Basic Diving Physics for understanding .pptx

DIVING PHYSICS FOR DIVERS

1. Diving Physics DR MUHAMMED RISHAM MD MARINE AND HYPERBARIC MEDICINE INDIAN NAVY

2. Introduction In this presentation we will be covering the following topics: Units of measurement Gas laws Contents of cylinders and quads Consumption and duration of cylinders and quads Partial pressures Temperature and temperature changes Analyser readings.

3. Units of Measurement There are two systems of measurement employed in diving: Metric and Imperial The basic units of both systems respectively are: Length: Metres and feet Volume: Litres and gallons Weight: Grammes or tonnes and pounds or tons Pressure: Bars and atmospheres or psi.

4. Units of Measurement (continued) In metric the basic units can be changed by adding a prefix: Smaller units – divide basic unit by 1000 and prefix with milli 1000 millibars = 1 bar Bigger units – multiply basic unit by 1000 and prefix with Kilo 1000 grammes = 1 kilogram There are other prefixes but milli and kilo are the most common.

5. Pressure Atmospheric pressure: This is the pressure at sea level and though it varies it is accepted to be: 1 bar, 1 atmosphere or 14.7psi To be absolutely precise: 1bar = 14.5psi but the difference is so slight it is usually ignored for practical purposes.

6. Pressure (continued) Gauge pressure: This is pressure exerted in excess of atmospheric pressure Gauges should be calibrated to read zero on the surface Absolute pressure: This is the total pressure Absolute pressure = gauge pressure + atmospheric pressure Hydrostatic pressure: This is the pressure in liquids 10 metres or 33ft of seawater will exert a pressure approximately equal to atmospheric pressure i.e. 1 bar At a depth of 20 metres the absolute pressure = 3 bar.

7. Density Density is defined as the mass of the unit volume of the substance. Density = mass ∕ volume Density is normally expressed in kilograms per cubic metre or pounds per cubic foot Typical written - 0.7 kg/m 3 Which in this example means that this particular substance has a density of 0.7 kilogrammes for each cubic metre of its volume or 1 cubic metre volume of the substance has a mass of 0.7kg. Which for most practical purposes will be the same as its weight!

8. Heat Heat is a measure of the kinetic energy of the molecules of the substance and can be transmitted in three ways: Convection Conduction Radiation.

9. Convection Convection takes place when gas or water is heated and then rises Cold gas or water moves in to fill the space vacated by the rising gas or water. This in turn is heated and rises In this way a single heat point can heat a chamber Of course the reverse is possible, cold air or water sinks and is replaced by warmer air or water. In this way convection currents are formed.

10. Conduction With conduction, heat is passed on by contact with another substance Heat will travel from a heat source through anything that is in direct contact with it In this context your body is the heat source and the water, metal hull of the chamber are the substances your body heat will be conducted through Some materials and substances are better conductors of heat than others Metal and water are very good conductors of heat .

11. Radiation Heat can be radiated through space to another substance or body Examples of this are the sun or an electric fire The wavelengths of radiated heat are normally the infrared frequency About 50% of the sun’s energy is infrared Infrared frequencies are tremendously high, being in the frequency/wavelength range immediately before microwaves: 300GHz to 430THz at wavelengths of 1mm to 700nm – (nanometres) respectively.

12. Heat Loss The human body loses heat by all three methods We are designed to breath air and live in a dry, warm environment When immersed in water we lose heat around 25 times faster than in air We therefore have to wear protective clothing whenever heat loss is a possibility.

13. Temperature Temperature is measured in two scales: Fahrenheit - F° Celsius (Centigrade) - C° Freezing point: 0°C and 32°F Boiling point: 100°C and 212°F To convert from one scale to the other: Fahrenheit to Celsius °C = (°F – 32)/1.8 Celsius to Fahrenheit °F = (1.8 x °C) + 32.

14. Absolute Temperature For certain calculations it is necessary to use absolute temperature Which is the lowest temperature that can possibly be reached To convert Fahrenheit to absolute temperature – called Rankine - °R : °R = °F + 460 To convert Celsius to absolute temperature - called Kelvin - °K : °K = °C + 273.

