CHAPTER 13 FLUIDS • Density ! Bulk Modulus ! Compressibility

Liquids CHAPTER 13 FLUIDS FLUIDS Gases

• Density To begin with ... some important definitions ... ! Bulk modulus ! Compressibility Mass m DENSITY: , i.e., ρ = • Pressure in a fluid Volume V ! Hydraulic lift [M] ! Hydrostatic paradox Dimension: ⇒ 3 [ L] • Measurement of pressure −3 Units: ⇒ kg ⋅ m ! Manometers and barometers

• Buoyancy and Archimedes Principle ! Upthrust Force F ! Apparent weight PRESSURE: , i.e., P = Area A • Fluids in motion [M] ! Continuity Dimension: ⇒ [L][T]2 ! Bernoulli’s equation Units: ⇒ N ⋅ m−2 ⇒ Pascals (Pa) Mass = Volume × Density

Question 13.1: What, approximately, is the mass of air Volume of room ≈ 15 m ×10 m × 3 m −3 in this room if the density of air is 1 .29 kg ⋅ m ? 3 −3 ∴ M ≈ 450 m ×1.29kg ⋅ m = 580 kg (over half-a-ton!) 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 ft3. 3 3 3 ∴ 18 ft ⇒ 18 × (0.305 m) = 0.51 m . 3 But 1 m of air has a mass of 1 .3 kg, so the mass of air in the refrigerator is Question 13.2: How does the mass of a medium sized 0 .51×1.3 kg = 0.66 kg, apples compare with the mass of air in a typical i.e., approximately the mass of 6 apples! 3 refrigerator? the density of cold air is ~ 1.3 kg/m . 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? definitions (continued) ... 5 Atmospheric pressure: P o = 1×10 Pa ∴ Force on the ceiling from floor of room above is ΔP BULK MODULUS: B = pressure area 1 105 (15 m 10 m) ΔV × ≈ × × × ( V) ≈ 1.2 ×107 N. [M] Dimension: ⇒ 2 [L][T] Units: ⇒ N ⋅ m−2 Compressibility ⇒ B−1

Δ P Why doesn’t it collapse under that weight ... ? Δ V Because pressure operates equally in all directions!

! Why? F

Δ P Δ P When air molecules “bounce” off V the walls they produce an impulse: F Δt = Δp ⇒ pressure Δ P Since the molecules are traveling Gases are easily compressed (B ⇒ very small). with equal speeds in all directions Liquids and solids much less compressible. ... the pressure is the same ! 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 Question 13.3: A balloon has a radius of 10 cm. By how the surface of the fluid. At equilibrium there is no net much does the radius change if the balloon is pushed force acting on the surfaces of the cylinder. down to a depth of 10 m in a large tank of water? (The bulk modulus of air is 2 ×105N⋅ m−2.) ∴ ∑Fy = 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? 3 3 2 5 ρ gh = 1×10 kg/m × 9.81 m/s ×10 m ≈10 Pa. 5 P ! = Pat ≈1.01×10 Pa. ∴ P ≈ 2Pat. The pressure difference is What’s the difference in the air pressure from the r Δ P = hρg ceiling to the floor in this room ... ? ΔP But B = . ΔV h ( V) The room height is ~ 3m so the pressure difference is: ΔP = hρg = 3 ×1.29 × 9.81 ≈ 38Pa Δ r ΔV ΔP hρg ∴ = = . ΔP 38 V B B ∴ ≈ ≈ 3.8 ×10−4 (0.038%) P 1 105 For an air-filled balloon at a ! × depth of 10m, we have i.e., negligible. ΔV 10 m ×1×103 kg/m3 × 9.81 m/s2 = ≈ 0.5 (50%). V 2 ×105 N/m2 Pascal’s principle ... Assuming the balloon is spherical, P ! + ΔP 4 If additional pressure ( Δ P) is V = πr3 so Δ V = 4πr2 × Δr. 3 applied, it is transmitted ΔV 3Δr through the whole fluid: ∴ = ≈ 0.5, h 1 h V r 2 ∴ P1 = P! + h1ρg + ΔP P 1 so, for an initial radius r = 0.1 m, and P 0.5r 2 P = P + h ρg + ΔP. Δr ≈ = 0.017m (1.7 cm). 2 ! 2 3 Blaise Pascal (1623-1662) Hydraulic lift: F 1 Area A 1 Area A 2 Δ x2 F1 Δx Area A Area A 1 1 2 F 2

