DIVING PHYSICS EXAMPLE QUESTIONS

Transcription

1 DIVING PHYSICS EXAMPLE QUESTIONS PLEASE NOTE: 1 bar = 10 Meter in Salt water 1 bar = 10.2 Meter in Fresh water. Will be GIVEN to you for calculations. 10m in Salt water = 1 bar 10m in Fresh water = 0.98 bar (1/10.2 = 0.98) 1 Litre Salt Water = Kilograms 1 Litre Fresh Water = Kilograms 1. Q: For every meter of depth in fresh water, pressure increases by: A: From the info given, we know that 1 bar in fresh water is reached at 10.2m depth. Therefore, 1 / 10.2 = 0.98 bar per 10 meter. From this we take 0.98 / 10 and our answer is that for every meter of depth in fresh water, pressure increases by bar. (0.98 / 10) 2. Q: The absolute pressure at a depth of 40m in salt water is: A: We know that for every 10metres of salt water, pressure increases by 1 bar. Taking into account the 1 bar atmospheric pressure; we calculate pressure in salt water, as follows: (Depth / 10) +1 Therefore pressure would be at 10m: (10/10) + 1 = 2bar 20m: (20/10) + 1 = 3bar 30m: (30/10) + 1 = 4bar 40m: (40/10) + 1 = 5bar Our answer is therefore at 40msw the pressure is 5 bars 3. Q: A known volume of air in a flexible container expands the most during a salt water ascent from: A: Pressure changes from 5bar to 4bar when ascending from 40msw to 30msw. 5/4 = 1.25 Pressure changes from 4bar to 3bar when ascending from 30msw to 20msw. 4/3 = 1.33 Pressure changes from 3bar to 2bar when ascending from 20msw to 10msw. 3/2 = 1.50 Pressure changes from 2bar to 1bar when ascending from 10msw to surface. 2/1 = 2.00 THEREFORE the largest pressure difference is from 10msw to the surface.

2 4. Q: An object with a volume of 117 litres, weighing 145kg is located at a depth of 30m in the ocean. How many 25 litre lift bags weighing 2.5kg each would it take to bring the object to the surface? A: According to Archimedes Principle, a body submersed in water is buoyed up by a force equal to the weight of the water it displaces. Therefore, the object weighing 145kg is buoyed up by (117 x 1.025kg which is the weight of salt water, = kg lift (buoyancy) = kg needed to lift the object. Our lift bags have a capacity of 25 litres each and weigh 2.5kg each therefore the lift capacity is only (25*1.025)-2.5 = kg lift capacity PER BAG? = 1.95 therefore, 1 lift bag is not enough. We need TWO lift bags for this exercise. Q: Air consumption: a. Decreases as depth increases b. Increases as depth increases c. Remains the same as depth increases d. Varies inversely with depth 5. A: The air delivery system supplies air to the diver at ambient pressure. As the pressure increases during descent, the air delivery system automatically adjusts and supplies more air from the cylinder. A diver s usage will vary directly with depth. The diver will consume twice as much air with each breath at 10msw (2 bar) than he will at sea level (1bar), Three times as much at 20msw (3bar) Four times as much at 30msw (4 bar) 6. Q: What would be the surface air consumption rate (SAC) of a diver with a 200bar, 15 litre? cylinder who uses 15bar in 10 minutes at a depth of 5 meters in the ocean? SAC = AIR USED AIR USED = START PRESSURE END PRESSURE WHERE USED? A: SAC = (AIR USED) X CYLINDER SIZE DEPTH (BAR) X TIME (MINUTES) SAC = (15bar) x 15l 1.5bar x 10 minutes SAC = 225 / 15 = 15 litres per minute.

3 7. Q: The ocean depth at which the partial pressure of oxygen in air equals 1.6bar is meter. Assuming O2 = 20%. A: 1.6 bar / 0.2 (Oxygen fraction) = 8. 8 bar = 70msw 8. Q: Breathing 0.5% carbon monoxide at a depth of 40m in the ocean is equivalent to breathing what % of carbon monoxide at the surface. A: 0.5% x 5bar (40msw = 5bar) = 2.5% 9. Q: An object that is 1.5m square and high (1.5 x 1.5 x 1.5) is floating in the ocean. 20cm of the object is above the waterline. What is the minimum line strength? that would be needed to lift the object out of the water? A: 1.5 x 1.5 x 1.5 = 3.375m3 20cm is above the waterline therefore 1.5 x 1.5 x 0.2 is above the waterline. 1.5 x 1.5 x 0.2 = 0.45m3 Thus = 2.925m3 is submersed 1.5 x 1.5 x 1.3 = 2.925m x (because ocean water weighs 1.025kg per litre) = Therefore we would need a 3.000kg line to lift/move this object. 10. Q: If a diver and his equipment weighs 87 kg when neutrally buoyant in salt water, the diver needs to add / remove kg to be neutrally buoyant for freshwater diving with the same equipment. A: 87kg x = = 2.18 The diver would have to REMOVE 2.2KG in order to be neutrally buoyant in fresh water. 11. Q: If 2 litres of air is needed in your buoyancy compensator to maintain neutral buoyancy 10m in the ocean, how much will your buoyancy increase if you ascend 3m? A: 2litres of air 1.7 bar (at 7msw) x 2 bar (diver was initially at 2 bar at 10msw) 2 / 1.7 = x 2 = = 0.35 Buoyancy will change by 0.35l.

