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Unread 29-06-2005, 00:26
sanddrag sanddrag is offline
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Pressure physics problems

When doing pressure physics problems like finding the pressure at the bottom of a swimming pool, do you have to add in the atmospheric pressure (101,325 Pa)? Or is pressure at the surface of the water typically taken to be zero?

EDIT: Atmospheric pressure is always taken into consideration in open containers. So in the swimming pool, yes you would add the atmospheric pressure to the pressure due to the depth of the water. In something like a dam, there is atmospheric pressure pressing on the top of the water and on the dry side of the dam, so it cancels out and does not need to be considered.
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Last edited by sanddrag : 29-06-2005 at 03:07.
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Unread 29-06-2005, 03:28
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Re: Pressure physics problems

Sand,

Yes, atmospheric pressure is added in.

As a rough number, 1 atmosphere of pressure for each 33 feet of seawater (this is from my days as a scuba diver). So that at 100 feet of depth, the pressure is about 4 atmospheres. 1 liter of compressed air inhaled into the lungs at 100 feet will expand to 2 liters at 33 feet (2 atmospheres) and 4 liters of air at the surface (1 atmosphere). This is why it is absolutely essential for divers to constantly exhale during ascents and most "damage" from someone holding their breath while surfacing occurs in the last 10 feet...

Regards,

Mike
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Unread 29-06-2005, 09:30
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Re: Pressure physics problems

Sanddrag,

Read the problem carefully and see what they are asking for. Pressure is generally reported two ways: 1) absolute pressure (i.e. add in atmospheric pressure); or 2) relative pressure (i.e. how much pressure above atmospheric).

Some problems require absolute pressure and some require relative. For example, tire pressure is relative pressure. Measuring pressure altitude with an altimeter is absolute pressure.
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Unread 29-06-2005, 11:43
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Re: Pressure physics problems

BY the way, relative pressure is often refered to as "guage" pressure. It is often indicated in the unit notation by adding a "g" at the end. Psi and psig are not the same thing.
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Unread 30-06-2005, 15:07
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Re: Pressure physics problems

Quote:
Originally Posted by Mike Betts
Sand,

Yes, atmospheric pressure is added in.

As a rough number, 1 atmosphere of pressure for each 33 feet of seawater (this is from my days as a scuba diver). So that at 100 feet of depth, the pressure is about 4 atmospheres. 1 liter of compressed air inhaled into the lungs at 100 feet will expand to 2 liters at 33 feet (2 atmospheres) and 4 liters of air at the surface (1 atmosphere). This is why it is absolutely essential for divers to constantly exhale during ascents and most "damage" from someone holding their breath while surfacing occurs in the last 10 feet...

Regards,

Mike
Mike -

Just a clarification. I seem to remember from my scuba (and a few other) classes) that overburden pressure increases of incompressible materials (e.g. pressure increases associated with increasing water depth) is linear. So in the example cited above, would it not be that 1 liter of compressed air at 100 feet will expand to 2 liters at 66 feet, and 3 liters at 33 feet, and 4 liters at the surface?

-dave
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Unread 30-06-2005, 15:21
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Re: Pressure physics problems

I also have some scuba classes under the belt, and I agree with you Dave, but I believe that he may have just misstated. I think he may have just meant 33 ft higher than 100, or 66ft.
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Unread 30-06-2005, 17:33
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Re: Pressure physics problems

Quote:
Originally Posted by dlavery
Mike -

Just a clarification. I seem to remember from my scuba (and a few other) classes) that overburden pressure increases of incompressible materials (e.g. pressure increases associated with increasing water depth) is linear. So in the example cited above, would it not be that 1 liter of compressed air at 100 feet will expand to 2 liters at 66 feet, and 3 liters at 33 feet, and 4 liters at the surface?

-dave
Dave,

I stand by my numbers...

1 atmosphere at sea level
2 atmospheres at 33 ft
3 atmospheres at 66 ft
4 atmospheres at 99 ft

PV=nRT and, assuming that nRT is a constant, P and V are complementary (have an inverse relationship). The pressure at 33 ft is 1/2 of the pressure at 99 ft so that the volume in 2X the volume at 99 ft. Likewise, the pressure at sea level is 1/4 of the pressure at 99 ft such that the volume at sea level will be 4X the volume at 99 ft and 2X the volume at 33 ft.

Like I said, the greatest danger is the last 10 feet. The esophagus is a very powerful muscle. Holding your breath in that last segment of an ascent will severely damage the lung tissues...

Regards,

Mike
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Unread 30-06-2005, 18:22
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Re: Pressure physics problems

Because I had two very learned men question my numbers, I went back to the physics book (see attachment).

Note that the plot of figure 17-4 (page 375 of Physics, 3rd edition, Halliday & Resnick, 1978) shows a linear (constant) P versus Y graph for depths below sea level.

Mike
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Unread 01-07-2005, 10:26
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Re: Pressure physics problems

Quote:
Originally Posted by dlavery
Mike -

Just a clarification. I seem to remember from my scuba (and a few other) classes) that overburden pressure increases of incompressible materials (e.g. pressure increases associated with increasing water depth) is linear. So in the example cited above, would it not be that 1 liter of compressed air at 100 feet will expand to 2 liters at 66 feet, and 3 liters at 33 feet, and 4 liters at the surface?

-dave
Pressure is linear with depth, but the volume of a given quantity of gas varies as the inverse of the pressure. 4/4 liter at 99 feet is 4/3 liter at 66 feet is 4/2 liter at 33 feet is 4/1 liter at the surface.
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