CHAPTER 12
mixed gas diving theory:
12-1 INTRODUCTION
12-1.1 Purpose. The fundamental laws and concepts of underwater physics presented in Chapter 2 (Volume 1) are basic to a proper understanding of mixed-gas diving techniques. In mixed-gas diving, calculations requiring the use of the various gas laws are vital to safe diving. A thorough working knowledge of the application of the gas laws is mandatory for the mixed-gas diver. This chapter reviews the gas laws.
12-1.2 Scope. This chapter discusses the theory and techniques used in mixed-gas diving.
12-2 BOYLE’S LAW
Boyle’s law states that at constant temperature, the absolute pressure and the volume of gas are inversely proportional. As pressure increases, the gas volume is reduced; as the pressure is reduced, the gas volume increases.
The formula for expressing Boyle’s law is:
C=P × V
Where: C is constant P is absolute pressure V is volume
Boyle’s law can also be expressed as:
P1V1 = P2V2
Where:
P1 = initial pressure V1 = initial volume P2 = final pressure V2 = final volume
When working with Boyle’s law, absolute pressure may be measured in atmo spheres absolute. To calculate absolute pressure using atmospheres absolute:
Depth fsw + 33 fsw psig + 14.7 psi
= ---------------------------------------------------or = -------------------------------------
Pata Pata
33 fsw 14.7 psi
Sample Problem 1. The average gas flow requirements of a diver using a MK 21 MOD 1 UBA doing moderate work is 1.4 acfm when measured at the depth of the diver. Determine the gas requirement, expressed in volume per minute at surface conditions, for a diver working at 132 fsw.
1. Rearrange the formula for Boyle’s law to find the initial volume (V1):
P2V2
V1 = -------------- P1
2. Calculate the final pressure (P2): 132 fsw + 33 fsw
P2 = -------------------------------------------
33 fsw = 5 ata
3. Substitute known values to find the initial volume (V1):
5 ata × 1.4 acfm V1 =
1 ata = 7.0 acfm
4. The gas requirement for a diver working at 132 fsw is 7.0 acfm.
Sample Problem 2. Determine the gas requirement, expressed in volume per minute at surface conditions, for a diver working at 231 fsw.
1. Rearrange the formula for Boyle’s law to find the initial volume (V1):
P2V2V1 = P1
2. Calculate the final pressure (P2): 231 fsw + 33 fsw
P2 =
33 fsw = 8 ata
3. Substitute the known values to find the initial volume (V1):
8 ata × 1.4 acfm
V1 = -----------------------------------------
1 ata = 11.2 acfm
The gas requirement for a diver working at 231 fsw is 11.2 surface acfm.
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Sample Problem 3. Determine the gas requirement, expressed in volume per minute at surface conditions, for a diver working at 297 fsw.
1. Rearrange the formula for Boyle’s law to find the initial volume (V1):
P2V2V1 = P1
2. Calculate the final pressure (P2): 297 fsw + 33 fsw
P2 =
33 fsw = 10 ata
3. Substitute the known values to find the initial volume (V1):
10 ata × 1.4 acfmV1 =
1 ata = 14.0 acfm
The gas requirement for a diver working at 297 fsw is 14.0 surface acfm.
Sample Problem 4. An open diving bell of 100-cubic-foot internal volume is to be used to support a diver at 198 fsw. Determine the pressure and total surface equivalent volume of the helium-oxygen gas that must be in the bell to balance the ambient water pressure at depth.
1. Calculate final pressure (P2): 198 fsw + 33 fsw
P2 =
33 fsw = 7 ata
2. Rearrange the formula to solve for the initial volume (V1):
P2 V2V1 = P1
3. Substitute the known values to find the initial volume (V1):
3
7 ata × 100 ftV1 =
1 ata 3
= 700 ftThere must be 700 ft3 of helium-oxygen gas in the bell to balance the water pres sure at depth.
Sample Problem 5. The open bell described in Sample Problem 4 is lowered to 297 fsw after pressurization to 198 fsw and no more gas is added. Determine the gas volume in the bell at 297 fsw.