15. Humidity T he following points regarding humidity are important and should be noted: The body sweats to reduce heat This system progressively fails as the atmospheric humidity approaches 100% and fails completely when it does reach 100% This is because the sweat cannot evaporate at such high levels of humidity The body will therefore overheat and hyperthermia may be the result.

16. Humidity Humidity is the amount of water vapour in the atmosphere It is measured in two ways: Absolute Relative Absolute humidity is the amount of water vapour in a given volume of gas This is often used to take gas samples The British standard defines the amount of water vapour in air as 0.5 mg/m³ Relative humidity is given as a percentage of the maximum amount of water vapour that the gas can hold Therefore 100% is the saturation point This is the type of measurement used in the chamber.

17. The Gas Laws Gases are subject to three closely related factors: Temperature Pressure Volume The relationship between these factors gives us three main gas laws: Boyle's Law Charles’ Law Dalton's Law Henry’s Law Though not one of the main gas laws in this presentation, it explains our need for decompression, so is obviously important to know about!

18. Boyle's Law States that provided the temperature is constant, the volume of a given mass of gas will vary inversely as its absolute pressure changes This means that if a mass of gas at a certain pressure and volume is compressed until the pressure is doubled then the volume will be halved This can be expressed in the formula: P1 V1 = P2 V2 More often written as: Free Gas Volume (FGV) = Floodable Volume (FV) x Pressure (P) or P x V = Consumption This formula is used for calculating divers’ breathing gas consumption and gas quantities.

19. Charles’ Law States that the change in either volume or pressure of a given mass of gas is directly proportional to the change in absolute temperature This means that if you half the absolute temperature either the pressure or the volume will half This can be expressed in the formula: V1/T1 = V2/T2 Since gas is kept in rigid cylinders and chambers the volume is normally constant and it is the pressure that changes, therefore the following formula applies: P1/T1 = P2/T2 or P1 T2 = P2 T1 It is important to remember that absolute temperature must be used for these calculations.

20. Dalton's Law Dalton's Law or the Law of Partial Pressures: States that in a mixture of gases each gas exerts a pressure as if it alone filled that space and the sum of all partial pressures equals the total pressure This means that if we have a mixture of gases, for example air, we can calculate the partial pressure of each gas Air is composed of 21% oxygen and 79% nitrogen Therefore the pressure exerted by the oxygen at the surface is 21% of the total pressure or 21% of one bar This can be read as: ppO 2 = 21 x 1/100 = 0.21 bars The oxygen exerts 21% of the pressure or 0.21 bars or 210mb.

21. Dalton's Law (continued) A formula can be constructed to find any partial pressure and is written as: ppO 2 = % x AP/100 Where AP equals absolute pressure and is the total pressure of the gases Example: What is the ppO 2 of air at 30msw? ppO 2 = % x AP/100 – therefore: ppO 2 = 21% x 4/100 = 0.84bar.

22. Henry's Law Henry's Law states that at a constant temperature the amount of gas that will dissolve in a liquid is almost directly proportional to the partial pressure of that gas Example: A liquid at the surface has 1L of nitrogen dissolved in it If the pressure of the gas above the liquid is increased to an equivalent of 30m of seawater (4bars absolute) then 4L of nitrogen could be dissolved in that liquid.

23. Henry's Law (continued) It is important to realise however that this will take some time Also; if the pressure is released then the gas will have to come out of the solution The rate of release of the gas will determine the size of the bubbles that have formed The liquid can be likened to a diver; gas dissolves into the diver’s body, which is mainly liquid, as he goes under pressure and is released as he ascends to the surface This law explains why decompression must be carefully controlled.

24. Contents of a Cylinder A question like this is based on Boyle’s Law The amount of gas required is expressed as a volume at atmospheric pressure (1bar) Commonly referred to as F ree G as V olume – FGV We use gauge pressure rather than absolute pressure for this kind of calculation because we can’t get the last bar out of the cylinder Example: What quantity of gas is contained in a cylinder that has a volume of 10L and is pressurised to 200bar? The formula to use is: FV x P = FGV Therefore: 10 x 200 = FGV of: 2000L or 2m 3.

