Scuba Horsepower

The regulator of a SCUBA apparatus operates in two stages. The first stage lets high pressure air leave the tank to enter a small inter-stage chamber. When the diver attempts to draw a breath through the mouthpiece, the valve exiting the second stage opens to supply the breathing air. A dive tank contains 2.6 kilograms of air at 300 K and 20 MPa. The average inhale volume of a diver is 500 cubic centimeters at an average 20 breaths per minute. Estimate the possible duration a dive.

Basic Thermodynamics ~ J. Pohl ©

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Scuba Horsepower

The regulator of a SCUBA apparatus operates in two stages. The first stage lets high pressure air leave the tank to enter a small inter-stage chamber. When the diver attempts to draw a breath through the mouthpiece, the valve exiting the second stage opens to supply the breathing air. A dive tank contains 2.6 kilograms of air at 300 K and 20 MPa. The average inhale volume of a diver is 500 cubic centimeters at an average 20 breaths per minute. The diver, as a member of a group, will explore a sunken ship. The average depth of operation of the SCUBA will be 20 meters.

Estimate the possible duration a dive.
♦  In breathing, the air that enters the lungs of the diver must have a pressure essentially equal to the pressure of the surrounding sea water at that depth. (Were it not for the air pressure, the diver would be unable to expand his chest to inhale). While diving at a depth of 20 meters, the ambient pressure in sea water will be:

Assume the sea water temperature to be 15°C. Each breath will draw a small mass of air from the tank. That mass is calculated to be:

The mass equation is applied to the air in the tank. Obviously there will be some air in the tank that cannot be breathed. However, for the moment, lets assume ALL of the air can be breathed.

The air in the tank at the second time (upper limit or end of the dive) is assumed equal to zero; m(t*)_tank = 0. The solution of the mass equation will give an approximate duration of a dive.

Our steps above change the mass equation from its "continuous rate" form to and "incremental rate" form. Each breath, taken in then exhaled, constitutes an event. The sum of the events exhausts all air from the tank.

We find a time of 72 minutes. But we made approximations; usually approximations give rosy or better-than-fact answers.

Approximately what horsepower is supplied by the compressed air to the breathing process?
♦   The system components are the diver and the apparatus or "SCUBA." The energy equation in rate form is written below. There will be negligible changes of kinetic or potential energy and negligible heat. Work of the system is expressed as the product of ambient water pressure times the volume increase of the diver with each breath.

Thus we obtain the differential equation, i) below. Steps to solve this equation are listed after it. First, we separate variables to obtain: ii). This step eliminates the dt's. Step iii) uses the fact that the differential operator d is the inverse of the integration operator, ∫. Consequently d ∫= 1. Equat

Equation ii) has "variables separated." Notice the differentials, dt eliminate themselves. Form iii) is obtained from ii) because d operating on ∫ equals 1. The pressure on the system boundary, p_B equals the ambient pressure at the depth. Next we integrate over one cycle of breathing.

The energy equation shows that the work for a breath equals a decrease of internal energy of the system.

Next replace the single cycle with the cycle rate.

Converted to horsepower we have.