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

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1 CHAPTER 13 FLUIDS FLUIDS Liquids Gases 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 To begin with... some important definitions... DENSITY: Dimension: [M] [L] 3 Units: kg m 3 PRESSURE: Dimension: Mass Volume, i.e., ρ = m V Force Area, i.e., P = F A [M] [L][T] 2 Units: N m 2 Pascals (Pa)

2 Mass = Volume Density Question 13.1: What, approximately, is the mass of air in this room if the density of air is 1.29 kg m 3? Volume of room 15 m 10 m 3 m M 450 m kg m 3 = 580 kg (over half-a-ton!)

3 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 13.2: How does the mass of a medium sized apples compare with the mass of air in a typical refrigerator? 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?

4 definitions (continued)... BULK MODULUS: B = Dimension: Units: N m 2 [M] [L][T] 2 ΔP ( ΔV V ) Compressibility B 1 ΔP ΔV Atmospheric pressure: P o = 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!

5 Pressure at a depth in a fluid... P! Area = A P! h w = mg P 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. 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. What s the pressure in water at a depth of 10m, say? ρgh = kg/m m/s 2 10 m 10 5 Pa. Question 13.3: 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? (The bulk modulus of air is N m 2.) P! = P at Pa. P 2P at.

6 r Δr The pressure difference is But B = ΔP = hρg ΔP ( ΔV V ). ΔV V = ΔP B = hρg B. For an air-filled balloon at a depth of 10m, we have ΔV V = 10 m kg/m m/s N/m (50%). Assuming the balloon is spherical, h V = 4 3 πr3 so ΔV = 4πr 2 Δr. ΔV V = 3Δr r 0.5, so, for an initial radius r = 0.1 m, Δr 0.5r = 0.017m (1.7 cm). 3 What s the difference in the air pressure from the ceiling to the floor in this room...? The room height is ~ 3m so the pressure difference is: i.e., negligible. ΔP = hρg = Pa ΔP P! (0.038%) Pascal s principle... P 2 h 2 P! + ΔP 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 ( )

7 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

8 Here s something surprising... no matter the shape of a vessel, the pressure depends only on the vertical depth. It is known as the hydrostatic paradox. We can use these ideas to measure pressure: P! P = 0 h P y 2 h P! P! y 2 y 1 y 1 P! P! P! P! Manometer Barometer h P! + ρgh 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

9 Barometer h P! P! P = 0 h = Using water: P! = ρgh, i.e., h = P! ρg P! Pa ρ = kg m 3 h = ~ 10m. Using mercury: ρ = kg m ~ 0.76m Standard pressure is 760mm of Hg. DISCUSSION PROBLEM 13.1: #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.

10 Buoyancy and the concept of upthrust 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). w = mg ρ s V s g = ρ L V L g V L = ρ s ρ L V s. Example: what volume of an iceberg is submerged? ρ s = kg m 3. ρ L = kg m 3. V L = ρ s = V s ρ L = 0.89, i.e., 89% of an iceberg is submerged! What... you don t believe me... B V L

11 Question 13.4: On Earth, an ice cube floats in a glass of water with 90% of its volume below the level of the water. If we poured ourselves a glass of water on the Moon, where the acceleration due to gravity is about 1 th 6 of that on Earth, and dropped in an ice cube, how much of the ice cube would be below the level of the water?

12 the ice cube below the surface. When an ice cube floats, the weight of the ice cube ( = ρ sv s g) equals the weight of the water displaced ( = ρ LV L g), 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 13.5: A piece of copper of mass 0.50 kg is suspended from a spring scale. If it is fully submerged in water, what is the reading (in N) on the spring scale? (The density of copper is kg m 3.)

13 B 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 But ρ L ρ s = kg/m kg/m 3 = T = ( ) = 4.36 N (0.444kg). In air instead of water... ρ s ρ air 1.29 kg/m 3 = ρ s kg/m 3 = So, the mass of the block would be mg less than in vacuum (0.50 kg). T (apparent weight) w = mg = V s ρ s g T Question 13.6: A beaker containing water is placed on a balance; its combined mass reading on the balance is kg. In (a) below, a copper block is hanging freely from a spring scale, which has a mass reading of 0.20 kg. If the copper block is totally immersed in the water, as shown in (b), what are the readings on the balance and the spring scales? (The density of copper is kg m 3.) 0.20kg?? 1.200kg?? (a) (b)

14 0.20kg 1.200kg (a) T 1 mg (a) Initially, the spring scale registers the weight of the copper block and the balance registers the weight of (water + beaker). (b) When the block is lowered into the water, the water exerts an upward buoyant force (the upthrust) on the copper block. Then T 2 + B = mg, i.e., T 2 = mg B, where B is the upthrust, which is equal to the weight of water displaced, i.e., B = ρ w V s g. The submerged volume is V s = m s 0.20 kg = ρ s kg/m 3 = m 3???? (b) T 2 B mg T 2 = ( m ρ w V s )g = ( 0.20 kg kg/m m 3 )g = 0.178g, i.e., the reading on the spring scale is kg. The upthrust B is the force on the water on the block; by Newton s 3rd law the block must exert an equal and opposite force on the water. Consequently, the reading on the balance will increase by kg, i.e., it will read kg when the bock is suberged. 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! Image from Paul Hewitt s Conceptual Physics website:

15 (a) (b) Question 13.7: A barge, loaded with steel cannisters, is floating in a lock. If the cargo is thrown into the water, what happens to the level of water at the side of the lock? Does it rise, stay the same, or fall? With cargo on the barge, as in (a), the weight of water displaced is equal to the weight of the barge plus the cargo, i.e., w w (a) = w b + w s = w b + NV s ρ s g, where w b is the weight of the barge, N is the number of blocks, V s is the volume of each block and ρ s their density. When the blocks 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 blocks, i.e., w w (b) = w b + NV s ρ w g. Since ρ w < ρ s, w w (b) < w w (a), i.e., less water is displaced in (b) than in (a). Since the volume of the lock is the same, the depth of water is less in (b) than in (a), i.e., the water level falls.

