Calculating an orifice opening size - air flow

[Shrimpfork]

I need some help with this one. It has been a while since I played around with fluid dynamic and differential equations.

I have an air tank with a free air volume of 10.7 ft^3. I would like to know what size of an orifice is required to change the pressure in this tank from 5 PSF to 25 PSF in exactly 1.5 seconds. Is this possible to figure out or am I missing a variable?

 

[chicopee]

The equation presented is for a tank discharging,however, I believe that it can be used when pressurizing it.
For non-choked flow:
Q=C*A*P2*(2gM/RT)^.5*(k/(k-1))^.5*((P1/P2)^2k-(P1/P2)^((k+1)/k))^.5
Q=mass flow rate,lbm/s
A=orifice size,ft^2
C=coefficient, assume .75
P2=upstream pressure,lbf/ft^2 absolute
P1=downstream pressure,lbf/ft^2 absolute
k=Cp/Cv
M=molecular weight
R=universal gas constant,1545 ft-lbf/lbmole-dR
T=gas temp dR
In your case I would have P2=25 lbf/ft^2 and P1=5 lbf/ft^2 and I would consider the compression to be isothermal.
I don'thave time to check out the result and if it seems unreasonable interchange the values of P1 and P2.

 

[Shrimpfork]

I'm not sure if this is what I am looking for. I don't see if how this will tie into the time to discharge other then the mass flow rate. If so I'm not sure how I get the mass of the air that was moved between this time period of 1.5 seconds.

 

[vpl]

Shrimp

No offense, but that's what can happen when you ask someone to solve your problem for free. People only do what they have time and interest to do -- after all, you're not paying them.

BTW, I believe you're missing at least one variable, if not more. What is your charging source? Why do you believe that an orifice can control the charging rate (as it is more like a function of the pressure of the charging source rather than the receipt tank)?

Patricia Lougheed

******

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[israelkk]

What are you trying to achieve? Filling the tank in 1.5 sec will cause the air to heat. Therefore, even if you will have 25 psi after 1.5 sec, few second later the pressure will drop when the air is cooled. If you know the charging pressure an orifice diameter can be calculated. However, since the heat transfer properties of the vessel are not known a test or tests will be needed to refine the orifice diameter.

 

[Shrimpfork]

The pressure we are talking about is not very much. It is in PSF, not PSI. I'm not sure what I'm going to have for charging pressure yet. I am simulating a Mother Nature wind gust and I'll need to cycle between 5 PSF to 25 PSF and back to 5 PSF in a 3 second time period all inside a cavity of a wall.

Let's look at discharging the tank then. Assume there is 25 PSF in the tank, what is the orifice size required to discharge the tank to 5 PSF? This is what I need to get figured out.

 

[israelkk]

What is PSF?

 

[israelkk]

Charging and discharging times are different with same diameter orifice.

 

[Shrimpfork]

PSF = Pounds per Square Foot

 

[chicopee]

So far, I have not seen any other equations presented by the above responders.
In so far as the equation that I presented, rearange the terms to get A; also Q=mass/time; also this equation is for isothermal conditions and considering the relatively short time span the error should not be significant.

 

[Shrimpfork]

Chicopee,
I think I follow your post. However, I am unclear on how I am going to calculate the mass of air that is moved in that period of time to suffice the Q variable. Also I am not 100% clear for the values I need to calculate k.

Can you help me here?

PS: I also found this article that I am looking through on the same subject:

http://rothfus.cheme.cmu.edu/tlab/compflow/projects/t6s06/t6_s06_r3.pdf

 

[sailoday28]

If you neglect heat transfer,
mu - mu(initial)= integral of hodm
where m mass in receiver
u specific internal energy
ho stagnation enthalpy of source
dm represents change in mass flow
integral of dm = m - m(initial)

Does ho, the source stagnation enthalpy vary?
What is initial temp of air in the receiver?

Classic problem. Filling an evacuated tank with a perfect gas and constant specific heat from a fixed source.
mu =ho*m
CvT= CpTo
T=gamma * To gamma =ratio of specif heats, T= abs temp