1 / 51

Diving Physics Practical

Diving Physics Practical. We will cover. Equations Cross-Multiplication Pressure and Depths Universal Gas Law Gauge Pressure Decanting Air. We will cover. Air Available At Depth Air Consumption Nitrox Theory Volumes Relative Density Buoyancy. Equations. Implies two things are equal

cyma
Download Presentation

Diving Physics Practical

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Diving Physics Practical

  2. We will cover • Equations • Cross-Multiplication • Pressure and Depths • Universal Gas Law • Gauge Pressure • Decanting Air

  3. We will cover • Air Available At Depth • Air Consumption • Nitrox Theory • Volumes • Relative Density • Buoyancy

  4. Equations • Implies two things are equal • Equations hold true for the following: • The same quantity can be added to both sides: • (2 + 3) + 4 = 5 + 4 • The same quantity can be subtracted to both sides: • (2 + 3) - 4 = 5 - 4 • The Same quantity can be Multiplied to both sides: • (2 + 3) x 4 = 5 x 4 • The same quantity can be divided to both sides • (2 + 3) / 4 = 5 / 4

  5. a c = b d a c (bd) (bd) = b d Cross-Multiplication Take this Equation: Multiply Each side by the product of the denominators b’s cancel out on the left d’s cancel out on the right This gives is this equation: ad = cb

  6. a c = d b ad c = b Cross-Multiplication You can skip the middle step by crossing the denominator on one side over to the numerator on the other side. ad = bc

  7. Pressure and Depths • Atmospheric is normally 1 bar • Fresh Water Pressure is 1 bar for every 10m depth. • Unless it’s a specific relative denisty question assume: 10m Seawater = 1 bar • P stands for Pressure. D stands for Depth • P = ( D / 10 ) + 1 • D = ( P – 1 ) x 10

  8. Pressure and Depths • Example: What is the absolute pressure at 37m of sea water? P = ( 37 / 10 ) + 1 = 4.7 bar • Example: What is the depth of sea water at 3.7 bar D = ( 3.7 – 1 ) x 10 = 27m

  9. Universal Gas LawSimplified P V P V • P = Pressure in bar • T = Temperature in Kelvin. 273°K = 0°C • V = Volume in Litres 1 1 2 2 = T T 1 2

  10. Universal Gas Law • Example: A 12 litre diving cylinder is filled to 200 Bar (P1), and the temperature is 10°C. It is left in the boot of a car on a hot summer's day and the temperature rises to 40°C. What is the pressure in the cylinder now? (P2) 40°C = 273 + 40 = 313 Kelvin. 10 °C = 273 + 10 = 283 Kelvin 200 x 12 = P2 x 12 283 313  Using cross-multiplication we get:  P2 = 200 x 12 x 313 12 x 283 P2 = 221.2 bar

  11. Universal Gas Law Example:A Balloon with a volume of 10 units will be reduced in size to what volume of units at 30m. Formula: P1V1 = P2V2 P1 = 1 bar. V1 = 10 units. P2 = 4 bar Substituting the values in to the formula: 1 x 10 units = 4 x V2 V2 = (10 / 4) = 2.5 units.

  12. Universal Gas Law Example: A collapsible container has a volume of 3.25 litres at a depth of 30 metres. The container will burst if filled to 5 litres. At what depth will the container burst? Formula: P1V1 = P2V2 P1 = 4 bar. V1 = 3.25 l. V2 = 5 l P2 = ( P1V1 ) / V2 P2 = (4 x 3.25) / 5 P2 = 2.6 bar = 16m depth

  13. Universal Gas Law Example: An ABLJ cylinder has a volume of 0.4 ltrs, andis filled to 200 Bar. Assuming the ABLJ willtake 16 litres of air to fill it at the surface, howmany fills will be available at 30 metres ? Total contents in cylinder:200 x 0.4 = 80 litres P1V1 = P2V2: P1 = 1 bar. V1 = 80 units. P2 = 4 bar Substituting the values in to the formula: 1 x 80 = 4 x V2 V2 = (80 / 4) = 20 litres The jacketrequires 16 litres so there will be: 20 / 16 = 1.25 fills.

