Snap, Crackle, Pop! Submarine Buoyancy, Compression, and Rotational Equilibrium Bill Sanford, Physics Teacher, Nansemond Suffolk Academy, Suffolk 2012 Naval Historical Foundation STEM-H Teacher Fellowship INSTRUCTIONAL GOAL: Explain and perform calculations regarding the buoyant force on a submarine, how the buoyant force on a submarine varies as its hull compresses, and how a submarine can maintain neutral buoyancy and rotational equilibrium by pumping water from tank to tank and on/off the boat. In an extension activity (#3), calculate the deformation of the submarine due to water pressure at various depths. NGSS References: HS-PS2-1 Analyze data to support the claim that Newton s second law of motion describes the mathematical relationship among the net force on a macroscopic object, its mass, and its acceleration HS-PS2-4 Use mathematical representations of Newton s Law of Gravitation to describe and predict the gravitational forces between objects BACKGROUND: Controlling Depth and Angle of the Sub: o A sub can adjust its density and thus the net vertical force it experiences by pumping water on or off. Its volume stays the same, but its mass changes, thus changing its density. o It can also control its vertical motion by changing some of its horizontal motion into vertical motion by angling its planes. o It can also pump water from a tank in the front to an aft tank to help control the balance of the ship. Does the buoyant force on an object change as the object is taken deeper in a fluid? o We assume it does not, since the buoyant force is determined by how much fluid the object displaces, which is determined by the object s volume, which we often assume to be constant... o But... if the object is compressible, as it goes deeper into the fluid, it undergoes what we call Bulk Deformation, such that it takes up less space... occupies a smaller volume. o Compression: the decks of a submarine float on the hull so they do not buckle when the hull compresses. This can create creaking noises as the submarine proceeds deeper, unless the deck/hull junctions are properly greased Note1: A special alloy is used for the hull... The trick for the metallurgist is to strike a compromise and to use the correct ratio of alloy elements to gain a hull plate that maximizes its resistance to pressure through high compression strength, but yields enough to allow the hull to bend instead of break. Note2: See the sketches on the next page, regarding how a submarine is affected by compression.
RESOURCES: floating decks will not buckle when the submarine is compressed decks of a compressed submarine if they were rigidly fixed to the hull Effects of Compression as a Submarine Goes Deeper Or Goes Too Deep
ACTIVITY 1: Buoyancy Problems OBJECTIVE: Practice and Reinforce concepts related to Fluid Pressure, primarily Buoyancy MATERIALS: How A Submarine Submerges - Video https://www.youtube.com/watch?v=aabu2iwnwwk This clip Buoyancy: Take Er Down and many others are available on the US Navy History Museum's You Tube Channel (https://www.youtube.com/user/usnavyhistory/videos?query=submarine ). INSTRUCTIONS: Submarines are basically teardrop shaped, with a sail : a rectangular protrusion on the top, from which the periscopes and antennas emerge. The image depicts a submarine that is moving to the right. Submarines have two sets of planes that can be angled so that the water flowing past them pushes them up or down. In this way, they are used to control the forward/aft angle of the submarine (its pitch). Similar to the wing flaps of an airplane, they are also used to change forward motion into vertical motion. One set is near the rear (aft) end of the submarine, or its stern. The other is either near the front (bow), or protruding from the sail. Two ways to control the vertical position of a submarine: o adjust its own density and thus the net vertical force it experiences by pumping water from its tanks into the ocean, or allowing some ocean water into its tanks. o changing some of its horizontal motion into vertical motion by angling its planes. Does the buoyant force on an object change as the object is taken deeper in a fluid? o Normally we assume it does not, since the buoyant force is determined by how much fluid the object displaces, which is determined by the object s volume and the density of the fluid, which we often assume to be constant, but o If and object is made of a compressible material, as it goes deeper into the fluid, it undergoes what we call Bulk Deformation, such that it takes up less space (occupies a smaller volume), so Yes, it experiences a weaker buoyant force as its volume decreases. A submarine s hull actually compresses (yes, it gets measurably smaller! not noticeably smaller, but measurably) due to the water pressure it experiences. Activity 3 of this lesson plan explores this concept farther. To prove this, some crews have tied a string tightly between the bulkheads (walls) while on surface. After diving to a deep depth, the string hangs loosely! o Also, although liquids are basically incompressible (you cannot make a defined amount of liquid take up less space by squeezing it, AKA its density does not change as you increase the pressure on it). Nevertheless, if the pressure is extreme, you can vary the density of a liquid. o Although pressure does have a small effect on the density of water, temperature has a larger effect, and the temperature of seawater drops sharply as you go deeper. Overall, the compression of the hull has a larger effect than either of these effects on the density of the seawater.
