size pipe would you recommend between the two air receivers to allow equalization to occur within 5 to 10 minutes ?

Graph 5

Shows for the conditions as you request how the pipe internal diameter influences the time to end pressure. The diameter is in inches and the 2 scales are logarithmic.

A diameter of 3.5" would have already allowed around 300s=5 minutes.

5- How long it would take to get the air pressures to be within 10 PSIG between the two tanks with the smaller 30,000 tank being the higher pressure ?

Graph 6

Here you see the pressure difference versus time. The left scale is in N/mm² and the right in ppsi.

You see that the 10 psi difference will be reached around 120" = 2 minutes after opening the valve.

Remark:

The opening time of the valve is considered as very small with respect to the whole process so that the valve was considered from the start (t=0) as fully open.

I hope that now all your questions have been correctly answered.

In fact there is a source as it is very often recommended : the engineering forum but the equations are complex and limited for single pressure difference so that even saying where in the wikipedia you can find the equations it would not help much.

With respect to the computations I shall see if I bring them on forum for every one use nut not yet.

Any way you have the data you were interested to have so that your problem is solved.

Regards

Nick Name

PS.: Sorry for the delay but I was busy with other problems.

Nick Name,

Thank you for your response to this problem.

Would it be possible to obtain a copy of the excel spreadsheet you used to solve this problem?

Thanks,

CWALTER

I have it no more, in fact I was a bit frustrated that no positive comment was made, no appreciation of the work which was not done in 5 minutes so that I send the progam to the waste corb.

Sorry to hear that Nick, excellent work on this problem.

We use a value of Cv to describe the friction losses from the pipe.

The website http://www.pressure-drop.com/ allows for the calculation of the Cv using the pressure drop associated with an assumed flow rate through a straight pipe.

I came up with a Cv of 162 for 150 feet of 4 inch diameter hose.

Using a similar excel spread sheet results in the following graph which is pretty similar to the results you obtained.

Thank you for the appreciation. It is good that two different approaches lead to same result this is a confirmation for both.

I am now not any more sensitive to such reactions, CR4 had a couple of participants who were interested in such calculations. It is not any more the case.

I still follow it because it is from engineering point of view to see all problems which could appear.

Nick Name,

great work on this.

Please let me know what is k and v10 in the equations you used. is v10 the initial volume? you assumed that it's a constant correct?

dp1/dt = - (p1o*v1o^k)/V1*m'

dp2/dt = +(p2o*v2o^k)/V2*m'

Initial values and adiabatic exponent.

Curious many consider the work as good or even very good but no one gives apositive rate.

This is the reason I do not do similar works for CR4. People do not appeciate it.

Some were even frustrated because I made a simulation and they believe only in "cut and try" approach old and expensive.

sorry, I did not realize that (new to CR4). I just rated your response as good answer.