SCUBA Horsepower

A self-contained underwater breathing apparatus (scuba) delivers air to a person diving at depth for respiration along with the power needed to breath. The apparatus consists of a steel tank containing air compressed to high pressure and a mechanical regulator which releases the air from the tank to the diver periodically and safely in accord with his or her respiration demands.

The regulator has two, valved stages. The first valve admits high pressure from the tank into the inter-stage chamber. When the diver draws a breath through the mouthpiece, the second valve opens to supply an increment of air into the breathers mouth, throat and lungs. Air admitted has oxygen for respiration and sufficient pressure to inflate the diver's chest against the ambient sea pressure. The expansion of the air to inflate the lungs constitutes a "work effect" of the air. A calculation of the work effect over the breathing times permits an "effective" SCUBA Horsepower to be calculated.

Dive tanks charged to 20MPa (at 300K) contain about 2.6 kilograms of air. The tank volume is about 0.011m³. For an average diver the tidal volume (full inhale to full exhale) is about 500 cubic centimeters. A full breath is taken, on average, 20 times per minute. The usual, prolonged depth of our operations is 20 meters of sea water. With this information calculate the approximate duration of a dive.

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

(1)

Open system mass equation.

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

(2)

Exiting mass represented incrementally.

(i) The final mass of air in the tank is m_final. When this air mass is attained the diver must surface. (Some breaths become available upon ascent, fortunately). The volume of the final mass is 0.0011m³. Its pressure will be "at depth." We use the hydrostatic equation to obtain 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) and assuming the sea water temperature to be 15°C... Consequently the 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 "increment" 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.

Eqn-7 shows we need the second (at depth) volume of the air. This is the collective volume of the bubbles (at depth) that stream upward.

(8)

The working volume of air breathed.

Having the "working volume" by 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.

We find the horsepower is negative. This is correct. The system is the air and during the events it expands. Our 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.