DATA: density of air = 1.25 kg/m 3 density of seawater = 1025 kg/m 3 1m = 3.281 feet volume of submarine = 8650 m 3 air pressure at sea level = 1ATM = or 1.013 x 10 5 Pa QUESTIONS: 1. A) What water pressure would a submarine experience when it is 400 feet below the surface of the ocean? Recall that oceans consist of saltwater... J B) What is the absolute pressure that the submarine experiences at this location (combination of the water pressure with the pressure caused by the air above the water)? C) What percent of the absolute pressure at 400 ft is contributed by the miles of air in our atmosphere? 2. A) What is the difference in air pressure when an airplane increases its altitude by 400 feet? B) Is this a significant percentage of regular atmospheric pressure, such that a human would notice or be affected? Calculate it and comment. 3. A) Some Navy aircraft, like the F- 35 Joint Strike Fighter, can fly at altitudes over 40,000 ft. To understand why pilots of the F- 35 need to have an oxygen system to help them breathe, calculate the air pressure at 10,000 ft, where many commercial aircraft fly. B) What percentage of sea level air pressure is this? 4. How much water would you need to be under, such that the absolute pressure (air and water pressure) is double the regular atmospheric air pressure you experience at sea level? 5. A submarine has a mass of 6.97million kg and does not change its mass by adding or removing any water during this problem. Assume that: The density of seawater is 1025 kg/m 3 and does not change appreciably with depth or temperature. The volume of the submarine at the surface is 6758m 3 The volume of the compressed submarine at 800ft is 6400m 3 A) When it is completely submerged just below the surface, calculate its density and its vertical acceleration, based solely on its weight and the buoyant force it is experiencing. Ignore any forces exerted by the water on the planes. Your solution should include a free body diagram.
B) When it is at 800ft, calculate: i. the density of the submarine ii. the buoyant force exerted on the submarine iii. the sub s new vertical acceleration. o Your solution should include a free body diagram with arrows drawn to the same scale as those in part A. C) Draw a simple submarine with its planes in positions that you think would help maintain its depth in this situation. Explain your choice of positions. Air Pressure: Conversion / Volume of a Cylinder / Pressure vs. Volume 6. A) When the submarine compresses due to the high water pressure, does the air pressure in the sub increase? B) If so, by how much, and would it be noticeable? You may assume the following: When the submarine is at the surface, the air inside it is at sea level atmospheric pressure Recall that PV=nRT o assume n (#molecules), R(gas constant), T(Temperature) are constant. The submarine is a simple cylinder. radius at the surface (before compression) = 198inches radius after compression = 197inches uncompressed length = 360 feet compressed length = 359 feet 9 inches NOTE: Solutions provided after Activity 3 questions.