25. Gas Volumes A gas volume example: A typical 16 cylinder (16 x 50 litres) quad is pressurised to 180bar. How much gas does it contain? Formula to use - FV x P = FGV FV = 16 x 50 = 800L – because the quad has 16 cylinders, each with a floodable volume of 50L P = 180bar – is the pressure of the gas in the cylinders Therefore: 800 x 180 = FGV of : 144,000 litres or 144m ³ Divide the number of litres by 1000 to get cubic metres.

26. Gas Volumes How much gas is used to charge a 16 x 50 litre quad from 50bar to 200bar? One method: There is already 50bar in the quad so we take that away from what we need: 200 – 50 = 150 We multiply the 150 by the FV of each cylinder multiplied by the number of cylinders in the quad, which is 16 Therefore: FGV = FV x P FGV = (200 – 50) x (50 x 16) = FGV = 150 x 800 = 120,000L or 120m 3. 200 bar -50 bar = 15 0 bar to add

27. Available Gas Not all the gas in a cylinder or quad can be used Therefore: For surface supply: Change over at a certain pressure, say 40 or 50bar and then calculate the FGV For emergency supplies such as bailout cylinders and onboard gas cylinders where you need as much as is actually available: Remove depth in bars plus regulator pressure and then calculate the FGV Decide which you need to use and then do the calculation Don’t get them mixed up!

28. Estimating Gas Consumption Divers’ breathing rates vary depending on personal consumption and activity, therefore the following is based on IMCA guidance There are other guidelines but the Company Diving Manual will tell you what breathing rates should be used for breathing gas calculations A commercial diver uses 35 litres per minute (1.25ft 3 /min) In an emergency this increases to 40 litres per minute (1.5ft 3 /min) These figures are averages used for calculation purposes Examination questions will tell you what rate to use The 35L per minute or the 40L per minute needs to be multiplied by the absolute pressure to get consumption; the volumes actually breathed at depth. The quantity is converted to a FGV at surface pressure The absolute pressure should be in bars.

29. Examples of Gas Consumption If a diver goes under pressure to 10 metres he will still be breathing as a rate of 35L/min but the pressure will be double - 10 msw = 2bars absolute Formula: Consumption = Absolute Pressure (AP) x Volume (V) This is essentially the same formula as that used to calculate available gas but this time the pressure used must always be absolute AP = 2bar V = 35L Therefore: Consumption = 2 x 35 = 70L/min.

30. Examples of Gas Consumption A diver breathing 35L/min is working at 30 metres. What is the diver’s breathing gas consumption at that depth? Formula used: Consumption = AP x V AP = 4bar V = 35L Therefore: Consumption = 4 x 35 = 140L/min The diver is consuming gas at a rate of 140L/min at 30msw.

31. Estimating Breathing Gas Duration Estimating the duration of a bailout cylinder: Because a bailout will only be used in an emergency the breathing rate used for estimating its duration is 40L/min In these calculations we assume that all the gas in a bailout cylinder may be needed i.e. it may be breathed completely empty Therefore we must make allowance for the regulator driving pressure and water pressure at depth In examination questions you will be told what regulator driving pressure to use We use gauge pressure because the one bar absolute left in the cylinder is not available to the diver.

32. Estimating Breathing Gas Duration Bailout cylinder duration example: A diver is working at a depth of 80msw. What is the estimated duration of his 10L bailout cylinder at depth if it is charged to 200bars? You will need to perform three calculations: Find out his emergency gas consumption at 80msw Find out how much gas is available to him at 80msw Divide answer one into answer two to find the duration of the cylinder in a worse case scenario.

33. Estimating Breathing Gas Duration Bailout cylinder duration example continued: The formula to use for the first calculation: Consumption = P x V P = 9bars absolute V = 40L Therefore: Consumption = 9 x 40 = 360L/min The formula to use for the second calculation: FGV = FV x P FV = 10L P = 200 - 8 - 10 = 182 bar – the -8 is for the water depth and the -10 is for the regulator Therefore: FGV = 10 x 182 = 1820L this is the FGV of the gas available to him at 80msw.