F h 2

Get a larger force OUT than you put IN? Too good to be true? same pressure No, not really, because, to do work (like lift something If a force F 1 is applied to the left hand piston, the heavy) the force F is applied through a distance x . F1 2 Δ 2 additional pressure, P1 = , is transmitted through the A1 But by conservation of energy A2 whole fluid. Therefore, on the surface of the right hand F2Δx2 = F1Δx1 ⇒ Δx1 = Δx2. A1 F2 piston, P2 = = P1. A2 So, although F 1 < F2, it is applied through a greater distance Δx1 > Δx2. F1 A1 ⎛ A2⎞ ∴ = , i.e., F2 = ⎜ ⎟ F1. Examples: F2 A2 ⎝ A1⎠ • lifts Wow ... the force is amplified!! • dentist’s chair Mechanical advantage • hydraulic brake systems We can use these ideas to measure pressure: Here’s something surprising ... no matter the shape of a vessel, the pressure depends only on the P ! P = 0 vertical depth. It is known as the hydrostatic paradox.

h h y y 2 P 2 P ! P !

y 1 y 1

P ! P ! P ! P ! Manometer Barometer

h P + ρgy1 = Po + ρgy 2 0 + ρgy 2 = Po + ρgy1 This is absolute pressure

P ! + ρgh i.e., P − P! = ρgh . i.e., P ! = ρgh . This is the Gauge pressure Atmospheric pressure DISCUSSION PROBLEM 13.1: Barometer

#2 The drawing shows two P P = 0 ! pumps, #1 and #2 to be used P ! = ρgh , i.e., h = ρg for pumping water from a very 5 P ! ≈ 1.01×10 Pa deep well (~30 m deep) to Using water: ground level. Pump #1 is h 3 −3 submerged in the water at the ρ = 1×10 kg ⋅ m #1 5 bottom of the well; the other P P 1.01×10 ! ! ∴h = ~ 10m. pump, #2, is located at ground 1×103 × 9.81 level. Which pump, if either, can be used to pump water Using mercury: to ground level? 3 −3 ρ = 13.6 ×10 kg ⋅ m 1.01×105 A: Both pumps #1 and #2. ∴h = ~ 0.76m 13.6 ×103 × 9.81 B: Pump #1. ‘Standard pressure’ is 760mm of Hg. C: Pump #2. D: Neither pump #1 nor pump #2. If the object is floating then ... w = B. Buoyancy and the concept of upthrust B

B V s V L w = mg V s w = mg Weight of object is w = mg = ρsVs g. If V L is the volume submerged, then the weight of liquid displaced is Archimedes Principle : when an object is partially or ρ LVL g. But according to Archimedes principle, this is wholly immersed in a fluid, the fluid exerts an upward equal to the upthrust (B). force ... upthrust ... (or buoyant force, B) on the object, ρs ∴ρsVs g = ρLVL g ⇒ VL = Vs. which is equal to the weight of fluid displaced. ρL Example: what volume of an iceberg is submerged? Submerged: w > B 3 −3 ρ s = 0.92 ×10 kg ⋅ m . 3 −3 ρ L = 1.03×10 kg ⋅ m . Weight of object w = mg = ρsVs g 3 Weight of liquid displaced = ρ V g VL ρs 0.92 ×10 L s ∴ = = 3 = 0.89, Vs ρL 1.03×10 If ρ s > ρL the object will sink. i.e., 89% of an iceberg is submerged! What ... you don’t believe me ... 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? When an ice cube floats, the

weight of the ice cube ( = ρsVsg) equals the weight of the water

displaced ( = ρLVLg), which is proportional to V L, the volume of the ice cube below the surface.