  14. Air in a Cylinder Size of a Cylinder refers to it’s Water Capacity Calculating Air in Cylinder is really calculating how much air is available at 1 bar. Basically a Gas Law equation. P1V1 = P2V2 As P1 = 1 bar implies : V1 = P2V2 or Cylinder Capacity = Cylinder Volume x Fill Pressure

  15. Air in a Cylinder Example: How many litres of air at 1 Bar is contained in a15 litre cylinder pumped to 232 Bar ? Capacity = 232 x 15 = 3480 litres Example: You have a 12.2L cylinder with 232 bar cylinder pressure. How much air is contained in the cylinder? Capacity = 232 x 12.2 = 2830.4 litres Example: You have a 15L cylinder which contains 3,315 Litres of air. What is the pressure of the cylinder? 3,315 = P2 x 15 P2= 3,315 / 15 = 221 bar

  16. Gauge Pressure Pressure Gauge Reads Zero on the Surface Absolute Pressure = Gauge Pressure + 1 bar Example: A diver contents gauge reads 220 bar in a 12.2L cylinder. How many litres of air are contained in the cylinder? Cylinder Pressure = 220 + 1 = 221 bar. Capacity = 221 bar x 12.2 litres = 2,696.2 Litres Example: A diver calculates that there is 230 bar cylinder pressure. What will the gauge read? Gauge Pressure = 230– 1 = 229 bar.

  17. Decanting Air Filling Mini Cylinder from Main Cylinder Mini Cylinders Would Include DSMB Cylinder and BCD Cylinder. Pressure in both cylinders calculated by: Calculating Total Volume of Air contained in Both Cylinders First. Divide by the Total Volume of Both Cylinders

  18. Decanting Air Example: A diver has a 12.2L cylinder containing 220 bar gauge pressure and an empty 0.4L mini cylinder. What is the final gauge pressure in both cylinders after filling the mini cylinder from the main cylinder. Cylinder Pressure = 220 + 1 = 221 bar Capacity: 221 x 12.2 = 2,696.2 L Mini Cylinder Pressure = 1 bar Capacity: 1 bar x 0.4L = 0.4L

  19. Decanting Air Total Air : 2,696.2L + 0.4L = 2,696.6L Total volume of Cylinders: 12.2L + 0.4L = 12.6L Capacity = Cylinder Volume x Fill Pressure Absolute pressure in both cylinders = 2,696.6 L / 12.6 L = 214 bar Gauge Pressure will read: 214 bar – 1 bar = 213 bar.

  20. Available Air at Depth Regulator Reduces Cylinder Pressure to Ambient. Amount of Air Available at Depth equals Cylinder Pressure minus Ambient Pressure Example: How much air is available to a diver at 30m depth with a 12.2 litre cylinder with Cylinder pressure of 232 bar. Ambient Pressure at 30m = 4 Bar Available air = (232 bar – 4 bar) x 12.2 litres = 2,781.6 litres

  21. Surface Air Consumption Surface Air Consumption is a measure on how many litres a minute a diver can breathe on the surface. To Calculate SAC a diver Fins at a constant depth for a fixed period of time. Calculate Gas used: Drop in Gauge Pressure x Cylinder size. Adjust back to surface by dividing by absolute pressure. Divide by Time in Minutes to get SAC.

  22. Surface Air Consumption Example: Fin at 10m (2 bar) for 10 minutes with contents reading 202 bar at start, 175 bar at end using a 12 litre cylinder. Gas Used: = (202 bar – 175 bar) x 12 litres = 324 litres Adjust to surface: 324 litres / 2 bar = 162 litres SAC = 162 litres / 10 minutes = 16.2 L/min.

  23. Gas Consumption Gas Consumption = SAC x Ambient Pressure x Dive Time Example: Calculate the amount of air needed to dive to 30m for 17 minutes when SAC = 25L/min: Gas consumption = 25l/min x 4 bar x 17 minutes = 1,700 litres

  24. Gas Consumption Example: Estimate the duration of a 15 litre bottle filled to 200 bar at a depth of 20m. Assume a surface air consumption of 20 l/min and a reserve of 50 bar. Ignore Ascent and Descent times. Available gas = (200 bar – 50 bar) X 15 litres = 2,250 litres Gas Consumption = SAC x Ambient Pressure x Dive Time 2,250 litres = 20l/min x 3 bar x Dive time Therefore, Dive Time = 2,250 / (20 x 3) = 37.5 minutes.

  25. Gas Consumption Example: Estimate the duration of a 12 litre bottle filled to 232 bar at a depth of 25m. Assume a surface air consumption of 25l/min and a reserve of 50 bar. Ignore Ascent and Descent times. Available gas = (232 bar – 50 bar) X 12 litres = 2,184 litres Gas Consumption = SAC x Ambient Pressure x Dive Time 2,184 litres = 25l/min x 3.5 bar x Dive time Therefore, Dive Time = 2,184 / (25 x 3.5) = 24.96 minutes.

