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Weight of gas left in vessel, non choked conditions

Weight of gas left in vessel, non choked conditions

Weight of gas left in vessel, non choked conditions

(OP)
I know how to figure out the amount of gas left in a vessel during choked flow conditions through an orifice at a given time. http://www.air-dispersion.com/feature2.html explains this.  

I know how to figure out the instantaneous non-choked flow from a vessel through at a given pressure.  

I was wondering what you do to calculate the amount of gas that is left in the tank, or the pressure in the tank, during non choked conditions at a given time?  I guess a conservative answer would be to use the minimum choked flow rate and treat that as constant after using the choked flow method to figure out the amount released up until that the flow becomes non choked.  Or in the case that there was never choked flow to begin with, use the initial non choked flow rate and treat that as constant.  There must be a more accurate way to describe the flow change with respect to time during these conditions in order to have a more accurate release calculation.  

This is assuming that the vessel doesn’t quite reach atmospheric conditions so that there is still some inventory left in the vessel.  
Possible scenario: partially opened valve that was found and shut or something like that.

I was just wondering this after stumbling upon and looking through www.air-dispersion.com, which btw is a great site.

/first posted question
//hope I’m not missing something obvious
///sorry ‘bout the length

RE: Weight of gas left in vessel, non choked conditions

Dezirak

For any problem like this you can usually get a first order idea by using the standard growth/decay equation
Q = Qo * exp(-T/k)
or Q = Qo * [1- exp(-T/k)]
{Plot them to see which way they go}

Qo is the quantity of "stuff" (or property) at the beginning of the timed period (T = 0)
Q is the value at time = T
k is the time constant

You will see than when T/k = 5 you hit 99.3% of the final condition.
When T/k = 6 you hit 99.75% of the final condition.
When T/k = 10 - 99.9955% etc

This allows you to "fudge" a value of k from the initial rate of change dQ/dT, which is the instantaneous rate of change (flow rate?) at the known pressure condition when T = 0.

If you assume that your vessel goes completely to atmospheric over the time T/k = 6 (for example)
then k = [6 * Qo] / [dQ/dt]

Plugging that value of k back into the formula allows you to find the residual at any other time span.

This works for a lot of different engineering problems as long as you don't mix units and you understand your time basis and you don't have any variable damping in the system (in which case it becomes 2nd order and I can't explain it too well here - I probably couldn't explain it at all but that's another story)

In the choked condition you have to use a choked flow relationship to find the initial rate of flow and the formula would only be useable in the choked range.  At the critical pressure you would have to start the clock again using the non-choked formula, (which is effectively the friction/velocity pressure formula for contraction and expansion though the hole.)

Remember that, if bleeding down to atmospheric, the formula only sees the differences so you would have to calculate the corresponding absolute values if that is what you need.
You need to think it through before you do the calc.

If this is too rough a solution for you, I seem to recall that somewhere on this site (I'm not sure what keyword you would need) there is a link to an erudite treatise which gives you a more accurate solution (it's a few years ago !!).  It all depends on what sort of accuracy you are using overall and whether you need a scientific or an engineering answer.
Perhaps other contributors may recall the link and be able to help out on this.

Good luck
David

RE: Weight of gas left in vessel, non choked conditions

dezirak:

First let me thank you for your kind words about my website. If you will contact me there, I can tell you one way to go about what you want to do.

Milton Beychok
(Visit me at www.air-dispersion.com)
.

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