## Assignment 12 Due: 11:59pm on Friday, May 9, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 15.6 A 0.600 -diameter vat of liquid is 2.30 deep. The pressure at the bottom of the vat is 1.30 . Part A What is the mass of the liquid in the vat? Express your answer with the appropriate units. ANSWER: Correct Problem 15.8 A 90-cm-thick layer of oil floats on a 160-cm-thick layer of water. Part A What is the pressure at the bottom of the water layer? Express your answer with the appropriate units. ANSWER: Correct m m atm 876 kg p = 1.25×105 Pa Problem 15.9 A research submarine has a 40.0 -diameter window 9.00 thick. The manufacturer says the window can withstand forces up to 1.20×106 . What is the submarine’s maximum safe depth? Part A The pressure inside the submarine is maintained at 1.0 atm. Express your answer with the appropriate units. ANSWER: Correct Problem 15.13 Part A What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 12 dog off the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 15.40 The 78 student in the figure balances a 1100 elephant on a hydraulic lift. cm cm N 947 m kg d = 3.8 cm kg kg You may want to review ( pages 415 – 419) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Part A What is the diameter of the piston the student is standing on? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Given that the height of the fluid on the two sides is the same in the figure, how is the pressure of the fluid on the two sides related? What is the definition of pressure? What is the area of the right cylinder? What is the force exerted by the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? What is the additional pressure above atmospheric pressure under the student in the left cylinder? What is the force exerted by the student on the left cylinder? What is the area of the left cylinder? ANSWER: Correct Part B d = 0.53 m When a second student joins the first, the piston sinks 40 . What is the second student’s mass? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the definition of pressure? How is the height difference between the left and right cylinders related to the pressure difference in the two cylinders? What is the standard value for the density of the oil given in the text? What is the force due to the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? Given the height difference between the two cylinders and the pressure in the right cylinder, what is the pressure above atmospheric pressure in the left cylinder? What is the force due to both students on the left cylinder? What is the sum of the masses of the students? What is the mass of the second student? ANSWER: Correct Enhanced EOC: Problem 15.17 A 6.80 rock whose density is 4900 is suspended by a string such that half of the rock’s volume is under water. You may want to review ( pages 419 – 423) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string? Express your answer with the appropriate units. Hint 1. How to approach the problem cm m = 80 kg kg kg/m3 What are the three forces acting on the rock? Draw a picture indicating the direction of the forces on the rock and an appropriate coordinate system indicating the positive direction. How is volume related to mass and density? What is the volume of the rock? What is the buoyant force on the rock given that half of the rock is underwater? What is the gravitational force on the rock? Given that the rock is suspended, what is the net force on the rock? Now, determine the tension in the string. ANSWER: Correct Problem 15.15 A block floats in water with its long axis vertical. The length of the block above water is 1.0 . Part A What is the block’s mass density? Express your answer with the appropriate units. ANSWER: Correct Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when 59.8 N 2.0 cm × 2.0 cm × 7.0 cm cm 857 kg m3 Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, , and the apparent weight of the crown when it is submerged in water, . See whether you can follow in Archimedes’ footsteps. The figure shows what is meant by weighing the crown while it is submerged in water. Part A Take the density of the crown to be . What is the ratio of the crown’s apparent weight (in water) to its actual weight ? Express your answer in terms of the density of the crown and the density of water . Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume . Find the actual weight of the crown. Express in terms of , (the magnitude of the acceleration due to gravity), and . ANSWER: Wactual Wapparent c Wapparent Wactual c w V Wactual Wactual V g c Wactual = cV g Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume , and take the density of water to be . Find the apparent weight of the crown submerged in water. Express your answer in terms of , (the magnitude of the acceleration due to gravity), , and . Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity. Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude of the buoyant force on the crown when it is completely submerged in water. Express your answer in terms of , , and the gravitational acceleration . ANSWER: ANSWER: ANSWER: Correct Part B Imagine that the apparent weight of the crown in water is , and the actual weight is . Is the crown made of pure (100%) gold? The density of water is V w Wapparent V g w c Fbuoyant w V g Fbuoyant = wV g Wapparent = (c − w)gV = Wapparent Wactual 1 − w c Wapparent = 4.50 N Wactual = 5.00 N grams per cubic centimeter. The density of gold is grams per cubic centimeter. Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for , what should the ratio of weights be if the crown is made of pure gold? Express your answer numerically, to two decimal places. ANSWER: ANSWER: Correct For the values given, , whereas for pure gold, . Thus, you can conclude that the the crown is not pure gold but contains some less-dense metal. The goldsmith made sure that the crown’s (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been. Problem 15.23 A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 9.0 . Part A What is the speed in the second segment? Express your answer with the appropriate units. w = 1.00 g = 19.32 Wapparent Wactual = 0.95 Wapparent Wactual Yes No = 4.50/5.00 = 0.90 Wapparent Wactual = 1 − = 0.95 Wapparent Wactual w g m/s ANSWER: Correct Part B What is the speed in the third segment? Express your answer with the appropriate units. ANSWER: Correct Part C What is the volume flow rate through the pipe? Express your answer with the appropriate units. ANSWER: Correct Understanding Bernoulli’s Equation Bernoulli’s equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli’s equation is an essential tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes. 2.25 ms 36.0 ms 7.07×10−4 m3 s For a fluid element of density that flows along a streamline, Bernoulli’s equation states that , where is the pressure, is the flow speed, is the height, is the acceleration due to gravity, and subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli’s equation becomes clearer if we rearrange the terms of the equation as follows: . The term on the left-hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right-hand