  26. ppg Fg P Nitrox Theory The Pressure T: • Partial pressure of gas – ppg • Decimal fraction of gas in mix – Fg • Pressure – P • Use Pressure ‘T’ ppg = Fg x P P = ppg/Fg Fg = ppg/P

  27. Nitrox Theory CFT Recommendations: 1.4 bar - max ppO2 for Recreational diving 1.6 bar - max ppO2 for deco stop

  28. ppg 1.4 Fg 0.32 P P Nitrox Theory Examples…  • On a dive holiday you are given a 32% mix (EAN32) What depth is this safe to dive to? • To find the MOD you need to use the ppg of 1.4 for O2 and the % of the mix in your cylinder ~ 0.32 P= 1.4 / 0.32 = 4.4 bar therefore the max depth for this mix is 34m

  29. ppg ppg Fg .32 P 4.2 Nitrox Theory Examples…  • You have a 32% Nitrox mix. You want to dive to 32m. Is this a good mix for this depth? • You know percentage of O2(32%) & P of the dive ~ depth 32m (4.2 bar),therefore you need to find the ppg ppg= .32 x 4.2 =1.34 bar therefore a mix of 32% is good for a 32m dive

  30. Nitrox Theory Example: A diver has a 40% mix. What are the two maximum operating depths for this mix according to CFT Recommendations? Recreational ppO2 = 1.4 bar. FO2 = 0.4 Pressure = 1.4 / 0.4 = 3.5 bar. MOD = 25m Deco schedule ppO2 = 1.6 bar. FO2 = 0.4 Pressure = 1.6 / 0.4 = 4 bar. MOD Deco = 30m

  31. Nitrox Theory Example: What is the Partial pressure of Nitrogen in EAN28 at a depth of 25 metres? Warning: Don’t calculate ppO2!! FN2 = 1 - 0.28 = 0.72. P = 3.5 bar ppN2 = 0.72 x 3.5 = 2.52 bar

  32. Nitrox Theory Equivalent Air Depth: Equivalent Depth where ppN2of Nitrox Mix is the same as an air. EAD (bar) = FN2in mix xP 0.79 (FN2 in air) Note, this may give you an intermediate depth. For EAD you use the next greatest depth on the dive tables.

  33. Nitrox Theory Example: What is the EAD for EAN32 at 33m? FN2 = 0.68; P = 4.3 bar EAD = (0.67 x 4.3) = 3.7 Bar 0.79 3.7 bar = 27m

  34. Volumes Volume of a sphere = 4πR3 3 Volume of a cylinder = πR2H Volume of a block = L x B x H 1 m3 = 1000 litres 1 litre = 1000 cm3

  35. Volumes Example: Calculate the volume of a block measuring 1m x 1.5m x 0.5m. Volume = 1 x 1.5 x 0.5 = 0.75 m3 1 m3 = 1,000 litres Volume = 0.75 x 1,000 = 750 litres

  36. Volumes Example: Calculate the volume of the cannon below: Note, lengths refer to the total length of the cannon (internal and external) and not just the length of the cylinder portion.

  37. Volumes A hemisphere is half a sphere. To solve this you’ll need to do work out four volumes. Outer hemisphere. Inner hemisphere, outer cylinder and inner cylinder. Subtract the inner volume from the outer volume. Therefore applying the formulae for cylinders and spheres: Vol ( large cylinder + hemisphere) minus vol (small cylinder + hemisphere) (πR2H+ 2/3πR3) - (πR2H+ 2/3πR3)

  38. Volumes Cylinder length = Cannon length - Hemisphere radius Outer Cylinder Length = 180cm - 10 cm = 170cm (Outer Cylinder length = Inner Cylinder Length) {[3.14 ×(10)2 x 170] + [2/3×3.14 x (10)3]} - {[3.14 ×(5)2 x 170] + [2/3 × 3.14 x (5)3]} { [ 53380 ] + [ 2093 ]} - {[ 13345 ] + [ 262 ]} = 41,866 c.c. or 41.9 litres

  39. Relative Density Ratio of density of a substance to that of fresh water at 4˚C. Relative density of 2 implies an object is twice as dense as Fresh Water. Fresh-water = 1, Sea-water = 1.027 1 litre of fresh water weighs 1 kg.