1. Calculate the final pressure (P2):
297 fsw + 33 fswP2 =
33 fsw = 10 ata
2. Rearrange the formula to solve for the final volume (V2):
P1V1V2 = P2
3. Substitute the known values to find the final volume (V2):
3
7 ata × 100 ftV2 =
10 ata 3 = 70 ft
The gas volume in the bell at 297 fsw is 70 ft3.
12-3 CHARLES’/GAY-LUSSAC’S LAW
Charles’ and Gay-Lussac’s laws state that at a constant pressure, the volume of a gas is directly proportional to the change in the absolute temperature. If the pres sure is kept constant and the absolute temperature is doubled, the volume will double. If temperature decreases, volume decreases. If volume instead of pressure is kept constant (i.e., heating gas in a rigid container), then the absolute pressure will change in proportion to the absolute temperature.
The formula for expressing Charles’/Gay-Lussac’s law when the pressure is constant is:
V1T2V2 = T1
Where: V1 = initial volume V2 = final volume T1 = initial absolute temperature T2 = final absolute temperature
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The formula for expressing Charles’/Gay-Lussac’s law when the volume is constant is:
P1T2
P2 = T1
Where: P1 = initial absolute pressure = final absolute pressure
P2 T1 = initial absolute temperature T2 = final absolute temperature
Sample Problem 1. The on-board gas supply of a PTC is charged on deck to 3,000 psig at an ambient temperature of 32°C. The capsule is deployed to a depth of 850 fsw where the water temperature is 7°C. Determine the pressure in the gas supply at the new temperature. Note that in this example the volume is constant; only pressure and temperature change.
1. Transpose the formula for Charles’/Gay-Lussac’s law to solve for the final pressure:
P1T2
P2 = T1
2. Convert Celsius temperatures to absolute temperature values (Kelvin):
°K = °C + 273
T1 = 32°C + 273 = 305°K
T2 = 7°C + 273 = 280ºK
3. Convert initial pressure to absolute pressure:
3 000 psig+ 14.7 psi
P1 = , 14.7 psi
= 205 ata
4. Substitute known values to find the final pressure:
205 ata × 280° K
P2 = 305° K
= 188.19 ata
5. Convert the final pressure to gauge pressure:
P2 = (188.19 ata – 1 ata)× (14.7 psi) = 2 751.79 psig
,
The pressure in the gas supply at the new temperature is 2749 psig.
Sample Problem 2. A habitat is deployed to a depth of 627 fsw at which the water temperature is 40°F. It is pressurized from the surface to bottom pressure, and because of the heat of compression, the internal temperature rises to 110°F. The entrance hatch is opened at depth and the divers begin their work routine. During the next few hours, the habitat atmosphere cools down to the surrounding sea water temperature because of a malfunction in the internal heating system. Determine the percentage of the internal volume that would be flooded by sea water assuming no additional gas was added to the habitat. Note that in this example pressure is constant; only volume and temperature change.
1. Convert Fahrenheit temperatures to absolute temperature values (Rankine): °R = °F + 460 T1 = 110°F + 460 = 570°R
T2 = 40°F + 460 = 500°R
2. Substitute known values to solve for the final volume:
V1T2
V2 =
T1 500° R
= V1 × -------------
570° R = 0.88V1
3. Change the value to a percentage:
V2 = (0.88 × 100%) V1 = 88% V1
4. Calculate the flooded volume:
Flooded volume = 100% - 88% = 12%
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Sample Problem 3. A 6-cubic-foot flask is charged to 3000 psig and the temperature in the flask room is 72°F. A fire in an adjoining space causes the temperature in the flask room to reach 170°F. What will happen to the pressure in the flask?
1. Convert gauge pressure unit to absolute pressure unit:
P1 = 3 000 psig+14.7
,
= 3 014.7 psia
,
2. Convert Fahrenheit temperatures to absolute temperatures (Rankine):
° R = ° F + 460
T1 = 72° F460+
= 532° R T2= 170° F + 460 = 630° R
3. Transpose the formula for Charles’s/Gay-Lussac’s law to solve for the final pressure (P2):
P1T2
P2 = ----------- T1
4. Substitute known values and solve for the final pressure (P2): 3 014.7 psia× 630° R
P2 = ,
532° R 1 899 261
,,
=
532° R = 3 570.03 psia
,
The pressure in the flask increased from 3,000 psig to 3,570.03 psia. Note that the pressure increased even though the flask’s volume and the volume of the gas remained the same.
