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### Topic: Relating psi and g (gravitational)  (Read 6949 times)

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#### Arkcon

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##### Relating psi and g (gravitational)
« on: September 11, 2016, 12:33:27 PM »
Kinda a weird physics question, I can explain it, but I can't really find the solution.

Briefly, I want to replace some bottles rated to be used under a certain gas pressure with some other bottles, rated for use in a centrifuge, at a certain g.

For more specifics, the bottles filled with a solution and attached to a manifold under argon at a pressure of 4-6 psi.  (Sorry for non-SI units, we can fix that later.)  A valve releases liquid and the pressure allows dispensing.  This is, in fact, an oligonucleotide synthesizer.

I want to replace the standard glass bottles with a centrifuge bottle, because it has a conical bottom which saves reagents.  Your average 100 mL Kimax glass centrifuge bottle has a g rating of 2000xg to 3000xg, depending on wall thickness.

Any one of those should handle any measly psi rating.  So I'm done here, thread over, just buy it, Arkcon.  Well, not so fast.  I would like to be able to prove I haven't bought the wrong thing, and justify why I don't buy thicker walled for added strength, or do buy thinner walled, and save money.  And this is also a fun logical problem to solve.

I did some Googling, and I found something on Yahoo.  Basically, there is minimal connection between the two ways of measuring things.  The g is a measurement of acceleration.  PSI is easy, its a force per unit area.  Problematical.  OK, Yahoo says:

g=a

F=ma

psi=F*area

So that's how we will relate the two measurements.

Internal area I can try to figure out.  FWIW I bought a couple already, and when they arrive, I can try to accurately determine internal area.  But what will I use for m?  Mass, of what?  Of a certain volume of argon?  That's trivial.  Mass of the liquid, assuming its incomprehensible, under a certain psi of argon that will exert a force on the glass, is that it?  But again, how to relate that mass, under pressure to a Force?  This is getting circular.

Many times I've seen glassware rated for a certain psi just up and shatter unexpectedly.  Other people have told me these stories too.  So I really want to look at this carefully.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### Borek

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##### Re: Relating psi and g (gravitational)
« Reply #1 on: September 11, 2016, 04:34:06 PM »
Why not ρah? Assume ρ of 1/mL, height of the liquid in the centrifugal bottle is not difficult to estimate, "a" you know.
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#### Arkcon

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##### Re: Relating psi and g (gravitational)
« Reply #2 on: September 11, 2016, 05:39:56 PM »
So, ρ, or rho is what?  You mean mass density?  Using 1/ml would be generous for this solution, so OK.  But how to compare to other situations?   I can't figure out how to use ρah.

Consider:  A bottle is rated to 2000x g.  Filled with 50mL of solution under 6 psi pressure, it holds just fine without shattering.  But why?  How do I know?  How do I compute?  That same bottle, filled the same well, will fail and shatter at some pressurizing gas pressure.  OK, its unlikely my instrument will generate that pressure, but I want to know how close or how far I am.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### Borek

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##### Re: Relating psi and g (gravitational)
« Reply #3 on: September 11, 2016, 05:53:27 PM »
ρah is a pressure at the bottom of a column of a liquid with the density ρ, height h and under acceleration a (compare https://en.wikipedia.org/wiki/Pressure#Liquid_pressure)

Liquid in your bottle can be simply described using these three parameters, so you should be able to easily calculate what is the pressure working on the bottom of the centrifuge bottle during centrifuging. That's probably the best estimate of the pressure these bottles withstand when used.
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#### Arkcon

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##### Re: Relating psi and g (gravitational)
« Reply #4 on: September 11, 2016, 06:14:35 PM »
I'll take some measurements and see what I can compute.  Thanks.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### Borek

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##### Re: Relating psi and g (gravitational)
« Reply #5 on: September 12, 2016, 02:47:07 AM »
Assuming an inch of water and 2000 g I got 0.025 m × 1000 kg/m3 × 2000 m/s2 = 50000 kg/(ms2) or 50 kPa, so around half of the standard atmospheric pressure.
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#### mjc123

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##### Re: Relating psi and g (gravitational)
« Reply #6 on: September 12, 2016, 05:12:16 AM »
2000 g is not 2000 m/s2.

#### Borek

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##### Re: Relating psi and g (gravitational)
« Reply #7 on: September 12, 2016, 11:14:04 AM »
Argh, good point. Tenfold more, so about 5 atm.
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#### Enthalpy

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##### Re: Relating psi and g (gravitational)
« Reply #8 on: September 15, 2016, 07:06:53 PM »
If the liquid thickness isn't negligible as compared with the radius, you can increase the computing accuracy with
ΔP = 0.5ρ(V2-v2)
where V and v are the azimutal speeds (in m/s, not rad/s) at the bigger and smaller liquid's rotation radii.

Convert the above formula to angular speed or acceleration if you wish. The nice thing with squared linear speeds is that they relate with the stress in the rotating solid (and its density) and with the liquid or gas pressure achievable, and also with the kinetic energy that can be stored.

In other terms: for a given pressure, turbines and compressors demand strong materials, and possibly several stages, and this material constraint is the same on small and big machines.

One consequence is that gas turbines need expensive alloys.