Scuba Physics

First, when we are concerned with scuba diving physics, we need to address why water is able dissipate body heat faster than air, at what rate this occurs and what effect it has on the diver. To answer these questions we conclude that water conducts heat far more efficiently than heat because water is more dense than air and conducts heat 20 times faster than air. Next we need to know the behavior of light as it passes from an air/water interface and what effect this has on the diver. To answer this we have to know that this behavior of light is called refraction. Refraction is caused by the process of light traveling at different speeds as it passes through different substances. Also, objects tend to look larger underwater by a ratio of 4:3 and cause objects to appear magnified by 33%. Now we need to look at the visual reversal phenomenon and explain its effect. This phenomenom refers to an objects tendency to appear further away than its actual distance. The single most important factor affecting this phenomenom is turbidity. Once this is understand we need to know why sound travels faster underwater than in air, by how much and what effect this has on the diver. To answer this we need to know that light waves contain electromagnetic energy while sound waves are comprised of mechanical energy. We also know that sound travels 4x faster in water than it does in air and this effects divers because divers have a difficulty determining the direction of sound underwater due to an insufficient delay between the sound striking one ear before the other. With this information understood we get to Archimedes Principle, the principles of Buoyancy- "Any object wholly or partially immersed in a fluid bouyed up by a force equal to the weight of the fluid displaced by the object." (From this principle we can see whether or not an object floats or sinks according to the weight of the fluid. When working with the scuba diving we recognize Saltwater to weigh 64 lbs per cubic foot and freshwater to weigh 62.4 lbs per cubic foot and we also need to know pure water has a specific gravity of 1.0)

Example: Approximatley how much water must be displaced to bring a 900 lb object to the surface if the object displaces 10 cubic ft and lies in 132 of seawater.

To solve this we disregard the depth and focus on the weight of the object along with the displacement. To begin we must multiply the cubic ft of 10 by the weight of the water 64. We arrive at 640. Now we must subtract 640 from the weight of 900.
This equals 260. The final step is divide 260 by the weight of the water (64) to arrive at our answer of 4.06.

Once we master the concept of Archimedes prinicple we move on to Boyles Law.
To deal with Boyles Law we must know that for every 33 feet of descent in seawater add the pressure of 1 atm or 14.7. If you take a full breath at 33 feet your inhaling twice the number of air molecules as a full breath at the surface. Note this when accounting for the surface pressure when finding absolute values.

Example: What is the ambient pressure at 50 ft of Seawater? to solve this we multiply the number of ft (50) by the pressure exerted (.445 for seawater). We arrive at the answer of 22.25. To finish the answer we must apply 1 atm for surface pressure and add 14.7 to 22.25. The completed answer is 36.95.

If we fully understand this, we realize that 1 atmosphere weighs 14.7 lbs and exerts .432 lbs for freshwater and .445 lbs for seawater. Absolute and Ambient pressures require that we add 1 atm or 14.7 to our gauge pressure. Gauge pressure ignores the atmospheric pressure.

Now we are ready for Boyles Law. Boyles Law describes pressure volume relationships and states that "If the temperature remains constant, the volume of a given mass of gas is inversely proportional to the absolute pressure". The mathematical equation for Boyles Law is P1 x V1 = P2 x V2.

We will use Boyles Law to explain the relationship between pressure and volume on a flexible gas-filled container (Balloon or our lungs) as well as in-flexible container and calculate the changes that will occur to that container as it is raised and lowered in the water column.

You may find this chart useful:
Depth Pressure Volume
0 1 ata Full
33' 2 ata 1/2
66' 3 ata 1/3
99' 4 ata 1/4 .........

Example: A balloon (our lungs) containing 10 cubic feet of air at 25 ft of seawater is taken to a depth of 85 ft. What will be the exact volume of the balloon (our lungs) upon reaching 85 ft.

To solve this we will use the formula Boyles Law Formula:
P1 X V1 = P2 X V2.