ACTIVITY 2: Balance of Torques using Submarine Trim Tanks OBJECTIVE: Review and Practice the concepts of torque and rotational equilibrium by balancing the pitching (forward/aft) torques on a submarine. MATERIALS: How A Submarine Submerges - Video. This clip Buoyancy: Take Er Down and many others are available on the Naval Historical Foundation U- Tube Channel. INSTRUCTIONS: Submarines are basically teardrop shaped, with a sail : a rectangular protrusion on the top, from which the periscopes and antennas emerge (see the images throughout this lesson). Submarines have two sets of planes that can be angled so that the water flowing past them pushes them up or down. In this way, they are used to control the forward/aft angle of the submarine (its pitch). Similar to the wings of an airplane, they are also used to change forward motion into vertical motion. One set is near the rear (aft) end of the submarine, or its stern, on either side of its rudder. The other is either near the front (bow), or protruding from the sail. On some submarines, the Bow Planes are replaced by a set of planes sticking out from the sides of the sail, as seen on the picture on the next page. For many reasons, a submarine routinely pumps water to/from forward and aft tanks, to help control the angle of the ship. For instance, consider the sanitary tanks (Note: sanitary is not a fitting term for what fills up these tanks). These tanks are forward of the center of the submarine, so as they fill up, the submarine s bow is heavier and tends to cause the sub to tilt downward. This can temporarily be controlled using the bow planes and stern planes, but after a while, it is useful to compensate by moving water from tank to tank, or adding/removing some water to/from a tank that is forward or aft of centerline. This would balance the forward/aft torque, but the sub would be light or heavy overall... in other words, it would no longer be neutrally buoyant.
DATA: Density of Seawater = 1025 kg/m3 1 meter cubed = 264.172 gallons 1m = 3.281 feet A Free Body Diagram used for the purpose of calculating torques should consist of a line to represent the object being analyzed, a labeled axis of rotation, and labeled force arrows. 1. On the following image, label the Stern Planes, Bow Planes, and Sail, and draw in the Periscope. 2. Currently, the buoyant force on the submarine is 6.930x10 7 N. The mass of the submarine is 7.100 million kg. A) Draw a Free Body Diagram of the submarine. B) The Officer of the Deck needs the submarine to hover at a constant depth to shoot missiles, so he tells the Chief of the Watch to establish neutral buoyancy. What must the Chief of the Watch do: open a valve to allow seawater into one of the tanks, or pump seawater out of a tank back into the ocean? You may assume the tank is located near the center of gravity of the submarine, so rotational equilibrium is not affected. C) How many gallons of seawater must be transferred? D) Draw a new Free Body Diagram to represent the submarine after the water is transferred and comment on the vertical forces and whether or not the sub is in vertical equilibrium.
3. Assume that the submarine begins in a state of neutral buoyancy and zero net torque around the center of gravity. Then, over time, the Sanitary Tank level increases by 1250 gallons. A. Draw a submarine, label its bow and stern and its Sanitary tank center of gravity (approximately in its geometric Forward Trim Tank center), and a small rectangle to represent each of the Centerline tanks described in the table to the right. Don t Trim Tank measure just draw them in about the right place. Aft Trim Tank B. Define Rotational Equilibrium. C. Determine how many m 3 of liquid were added to the Sanitary Tank. D. On your Free Body Diagram of the submarine (label the weight of liquid in sanitary tank). Note: For several reasons you do Not need to draw arrows to represent its overall weight and the buoyant force it experiences (they are equal, and can both be assumed to act at the center of gravity, where we are locating our center of gravity). So only draw arrows for the weights of liquids in the tanks, since they cause torques. Note 1: assume all tanks are initially empty, except what we have said has been added to the sanitary tank. Note 2: For this problem, ignore forces that the flowing water exerts on the planes. E. What is the lever arm of the weight of the contents of the Sanitary Tank? F. Calculate the net torque (in Nm) on the submarine. G. Why is the submarine designed such that the Sanitary Tank is located close to the middle of the submarine (forward/aft)? For the next two questions, do not worry about the effect your actions would have on the submarine s vertical equilibrium or neutral buoyancy. H. Which of the following could the Chief of the Watch do to re- establish