questions re. SAC rate calc's

[Bubbletrubble]

I guess I just don't see how the values in the formula relate to each other.

For example in (D+33)/33 I see the depth plus 33 (for the surface atmosphere) divided by an atmospere to give the avg. atmosphere.

Pengwe's explanation of this is right on. I think the equation (D+33)/33 makes more sense if you convert it to (D/33) + 1.
(D/33) represents how many atmospheres of pressure contributed by the water. "1" represents surface atmosphere. Add (D/33) to "1" and you get the total atmospheric pressure.

But in JxH/I I don't quite understand the relationship between psi used, tank volume and tank service pressure. I keep thinking that the various starting psi pressures should be taken into account some how, that [I.] should be a variable. Maybe I'm just chasing a ghost though.

@DaleC: I'll attempt to clarify this point. Once again pengwe has done a good job so far.

To restate the "variables" and accompanying equations:
F. Start PSI
G. End PSI
H. Tank Capacity in Cu. ft.
I. Tank Service Pressure PSI
J. PSI used [= F-G]
K. Cu. ft. used [=JxH/I]

Let's discuss the stuff in the K = JxH/I equation. You state that "the various starting psi pressures should be taken into account somehow." I agree 100%.
J represents the PSI used.
J = Start PSI - End PSI = F - G
K = J x (H/I) is simply a way to convert the "PSI used" (which you can easily calculate by looking at your SPG) to "cubic feet used." This conversion requires information about the tank (its capacity at the service pressure). Some people call (H/I) a "tank conversion factor."
You need to convert the amount of gas used from psi to cubic feet because "cubic feet/minute" is the unit being used to describe your SAC or RMV rate.
"Cubic feet" is simply more descriptive of gas usage. If I tell you that I used 2000 psi on a dive, you might wonder: "Well, what kind of tank did he use?" For instance, 2000 psi in AL80 is a different amount of gas than 2000 psi in a steel HP100. Then there's the issue of how deep I was when I used that gas...

Hope this helps...

 

[Blackwood]

Dale -

I'd group it as follows:

DIVE INFO
A1 Initial Pressure (PSI)
A2 Ending Pressure (PSI)
A3 Bottom Time (min)
A4 Average Depth (FSW)
CYLINDER INFO
A5 Rated Pressure (PSI)
A6 Rated Volume (ft³)

Then, SAC Rate (ft³/min/ATM) is:
((A1-A2)/((1+A4/33)*A3))/(A5/A6)


Note, it's important to check that your formula is working. Fortunately, it's also easy to do if you set up a case to which you should know the answer. One suggestion is:

Set the Initial Pressure (A1) to the rated pressure of your tank,
Zero out the Ending Pressure (A2),
Set the Bottom Time (A3) to 1, and
Zero out the Average Depth (A4).

That simulates starting with a full tank and sucking it empty in one minute at the surface. If the resulting SAC Rate reads the Rated Volume of the tank - in your example 77.4 (CFM) - the formula is working.

As a second check, I'd suggest setting the Final Pressure (A2) to the Initial Pressure (A1) and ensuring that the SAC Rate reads 0.

 

[b1gcountry]

Thanks Lynne,
I guess I just don't see how the values in the formula relate to each other.

For example in (D+33)/33 I see the depth plus 33 (for the surface atmosphere) divided by an atmospere to give the avg. atmosphere.

But in JxH/I I don't quite understand the relationship between psi used, tank volume and tank service pressure. I keep thinking that the various starting psi pressures should be taken into account some how, that [I.] should be a variable. Maybe I'm just chasing a ghost though.

Using the US convention on tanks...

Rated Capacity/Rated Pressure = CF Used/PSI used


So with an AL 80 (rounding up)

80cf/3000psi = ??cf/1500psi

Solving for cf used...
80cf * 1500psi/3000psi = 40cf used

 

[DaleC]

You guys are great! Thanks all for your responses. I'm going to mess around in excel tonight to see how these different ideas work.

Marc, I'm going to chew on that formula for a bit. I like your example for testing the values as it was one thing I wondered about (how do I know if I'm getting the right rate or just some random number). That would answer it.