12-4 THE GENERAL GAS LAW
The general gas law is a combination of Boyle’s law, Charles’ law, and Gay Lussac’s law, and is used to predict the behavior of a given quantity of gas when pressure, volume, or temperature changes.
The formula for expressing the general gas law is:
P1V1 P2V2
=
T1 T2
Where: P1 = initial absolute pressure V1 = initial volume T1 = initial absolute temperature P2 = final absolute pressure V2 = final volume T2 = final absolute temperature
The following points should be noted when using the general gas law:
�
There can be only one unknown value.
�
If it is known that a value remains unchanged (such as the volume of a tank) or that the change in one of the variables will be of little consequence, cancel the value out of both sides of the equation to simplify the computations.
Sample Problem 1. A bank of cylinders having an internal volume of 20 cubic feet is to be charged with helium and oxygen to a final pressure of 2,200 psig to provide mixed gas for a dive. The cylinders are rapidly charged from a large premixed supply, and the gas temperature in the cylinders rises to 160°F by the time final pressure is reached. The temperature in the cylinder bank compartment is 75°F. Determine the final cylinder pressure when the gas has cooled.
1. Simplify the equation by eliminating the variables that will not change. The volume of the tank will not change, so V1 and V2 can be eliminated from the formula in this problem:
P1 P2
=
T1 T2
2. Multiply each side of the equation by T2, then rearrange the equation to solve for the final pressure (P2):
P1T2
P2 = ----------- T1
3. Calculate the initial pressure by converting the gauge pressure unit to the atmospheric pressure unit:
P1 =2 200 psig+14.7 psi
,
= 2 214.7 psia
,
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4. Convert Fahrenheit temperatures to absolute temperature values (Rankine):
° R= ° F + 460 T1= 160° F + 460 = 620° R T2= 75° F460
+ = 535° R
5. Fill in known values to find the final pressure (P2):
2 214.7 psia× 535° RP2 = ,
620° R = 1 911.07 psia
,
6. Convert final pressure (P2) to gauge pressure:
P2 = 1 191.07 psig
,
= 1 896.3 psig
,
The pressure when the cylinder cools will be 1896.3 psig.
Sample Problem 2. Using the same scenario as in Sample Problem 1, determine the volume of gas at standard temperature and pressure (STP = 70°F @ 14.7 psia) resulting from rapid charging.
1. Rearrange the formula to solve for the final volume (V2):
P1V1T2V2 = P2T1
2. Convert Fahrenheit temperatures to absolute temperature values (Rankine):
°R = °F + 460 T1 = 160°F + 460
= 620°R T2 = 70°F + 460 = 530°R
3. Fill in known values to find the final volume (V2):
2 214.7 psia,× 20ft3 × 530° RV2 = 14.7 psia × 620° R
= 2 575.79 ft, 3STP
Sample Problem 3. Determine the volume of the gas at STP resulting from slow charging (maintaining 70°F temperature to 2,200 psig).
1. Rearrange the formula to solve for the final volume (V2):
P1V1T2
V2 = ------------------ P2T1
2. Convert Fahrenheit temperatures to absolute temperature values (Rankine):
T1= 75° F460+
= 535° R
T2= 70° F460+
= 530° R
3. Substitute known values to find the final volume (V2):
2 214.7 psia,× 20ft3 × 530° RV2 = 14.7 psia × 535° R
= 2 985.03 ft, 3STP
Sample Problem 4. A 100-cubic-foot salvage bag is to be used to lift a 3,200 pound torpedo from the sea floor at a depth of 231 fsw. An air compressor with a suction of 120 cfm at 60°F and a discharge temperature of 140°F is to be used to inflate the bag. Water temperature at depth is 55°F. To calculate the amount of time required before the torpedo starts to rise (neglecting torpedo displacement, breakout forces, compressor efficiency and the weight of the salvage bag), the displacement of the bag required to lift the torpedo is computed as follows:
1. Calculate the final volume (V2):
3200 lbs
=
V2 3
64 lb / ft
3
= 50ft
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2. Calculate the final pressure (P2):
231 fsw + 33 fsw
P2 = -------------------------------------------
33 fsw = 8 ata
3. Convert Fahrenheit temperatures to absolute temperature values (Rankine):
° R= ° F + 460
T1= 60° F460
+ = 520° R T2= 55° F460
+ = 515° R
4. Rearrange the formula to solve for the initial volume (V1):
P2 × V2 × T1
V1 = ------------------------------
P1 × T2
5. Substitute known values to find the initial volume (V1):
3
8 ata × 50 ft× 520° RV1 =
1 ata × 515° R 3
= 403.8 ft
6. Compute the time:
Volume Required
Time = ----------------------------------------------------------------------
Compressor Displacement 3
403.8 ft
=
120 ft3 / min = :03::22
(Note that the 140°F compressor discharge temperature is an intermediate temperature and does not enter into the problem.)
