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I am a scuba diver, trying to understand some physics that occurs throughout a dive. Primarily I am trying to understand the surface air consumption (SAC) derivation.

For those unfamiliar, SAC is the change in pressure (PSI or bar) per unit time at the surface (1 atm). SAC is calculated by measuring the change in PSI at the dive depth and dividing it by the pressure at depth (expressed in atm).

$ SAC = \frac{\Delta PSI* P_{sea level}}{P_{depth}} $

Since the pressure at sea-level is 1 atmosphere the equation can be simplified by expressing the pressure at depth in atmospheres as $ P_{depth} = N_{atm}*1atm$. $N_{atm}$ is the number of atmospheres the pressure at depth is equivalent to.

In this case, SAC is expressed as

$SAC = \frac{\Delta PSI * 1atm}{N_{ATM}*1atm} = \frac{\Delta PSI}{N_{ATM}} $

I am trying to determine the assumptions/derivation to arrive at this equation.

So far, what I have done is assumed that

  • The volume of air inhaled per breath is constant regardless of depth
  • Breathing rate is constant regardless of depth
  • Temperature of the gas is independent of depth

Using these assumptions and Boyle's law, it is easy to show that one breath at depth would be equivalent to $ N_{ATM} $ breaths at sea level.

I am not sure about what steps to take next. I need to figure out how much my tank's PSI will change for a breath, given the volume of 1 breath, the ambient pressure, and the tank's volume. I also need to show the PSI change per breath is independent of the PSI of the cylinder. I am rusty on my thermodynamics and looking for advice on how to derive/prove these two quantities.

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  • $\begingroup$ You are mentioning too many units for pressure in ways that they contradict your equations. $\endgroup$
    – Themis
    Jun 29 at 21:13
  • $\begingroup$ @Themis can you clarify a bit how the different units are contradicting the equations? I realized I am using PSI and atmosphere which I agree is probably a bit confusing. I am going to edit to more clearly show that I assume the pressure at sea-level is 1 atmosphere $\endgroup$
    – CMH12
    Jul 7 at 20:30
  • $\begingroup$ I meant that you are referring to PSI, atm and bar at the same time. $\endgroup$
    – Themis
    Jul 9 at 19:07

1 Answer 1

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I need to figure out how much my tank's PSI will change for a breath, given the volume of 1 breath, the ambient pressure, and the tank's volume.

Consider applying the ideal gas law both in the tank and in the lungs, with the connection being the amount of gas $\Delta n$ transferred from the former to the latter:

$$\text{Tank, constant volume and temperature:}\;\Delta P_\text{tank}V_\text{tank}=\Delta nRT;$$

$$\text{Lungs, constant pressure and temperature:}\;P_\text{ambient}\Delta V_\text{breath}=\Delta nRT;$$

$$\Delta P_\text{tank}=\frac{P_\text{ambient}\Delta V_\text{breath}}{V_\text{tank}}.$$

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  • $\begingroup$ Hi Thank you for the answer it is very helpful. My only question is : is the ideal gas law still applicable at pressures in the thousands of PSI? Typically a scuba tank is used from 3000psi to around 1000 psi $\endgroup$
    – CMH12
    Jul 12 at 23:06
  • $\begingroup$ The ideal gas law can be safety applied if the compressibility factor is suitably close to 1 at the conditions of interest. $\endgroup$
    – Chemomechanics
    Jul 12 at 23:20
  • $\begingroup$ Awesome, thank you so much for the answer you really clarified these concepts for me $\endgroup$
    – CMH12
    Jul 15 at 11:02

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