CHAPTER 13. Liquids FLUIDS FLUIDS. Gases. Density! Bulk modulus! Compressibility. To begin with... some important definitions...


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1 CHAPTER 13 FLUIDS Density! Bulk modulus! Compressibility Pressure in a fluid! Hydraulic lift! Hydrostatic paradox Measurement of pressure! Manometers and barometers Buoyancy and Archimedes Principle! Upthrust! Apparent weight Fluids in motion! Continuity! Bernoulli s equation FLUIDS To begin with... some important definitions... DENSITY: Dimension: [M] [L] 3 Units: PRESSURE: Mass Volume, i.e., ρ = m V kg m 3 Dimension: Liquids Gases Force Area, i.e., P = F A [M] [L][T] 2 Units: N m 2 Pascals (Pa)
2 The weight of a medium apple is ~ 1 N, so the mass of a medium apple is ~ 0.1 kg. A typical refrigerator has a capacity of ~ 18 ft ft 3 18 (0.305 m) 3 = 0.51 m 3. Question 13.1: How does the mass of air inside a typical refrigerator compare with the mass of a medium size apple? The density of cold air is ~ 1.3 kg m 3. But 1 m 3 of air has a mass of 1.3 kg, so the mass of air in the refrigerator is kg = 0.66 kg, i.e., approximately the mass of 6 apples! We don t notice the weight of air because we are immersed in air... you wouldn t notice the weight of a bag of water if it was handed to you underwater would you?
3 In the previous chapter we defined the Young s modulus and the shear modulus. He is another modulus. BULK MODULUS: B = Dimension: Units: N m 2 [M] [L][T] 2 ΔP ( ΔV V ) Compressibility B 1 ΔP ΔV Atmospheric pressure P! = Pa. Force on the ceiling from floor of room above is pressure area (15 m 10 m) N. Why doesn t it collapse under that weight...? Because pressure operates equally in all directions! Why?! F ΔP ΔP V ΔP Gases are easily compressed (B very small). Liquids and solids much less compressible. When air molecules bounce off the walls they produce an impulse: FΔt = Δp pressure Since the molecules are traveling with equal speeds in all directions... the pressure is the same!
4 Pressure at a depth in a fluid P! Area = A Imagine a cylindrical body of the fluid with its top face at the surface of the fluid. At equilibrium there is no net force acting on the surfaces of the cylinder. P! h w = mg P F y = 0 at the lower face, i.e., P A = P! A + mg. But the mass of fluid in the cylinder is m = ρv = ρah. P = P! + ρgh. Question 13.2: A balloon has a radius of 10 cm. By how much does the radius change if the balloon is pushed down to a depth of 10 m in a large tank of water. Assume the balloon remains spherical. (The bulk modulus of air is N m 2.) What s the pressure on water at a depth of 10 m, say? ρgh = ( kg m 3 )(9.81 m/s 2 )(10 m) 10 5 Pa. P! = P at Pa. P 2P at.
