NAUI Master Scuba Diver 226 Search and Light Salvage pendent cylinder, not from the diver’s cylinders. The purpose of this is not to conserve air, but to ensure that the diver’s regulator or hoses do not become entangled in the lift as it begins to ascend. If a lift bag is inflated until the object begins to rise, there is probably too much air in the bag. It takes time to overcome inertia in water. Add air to the lift bag slowly and test the lift from time to time by raising it from the bottom. Your lifting will tell you how much additional air is required and will help overcome any inertia the lift may develop. When the lift begins to ascend, you and your buddy must clear the area beneath the lift and should ascend with it. One of you should control the dump valve on the lift bag, and the other should remain beside and in contact with the object being lifted to watch it and to help control it if necessary. If the lift bag being used is not equipped with a dump valve, a line should be attached to the top of the lift bag so it can be pulled to force air down and out of the bag as required. If a lift rises faster than it should and becomes uncontrollable, release it and swim away from it. If you release too much air from a lift and it begins sinking, let it go. Follow it to the bottom and start over. Swimming to support a sinking lift is dangerous because the lift becomes heavier as its volume decreases with increasing depth. When an object has been lifted to the surface, it can be pushed or towed to the exit point and pulled out. If the object is to be removed from deep water, a line from a boat or pier should be secured to it before lifting by the bag so it will not sink if it falls or the bag tears. English Lift Example You have offered to recover a sunken outboard motor. It rests in 35 feet of salt water, weighs 100 lbs on land, and displaces 0.5 ft3 when submerged. You have a lift bag which holds 5 ft3 and an extra 15 ft3 cylinder available to fill the bag. How much air (surface equivalent) will you need to add to the lift bag to make the motor neutrally byouant? Answer: The first thing to figure out is the weight to be lifted underwater. Since the motor displaces 0.5 ft3 of water, it will be buoyed up by the weight of the water displaced: 0.5 ft3 X 64 lbs/ft3 = 32 lb 100 lb - 32 lb = 68 lb to be lifted Next, translate the weight to be lifted to an equivalent amount of air: 68 lb / 64 lb/ft3 = 1.1 ft3 of air required at depth Now, if 1.1 ft3 of air is required at a depth of 35 feet, how much surface air is this? Using Boyle’s Law: P1V1 = P2V2 P1 = (35 ft X 1atm/33 ft) + 1 atm = 2.1 ata P2 = 1 atm V1 = 1.1 ft3 V2 = X Substituting and solving for X: The 1.1 ft3 of air at a depth of 35 feet will make the motor neutrally buoyant and will have a volume of 2.3 ft3 of air at the surface. The 15 ft3 cylinder will be ideal for the task. Metric Lift Example A motor rests in 15 meters of salt water, weighs 50 kg on land, and displaces 15 liters. Your lift bag holds 100 liters and an extra cylinder with capacity of 500 liters is available. Answer: The motor displaces 15 liters, and it will be buoyed by: 15 L X 1.0256 kg/L = 15.4 kg 50 kg - 15.4 kg = 34.6 kg to be lifted Translating this weight to an equivalent amount of air: 34.6 kg / 1.0256 kg/L = 33.7 L of air required at depth Surface equivalent: P1V1 = P2V2
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