  40. Relative Density Example: A concrete block with a volume of 750 litres has a relative density of 4.5. What is the weight of the block? 750 Litres of Water weighs 750kgs. Concrete block’s relative density = 4.5. The weight of the concrete block is: 750kgs x 4.5 = 3,375kgs

  41. Relative Density Example: A cannon with a volume of 41.9 litres has a relative density of 8. What is the weight of the block? 41.9 Litres of Water weighs 41.9kgs. Cannon’s relative density = 8. The weight of the concrete block is: 41.9kgs x 8 = 335.2kgs

  42. Weight of Water Displaced 1 litre of Fresh water = 1 kg 1 litre of Salt water = 1.027 kg Example: How much weight of fresh water and sea water will a 750 litre concrete block displace? Freshwater: 750 x 1 = 750kgs Seawater: 750 x 1.027 = 770.25kgs

  43. Buoyancy Archimedes Principle: Any object wholly or partially immersedin a liquid experiences an upthrust(apparent loss of weight) equal to theweight of liquid displaced. Buoyancy questions usually Follow these steps: Calculate the volume of the object. Calculate the weight of the object. Calculate the upthrust, which is the weight of the liquid displaced. Calculate the weight of the object in water by subtracting the weight of the object from the weight of the liquid displaced. Then calculate how much air is needed to make this object neutrally buoyant.

  44. Buoyancy Example: A concrete block measuring 1m x 1.5m x 0.5m isto be lifted from the sea-bed. Its relative densityis 4.5. How many litres of air at 1 Bar arerequired to effect neutral buoyancy at 30 metres ? Volume = 1 x 1.5 x 0.5 = 0.75 m3 x 1000 = 750 litres. Weight of block: 750 x 4.5 = 3,375 kgs Weight of displaced sea-water: 750 x 1.027= 770 kgs Effective weight at 30 metres:= 3,375 – 770 = 2,605 kgs One litre of air will displace 1.027 kgs of sea-water, therefore it will require 2605/1.027 litresof air to give neutral buoyancy:- 2,537litres Volume of air required at 30 metres = 2,537litres 2,537litres x 4 = 10,148litres of air at 1 Bar

  45. Buoyancy Example 2: You are required to lift a bronze cannon withwith a volume of 42 litres at a depth of 20 metres sea-water. How many 50-litre oil drums and howmany litres of air at 1 Bar will be necessary?The relative density of bronze is 8 and the oildrums each weigh 2 kgs in seawater. Cannon Weight = 8 x 42 = 336 kgs Displaces 42 litres of sea water, 42 x 1.027 = 43.1 kgs Cannon weight in water = 336 – 43.1 = 292.9 kgs We require to neutralise 293 kgs to effect neutral buoyancy.

  46. Buoyancy Each drum has a volume of 50 litres of air, which will give an upthrust of 51.35 kgs in seawater, less its own weight of 2 kgs, leaving an effective lifting power of 49.35 kgs per drum. Therefore we will require:293 = 5.9 drums required 49.35 Each drum will contain 50 litres of air at 3 Bar (20 metres). Therefore we will require 5.9 x 50 x 3 = 885 litresof air at one Bar to effect neutral buoyancy.

  47. Buoyancy Example: Calculate the air required to raise a 250kg sea mooring with an assumed relative density of 10. The depth of the mooring is 35m. Relative density of 10 implies mooring is 10 times denser than freshwater. 1 litre of freshwater = 1kg. Mooring Volume = 250 / 10 = 25 litres 25 litres of sea water displaces: 25 x 1.027 = 25.7kgs Apparent weight =250–25.7= 224.3kgs One litre of sea water displaces 1.027 kgs of water. 224.3 / 1.027 = 218.4 litres At 1 bar: 218.4 x 4.5 = 982.8 litres. 982.8 litres of air needed.

  48. Buoyancy Example: Calculate the air required to raise a 40kg sea mooring with an assumed relative density of 10. The depth of the mooring is 20m. Relative density of 10 implies mooring is 10 times denser than freshwater. Mooring Volume = 40 / 10 = 4 litres 4 litres of sea water displaces 4 x 1.027 = 4.1 kgs Apparent weight =40–4.1 = 35.9 kgs One litre of sea water displaces 1.027 kgs of water. 35.9 / 1.027 = 35 litres At 1 bar: 35 x 3 = 105 litres.

  49. Buoyancy Example: What is the apparent weight of a 1m3 block of concrete and steel weighing 7000kgs when immersed in fresh water. 1m3 = 1000 litres. 1 litre of fresh water weighs 1 kg. Weight of water displaced = 1,000 kgs Apparent weight = 7000 – 1000 = 6000kgs

  50. Summary Equations Cross-Multiplication Pressure and Depths Universal Gas Law Gauge Pressure Decanting Air

More Related