Calculation pressure loss rate

[Rattletrap]

Hi,

I would like to calculate the pressure drop in a chamber when there is an orifice with a certain flow rate.

The chamber volume about 950 ML.

Ambient temperature may vary between ambient and say 30 degrees C.

The pressure drop to the chamber is from 101 kPa to 20 kPa over a minute. The unit is subjected to a vacuum caused by piston effect in a tunnel.

I would like to keep the pressure difference between the ambient and the chamber at about 10 Kpa to prevent stresses that may cause failure to the part.

So I need to know the final flow rate that will give me the size of the orifice that I need to use.

Anyone out there with a solution to this?

Thanks in advance

 

[israelkk]

Is the volume 950 Milliliter (950 cm^3)?
I understand the initial pressure is 101 kPa (1 atm) and the after 1 minute the pressure in the chamber will be 20 kPa. Therefore, what do you mean by the "I would like to keep the pressure difference between the ambient and the chamber at about 10 kPa" statement. The ambient pressure is normally 101 kPa?

 

[Rattletraped]

Hi Israelk

The ambient is changing from 1 atmosphere to a vacuum of about 20 kPa. So by saying ambient, in this case it is the ambient that is changing. The chamber needs to follow.

That is from 101 kPa to 20 Kpa. so the chamber have the original pressure that is the 101 Kpa inside of it if the unit was sealed. I cant allow for this as the pressure difference needs to be not more than 15 Kpa between the inside and the outside of the chamber, as the chamber is not strong enough to hold it due to deformation of some plastic covers.

what I really need to know is how to calculate what the flow rate of a designed orifice should be to ensure that the pressure difference remain within 10 - 15 kPa.
I can see that the flow will be greater when the pressure difference if higher than what it will be when the pressure is close to equal.

The volume is nearly 1 liter yes.

Any ideas?



Thanks,

Stan

 

[austsa]

If I understand correctly, you have a 950mL chamber that starts at 1 atm pressure and ..room?.. temperature. You need to calculate the mass of its contents at this initial stage. At the final stage, the chamber needs to be 20kPa and ...?.. temperature. Calculate the mass of its contents at this final stage. Then calculate mass flow rate (loss) required and convert to volume flow rate based on an average pressure. This is assuming that your chamber is a rigid body.

 

[israelkk]

Assuming the ambient pressure decreases at a constant rate from 101kPa to 20kPa in 60 seconds, you need an orifice that under the 10-15kPa pressure difference between the pressure inside the chamber and the ambient pressure will quite follow same pressure decrease rate. Since the pressure difference is only 10-15kPa therefore, when the ambient pressure reached 20kPa the pressure in the chamber will 30-35kPa. Therefore, the whole time the flow will be in the non choked range of compressible air flow. The pressure change inside the chamber will not be at constant rate. To solve this problem you need to use a set of differential equations simultaneously numerically solved. It may be that using a constant size orifice will not be enough. After solving the equations you will know if a constant size orifice can keep the pressure difference at 10-15kPa all the way from 101kPa to 20kPa ambient pressure.

 

[prex]

Rattletrap or Rattletraped (?),
Let's call R the rate of change in time of the external pressure, assuming of course it is small enough to let us consider each state of the vessel as a steady state.
Now ΔV/V=ΔP/P and ΔV/VΔt=ΔP/PΔt or Q=VR/P where Q=volumetric flow rate through the orifice and P,V are pressure and volume in the vessel.
The volumetric flow rate is (nearly) proportional to √(dp/ρ) where dp is the pressure loss across the orifice, that is Q=K√(dp/P) where K is (nearly) a constant that depends mainly on orifice area and other (nearly) constant factors: the expression for K is easily derived from the orifice equation and the gas law.
Finally K=VR/√(Pdp). As V is constant, dp is a datum, and if R is constant across the pressure range (or higher at the higher pressures), then K should be smaller at the higher pressures.
In conclusion you should dimension the orifice for giving Q=VR/P by using P=101 kPa and a pressure loss of 10 (or 15) kPa. The pressure loss at the lower pressures (R remaining the same or smaller) will be smaller.

prex
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[Rattletraped]

Thanks all, I will have a go at the suggestions. All make sense to me now. I am also building a testing apparatus to compare theory against a physical test.



Thanks,

Stan