Chronicle of an older diver

The pressure of sea water December 10, 2008

Some time ago I wrote about a web page called Calchemy that uses dimensional analysis to do unit conversion, from simple arithmetic like metres to feet to complicated calculations in physics or electronics. The toughest, but best, part of using this calculator is getting the units to match. For instance, if you try to convert square metres to cubic feet you will get an error since the metric units have one less dimension than the imperial units (thus the name dimensional analysis).

I started thinking the other day about how to write decompression software, noting that many of the papers written by North American authors use feet of salt/sea water (FSW) to express ambient and tissue compartment pressures, even when it is below atmospheric pressure (e.g. a vacuum is -33 FSW because the pressure at a depth of 33 FSW is approximately one atmosphere).  For a while I was very confused by some literature because the pressures were expressed as absolute, so sea level was 33 FSW, and 100 FSW depth was called 133. It’s a little bizarre to think about things that way but it makes the calculations easier. So  diving in a mountain lake you may find yourself at -4 FSW at the start of the dive, for instance.

Anyone who was done diving calculations (say for the Enriched Air Specialty) should know that it is often necessary to convert gauge to absolute pressure by adding 33 feet to the depth, then convert back after the calculation from absolute to gauge by subtracting the same amount. It exactly the same rationale, but in diving education this is done explicitly, while in the scientific literature it’s buried in the numbers.

So in my quest to design decompression software, at least in my head, I decided that the best unit of pressure would be the Pascal. When weather forecasters in Canada bother to talk about atmospheric pressure to the public (Aviation is different, and still uses inches of Mercury), they use KiloPascals as their primary unit (standard pressure being 101.3 KPa)

In my previous post I noted I was having trouble converting the density of sea water (a built in unit in the program) to determine the depth of one atmosphere (another built in unit). I had similar trouble trying to use it to determine the conversion factor from FSW to Pa.

So finally I have figured it out. Pressure is force/area; and while we tend to associate this force with weight (i.e. pounds per square inch), weight varies from planet to planet so scientifically speaking we need to deal with weight as the force of gravity times the mass, not just the mass itself. So a Pascal is based on Newtons per square Metre, where a Newton is one kilogram metre per second per second. The acceleration of gravity is expressed as distance over time squared (metres per second per second) so force is simply mass times gravity, so pascals are kilograms times earth’s gravitational constant g. Standard Gravity (it varies by up to 1/2 percent in different parts of the Earth) is taken to be 9.80665 m/s²

The number of kilograms is simply the density (mass over volume, where volume is cubic centimetres, inches, or whatever) times the depth (simply measured in length). So by substitution the pressure is equal to the density times the depth times g.  So now I can put this into Calchemy without difficulty. All I do is enter the expression: dens_sea_water*meter*gn which is saying I want to convert the pressure exerted by 1 meter of sea water (gn is the term for g in Calchemy), and the result unit as Pascal. This yields the following result:

dens_sea_water*meter*gn ? pascal
= 10051.8 pascal

So the answer is that the pressure exerted by 1 metre of sea water is 10051.8 pascals, or 10.0518 kilopascals. You should recall from open water class that 1 atmosphere is about 1 bar is about 33 FSW is about 10 metres of sea water (MSW). 10 metres of sea water is 100.518 KPa, which is very close to the standard atmospheric pressure of 101.3 KPa so the answer is in the right zone.

For fresh water I get the following:

dens_water*meter*gn ? pascal
= 9806.65 pascal

This is a little less pressure than sea water, which you would expect. Finally, let’s ask Calchemy to do pounds per square inch for 10 metres of sea water. I change the Expression to 10*dens_sea_water*meter*gn to show I want 10 metres instead of 1, and change the result units to psi, to get:

10*dens_sea_water*meter*gn ? psi
= 14.5789 psi

So an atmosphere is about 14.7 PSI (when those of us still on Imperial units do calculations about scuba tanks, we use this number) so our answer for 10 metres of sea water is pretty close. If we want bar, we just change the result unit to bar and get:

10*dens_sea_water*meter*gn ? bar
= 1.00518 bar

Again, just what you’d expect. 1 bar is about 1 atmosphere which is about 14.7 PSI which is about 33 FSW.

What I really like about this little program is that the only units I really have to worry about are the ones which I use for input and output, namely depth (metres, feet, inches, etc.) and pressure (Pascals, PSI, bar, etc.). I didn’t have to worry about the units for water density or the gravitational constant.

The difficulty with pressure is that it is based on force. We users of the imperial system (Canada is particularly divided using imperial and metric in day-to-day life) use pounds as a measure of force as well as mass. Whether the term refers to force or mass depends on the context – it may be different from application to application.