34. Estimating Breathing Gas Duration Bailout cylinder duration example continued: So we now know that the diver would have 1820L of gas available to him at 80msw We also know that at 80msw his breathing rate is estimated to be 360L/min Therefore: Duration = FGV/Consumption FGV = 1820 Consumption = 360L/min Therefore: Duration = 1820/360 = 5 minutes.

35. Pressure Changes Due To Temperature According to Charles’ Law when the temperature of a gas drops its pressure will drop, assuming it is in a rigid container Likewise, when a cylinder or chamber are compressed the gas heats up and the pressure increases When it cools the pressure drops Under certain circumstances being able to calculate this relationship is useful The following formula can be used for rigid containers: P1/T1 = P2/T2 or P1T2 = P2T1 Note the following basic rules when carrying out these calculations: Absolute temperature = °C + 273 or °F + 460 When working with chamber pressures use absolute pressure When working with cylinder pressures gauge pressure can be used.

36. Pressure Changes Due To Temperature Charles’ Law cylinder example: A cylinder is charged to 200bars and is at a temperature of 40°C. What will its final pressure be when the temperature drops to 5°C? Formula transposed to find P2: P2 = P1 x T2/T1 P1 = 200bar - the pressure after charging T2 = 278°K 5 + 273 to give the absolute temperature after cooling T1 = 313°K 40 + 273 to give the absolute temperature before cooling Therefore: P2 = 200 x 278/313 = 177.6bars.

37. Charles’ Law chamber example: A chamber is pressurised to 55 metres and is at a temperature of 45°C. What will the depth be when the temperature stabilises at 25°C? Formula transposed to find P2: P2 = P1 x T2/T1 P1 = 6.5bar - the absolute pressure after pressurisation T2 = 298°K 25 + 273 to give the absolute temperature after cooling T1 = 318°K 45 + 273 to give the absolute temperature before cooling Therefore: P2 = 6.5 x 298/318 = 6.09bars absolute minus 1 and multiply by 10 to get 50.9 the depth in metres after cooling. Pressure Changes Due To Temperature

38. Calculating Partial Pressure To calculate partial pressures the following formula is used: PP = % x AP/100 Where: % = Percentage of the gas whose partial pressure we are calculating AP = Absolute Pressure in bars - If a depth, this value can be changed to depth absolute in metres instead of dividing by 100 Remember: If AP is bars absolute, then the answer will be in bars If dealing with a depth and you choose to enter the absolute depth in metres instead in bars, then the answer will be in absolute metres This will probably need to be converted to a gauge depth.

39. Some partial pressure calculation examples: If a diver is working at 80msw and is breathing 10/90 what is the partial pressure of the oxygen in the mix at that depth? Note: When a mix is given such as here, 10/90, the oxygen percentage is always the first value Formula: ppO 2 = % x AP/100 Where: % = 10 AP = 9bars Therefore: ppO 2 = 10 x 9/100 = 0.9bar. Calculating Partial Pressure

40. Partial pressure calculation examples continued: Although not strictly mathematically correct, most Supervisors and LSTs use metres and millibars . So the previous example would change a little: % would remain the same AP would be converted to depth absolute This will give the answer in millibars Therefore: PPO2 = 10 x 90 = 900mbs Used carefully this is a quick and convenient way of doing this type of calculation but be careful if being asked to calculate a depth Refer to the depth example given later for an explanation. Calculating Partial Pressure

41. Dalton’s Law Easy Formula Transposition PP % AP The square can be used if AP and pp are given in bars. Turn the square to find the top value, pp in this example i.e. pp = AP x % / 100. The triangle can be used if AP is given in depth absolute and pp is given in millibars . Cover the value to be found and calculate using the values left showing i.e. pp = % x AP.