P1 = 25 ft x .445 = 11.125 + 14.7 = 25.82
V1 = 10 ( # of cubic ft)
P2 = 85 ft x .445 = 37.82 + 14.7 = 52.53
V2 = x
Plug in variables
(P1)25.82 X (V1)10 = (P2)52.53 X (V2)x
258.2 = 52.53 X x
258.2 / 52.53 = 4.91
x = 4.91

Still dealing with Boyles Law we now need to explain the relationship between depth and the density of the air a diver breathes and calculate this relationship in increments of whole atmospheres. To determine this relationship we must determine the type of container being used being a balloon or lift bag or an inflexible container such as a scuba tank. When dealing with an inflexible container we know if pressure increases the volume must decrease and if pressure decreases volume must increase and the opposite when dealing with a balloon or flexible container. A flexible container will expand upon ascent, and reach the surface with the original quantity of gas times the number of atmospheres from which it was released. And we also know that a scuba tank or inflexible container is unaffected by the surrounding water pressure. Next we need to determine a divers air consumption rate at one depth and calculate how that consumption rate changes when depth changes.

Example: A diver has an air consumption rate of 3 cubic ft. per mintue at 66 ft of seawater. If all other factors but depth remain unchanged, what will his consumption rate be at 200 ft?

To answer this we must first determine the surface consumption rate. Which in this case the diver will consume one third the amount of air at the surface then he would at 66 ft (3ata).
Example: How much water is exerted at a depth of 200 ft? By dividing 33 feet into 200 ft we determine that 200 ft is 7 ata. (200/33 = 6.06. When accounting for atmospheric pressure we must add 1 to the gauge pressure. In this case the gauge pressure is 6.06. 6.06 + 1 = 7.06) We now know that the air consumption rate will increase to slightly more than 7 fold at 200 ft.

With all of this understood we now understand the Law and principles of Sir Robert Boyle and what happens to a volume of air when the pressure changes.

Moving on, we arrive at Charles Law which states, "the amount of change in volume of gas is directly proportional to the change in the absolute temperature at a constant pressure." Boyle's Law only dealt with the effects of pressure and volume. Charles Law deals with Temperature. Charles Law says that the amount of change in either volume or pressure of a given gas volume is directly proportional to the change in the absolute temperature.
The general guideline for Charles Law is the scuba tank pressure will change 5 psi for every 1 degree change. We may also combine the formula for Boyles Law:

P1/T1 = P2/T2

We also need to know that a flexible container such as a balloon or lift bag when placed in a freezer will decrease in volume because of less pressure in the balloon. As temperature decreases, the motion of the molecules decreases. As the motion decreases, the force of the impact of their collisions with each other decreases and since "the amount of change in volume of gas is directly proportional to the amount of change in the absolute temperature" we know that when pressure decreases volume decreases and when pressure increases volume increases. On the flip side when dealing with an in-flexible container such as a scuba tank the volume remains unchanged regardless of changes to the external pressure.

Example: A 80 cubic foot scuba tank is filled to 3225 psig at an ambient temperature of 78 degrees F. If the tank is then used in water temperature of 44 degreees F, what would the approximate tank pressure be?

To solve this we will need to remember the scuba tank pressure will change 5 psi for every 1 degree change. We will also need to use the formula, P1/T1 = P2/T2. Next we must determine what we know from this problem. The pressure the tank is filled to is 3225 and it asks us to determine what the pressure will become(P2). To predict the behavior of gases we must work in absolute terms, so we must add 15 (14.7) to the 3225 psig to arrive at 3240 psia or P1. The temperature of the tank is 78 degrees F and it will be used in 44 degrees F. When working in absolute terms we must convert degrees F into degrees Rankin. For this conversion we just add 460 degrees to the temperature of 78 degrees arriving at 538 degrees Rankin and 44 degrees F now becomes 504 degrees Rankin. Now we can plug these values into the equation.
3240/538 = x/504
504x3240/538 = x
x = 3035
3035 - 15 (14.7) = 3020.

If we understand the principles so far then we have a basic understanding of Charles Law and the principles in which apply including Boyle's Law.