5 The pressure difference is But B = For an airfilled balloon at a depth of 10 m, we have Since the balloon is spherical, ΔP = hρg. ΔP ( ). ΔV V ΔV V = ΔP B = hρg B. ΔV V = (10 m)(1 103 kg m 3 )(9.81 m/s 2 ) N m (50%). V = 4 3 πr3, i.e., ΔV = 4πr 2 Δr. ΔV V = 3 Δr r 0.5, So, for an initial radius r = 0.1 m, Δr 0.5r 3 = m (1.7 cm). What s the pressure difference from the ceiling to the floor in a typical room? Assume a room height of 3 m, so the pressure difference is ΔP = hρg = (3 m)(1.29 kg m 3 )(9.81 m/s 2 ) 38 Pa. Pascal s principle... P 2 h 2 ΔP P i.e., negligible. P! + ΔP 38 Pa (0.038%). Pa h 1 P 1 If additional pressure ( ΔP) is applied, it is transmitted through the whole fluid: and P 1 = P! + h 1 ρg + ΔP P 2 = P! + h 2 ρg + ΔP. Blaise Pascal ( )
6 Hydraulic lift F 1 Area A 1 Area A 2 F 1 Area A 1 Area A 2 Δx 2 Δx 1 F 2 h F 2 If a force F 1 is applied to the left hand piston, the additional pressure, P 1 = F 1 A 1, is transmitted through the whole fluid. Therefore, on the surface of the right hand piston, P 2 = F 2 A 2 = P 1. same pressure F 1 = A 1, i.e., F F 2 A 2 = A 2 F 1. 2 Wow... the force is amplified!! A 1 Mechanical advantage Get a larger force OUT than you put IN? Too good to be true? No, not really, because, to do work (like lift something heavy) the force F 2 is applied through a distance Δx 2. But by conservation of energy F 2 Δx 2 = F 1 Δx 1 Δx 1 = A 2 A 1 Δx 2. So, although F 1 < F 2, it is applied through a greater distance Δx 1 > Δx 2. Examples: lifts dentist s chair hydraulic brake systems
7 Hydrostatic Paradox Measurement of pressure P! P = 0 P! P! P! P! P! h h P y 2 h P! P! y 2 P! + hρg y 1 y 1 Manometer Barometer No matter the shape of a vessel, the pressure depends only on the vertical depth. P + ρgy 1 = P o + ρgy ρgy 2 = P o + ρgy 1 This is absolute pressure i.e., P P! = ρgh. i.e., P! = ρgh. This is the Gauge pressure Atmospheric pressure
8 Barometer P! = hρg, i.e., h = P! ρg. P Pa. Using water: ρ = kg m 3, Pa h = ( kg m 3 )(9.81 m/s 2 ) 10 m. Using mercury: ρ = kg m 3, Pa h = ( kg m 3 )(9.81 m/s 2 ) 0.76 m. Standard atmospheric pressure is defined as 760 mm of Hg. DISCUSSION PROBLEM #2 #1 The drawing shows two pumps, #1 and #2 to be used for pumping water from a very deep well (~30 m deep) to ground level. Pump #1 is submerged in the water at the bottom of the well; the other pump, #2, is located at ground level. Which pump, if either, can be used to pump water to ground level? A: Both pumps #1 and #2. B: Pump #1. C: Pump #2. D: Neither pump #1 nor pump #2.
9 Buoyancy and Archimedes Principle B V s w = mg Archimedes Principle : when an object is partially or wholly immersed in a fluid, the fluid exerts an upward force... upthrust... (or buoyant force, B) on the object, which is equal to the weight of fluid displaced. Submerged: w > B Weight of object w = mg = ρ s V s g Weight of liquid displaced = ρ L V s g If ρ s > ρ L the object will sink. If the object is floating then... w = B. V s Weight of object is w = mg = ρ s V s g. If V L is the volume submerged, then the weight of liquid displaced is ρ L V L g. But according to Archimedes principle, this is equal to the upthrust (B). B w = mg ρ s V s g = ρ L V L g V L = ρ s ρ L V s. Example: What fraction of an iceberg is submerged? ρ s = kg m 3 and ρ L = kg m 3. V L = ρ s = kg m 3 V s ρ L = 0.89, 3 kg m i.e., 89% of an iceberg is submerged! If you don t believe it V L
10 Question 13.3: On Earth, an ice cube floats in a glass of water with about 90% of its volume below the level of the water. If we poured ourselves a glass of water on the Moon, where the acceleration of gravity is about 16% of that on Earth, and dropped in an ice cube, how much of the ice cube would be below the level of the water?
11 the ice cube below the surface. When an ice cube floats, the weight of the ice cube ( = ρ s V s g) equals the weight of the water displaced ( = ρ L V Lg), which is proportional to V L, the volume of ρ L V L g = ρ s V s g, i.e., V L = ρ s ρ L V s, which is independent of g. So, the volume submerged would remain the same! Since both the weight of an object, which is floating, and the weight of a fluid it displaces are proportional to the local value of g, the submerged volume does not depend on g. Question 13.4: A block of copper with mass 0.50 kg is suspended from a spring scale. When it is fully submerged in water, what is the reading (in N) on the spring scale? The density of copper is kg m 3.