rotational equilibrium? Draw a new Free body Diagram, answer the question and also calculate the amount (gallons) of seawater. Allow seawater into the submarine and put it in the fwd trim tank. Allow seawater into the submarine and put it in the aft trim tank. I. As an alternative, which of the following could the Chief of the Watch do to re- establish rotational equilibrium? You may assume there is water in all the trim tanks. Draw a new Free body Diagram, answer the question and also calculate the amount (gallons) of seawater. Pump water from the fwd trim tank off the submarine into the ocean Pump water from the aft trim tank off the submarine into the ocean J. Assume that the Chief of the Watch chose one of the options in Part H (to allow seawater Into a trim tank). The submarine should now be in rotational equilibrium, but it is no longer neutrally buoyant. It will tend to sink or float. a. Which will it tend to do: sink or float? b. Explain what the Chief of the Watch could do to regain neutral buoyancy, while not affecting the sub s rotational equilibrium. NOTE: Solutions provided after Activity 3 questions. Tank Horizontal position (feet) from the submarine s center of gravity 15.0 forward 50.0 forward 0.0 90.0 aft
ACTIVITY 3: Bulk Deformation Problem OBJECTIVE: Practice and Reinforce understanding of Deformation as it applies to a submarine at various depths MATERIALS: None BACKGROUND: A submarine s hull actually compresses due to the water pressure it experiences. o To prove this, some crews have tied a string tightly between the bulkheads (walls) while on surface. After diving to a deep depth, the string hangs loosely. Similar to what is shown in the drawing below, the decks of a submarine float on the hull so they do not buckle when the hull compresses. As the submarine proceeds deeper and the cylindrical hull compresses, the decks stay the same width, so they tend to rub against their supports. Unless the deck/hull junctions are properly greased, this can cause creaking noises that could be detected by enemy submarines. The movies Das Boot and U- 571 have relatively realistic depictions of World War II submarines at war. Floating decks will not buckle when the submarine is compressed Decks of a compressed submarine if they were rigidly fixed to the hull
To demonstrate this concept, Navy researchers have taken Styrofoam cups down deep in the ocean with research submarines. To be clear, nothing inside the submarine is crushed (Refer to Activity 1 of this lesson plan for calculations that prove that the air pressure inside the submarine does not increase just because the hull is exposed to extreme water pressure). The Styrofoam cup must be exposed to the high water pressure. The researchers put the cups in a small compartment on the outside of the submarine that is enclosed enough that the cups are not swept away, but with small holes that let water in. The cup on the left is regular size. Why do you suppose the one on the right so much smaller than the one in the middle? NOTE: An alloy is a mixture of metals, or a metal and other elements. For instance, various amounts of carbon are alloyed with iron to obtain steel with various characteristics: strength, flexibility, etc. Steel alloys are used for submarine hulls. The trick for the metallurgist is to strike a compromise and use the correct ratio of alloy elements to create a hull plate with high compression strength (that maximizes its resistance to high water pressure, but also one that flexes (yields) enough to allow the hull to bend, not break.
If you have not covered them in class, look up Hooke s Law as it applies to the concepts of Compression and Bulk Deformation. In physics books, they most often in the chapter about Springs. Now, consider a cube of Styrofoam taken to 750 feet below the surface of the ocean. Be sure to have the correct units for each answer. DATA: 1m = 3.281 feet density of seawater = 1025 kg/m 3 Length of a side of the cube before being submerged: 6.0 inches Length of a side of the cube after being submerged at 750 ft: 2.0 inches 1. Explain which concept we should use to analyze what will happen to the Styrofoam: Compression or Bulk deformation? 2. Calculate the Stress in this situation. 3. Calculate the Strain. 4. Determine the bulk modulus of Styrofoam in Pascals. Note: This analysis sequence is a little misleading, as Hooke s Law models materials that rebound to their original shape as you remove the stress. In this situation, the Styrofoam has undergone permanent deformation, as a submarine would if it ventured too deep! Only a few special research submarines are designed to explore the deepest depths of our oceans (~36,000 ft, almost 7 miles), which are deeper than Mount Everest is tall (~29,000 ft)!!! NOTE: Solutions provided after Activity 3 questions.
