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Atm vs. Gauge Pressure 1

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rebel

Mechanical
Jun 14, 2002
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Can someone please tell me the difference between gauge pressure and atmospheric? Psig vs. Psia. In laymen terms.

Thank you
 
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It is gauge pressure vs. ABSOLUTE, not "atmoshperic".

The difference does have to do with the atmosphere: the weight of the air on an area 1 inch x 1 inch, from the ground (sea level) to the "top" of the atmosphere, is approximately 14.7 lbs. This pressure, 14.7 psia is the atmospheric pressure. It is imposed upon everything that is exposed to "open air". (This pressure is less at elevations above sea level.)

Sometimes it is convenient to deal with the net pressure in a situation where the atmospheric pressure acts on everything under consideration and therefore has no net effect itself. This is referred to as "gauge pressure".

You inflate you tires to 32 psig. This is the same as 46.7 psia, but the atmospheric pressure acts on the inside and the outside and therefore "doesn't count" in providing any inflation. So as a practical matter, we often measure the gauge pressure.

Typically in a factory or shop setting, when the pressure is given as "10 pounds", for example, this is sloppy (but commonly understood) shorthand for "10 pounds per square inch, gauge pressure".



 
psia takes in consideration the pressure of the air surrounding us at sea level.

psig= psia - 14.7, in other words the reference is 0 psia at sea level Steven van Els
SAvanEls@cq-link.sr
 
In calculation you should consider psig, because the vessel is under an internal pressure p and external pressure patm, and therefore, it's under pressure p-patm = pgage. In metric patm =1 bar.
 
Don't forget that the air is lighter in Europe! Must be something to do with the atmosphere being shoved around by the Alps. Or maybe a bar is atmospheric pressure at some reference elevation other than sea level... maybe at the top of a really tall column!

1 atm = 14.7 psia
1 bar = 14.5 psia

jt

 
I am confused jte.

I know 1 bar = 14.5 psi

Depending on the conversion, it can be barg or bara.

E.g. 2 barg = 29psig
and
2 bara = 29psia

I stand to be corrected.
 
There is perhaps a subtle distinction in designating pressures that has not been brought out in this thread. It is almost a point of grammar, rather than engineering, but here goes.

Common "suffixes" in designating pressure are:

"a" - absolute
"g" - gauge
"d" - differential

These suffixes may be, and are, used irrespective of whether the units are US customary, or "metric"/SI.

However, if an increment of pressure is referred to, and it is either outside of any context, or the context is understood (at your own risk of causing a MISunderstanding), the proper term is to simply speak of "psi" of "bar". For example, in a table of conversions, one ought to see something like

1 bar = 14.5 psi

Therefore, both of the examples that you gave to illustrate your confusion are correct "mucour", and "jte" really ought to have written "1 atm = 14.7 psi" and "1 bar = 14.5 psi" without the "a" suffix.

Of course, an exception is the (somewhat bastardized) units of "ata" = 14.223 psia, where "absolute" is part of the definition of the units.

BTW, There should be a similar distinction in talking about temperatures vs. temperature differences. For example to say "the boiling point of water is 212 degrees Fahrenheit at 14.7 psia" and "the steam superheat is 150 Celsius degrees".

In context it is often easily understood, but it is sometimes a useful distinction.
 
As I'm Cartesien I explain to mucour with my manner :
1 bar = 0.069 psi I take generally 7/100 of psi for quick calc.
Pcac. = p – patm ( p: internal pressure, pcac. : p for calculation)
Pcac. = p – patm , all in bar.
Pcac.(psi) = 14.504 * p(bar) – 14.696 as
Patm = 1.01325 bar = 14.696 PSI
And 1 bar = 14.5038 PSI.

To answer jte, Atmospheric pressure vary following law p1 = p2 + ro*g*(h1-h2). It's generally not convenient to take 1.013.. and I don't know how many fraction of decimals for calculation. We round to 1 bar and it's simple.
 
Holy moly Batman, you guys kill me. Most of you guys must be PHDs or something. Poor rebel is really F&%@ up now. Just kidding guys.

So for a BSc to answer rebel's question here goes.

- psia is the absolute pressure. It is a theoretical value for calculations purpose. It is P(absoulte) = P(gauge) + P(atmospheric).

- Gauge pressure is the pressure "difference" you read off of a gauge. Or, a book says "pressure gauge read the excess of the the test pressure over atmospheric pressure". It already accounted for the pressure from the atmosphere. Another word, P(gauge) = P(asoulte) - P(atm)

enough said. Ok, how did I do? rebel you be the judge.
 
Pressure is defined as force per unit area.

Atmospheric Pressure –
The weight of a column of air, one square inch in cross
section and extending from the earth to the upper level
of the blanket of air surrounding the earth. This air
exerts a pressure of 14.7 pounds per square inch at sea
level, where water will boil at 212 degrees F. High
altitudes have lower atmospheric pressure with
correspondingly lower boiler point temperatures.
Standard is an average atmospheric pressure at sea level, and is defined as 1 atmosphere on Earth, equal to 760 millimeters of mercury (760 Torr) and 101,325 Pascals and 1 PSIA.

Gauge pressure is simply the pressure without considering the 'atmospheric pressure'.

An example would be an empty baloon sealed at sea level, it would have 1 PSIA (0 PSIG) at both the inside AND outside surfaces. However, if were were to take that same sealed empty baloon and start lifting it, the balloon would start to 'inflate' (decreasing PSIA external, increasing PSIG internal) as it increased in altitude, until it got to the edge of the earth's atmosphere.... where it would have 0 PSIA on the outside and 1 PSIG internal pressure (without adjustments for temperatures).
 

It cannot be determined what the inside pressure of the balloon is "at the edge of the earth's atmosphere..." The balloon expands, so the pressure is ultimately less than 1 psia. But the final value can't be determined because the balloon itself supports a certain pressure drop, roughly analogous to surface tension. So, unless the (non-linear) properties of the balloon's rubber were specified, you can say only that it is greater than the external pressure and less than its initial pressure.

All of which is off the subject of the original question -- except to say that there is nothing magical (or in this case, correct) about the value of "1 psig" or "1 psia".

 
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