Here's some questions for you:
1. If you fill your tank to 3000 psi, how much air is in it?
2. If you fill the tank to 2000 psi, how much air is in it?
3. If you fill it to 2800psi, how much air is in it?
4. Then you breathe it down to 700psi, how much air is left?
5. How much air did you use?

1. 77.4
2.51.6
3. 72.24
4. 2100
5. 54.18

I see what's happening now. I was looking for something that wasn't there.

I'm afraid of where this will all end - I'm even on the hunt for a basic BT/DG which was something I never thought I'd want just awhile ago. Down the rabbit hole I go.

 

[Blackwood]

Marc, I'm going to chew on that formula for a bit.

The way I structure data is often different than the way others do it so you'd be well served to use whichever format makes the most sense to you.

I think this whole exercise would be easier if you put units to you variables. You know what you start with (PSI, feet and minutes) and what you want to end up with (CF, atmospheres and minutes), and writing out the units makes it a lot easier (for me, anyway) to figure out how to get there.

I don't often memorize formulae. On the rare instances I may be called upon to solve for END or MOD or SAC Rate or whatever, I derive the formula. That way I know I got it right.

I'll try to explain the flow of my example.

A1-A2 = how much gas I used in PSI. Simple.

If I divide that by A3, I get PSI/min. Again, simple.

Now I need to normalize for depth, which I'll do by dividing the above by my pressure depth which is 1+A4/33 in ATM .

So now what do I have? I have PSI/min/ATM. Almost there, but if I want to go from one tank to another, I should go to volume rather than pressure.

Relating PSI to CF is pretty easy given the tank specifications of 77.4CF per 3000PSI.

Simple algebra, if I have PSI, I can multiply it by CF/PSI to get CF. That operation is exactly the same as dividing PSI by PSI/CF, which is how I wrote it above.

PSI/min/ATM divided by PSI/CF = CF/min/ATM (which is SAC rate where ATM = 1, i.e. at the Surface).

 

[TSandM]

I think this whole exercise would be easier if you put units to you variables. You know what you start with (PSI, feet and minutes) and what you want to end up with (CF, atmospheres and minutes), and writing out the units makes it a lot easier (for me, anyway) to figure out how to get there.

I don't often memorize formulae. On the rare instances I may be called upon to solve for END or MOD or SAC Rate or whatever, I derive the formula. That way I know I got it right.

Dimensional analysis and formula derivation . . . Mr. Blackwood, you and I are kindred spirits!

 

[SparticleBrane]

Yay for dimensional analysis! Always better to teach concepts rather than rote memorization...

 

[DaleC]

Just in case someone ever asks this question again:
What was confusing me about the psi consumed converted to cuft. consumed formula was the value "3000". I kept thinking that, sometimes I started with 2800psi or 3200 psi so that value should be variable. Now I see that the 77.4/3000 is simply determining a cuft. value for each psi (.0258). The consumed psi is then multiplied against that value to give me the cuft. consumed.
The value 77.4 changes, depending on the tank size and the 3000 can change, depending on the rated fill pressure for the tank.
Also, Marc, I ran those test values and the fomula works out

 

[lamont]

Just in case someone ever asks this question again:
What was confusing me about the psi consumed converted to cuft. consumed formula was the value "3000". I kept thinking that, sometimes I started with 2800psi or 3200 psi so that value should be variable. Now I see that the 77.4/3000 is simply determining a cuft. value for each psi (.0258). The consumed psi is then multiplied against that value to give me the cuft. consumed.
The value 77.4 changes, depending on the tank size and the 3000 can change, depending on the rated fill pressure for the tank.
Also, Marc, I ran those test values and the fomula works out

the 77.4/3000 ratio is also a measurement of the size of the tank. bigger tanks take more cu ft of gas added to them to create the same difference in pressure because the volume of the tank itself is bigger. a higher pressure tank of the same physical volume will have the same ratio, but both the cu ft and the pressure increase (assuming linear ideal gasses, which is close enough for our purposes).

in metric it gets easy.

if you have a 10L tank (volume at 1 bar), filled to 200 bar, then you have 2000L of gas in the tank. in this case the 77.4 cu ft / 3000 psi is replaced by a 10 L / 1 bar ratio, but the tanks are just sold as "10L" tanks.