Activity 1: Buoyancy Problems Activity Answers density of air = 1.25 kg/m 3 density of seawater = 1025 kg/m 3 1m = 3.281 feet volume of submarine = 8650 m 3 Air pressure at sea level = 1ATM = or 1.013 x 10 5 Pa 1. A) What water pressure would a submarine experience when it is 400 feet below the surface of the ocean? Recall that oceans consist of saltwater... J Convert 400 ft to meters: 400 ft (1m / 3.281 ft) = 121.914m Difference in Fluid Pressure = density of fluid * g * difference in depth ΔP due to layers of water above = 1025kg/m 3 * 9.8m/s 2 * 121.914m = 1224626 Pascals B) What is the absolute pressure that the submarine experiences at this location (combination of the water pressure with the pressure caused by the air above the water)? Air pressure at sea level = 1 ATM = 1.013x 10 5 Pa Absolute pressure (due to water And air) = 1224626 Pascals + 1.013x 10 5 Pa Absolute Pressure = 1325926 Pa C) What percent of the absolute pressure at 400 ft is contributed by the miles of air in our atmosphere? #1 C) While underneath merely 400 feet of ocean water, the contribution of air pressure from the miles of air above the ocean is less than 10%... 1.013x10 5 Pa / 1.326x10 6 Pa =.076 or 7.6% 2. A) What is the difference in air pressure when an airplane increases its altitude by 400 feet? ΔP due to layers of fluid = density of fluid * g * difference in altitude ΔP due to layers of fluid = 1.25kg/m 3 * 9.8m/s 2 * 121.914m ΔP due to changing altitude by 400 feet = 1493.44 Pa B) Is this a significant percentage of regular atmospheric pressure, such that a human would notice or be affected? Calculate it and comment. 1493.44 Pa / 1.013x10 5 Pa =.0147 or 1.5% A human would barely notice this. 3. A) Some Navy aircraft, like the F- 35 Joint Strike Fighter shown, can fly at altitudes over 40,000 ft. To understand why pilots of the F- 35 need to have an oxygen system to help them breathe, calculate the air pressure at 10,000 ft, where many commercial aircraft fly. Convert 10000 ft to meters: 10000 ft (1m / 3.281 ft) = 3047.9m ΔP due to layers of fluid = density of fluid * g * difference in altitude ΔP due to layers of air = 1.25kg/m 3 * 9.8m/s 2 * 3047.9m ΔP due to changing altitude by 10000 feet = 37336.2 Pa P at 10,000 ft = P sea Level - ΔP due to changing altitude by 10000 feet = 1.013x10 5-37336.2 = 63963.8 Pa B) What percentage of sea level air pressure is this? 63963.8 Pa / 1.013x10 5 Pa =.6314 or 63% Note: This is only a very rough estimate, since as you increase in altitude, the density of the air decreases significantly from the sea level value of 1.25kg/m 3. Thus, the decrease would be less than we calculated. The air pressure at 10,000 ft is closer to 70,000 Pa.