12-5 DALTON’S LAW
Dalton’s law states that the total pressure exerted by a mixture of gases is equal to the sum of the pressures of the different gases making up the mixture, with each gas acting as if it alone occupied the total volume. The pressure contributed by any gas in the mixture is proportional to the number of molecules of that gas in the total volume. The pressure of that gas is called its partial pressure (pp), meaning its part of the whole.
The formula for expressing Dalton’s law is:
= ppA+ppB+ppC+ …
PTotal Where: A, B, and C are gases and
× %VolA
PTotal
ppA = --------------------------------------
100%
Sample Problem 1. A helium-oxygen mixture is to be prepared which will provide an oxygen partial pressure of 1.2 ata at a depth of 231 fsw. Compute the oxygen percentage in the mix.
1. Convert depth to pressure in atmospheres absolute: 231 fsw + 33 fsw
=
PTotal
33 fsw = 8 ata
2. Calculate the oxygen percentage of the mix. Since:
%VolA
ppA = × ----------------
PTotal
100% Then:
ppA
%VolA = --------------× 100% PTotal
1.2 ata
= -----------------× 100%
8 ata = 15% oxygen
The oxygen percentage of the mix is 15 percent.
Sample Problem 2. A 30-minute bottom time dive is to be conducted at 264 fsw. The maximum safe oxygen partial pressure for a dive under normal operating conditions is 1.3 ata (Table 14-4 ). Two premixed supplies of HeO2 are available: 84/16 percent and 86/14 percent. Which of these mixtures is safe for the intended dive?
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1. Convert depth to pressure in atmospheres absolute:
264 fsw + 33 fsw
=
PTotal
33 fsw = 9 ata
2. Calculate the maximum allowable O2 percentage:
ppA
%VolA = --------------× 100% PTotal
1.3 ata
= -----------------× 100%
9 ata = 14.4% oxygen Result: The 14 percent O2 mix is safe to use; the 16 percent O2 mix is unsafe. 14%
The pp ofthe 14% mix = 9 ata × -------------
100% = 1.26 ataO2
1.26 ata O2 is less than the maximum allowable. 16%
The pp ofthe 16% mix = 9 ata × -------------
100% = 1.44 ataO2
Use of this mixture will result in a greater risk of oxygen toxicity.
Sample Problem 3. Gas cylinders aboard a PTC are to be charged with an HeO2 mixture. The mixture should provide a ppO2 of 0.9 ata to the diver using a MK 21 MOD 0 helmet at a saturation depth of 660 fsw. Determine the oxygen percentage in the charging gas, then compute the oxygen partial pressure of the breathing gas if the diver makes an excursion from saturation depth to 726 fsw.
1. Convert depth to pressure in atmospheres absolute: 660 fsw + 33 fsw
=
PTotal
33 fsw = 21 ata
2. Calculate the O2 content of the charging mix:
0.9 ata
%VolO2 = -----------------× 100%
21 ata = 4.3% O2
3. Convert excursion depth to pressure in atmospheres absolute: 726 fsw + 33 fsw
=
PTotal
33 fsw = 23 ata
4. Calculate the O2 partial pressure at excursion depth: 4.3% O2
ppO2 = 23 ata × --------------------
100% = 0.99 ata
12-6 HENRY’S LAW
Henry’s law states that the amount of gas that will dissolve in a liquid at a given temperature is almost directly proportional to the partial pressure of that gas. If one unit of gas is dissolved at one atmosphere partial pressure, then two units will be dissolved at two atmospheres, and so on.
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