Converting BAR to PSI

[RockinWagin]

Ok, I am looking at cylinder compression figrues from the manua. The compression should check at 24-30 bar but I haven't found a conversion table for BAR to PSI. Anyone seen one? Thanks.

[ericnguyen]

Hi RockinWagin:

Multiple the value in BAR by 14.5 to get the equivalent in PSI.

24 -> 30 BAR = 348 -> 435 PSI

Eric

[dmorrison]

Download this converter and you dont' have to worry about it.

http://joshmadison.com/software/convert/

Dave

[TomJ]

Originally posted by ericnguyen
Hi RockinWagin:

Multiple the value in BAR by 14.5 to get the equivalent in PSI.

24 -> 30 BAR = 348 -> 435 PSI

Eric

14.50326 to be exact (or more so anyway)

[ericnguyen]

Dear TomJ:

You are very wrong. If you want to talk in a scientific way, please be accurate in your listed values . RockinWagin was just asking how to get a quick and approximate answer, so 14.5 is good enough for him.

Let me show you how to get the exact value for the conversion factor between Bar and PSI.

The standard acceleration of gravity = 9.80665 m*s^(-2)

1 lb = 0.45359237 kg

1 inch = 2.54 cm = 0.0254 m

According to Isaac Newton's second law of classical mechanics, the force is a product of the mass and uniform acceleration, i.e.

F = m*a

The pressure is defined as Force per Area unit, i.e.

P = F / area

On the Earth, a = g = 9.80665 m*s^(-2)

By definition, 1 Pascal (Pa) is the a pressure of 1 Newton (N) / square meter, i.e. 1 Pa = 1 N*m^(-2)

Now let's try to convert PSI into Pascal:

1 PSI = F / area
= (0.45359237 kg) * (9.80665 m*s^(-2)) / [(0.0254 m)^2]
= 6894.75729 N*m^(-2)
= 6894.75729 Pa

(The remaining decimal places are truncated because the precision of "g" is accurate only to 5 decimal places.)

Also, by definition: 1 Bar = 100,000 Pa

If 1 Bar = 100,000 Pa and 1 PSI = 6894.75729 Pa, the conversion from 1 Bar to its PSI equivalent will be:

1 Bar = (100,000) / (6894.75729) = 14.50377 PSI

If you read anywhere else that shows a conversion factor that is more accurate than 5 decimal places (e.g. 14.50377377...) as in the final result shown above, it is just plain wrong (mathematically speaking.)

So the most accurate conversion factor from Bar to PSI is:

14.50377

[RockinWagin]

Yep, I pulled out my old physics book and found the conversion. Not really, I skipped all the hard science stuff. I'm just relying on the experts here. Thanks. 14.5 is close enough for a 20 year old diesel engine

[Habanero]

14.50377 isn't actually correct, at least according to the significant digit theory I was taught. Since you have gravity as 9.80665 (6 significant digits), your answer should actually be 14.5038 (also 6 significant figures). At least that was what they taught me.

[sublettm]

Eric,
One small problem with your explanation...if I am not mistaken, and I know I'm not, the gravity you listed as 9.80665 is standard gravity. Standard gravity is the gravity in Grenich, England. Gravity changes depending where you are on Earth...for instance, the gravity in my lab here in Houston is 9.79280. Gravity changes depending on your altitude and your latitude. So, your calculation could be a little off if you were in Denver, on parts of Louisiana where they are below sea level.

A scary thought for you folks who may fly to Venezuala... The airlines use gravity corrections to derive lift tables for airplanes. They typically know exactly what their local gravity is at their airport. I had one of my techs down there working at one of the oil refineries a while back. He needed to know what their local gravity was so he could make corrections for one of our pressure standards. He made a call to the airport and asked what the local gravity was. They stated 9.80665. This was cause for serious alarm. As stated above, this is the gravity for England which is way north of Veneula. I am surprised they could even get planes off the ground. This was also cause for alarm because the boys in the refinery were also using this gravity number for the pressure calculations they were doing. It was a huge error. For instance, if you are sourcing pressure out of a standard whose weights were cut to standard gravity (9.80665) in Houston, your 1000psi would actually be 998.5psi. This may not seem like that gross of an error, but when you are testing items with a tolerance of 0.025%, it is quite large.

Anyway, just my take on your comments.


Mike

[ericnguyen]

Dear sublettm:

You must distinguish between ADOPTED scientific constants and
UNIVERSAL scientific constants. ADOPTED ones are based on a consensus established by the International Committee on Data for Science and Technology (CODATA). You are not right in saying "Standard gravity is the gravity in Grenich, England". The standard acceleration of gravity of 9.80665 m*s^(-2) has nothing to do with the location of Greenwich, England. Greenwich, England only represents the Greenwich Mean Time (GMT), and the Greenwich Meridian Line (Plane) corresponding to 0° longitude (note: this is not latitude).

The standard acceleration of gravity is just an internationally (by CODATA) adopted value for standardized use in all geodesic, geological, meteorological, astronomical etc... computations. Roughly speaking, this specific value gives the approximate acceleration of gravity for any locations situated at about 45° latitude at sea level. In fact, you can also say that this is kind of a hand-picked value to represent the average acceleration of gravity for all locations on the Earth in general, and not for any specific location of the Earth. It is just an adopted standardized value for the sake of having a standard value for acceleration of gravity.

When there's a need for more accurate value of acceleration of gravity at some specific location on the Earth, a carefully measured value is to be stated. As a matter of fact, by taking into account various contributing factors (e.g. Earth shape, flattening, core material distribution, speed of rotation, centrifugal force etc....) and experimental data from many locations on the Earth, scientists have been able to derive an approximate formula that gives the value of Earth's acceleration of gravity "g" at a given latitude (°L) and supposedly at sea level (and at any longitude). Because you can clearly see that this formula is not dependent upon longitude, it's obvious that this formula is just more or less an empirical formula in a whole. However, this formula is quite good in predicting the approximate acceleration of gravity for many places on the Earth, by just knowing its latitude.

-------------------------------------------------------------------------------
g= 9.780495*{1 + 0.0052892*[sin(L)]^2 - 0.0000073*[sin(2L)]^2} m*s^(-2)
-------------------------------------------------------------------------------

For example, Washington DC has a latitude of about 38.9° (North). By substituting this value into the above formula, you can get the approximate acceleration of gravity in Washington DC:

g=9.780495 * 1.002078 ~ 9.800819 m*s^(-2)

If you take a look at my calculations of the conversion factor between BAR and PSI in the previous post, you will see that all the values and definitions I used are also adopted scientific values and definitions. Therefore, my use of the standard acceleration of gravity value of 9.80665 m*s^(-2) for those calculations is correct.

Sorry for my ESL. I hope my explanations are clear enough.

Eric

[dmorrison]

RockinWagin

Here

http://joshmadison.com/software/convert/


ericnguyen, sublettm, Habanero,
When a man asks you what time it is. He's not asking you to tell him how to build a watch. Just what time it is! This reminds me of my College professors.

Dave