Air Flow Formula Help: Calculating % of Volume Lost in 2.8 Seconds

[Mark Selvey]

Hello

New to the forum and have a fairly simple problem.

I am looking for help in building an equation for volume lost in a container.

Example.

Known variables:

Container has a volume of 44 fl oz
Container will be pressurized to 6" of water column.
Container has an orifice with a .010" dia.
Once container has reached its 6.0" W.C. target pressure, the orifice will be opened for 2.8 seconds.
These are the only variables that can be determined.

How can I determine the % of volume of air lost in that 2.8 secs?
Can it be accomplished with information given?

I only need to find a close estimate so I am not really concerned with other variables such as Altitude, temp etc. I would prefer a clean/simple formula.

Thanks for any help.

 

[BruceW]

Welcome to the forum :)
Sorry, I'm having trouble understanding what the set-up is.
At the moment I am imagining a container full of air that someone is holding at a fixed depth (6 inches) under a lake. And that the container has a small hole in it, so it is filling with water.
And what does "once container has reached its 6 inches W.C. target pressure" mean?

 

[boneh3ad]

I think mass is a better measure for you here since air is highly compressible.

There are a few questions though. When you open your valve, are you keeping your pressure constant and allowing the volume to change (or feeding more ai into th box) or are you holding volume constant and letting pressure change?

 

[Mark Selvey]

Bruce
The container is pressurized to a target of 6" WC. Once the transducer sees the target has been reached, the fill valve closes and the orifice opens. 2.8 seconds later, the transducer takes another measurement and compares the 2 to determine pressure drop.
My problem is, I do not have access to the pressure readings in order to make a conversion from pressure lost to volume lost.

 

[Mark Selvey]

Boneh3ad.
No air is added to the container after the target pressure is reached.
When the orifice is open, the pressure will drop along with volume dropping.

 

[Mark Selvey]

Maybe a better description will help.

This involves a machine that is used to detect a leak in a container on a production line.
The machine needs to have test parameters ( Pass/Fail ) setup in order to detect holes.
The standard is the machine must detect a .010 dia hole.
It has a setpoint that is called "leak limit %" or in other words, how much air will it allow to be lost before it determines the container has a leak.
When a new product size is setup, the setup guy needs a starting point for lhis "leak limt %"
setpoint. I was hoping to be able to calculate that based on Container volume, pressure, orifice size and time.

 

[boneh3ad]

No, you have enough information. You can use Bernoulli's equation to get the velocity out of the orifice. That velocity will give you a mass flow rate because you will know the density based on your pressure and volume (ideal gas law). The problem is that your pressure inside the container is not steady.

At any rate, reducing Bernoulli's equation will give you the following:

$$p_{\textrm{res}} = p_{\textrm{atm}} + \frac{1}{2} \rho v_{\textrm{out}}^2$$

$$v_{\textrm{out}} = \sqrt{ \frac{2 \left(p_{\textrm{res}} - p_{\textrm{atm}} \right)}{\rho}}$$

$$\dot{m}_{\textrm{out}} = - \frac{dm}{dt}= \rho v_{\textrm{out}} A_{\textrm{out}}$$

$$-\frac{dm}{dt} = \frac{\pi D_{\textrm{out}}^2}{4}\sqrt{ 2 \rho \left(p_{\textrm{res}} - p_{\textrm{atm}} \right)}$$

Of course, we know that $p_{\textrm{res}} - p_{\textrm{atm}}$ is just the gage pressure read out on your manometer (since you are using inches of water, I just assumed that is how you were measuring it. Unfortunately, though, it won't help us because we have to use the absolute pressure in the box if we want to relate it to mass, not the gage pressure.

You know from the ideal gas law that $p_{\textrm{res}} = \rho \bar{R} T$ which can be plugged into our previous equation:

$$- \frac{dm}{dt} = \frac{\pi D_{\textrm{out}}^2}{4}\sqrt{ 2 \rho \left(\rho \bar{R} T - p_{\textrm{atm}} \right)}$$

We know the definition of density, so can plug $\rho = m/V_{\textrm{res}}$ into the equation.

$$- \frac{dm}{dt} = \frac{\pi D_{\textrm{out}}^2}{4}\sqrt{ 2 \frac{m}{V_{\textrm{res}}} \left(\frac{m}{V_{\textrm{res}}} \bar{R} T - p_{\textrm{atm}} \right)}$$

$$\frac{dm}{dt} = -\frac{\pi D_{\textrm{out}}^2}{4}\sqrt{ 2 \left( m(t)^2 \frac{\bar{R} T}{V_{\textrm{res}}^2} - m(t) \frac{p_{\textrm{atm}}}{V_{\textrm{res}}} \right)}$$

Unfortunately, that is pretty much as far as it goes. From there you have to solve the differential equation, which isn't separable or integrable, so you would need to do it numerically. That would give you a complete time history of the mass in the container for whatever time period you happen to want.

If you tried earlier to work with $p_{\textrm{gage}}$, you would end up with:

$$\frac{dm}{dt} = -\frac{\pi D_{\textrm{out}}^2}{4}\sqrt{ 2 \left( m(t) \frac{p_{\textrm{gage}}}{V_{\textrm{res}}} \right)}$$

This looks a lot nicer at first, but then you have to realize that $p_{\textrm{gage}}$ changes with time, and the only way to relate this to what is physically happening is through the definition of gage pressures, ending in the same place as before.

So again, that gives us the final answer of:

$$\boxed{\frac{dm}{dt} = -\frac{\pi D_{\textrm{out}}^2}{4}\sqrt{ 2 \left( m(t)^2 \frac{\bar{R} T}{V_{\textrm{res}}^2} - m(t) \frac{p_{\textrm{atm}}}{V_{\textrm{res}}} \right)}}$$

Which needs to be integrated numerically. The initial conditions are given from converting $p_{\textrm{gage}}$ into an initial mass, and an initial mass loss rate of zero.

 

[Mark Selvey]

Thanks Boneh3ad
I will try your final equation and see if the results are consistent with how the machine is responding .

Thanks for the help

 

[boneh3ad]

Yeah maybe someone else can check my math, but the concept is sound. You will definitely need to use a computer to integrate what I came up with so hopefully I am not leading you down the wrong path because of some stupid math error I made in haste somewhere... haha

 

[BruceW]

boneh3ad's analysis looks good to me. But you don't necessarily have to use numerical integration. If the change in mass is small compared to the initial mass, then you can just use the initial mass instead of time-dependent mass in boneh3ad's final equation.

 

[boneh3ad]

boneh3ad's analysis looks good to me. But you don't necessarily have to use numerical integration. If the change in mass is small compared to the initial mass, then you can just use the initial mass instead of time-dependent mass in boneh3ad's final equation.

True, that would give you a reasonable approximation. The container would have to be quite large and/or the gage pressure quite low. Given that the gage pressure here is about 6" of water, that comes out to something like 0.2 psi, which is pretty small and may allow for the constant mass approximation to be made.

 

[Mark Selvey]

I think constant mass will work for me also. Outside presure on container will be Atmospheric pressure, inside will be atmospheric plus .216 psi.
Thanks again guys