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A SCUBA regulator controls air flow from the SCUBA. The flows through two valved stages to the breathers mouth, throat and lungs. The first stage admits high pressure from the tank to a small, lower pressure inter-stage chamber. When the diver draws a breath through the mouthpiece, an exiting valve of the chamber opens to supply an incremental mass of air at the diving depth pressure. The increment of air contains sufficient oxygen needed for the diver's respiration. Also the pressure of the addmitted air inflates the diver's chest - permitting the diver to inhale. The expansion of the air to expand the lungs constitutes a "work effect" bu the SCUBA apparatus. A calculation of this work effect rated over the vreathing times will permit an effective, "SCUBA Horsepower" to be calculated. Some available data are:

Charged dive tanks contain about 2.6 kilograms of air at 300K and 20MPa. The tank volume is about 0.011m³ (please check these numbers). On average, the tidal volume (full inhale to full exhale) of a diver is 500 cubic centimeters. Breathing happens an average of 20 breaths per minute (at all depths of our jobs). The usual, prolonged depth of our operations is 20 meters of sea water. With this information please calculate the approximate duration of a dive. It would seem some manner of "effective horsepower" appertains. Please explain and calculate a number.

**Duration:** For the system we select the air in the tank (and lungs). An open system model will be used. The mass equation is written below, left, Eqn-1. Below right shows variables separated, integration operator applied with limits. Eqn-1.

(1) |

Integration of Eqn-1 yields three components. We list these and address them in order.

(2) |

Exiting mass represented incrementally.

**(i)** m_{final}, is the air remaining in the tank having a pressure "at depth." When this mass is attained, the dive is over. No further air is available at the design depth. (Some breaths are available upon ascent, fortunately). The hydrostatic equation yields the pressure, Eqn-3.

(3) |

Ambient pressure at the depth.

Now, knowing the pressure of that final "minimum amount of air" that will not come from the tank because the surrounding, ambient pressure equals 302.4kPa... Assume the sea water temperature to be 15°C. Consequently air mass within the tank at the end of the dive is:

(4) |

Air mass remaining in tank which cannot be breathed.

**(ii)** Each breath extracts a small incremental mass of air, a lung-full, from the tank. That extracted "mass per breath" is calculated to be:

(5) |

Mass of "air-out" with each "breath" event.

Having numbers for (i) and (ii) the mass equation solves to give an approximate duration of a dive Eqn-6.

(6) |

An estimate: Duration of dive time.

Our steps above cast mass in an "incremental rate" form. Each breath, taken then exhaled, constitutes an event. The major event (the dive) is a sum of "breath" events which continue as air exhausts the tank till the pressure or remaining air is too low, terminally low.

**Horsepower:** Upon thought, one realizes there might be a horsepower aspect, call it an "effective horsepower" of SCUBA operation. Air (the system) is the working substance. The energy of compressed air decreases at a rate - the horsepower will be a negative number. As air leaves the tank, it is controlled to enter the lungs of a human in a work-like fashion. The air expands the lungs, filling them ~ a volume increase (displacement) against the boundary force. Immediately, the air displaces the complicated surface of the diver's inner chest. More visually, the inner and outer chest of of the diver are "one." The outer chest "displaces" in turn against the hydrostatic resistance of the surrounding sea water. Obviously a work effort happens intermittently with each filling of the diver's lungs over the dive duration. Work happening over time is power.

Our calculation is reasonably easy. Work rate for constant pressure is:

(7) |

Work rate of a simple compressible working substance.

Using the ideal gas equation, we calculate the "working volume" of the gas. The final, expanding volume of that inflates the chest each "breath" event. These volumes, collectively, are the bubbles that stream upward.

(8) |

The working volume of air breathed.

Having the "working volume," Eqn-8, we apply it to Eqn-7. Then tidy-up the units.

(9) |

Horsepower as negative is correct. The system is the air. The energy rate of the air is that its energy is less with time - decrases.

This solution omits the "pause" time most of us have while breathing. The basic idea is about right. Mass equations with "incremental terms" desereve better treatment than given in this example.