∴ ρLVLg = ρsVsg, ρs i.e., VL = Vs, ρL which is independent of g. So, the volume submerged Question 13.5: A piece of copper of mass 0.50 kg is would remain the same! suspended from a spring scale. If it is fully submerged in water, what is the reading (in N) on the spring scale? Since both the weight of an object, which is floating, and 3 −3 (The density of copper is 9 .0 ×10 kg ⋅ m .) the weight of a fluid it displaces are proportional to the local value of g, the submerged volume does not depend on g. T (apparent weight) T Question 13.6: A beaker containing water is placed on a

B w = mg = Vsρsg balance; its combined mass reading on the balance is 1.200 kg. In (a) below, a copper block is hanging freely from a spring scale, which has a mass reading of 0.20 kg. Identify the forces acting on the block. If the copper block is totally immersed in the water, as At equilibrium ∑Fy = T + B + (−w) = 0. shown in (b), what are the readings on the balance and ⎛ ρL⎞ ∴T = w − B = ρsVsg − ρLVsg = ρsVsg⎜1 − ⎟ . the spring scales? ρ ⎝ s ⎠ 3 −3 (The density of copper is 9 .0 ×10 kg ⋅ m .) True weight = mg . Upthrust 3 3 ρL 1×10 kg/m But = = 0.111. 0.20kg ρ 9.0 ×103 kg/m3 s ?? ∴ T = 0.5 × 9.81× (1− 0.111) = 4.36 N (0.444kg). In air instead of water ... 3 ρair 1.29 kg/m −4 1.200kg ?? = 3 3 = 0.143×10 . ρs 9.0 ×10 kg/m (a) (b) So, the mass of the block would be 0 .9998mg less than in vacuum (0.50 kg). ∴ T2 = (m − ρwVs)g = 0.20 kg −1×103 kg/m3 × 2.22 ×10−5m3 g 0.20kg ( ) ?? T 1 = 0.178g, i.e., the reading on the spring scale is 0 178 kg. T 2 B . mg mg The upthrust B is the force on the water on the block; by 1.200kg ?? (a) (b) Newton’s 3rd law the block must exert an equal and opposite force on the water. Consequently, the reading (a) Initially, the spring scale registers the weight of the on the balance will increase by 0.178 kg, i.e., it will read copper block and the balance registers the weight of 1.378 kg when the bock is suberged. (water + beaker). (b) When the block is lowered into the water, the water exerts an upward buoyant force (the So, when you dip a teabag into your cup, the weight of upthrust) on the copper block. Then the teabag is reduced, but the weight of the cup (and T 2 + B = mg , i.e., T 2 = mg − B, contents) is increased! where B is the upthrust, which is equal to the weight of water displaced, i.e., B = ρwVsg. The submerged volume is ms 0.20 kg −5 3 Image from Paul Hewitt’s Vs = = 3 3 = 2.22 ×10 m Conceptual Physics website: ρs 9.0 ×10 kg/m http://www.arborsci.com/ConceptualPhysics/ (a) (b)

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) = wb + ws = wb + NVsρsg, 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 Question 13.7: A barge, loaded with steel cannisters, is density. When the blocks are thrown into the water, as floating in a lock. If the cargo is thrown into the water, in (b), the weight of water displaced is equal to the what happens to the level of water at the side of the weight of the barge plus the weight of the water lock? Does it rise, stay the same, or fall? displaced by the blocks, i.e., w w(b) = wb + NVsρwg. Since ρ w < ρs, w w(b) < ww (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. 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 Question 13.8: A block of balsa wood with a rock tied to displaced is the same in both cases. If the volume of the it floats in water. When the rock is on top, exactly one- block is V b and the volume of the rock is V r, when the half of the wood is submerged below the water line. If rock is on top the volume of water displaced is 0 .5Vb. If the block is turned over so that the rock is now the fraction of the wood submerged when the rock is underneath and submerged, what can you say about the underneath is xV b , then, since the volume of water amount of the block below the water line? Is it less than displaced with the rock underneath is the same as with one-half of the block, one-half of the block, more than the rock on top, one-half of the block. 0 .5Vb = xVb + Vr, i.e., x < 0.5, so less than one-half of the wood is submerged. Fluids in motion; mass continuity

v 1 v 2

Area A1 Area A2 v 1 v Δt 2 v 2 v 1Δt Area A1 Area A2 We assume the fluid is incompressible, i.e., a liquid, so there is no change in density from 1 → 2. Then, the mass v 2Δt v 1Δt and volume must be conserved from 1 → 2 as the liquid flows down the pipe, If the density changes (from ρ 1 ⇒ ρ2) then, since mass is i.e., V 1 = V2. conserved we have ... Then in time Δ t, we have m 1 = m2 A 1v1Δt = A2v2Δt, i.e., ( A1v1Δt)ρ1 = (A2v2Δt)ρ2 i.e., A 1v1 = A2v2 ⇒ constant. ∴ A1v1ρ1 = A2v2ρ2 This is called the continuity equation for an This is the mass continuity equation. incompressible liquid. The conserved quantity ... Area × velocity ⎛ dV ⎞ ... is called the volume flow rate ⎜ (m3 ⋅s−1)⎟ . ⎝ dt ⎠ Bernoulli’s equation The continuity equation in everyday life ... Area A 2