12 T (apparent weight) T B w = mg = V s ρ s g Identify the forces acting on the block. At equilibrium F y = T + B + ( w) = 0. T = w B = ρ s V s g ρ L V s g = ρ s V s g 1 ρ L. True weight = mg. Upthrust ρ L = kg m 3 ρ s = kg m T = (0.5 kg)(9.81 m/s 2 )( ) = 4.36 N (0.444 kg). In air instead of water: So, the mass of the block would be 71.5 mg less than in vacuum (0.50 kg). ρ air 1.29 kg m 3 = ρ s kg m 3 = ρ s Question 13.5: A beaker containing water is placed on top of a weighing scale and the reading is kg. In (a), a copper block is hanging freely from a spring scale, which has a mass reading of kg. When the copper block is totally immersed in the water, as shown in (b), what are the readings on the two scales? (The density of copper is kg m 3.)
13 T 2 = (m ρ w V s )g ( ) (9.81 m/s2 ) = 1.74 N, i.e., the reading on the spring scale is kg. = (0.20 kg) ( kg m 3 )( m 3 The upthrust B is the force of the water on the block; by (a) Initially, spring scale registers the weight of the copper block and the weighing scale registers the weight of the beaker plus water. (b) When the block is lowered into the water, the water exerts a buoyant force, B, (upthrust) on the copper block. Then, B is equal to the weight of water displaced, i.e., B = ρ w V s g, where the submerged volume, i.e., the volume of the copper block, is T 2 + B = mg, i.e., T 2 = mg B. V s = m s 0.20 kg = ρ s kg m 3 = m. Newton s 3 rd Law, the block must exert an equal and opposite force on the water. Consequently, the reading on the weighing scale will increase by kg, i.e., it will read kg when the block is submerged. So, when you dip a teabag into your cup, the weight of the teabag is reduced, but the weight of the cup (and contents) is increased!
14 With the cargo on board, as in (a), the weight of water displaced is equal to the weight of the barge plus cargo, i.e., w w (a) = w b + w c = w b + NV c ρ c g, Question 13.6: A barge, loaded with steel canisters is floating in a closed lock. If the cargo is thrown over the side, what happens to the level of water in the lock? Does it rise, stay the same, or fall? where is the weight of the barge, N is the number of w b Vc ρc canisters, is the volume of each canister and their density. When the canisters are thrown into the water, as in (b), the weight of water displaced is equal to the weight of the barge plus the weight of the water displaced by the canisters, i.e., w w (b) = w b + NV c ρ w g. Since ρ w < ρ c, w w (b) < w w (a), i.e., less water is displaced in (b) than in (a). The volume of water is unchanged, so the depth of water is less in (b) compared with (a), i.e., the water level falls.
15 Fluids in motion v 1 v 2 Area A 1 Area A 2 v 1 Δt Consider an incompressible fluid (a liquid) flowing down a tube of nonuniform size. In time Δt, the mass of fluid in the lefthand shaded volume is m 1 = ρa 1 v 1 Δt and the mass of fluid in the righthand volume is m 2 = ρa 2 v 2 Δt. Note that because the fluid is incompressible, the density remains constant. If the flow is steady, the mass that crosses A 1 must equal that crossing A 2, i.e., m 1 = m 2. A 1 v 1 Δt = A 2 v 2 Δt, i.e., A 1 v 1 = A 2 v 2 constant. This expression is called the continuity equation for an incompressible fluid. The conserved quantity, A v, is called the volume flow rate, Q (m 3 /s). v 2 Δt If the density changes (from ρ 1 ρ 2 ) then, since mass is conserved we have... v 1 v 2 Area A 1 Area A 2 v 1 Δt m 1 = m 2 i.e., ( A 1 v 1 Δt)ρ 1 = ( A 2 v 2 Δt)ρ 2 A 1 v 1 ρ 1 = A 2 v 2 ρ 2 This is the mass continuity equation. v 2 Δt
16 The speed of the water from a faucet increases as it falls because of gravity. The continuity equation tells us that the cross sectional area will decrease as the speed increases. The speed of the water from a garden hose increases as you reduce the area by putting your thumb over the end of the hose. So, the water squirts further. Question 13.7: A garden hose with an inside diameter of 16 mm fills a 10 liter bucket in 20 s. (a) What is the speed of the water out the end of the hose? (b) What diameter nozzle would increase the speed by a factor of two? (c) How long would it take to fill the same bucket with the nozzle referred to in part (b)? Other examples include lanes at highway tolls (increasing the number of lanes in an attempt to maintain traffic flow).