42. Calculating other values: In some instances we are given two of the variables and need to calculate the third. The following is a typical example: A dive is to be made to 80msw and company policy dictates that the ppO 2 of the breathing gas should be 800mbar, what gas mix should be provided? We transpose the formula to find % % = ppO 2 / AP or % = ppO 2 x 100 / AP Where: ppO 2 = 800mbar or 0.8bar AP = 90 or 9bar Decide which version of the formula you want use and then: % = 800/90 = 8.9% or % = 0.8 x 100/9 = 8.9% It would be hard to produce exactly that percentage therefore the mix nearest to that figure, within limits, would be the gas used It is normal for company manuals to give ranges rather than a single figure. Dalton’s Law - Calculating Other Values

43. A diver is diving from a bell breathing 8% oxygen that has a calculated ppO 2 of 850mbar, what depth is the diver working at? Transpose the formula to find AP AP = pp / % or pp x 100/% PP = 850mbar or 0.85bar % = 8 Therefore: AP = 850 / 8 = 106.25 minus 10 is 96.25msw or AP = 0.85 x 100 / 8 = 10.625bar abs. minus 1 x 10 is 96.25msw It doesn’t matter which version of the formula you use but don’t mix them up! Calculating Other Values (continued)

44. When a chamber or bell is initially blown down from the surface the percentage of gases at the end of the blowdown is not the same as that used for the blowdown itself This is because there was air in the chamber or bell before the blowdown started The final ppO 2 will be a mixture of the two gases What will the final ppO 2 of the bell be after it is blown down from the surface to 100m using 11 / 89? ppO 2 of gas is: 100 x 11 = 1100mbar ppO 2 of air is: 210mbar Therefore: 1100 + 210 = 1310mbs or 1.31bar is the final ppO 2. Chamber Partial Pressure

45. Chamber Blowdown It will be necessary to estimate the amounts of air needed to compress chamber, transfer locks etc. The following is an example: “How much air will be needed to compress a 17.5m 3 chamber and 6m 3 transfer lock to a pressure equivalent depth of 35msw?” As the chamber and transfer lock are to be compressed the same distance we can start by simply adding the two volumes together: 17.5 + 6 = 23.5m 3 Now apply the FGV = V x P formula to find the answer FGV = 23.5 x 3.5 = 82.25m 3 of air will be needed “What will the ppO 2 be at 35m?” pp = % x AP / 100 ppO 2 = 21 x 4.5 / 100 = 0.945bar Chronic oxygen toxicity is a potential problem for anyone breathing such a high ppO 2 for more than a few hours “What might be a solution to this?” Filling the chamber with a different gas.

46. Analyser Reading Examples If a chamber oxygen analyser is reading 6.52% and the depth gauge indicates a depth of 55m. What is the ppO 2 ? Use the Dalton’s Law formula: PP = AP x % AP in this case is in metres absolute Change the depth to metres absolute by adding 10 55+10 = 65 Multiply by 6.52 65 x 6.52 = 423.8mbs If the reading is given in ppm then this has to be changed to a percentage by dividing by 10,000.

47. Surface Equivalent Value - SEV Sometimes the partial pressure of the limits you are looking for are expressed as a surface equivalent i.e. the limit for CO 2 for air diving is given in the U.S. Navy manual as 1.5% SEV To calculate SEV from the analyser reading, multiply by the absolute depth to get them in bars or atmospheres A reading of 0.08% CO 2 at 90 m would therefore equate to 10 × 0.08 or 0.8% SEV Literally the equivalent of breathing 0.8% CO 2 on the surface If the analyser has its sensor in the chamber the analyser should read directly in partial pressure and no conversions are required If the analyser readout is a percentage then multiply by 10 to convert the reading to millibars Therefore for a 0.5% reading: 0.5% x 10 = 5mbs Divide by 100 to convert the reading into bars 0.5 / 100 = 0.005bar Multiply by 1000 to get back to millibars 0.005 x 1000 = 5mbs.

48. Summary In this section we have looked at: Units of measurement Units of pressure Heat and its effect on divers, bells and chambers Temperatures and absolute temperatures Surface Equivalent values – SEV Analyser readings. The Gas Laws We have learnt how to use the gas law formulas to: Calculate the contents of cylinders Estimate gas consumption Estimate the duration of cylinders Calculate pressure changes due to temperature changes Calculate partial pressures Calculate oxygen consumption and CO 2 production. Keep trying and practicing the calculations until they are second nature However, remember they are just guides and safety margins must be applied and adhered to.

49. 49 Are there any questions?