4. How much water would you need to be under, such that the absolute pressure (air and water pressure) is double the regular atmospheric air pressure you experience at sea level? ΔP due to layers of water = Regular atmospheric air pressure = 1.013x10 5 Pa. ΔP due to layers of fluid = density of fluid * g * difference in depth 1.013x10 5 Pa= 1025kg/m 3 * 9.8m/s 2 * depth depth = 10.1m or ~ 33.1 ft 5. A submarine has a mass of 6.97million kg and does not change its mass by adding or removing any water during this problem. Assume that: The density of seawater is 1025 kg/m 3 and does not change appreciably with depth or temperature. The volume of the submarine at the surface is 6758m 3 The volume of the compressed submarine at 800ft is 6400m 3 A. When it is completely submerged just below the surface, calculate its density and its vertical acceleration, based solely on its weight and the buoyant force it is experiencing. Ignore any forces exerted by the water on the planes. Your solution should include a free body diagram. Just below the surface: Density = Mass / Volume Weight = mass * acceleration due to gravity B = weight of displaced fluid Vertical Acceleration of sub = Net Vertical Force exerted on the sub / mass submarine Accel. = B - W submarine / 6.97 E 6 = (6.79E7 6.83E7) / 6.97 E 6 = -.0574 m/s 2 or.06 m/s 2 downward B. When it is at 800ft, calculate: iv. the density of the submarine v. the buoyant force exerted on the submarine vi. the sub s new vertical acceleration. o Your solution should include a free body diagram with arrows drawn to the same scale as those in part A. At 800 ft Vert. accel. sub = ΣF on the sub / m sub Accel. = B - W submarine / 6.97 E 6 = (6.43E7 6.83E7) / 6.97 E 6 = -.574 m/s 2 or.6 m/s 2 downward Quantity Submarine Displaced Fluid (seawater) Weight (N) 6.831x10 7 6.79 E 7 Mass (kg) 6.97x10 6 6.93 E 6 Volume (m 3 ) 6758 6758 Density (kg/m 3 ) 1031 1025 Quantity Submarine Displaced Fluid (seawater) Weight (N) 6.831x10 7 6.43 E 7 Mass (kg) 6.97x10 6 6.56 E 6 Volume (m 3 ) 6400 6400 Density (kg/m 3 ) 1089 1025
C.Draw a simple submarine with its planes in positions that you think would help maintain its depth in this situation. Explain your choice of positions. This would be acceptable, if explained that as water collided with each set of planes, it would exert upward forces on both of them, creating zero net torque, but an overall upward force that would be able to overcome the problem that the weight of the submarine is presently greater than the buoyant force it is experiencing. Notice that the angle on the bow planes is greater than the angle on the stern planes, since the bow planes are closer to the center of gravity of the sub and thus its axis of rotation. The stern planes have a longer lever arm, so we do not need as strong a force on them, or the submarine would begin to point downward. Staying level is preferred, unless using the angle of the entire ship to help change depth. and this, although a more aggressive solution and only necessary if the downward acceleration were excessive, would be acceptable, since it would cause a counter clockwise torque on the submarine, causing the entire sub to point toward the surface, making the entire bottom of the sub into a plane type surface that water would hit and create an upward force that could support the sub, keeping it from continuing to accelerate downward. This would be a temporary situation, held only until the submarine achieved an appropriate upward angle. Notice that the stern planes are only angled downward a little bit, since they have a long lever arm and do not need to exert as strong a force to have an affect on the angle of the sub. Air Pressure: Conversion / Volume of a Cylinder / Pressure vs. Volume 6.A. When the submarine compresses due to the high water pressure, does the air pressure in the sub increase? The bottom line is Yes, the air pressure does increase, but only a little bit, such that it would be barely noticeable This is because gases are compressible, whereas solids and for the most part, liquids are not B. If so, by how much, You may assume the following: When the submarine is at the surface, the air inside it is at sea level atmospheric pressure Recall that PV=nRT o assume n (#molecules), R(gas constant), T(Temperature) are constant. The submarine is a simple cylinder. radius at the surface (before compression) = 198inches radius after compression = 197inches uncompressed length = 360 feet compressed length = 359 feet 9 inches
First determine how much the volume of the submarine changes. delta radius = 1in (.025399m) out of 198in (5.02895m) delta length = 3in (.076196m) out of 4320in (360feet, 109.7226m) initial volume = πr 2 h = 8717.67m 3 compressed volume = 8623.8447m 3 delta volume = 93.825m 3 Note, this delta Volume is not used in the rest of the problem. Then, applying the general rule for how the pressure of a sample of gas is related to its temperature and pressure PV=nRT assuming n, T are constant, we can say that P = constant (1 / V) P original = constant (1/V original ) 101300 = constant (1/8717.67) constant = 883099971 P final = constant (1/V final ) P final = 883099971 (1/ 8623.8447) P final = 102402 delta p = 1102Pa and would it be noticeable? Difference in Fluid Pressure = density * g * h 1102 Pa = 1000 (9.8) h h =.11m or 11cm o the air pressure change would be about the same as going under 11cm of water, which a human could barely notice.