Now we are ready to move on to Daltons Law which states, "the total pressure exerted by a mixture of gases is equal to the sum of the pressure of each of the different gases making up the mixture - each gas acts as if it alone were present and occupied the total volume." In essence, this means that each gas within a gas mixture acts independently of the others. This independent pressure is referred to as the partial pressure. When dealing with Dalton's Law we are referring to partial pressures and when dealing with partial pressures we are explaining the effects of breathing contaminated air mixtures at depth and calculating the equivalent effect such contamination would have upon the diver at the surface. We also must remember once the tank is filled, the percentages of gases within it cannot change.

Example:
Breathing from a contaminated air source with 1.5% carbon monoxide at a depth of 300 feet of seawater would have the same effect as breathing approximately what percentage of carbon monoxide at the surface?

We must first determine that 300 feet is 10 ata. We then multiply 1.5% by the ata of 10 to arrive at 15%.

Dalton's Law is important for divers because it deals with individual gases as well as surface equivalency which concerns us when we deal with toxic contaminates in our breathing gases.
Example: On the surface 0.5% of carbon monoxide is not toxic but at 5 ata, it is.

Once we understand Daltons Law and partial pressures we are ready for Henrys Law. The most common example of Henry's Law is a carbonated beverage such as a coke. When you open the beverage it foams and fizzes as carbon dioxide, dissolved in the liquid, comes out of solution. This demonstrates that liquids dissolve gases, and that if conditions change, the amount of gas that can stay in solution changes. This is true when dealing with gas in our bodies, such as nitrogen. Also, we should note that the pressure exerted from inside a liquid by a particular gas in solution is called gas tension. The difference between the partial pressure of gases in contact with a liquid and the gas tension within the liquid is referred to as the pressure gradient. When the gas tension within a liquid reaches equilibrium with the partial pressure of the gas in contact with the liquid, no more net exchange of the gas occurs. At this point the liquid is said to be saturated with that gas. Henry's Law describes supersaturation and the effects it has on a diver. Without Henry's Law we wouldn't have the dive tables and computers that allow us to minimize the risk of decompression sickness by providing no stop limits and/ or decompression stops. Now we need to explain what will occur to a liquid saturated with a gas at high pressure when the pressure of the gas in contact with the liquid is quickly reduced. To understand this we must know there is a tendency for a state of equilibrium to exist between the pressure within the liquid (gas tension), and the pressure of the gas in contact with that liquid. This equilibrium will be maintained until the pressure in contact with the liquid changes. Thus concluded, "the amount of gas that will dissolve into a liquid is almost directly proportional to the partial pressure of that gas."
Example: If a glass of water is placed in a vacuum for several days , no longer containing any gases, if it is then placed in a pressure pot and pressurized to 2 ata for several days what will be the gas pressure within the liquid? And if this vacuum is created, how will pressure of the gas, inside the liquid, change?

Since the amount of gas that will dissolve into a liquid is almost directly proportional to the partial pressure of that gas, the gas pressure within the liquid is 2. If the vacuum is created the pressure will decrease. A vacuum would represent zero pressure in contact with the liquid. Therefore, the tendency would be for any gas contained in the liquid to come out. So, the pressure will decrease.

Once we understand how a saturated liquid at high pressure will react when the pressure of a gas in contact with the liquid is quickly reduced, we are ready to learn about "supersaturation" and what conditions are necessary for gas bubbles to form in a supersaturated liquid. We use supersaturation to predict decompression outcomes with very high reliability. While diving, nitrogen gets absorbed in our tissues. Although at different rates, our tissues become saturated with nitrogen and when we surface our tissues become supersaturated with nitrogen. While diving we plan for no stop limits and decompression stops so that we don't end up with an excessive pressure gradient that results in decompression sickness. During these stops nitrogen is released out of solution (our tissues) and when the pressure gradient has declined enough you may move on the next stop.

To learn more about how our body deals with dissolved gases please see our blog on Diver Physiology.

I hope you have enjoyed our lesson on Scuba physics and hope you will visit us in San Diego. Thank you for visiting www.GetWetSanDiego.com and we hope you check back often for our latest blogs.