17 (a) The volume flow rate is Q = 10 L 20 s = 10( m 3 ) = m 3 /s. 20 s v = Q A = Q m 3 /s = 2 πr π(0.008 m) 2 = 2.5 m/s. (b) Since Q = Av remains constant, if v is increased by a factor of 2, then A must be reduced by a factor of 2. But A r 2 so the radius must be reduced by a factor of 2. So, the nozzle diameter should be 11.3 mm. (c) Since Q, the volume flow rate, i.e., the volume of water delivered each second, remains constant, it takes the same time (20 s) to fill the bucket with the nozzle as without the nozzle.
18 A 2 A 1 Question 13.8: A large tank of water has an outlet a distance h = 3.0 m below the surface of the water. Initially, the tank is filled with water to a depth y 2 = 4.0 m. (a) What is the speed of the water as it flows out of the hole? (b) What is the distance x reached by the water flowing out of the hole? You may assume the tank has a very large diameter so the level of the water remains constant. Also, you can model the water leaving the hole as a projectile. (a) We apply Bernoulli s equation to points and. But P 1 = P 2 = P! since both the hole and the surface of the water in the tank are at atmospheric pressure. Since v 2 = 0, then P 1 + ρgy ρv 1 2 = P 2 + ρgy ρv ρv 1 2 = ρg(y 2 y 1 ) ρv 2 2 = ρgh ρv 2 2. (As though in free fall!) 1 2 ρv 1 2 = ρgh, i.e., v 1 = 2gh = 2(9.81 m/s 2 )(3.0 m) = 7.67 m/s.
19 A 2 A 1 (b) If we model the water leaving the hole as a projectile, then the time to reach the ground is given by but v yi = 0, as the water emerges horizontally. The range is x = v xi t, where v xi = v 1, which remains constant in projectile motion. y 1 = v yi t 1 2 gt2, t = 2y 1 g = 2(y 2 h) g = 2(1.0 m) (9.81 m/s 2 ) x = v 1 t = (7.67 m/s)(0.45 s) = 3.46 m. = 0.45 s. Question 13.9: A siphon is a device for transferring a liquid from one container to another. The tube must be filled with liquid to start the siphon. (a) Derive an expression for the speed that water would flow through the tube. (b) What is the pressure at the highest section of the tube?
20 (a) We apply Bernoulli s equation to the surface of the liquid in the left hand container and the liquid at C. Then P! ρv s = P! ρv c 2 h c ρg, where the velocity at the surface v s = 0 as the surface area of the container is much greater than the area of the tube. we find P! = P B + h b ρg ρv 2 B. P B = P! h b ρg 1 2 ρv 2 B. But, since the tube has a constant cross sectional area, v A = v B = v C. Therefore, substituting for v B = v C = 2gh c, P B = P! (h b + h c )ρg. 1 2 v c 2 = h c g, i.e., v c = 2gh c. So the velocity depends only on the height difference between the surface of the liquid in the reservoir and the drain point. (b) Applying Bernoulli s equation to the surface in the Note that if (h b + h c ) = P! ρg, then P B = 0, which represents the longest length for the siphon tube. With water as the liquid, (h b + h c ) 10 m. reservoir and the point B we find
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