Activity 2: Balance of Torques using Submarine Trim Tanks 1. stern planes periscope bow planes 2. A) 6.930 x 10 7 N B) 6.958 x 10 7 N RIGHT NOW, THE SUBMARINE IS HEAVY. The downward gravitational force exerted by the earth on the submarine is stronger than the upward buoyant force exerted by the water on the submarine. The COW must PUMP WATER OFF THE SUBMARINE INTO THE OCEAN C) Net Force on submarine = 6.958 x 10 7 6.930 x 10 7 = 280000N downward, so water that weighs that much must be pumped off. mass = Net Force / g = 28571 kg of water must be pumped off volume = mass / density = 28571kg / 1025 kg/m 3 = 27.8946 m 3 pumped off 27.8946 m 3 (264.172 gallons/m 3 ) = 7363.7 gallons, or 7364 gallons pumped off (4 digits of precision) D) Net Vertical Force = 0 Vertical Equilibrium Vertical Acceleration = 0 6.930 x 10 7 N 6.930 x 10 7 N
3. A. center of gravity B. Rotational Equilibrium is a state in which the net torque on an object is zero (no torques or all the torques exerted on the object cancel each other out). This results in the object having zero rotational acceleration. C. 1250 gallons (264.172 gallons/m 3 ) = 4.7318 m 3 of waste liquid were added. D. 15 feet or 4.5718m between cog and weight of SAN Tank E. See Diagram, above. 15 feet (1m / 3.281 ft) = 4.5718m 47530.6N F. torque = force * lever arm torque = weight of San. Tank * 4.5718m torque = 47530.6N * 4.5718m = 217299.3 Nm G. Having the Sanitary Tank close to the forward/aft centerline reduces the lever arm of the weight of its contents, minimizing the inevitable variations in torque as this tank repeatedly fills and is emptied. H. Allow seawater into the submarine and put it in the aft trim tank. 27.431m 4.5718m 7921.7 N Note: by convention, clockwise torque is defined as positive 47530.6N We want the sub back in rotational equilibrium, in other words, we want: Net torque = 0 torque of water in Aft Trim + torque of liquid in Sanitary Tank = 0 torque of water in Aft Trim = torque of liquid in Sanitary Tank 27.431 m * weight of water added to Aft Trim = ( ) 4.5718m * 47530.6N Weight of water that must be added to Aft Trim = 7921.7 N mass = Net Force / accel. = weight of water / g = 7921.7 N / 9.8 m/s 2 mass of water added to Aft Trim Tank = 808.344 kg Using conversions similar to those seen above, we find that.78863 m 3 or 208.35 gallons of water must be added to the Aft Trim Tank
I. J. Pump water from the fwd trim tank off the submarine into the ocean The torque caused by the weight of the Sanitary Tank is clockwise. In the previous solution, we balanced this torque with an equal but opposite (counter- clockwise) torque created by adding water to the Aft Trim Tank. In this solution, we analyze another way to counter the clockwise torque of the sanitary tank: removing water from the forward trim tank, thus reducing the clockwise torque caused by it. 15.239 m * weight of water removed from Fwd Trim = 4.5718m * 47530.6N Weight of water that must be removed Fwd Trim = 14259 N mass = Net Force / accel. = weight of water / g = 14259 N / 9.8 m/s 2 mass of water added to Aft Trim Tank = 1455 kg Using conversions similar to those seen above, we find that o 1.4195 m 3 or 375 gallons of water must be removed the Fwd Trim Tank Since the sub began in a state of neutral buoyancy, and the COW took on additional water, it will tend to Sink. In other words, the COW allowed water into a tank that was previously filled with air, so the submarine s mass has increased, while its volume remained the same, thus the submarine is more dense than it was before. To regain neutral buoyancy, the COW could pump water from the Centerline Trim Tank off the submarine, thus decreasing its weight, while not affecting its rotational equilibrium (since there is no affect on torque, since the tank is at the fwd/aft centerline of the sub, thus the lever arm of its weight is zero).