4 12. Q: A lift bag that weighs 1kg and contains 20 litres of air at a depth of 20m in the ocean would provide kg of buoyancy upon reaching the surface provided no air escaped during ascent. A: NOTE: The CAPACITY is not given therefore we assume the capacity is limitless. 20 litres of air at 20m will expand. 20litres at 3bar will expand to 3 x its original volume. Therefore 20 x 3 = 60 HOWEVER keep into account the weight bag weighs 1kg and therefore can only carry 59l of air. 60 1= Q: The pressure increases per meter of depth in saltwater by: A: For every 10metres, it is 1 bar. Therefore 1bar / 10msw = 0.1bar per 1meter salt water. 14. At an altitude of 5,500 meters, the atmospheric pressure is approximately HALF of the atmospheric pressure at sea level. WE USE THIS INFORMATION WHEN FLYING. 15. Q: If a scuba cylinder contains 12 litres of air at a pressure of 200bar, how much air does the cylinder contain at 150bar if the temperature remains constant? A: 12l cylinder x 150 bar = (12x150) 1800 bar litters. 16. Q: If a diver takes a sealed, rigid, cubical container measuring 50cm per side to a depth of 5m in to the ocean, the total crushing force (assume the wall thickness is zero) on the container would be about: A: Well, how much pressure is at 5msw? It s 1.5bar. It will remain 1.5bar no matter what you submerse. 17. Q: The partial pressure law requires the sum of the partial pressures to always equal the pressure. A: Dalton s law uses the term total pressure to describe the sum of the partial pressures of each gas in a mixture of gases. Total pressure is synonymous with absolute pressure. Always use absolute pressure, the term used in physics, for pressure calculations. The answer is TOTAL or ABSOLUTE.

5 18. Q: The absolute pressure at a depth of 20m in fresh water at an altitude of 1000m (atmospheric pressure is 0.9bar) is bar. A: FRESH WATER 1 bar is attained at 10.2 metres therefore 20mfw = 20/ /10.2 = 1.96 Atmospheric pressure is given as 0.9bar = 2.86 Therefore the answer is 2.86bar 19. Q: An object floats when it displaces water weighing the weight of the object.. A: A. As much as B. Less than. C. None of the answers are correct D. More than Let s think about this logically. An object is buoyed up by a force equal to the weight of the water it displaces. Coins, gold, and metals these are very dense and therefore sink to the bottom easily. Polystyrene is less dense, therefore floats easily. A small coin displaces less water, therefore it sinks. The inverse of this is an object that displaces MORE water than what it weighs, will FLOAT. So, if we force an empty, sealed plastic bottle into a bucket of water, the sealed bottle will displace more water than what the bottle weighs and therefore it will surface and float the moment we let go of it. If the weight of the water displaced by an object weighs more than the object itself, the object will be buoyed up by a force equal to the difference between the weight of the object and the weight of the water it displaces. 20. Q: Assuming a maximum partial pressure limitation of 1.4 ATA, the maximum depth for a 40% Nitrox mixture is m of seawater. A: PPO2 Depth (bar) EANx O2 Fraction PPO2 is given as 1.4. O2 Fraction is given as 0.40 (40%) 1.4/ 0.4= bar = 25msw

6 21. Q: What would be the Surface Respiratory Volume (SRV) of a diver with a 200 bar, 15 litre cylinder? who uses 15 bars in 10 min. at a depth of 5 msw in the ocean? A: (15bar x 15l cylinder) / (10 minutes x 1.5bar depth) 225 /15 15litres per minute 22. Q: An object with a volume of 30 litres, weighing 70 kg is embedded in the sediment at a depth of 30 meters in the ocean harbour. This object is best lifted to the surface by: A: 1. Two 40 litres lift bags weighing 3kg each Volume = 30l 30 x (weight of salt 2. A line tied to a boat at the surface water) = Three 250 litres lift bags weighing 20 kg each 1 lift bag provides (40-3) 37kg of lift. We 4. Three 500 litres lift bags and two 25 l lift bags we therefore need 2 x 40l lift bags. 23. Q: The ocean depth at which the partial pressure of oxygen in air equals 1.6 bars is assuming O2 = 20,9%. A: 1.6 / = 7.665bar 7.665bar 1bar atmospheric pressure = 6.665bar water pressure x 10 = 66.5msw