[1] The garden hose: v 2

Area A 1 If you squeeze the end of a y 2 garden hose the area is A 1 v 1 y 1 reduced and so the water v 1 velocity increases, since A 2 A 1v1 = A2v2 We state Bernoulli’s equation without proof. It relates v 2 ⎛ A1⎞ the pressure (P), elevation (y) and speed (v) of an i.e., v2 = ⎜ ⎟ v1. ⎝ A2⎠ incompressible fluid in steady (i.e., non-turbulent) flow If you don’t restrict the pipe too much, the volume flow down a pipe, rate remains constant, which means you will fill a bucket 1 2 1 2 P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2 , in the same time whether the end of the hose is restricted 2 2 1 or not! i.e.,P + ρgy + ρv2 = constant. 2 “potential energy” “kinetic energy” [2] Lanes at highway tolls: Note, if fluid is at rest v 1 = v2 = 0. ∴P − P = ρg(y − y ) ⇒ ΔP = ρgΔy, ~ See useful notes on web-site ~ 2 1 2 1 a result we obtained earlier. A 2 2

Question 13.9: A large tank of water has an outlet a h y distance h below the surface of the water. (a) What is 2 1 A the speed of the water as it flows out of the hole? (b) 1 y 1 What is the distance x reached by the water flowing out x of the hole? (c) What value of h would cause the water (a) We start with Bernoulli’s equation to reach a maximum value of x? You may assume that 1 2 1 2 the tank has a very large diameter so the level of the P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2 . 2 2 water remains constant. But, P 2 = P1 = P! since both the hole and the top of the tank are at atmospheric pressure. Then 1 2 1 2 1 2 ∴ ρv1 = ρg(y2 − y1) + ρv2 = ρgh + ρv2 . h 2 2 2 y 2 However, if A 2 >> A1, then v 1 >> v2; in fact, we were y 1 told that v 2 = 0. Then 1 x 2 ρv1 = ρgh , i.e., v 1 = 2gh . 2 (Does this seem familiar?) A 2 2

h y 2 1 A 1 y 1 (c) To find the maximum value of x, we take x = 2 y y − y 2 x ( 1 2 1 ) dx (b) To find x we model the water leaving the hole as a and set = 0. (Note: we were told that the water level dy projectile. The time it takes for a volume element of 1 water to exit the hole and reach the ground is given by, remains constant, so y 2 is constant.) dx 1 −1 2 1 2 2 y y y 2 2 y 2y 0. y1 = vyit − gt , ∴ = ( )( 1 2 − 1 ) ( 2 − 1) = 2 dy1 2 but v 0, as the water emerges horizontally. yi = This equation is satisfied if ( y2 − 2y1) = 0, 2y1 1 ∴t = . i.e., when y1 = y2. 2 g Thus, the hole should be halfway between the bottom of The “range” is x = vxit, but v xi = v1, i.e., in projectile the tank and the surface of the water. motion it remains constant.

2y1 ∴x = v1 = 2 hy1 . g Also, substituting for h we find x 2 y y y 2 . = ( 1 2 − 1 ) 2 h d 1 Question 13.10: A large tank of water has an outlet a distance d below the surface of the water. The outlet is Using Bernoulli’s equation we have curved so that the exiting water is directed vertically 1 2 1 2 P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2 . upward. Prove that the vertical height achieved by the 2 2 water, h, is equal to the depth of the outlet below the But P 1 = P2 = P!. Also, if the initial speed of an element surface of the water, d. of water is v 1, then 2 2 v 2 = v1 = 2(−g)h. But v 2 = 0. v 2gh . h d ∴ 1 = Substituting in Bernoulli’s equation we obtain 1 2 1 ρv1 = ρg(y2 − y1), i.e., ρ(2gh) = ρgd , 2 2

since ( y2 − y1) = d, then d = h.