ACTIVITY 3: Bulk Deformation Problem If you have not covered them in class, look up Hooke s Law as it applies to the concepts of Compression and Bulk Deformation. In physics books, they most often in the chapter about Springs. Now, consider a cube of Styrofoam taken to 750 feet below the surface of the ocean. Be sure to have the correct units for each answer. 1m = 3.281 feet density of seawater = 1025 kg/m 3 Length of a side of the cube before being submerged: 6.0 inches Length of a side of the cube after being submerged at 750 ft: 2.0 inches 1. Explain which concept we should use to analyze what will happen to the Styrofoam: Compression or Bulk deformation? Compression describes 2- dimensional deformation involving a percentage change in length, that is caused by a force directed along one axis, like stretching a piece of fishing line, or shortening your spine by carrying heavy objects on your head. Bulk deformation describes 3- dimensional deformation involving a percentage change in volume, caused by a change in fluid pressure, such as changing your depth in water. Taking the Styrofoam cup deep below the surface of the ocean will involve Bulk Deformation. 2. Calculate the Stress in this situation. Stress is what causes Strain. In the case of Bulk deformation, the Stress is the change in fluid pressure. Convert depth to meters: 725 ft * (1m / 3.281 ft) = 220.9692 m Determine difference in pressure between the surface and at 725ft: o Change in Pressure = density of sea water * g * difference in depth o Change in pressure = 1000kg/m 3 * 9.8m/s 2 * 220.9692m = 2165498 Pascals 3. Calculate the Strain. Strain is not the change in length or the change in volume, but the percentage change. In this case, the strain is the %change in volume of the cup. Convert original length to m: 6.0in * (1ft / 12in)(1m / 3.281 ft) =.152393m Convert final length to m: 2.0in * (1ft / 12in) (1m / 3.281 ft) =.050798m Original Volume = l 3 o = (.152393m) 3 =.003539088 m 3 Final Volume = = l 3 f = (.050798m) 3 =.000131077 m 3 percentage change in volume = [(V final V original ) / V original ] = [(.000131077-.003539088) /.003539088] =.629630571 o Note that the percentage change in volume is unitless and that it is negative. Being negative indicates that the volume of the cup decreased.
4. Determine the bulk modulus of Styrofoam in Pascals. Finding the Bulk Modulus Stress = Bulk Modulus * Strain Change in Pressure = Bulk Modulus * percentage change in volume Bulk Modulus = Change in Pressure / percentage change in volume Bulk Modulus = 2165498Pa /.629630571 Bulk Modulus = 3439315 Pa Note: This analysis is a little misleading, as Hooke s Law models materials that rebound to their original shape as you remove the stress. In this situation, the Styrofoam has undergone permanent deformation. Note: This analysis is a little misleading, as Hooke s Law models materials that rebound to their original shape as you remove the stress. In this situation, the Styrofoam has undergone permanent deformation, as a submarine would if it ventured too deep! Only a few special research submarines are designed to explore the deepest depths of our oceans (~36,000 ft, almost 7 miles), which are deeper than Mount Everest is tall (~29,000 ft)!!!