16 Question 13.8: A block of balsa wood with a rock tied to it floats in water. When the rock is on top, exactly onehalf of the wood is submerged below the water line. If the block is turned over so that the rock is now underneath and submerged, what can you say about the amount of the block below the water line? Is it less than one-half of the block, one-half of the block, more than one-half of the block. By Archimedes principle, the rock plus wood displaces its combined weight of water whether the rock is on top or underneath. So, the weight and volume of water displaced is the same in both cases. If the volume of the block is V b and the volume of the rock is V r, when the rock is on top the volume of water displaced is 0.5V b. If the fraction of the wood submerged when the rock is underneath is xv b, then, since the volume of water displaced with the rock underneath is the same as with the rock on top, 0.5V b = xv b + V r, i.e., x < 0.5, so less than one-half of the wood is submerged.

17 Fluids in motion; mass continuity v 1 v 2 Area A 1 Area A 2 v 1 Δt We assume the fluid is incompressible, i.e., a liquid, so there is no change in density from 1 2. Then, the mass and volume must be conserved from 1 2 as the liquid flows down the pipe, Then in time Δt, we have i.e., V 1 = V 2. A 1 v 1 Δt = A 2 v 2 Δt, v 2 Δt i.e., A 1 v 1 = A 2 v 2 constant. This is called the continuity equation for an incompressible liquid. The conserved quantity... Area velocity dv... is called the volume flow rate (m 3 s 1 ). dt 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

18 The continuity equation in everyday life... [1] The garden hose: Bernoulli s equation Area A 2 v 2 A 1 v 1 A 2 v 2 If you squeeze the end of a garden hose the area is reduced and so the water velocity increases, since A 1 v 1 = A 2 v 2 i.e., v 2 = A 1 v 1. A 2 If you don t restrict the pipe too much, the volume flow rate remains constant, which means you will fill a bucket in the same time whether the end of the hose is restricted or not! [2] Lanes at highway tolls: ~ See useful notes on web-site ~ Area A 1 y 2 y 1 v 1 We state Bernoulli s equation without proof. It relates the pressure (P), elevation (y) and speed (v) of an incompressible fluid in steady (i.e., non-turbulent) flow down a pipe, P 1 + ρgy ρv 1 2 = P 2 + ρgy ρv 2 2, i.e.,p + ρgy ρv2 = constant. potential energy kinetic energy Note, if fluid is at rest v 1 = v 2 = 0. P 2 P 1 = ρg(y 2 y 1 ) ΔP = ρgδy, a result we obtained earlier.

19 A 2 2 Question 13.9: A large tank of water has an outlet a distance h below the surface of the water. (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? (c) What value of h would cause the water to reach a maximum value of x? You may assume that the tank has a very large diameter so the level of the water remains constant. h y 2 1 A 1 y 1 x (a) We start with Bernoulli s equation P 1 + ρgy ρv 1 2 = P 2 + ρgy ρv 2 2. But, P 2 = P 1 = P! since both the hole and the top of the tank are at atmospheric pressure. Then y 2 x h y ρv 1 2 = ρg(y 2 y 1 ) ρv 2 2 = ρgh ρv 2 2. However, if A 2 >> A 1, then v 1 >> v 2 ; in fact, we were told that v 2 = 0. Then 1 2 ρv 1 2 = ρgh, i.e., v 1 = 2gh. (Does this seem familiar?)

20 A 2 2 y 2 (b) To find x we model the water leaving the hole as a projectile. The time it takes for a volume element of water to exit the hole and reach the ground is given by, y 1 2 = v yi t 1 2 gt2, but v yi = 0, as the water emerges horizontally. t = 2y 1 g. The range is x = v xi t, but v xi = v 1, i.e., in projectile motion it remains constant. 1 A 1 x h y 1 (c) To find the maximum value of x, we take ( ) x = 2 y 1 y 2 y 1 2 and set dx dy 1 = 0. (Note: we were told that the water level remains constant, so y 2 is constant.) dx = 1 dy 1 2 (2) y 1y 2 y 2 ( 1 ) 12 (y 2 2y 1 ) = 0. This equation is satisfied if (y 2 2y 1 ) = 0, i.e., when y 1 = 1 2 y 2. Thus, the hole should be halfway between the bottom of the tank and the surface of the water. x = v 1 2y 1 g = 2 hy 1. Also, substituting for h we find x = 2 y 1 y 2 y 2 ( 1 ).

21 Question 13.10: A large tank of water has an outlet a distance d below the surface of the water. The outlet is curved so that the exiting water is directed vertically upward. Prove that the vertical height achieved by the water, h, is equal to the depth of the outlet below the surface of the water, d. h d Using Bernoulli s equation we have P 1 + ρgy ρv 1 2 = P 2 + ρgy ρv 2 2. But P 1 = P 2 = P!. Also, if the initial speed of an element of water is v 1, then But v 2 = 0. 2 h d 1 v 2 2 = v 1 2 = 2( g)h. v 1 = 2gh. Substituting in Bernoulli s equation we obtain 1 2 ρv 1 2 = ρg(y 2 y 1 ), i.e., 1 ρ(2gh) = ρgd, 2 since (y 2 y 1 ) = d, then d = h.

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