side represent, respectively, the change in potential energy, , and the change in kinetic energy, , of the unit volume during its flow from point 1 to point 2. In other words, Bernoulli’s equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline. Part A Consider the portion of a flow tube shown in the figure. Point 1 and point 2 are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point 2. If the cross section of the flow tube at point 1 is greater than that at point 2, what can you say about the pressure at point 2? Hint 1. How to approach the problem Apply Bernoulli’s equation to point 1 and to point 2. Since the points are both at the same height, their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli’s equation, you will be able to determine the pressure at point 2 with respect to the pressure at point 1. Hint 2. Apply Bernoulli’s equation p1 +gh1 + = +g + 1 2 v21 p2 h2 1 2 v22 p v h g p1 − p2 = g(h2 −h1)+ ( − ) 1 2 v22 v21 p1 − p2 g(h2 − h1) 1 ( − ) 2 v22 v21 Apply Bernoulli’s equation to point 1 and to point 2 to complete the expression below. Here and are the pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use for elevation with the appropriate subscript, and use for the density of the fluid. Express your answer in terms of some or all of the variables , , , , , , and . Hint 1. Flow along a horizontal streamline Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words, when applying Bernoulli’s equation to any two points of the streamline, and they cancel out. ANSWER: Hint 3. Determine with respect to By applying the continuity equation, determine which of the following is true. Hint 1. The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It says that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections, or . Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice versa. ANSWER: p v h p1 v1 h1 p2 v2 h2 h1 = h2 p1 + = 1 2 v21 p2 + v2 2 2 v2 v1 $V A $t $V = = $t A1v1 A2v2 v2 > v1 v2 = v1 v2 < v1 ANSWER: Correct Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal fluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases. In other words, if , then and . Part B As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________. Hint 1. Effects from conservation of mass Recall that, if the cross section of the flow tube varies, the flow speed must change to conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section. ANSWER: Correct Part C Now assume that point 2 is at height with respect to point 1, as shown in the figure. The ends of the flow tube have the same areas as The pressure at point 2 is lower than the pressure at point 1. equal to the pressure at point 1. higher than the pressure at point 1. A2 < A1 v2 > v1 p2 < p1 A v decrease in speed increase in speed remain in equilibrium h the ends of the horizontal flow tube shown in Part A. Since the cross section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward point 2 from point 1 moves toward a region of lower pressure. In this case, what is the pressure drop experienced by the fluid element? Hint 1. How to approach the problem Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take into account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? ANSWER: Correct Part D From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? Hint 1. Physical meaning of the pressure drop in a tube As explained in the introduction, the difference in pressure between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow. equal to the pressure drop occurring in a purely horizontal flow. larger than the pressure drop occurring in a purely horizontal flow. p1 − p2 surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. ANSWER: Correct In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur. Venturi Meter with Two Tubes A pair of vertical, open-ended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density . The fluid rises to heights and in the two open-ended tubes (see figure). The cross-sectional area of the pipe is at the position of tube 1, and at the position of tube 2. A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy. larger decrease in kinetic energy. h1 h2 A1 A2 Part A Find , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the terms involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density. Hint 2. Simplified Bernoulli's equation For a fluid of uniform density that is not flowing, the pressure at a depth below the surface is given by , where is the pressure at the surface and is the magnitude of the acceleration due to gravity. ANSWER: Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube. Part B Find , the speed of the fluid in the left end of the main pipe. Express your answer in terms of , , , and either and or , which is equal to . Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of p1 g p h p = p0 + gh p0 g p1 = gh1 v1 h1 h2 g A1 A2 A1 A2 tube 1, and the fluid pressure and velocity at the bottom of tube 2. Hint 2. Find in terms of What is , the pressure at the bottom of tube 2? Express your answer in terms of , , and any other given quantities. Hint 1. Recall Part A Obtain the solution for this part in the same way that you found an expression for in terms of in Part A of this problem. ANSWER: Hint 3. Find in terms of given quantities Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , , , and . Hint 1. Find the pressure at the bottom of tube 2 Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , and . ANSWER: Hint 2. Find in terms of The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities and in terms of the geometry of the pipe. Find , the fluid velocity at the bottom p2 h2 p2 h2 g p1 h1 p2 = gh2 p2 p2 p1 v1 A1 A2 p2 p1 v1 v2 p2 = p1 + ( − ) 1 2 v1 2 v2 2 v2 v1 v1 v2 v2 of tube 2, in terms of . Your answer may include and , the cross-sectional areas of the pipe. ANSWER: ANSWER: ANSWER: Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves. Problem 15.39 The container shown in the figure is filled with oil. It is open to the atmosphere on the left. v1 A1 A2 v2 = A1 A2 v1 p2 = p1 + ( )(1 − ) 1 2 v1 2 ( ) A1 A2 2 v1 = 2g h1−h2 ( ) −1 A1 A2 2 −−−−−−−−−−−−−− Part A What is the pressure at point A? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the pressure difference between points A and B? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the pressure difference between points A and C? PA = 106 kPa PB − PA = 4.4 kPa Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 15.48 You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 32.4 . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17 . Part A What is the density? Express your answer with the appropriate units. ANSWER: Correct Problem 15.60 Water flows from the pipe shown in the figure with a speed of 7.0 . PC − PA = 4.4 kPa N N statue = 2100 kg m3 m/s Part A What is the water pressure as it exits into the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the height of the standing column of water? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.9%. You received 93.92 out of a possible total of 94 points. P = 1.0×105 Pa h h = 5.9 m

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