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Lecture 21. Goals:. Chapter 15 U nderstand pressure in liquids and gases Use Archimedes’ principle to understand buoyancy Understand the equation of continuity Use an ideal-fluid model to study fluid flow. I nvestigate the elastic deformation of solids and liquids. Assignment
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Lecture 21 Goals: • Chapter 15 • Understand pressure in liquids and gases • Use Archimedes’ principle to understand buoyancy • Understand the equation of continuity • Use an ideal-fluid model to study fluid flow. • Investigate the elastic deformation of solids and liquids • Assignment • HW9, Due Wednesday, Apr. 8th • Thursday: Read all of Chapter 16
Pressure vs. DepthIncompressible Fluids (liquids) • In many fluids the bulk modulus is such that we can treat the density as a constant independent of pressure: An incompressible fluid • For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! • p(y) = p0 - y g r = p0 + d g r • Gauge pressure (subtract p0, pressure 1 atm, i.e. car tires) F2 = F1+ m g = F1+ rVg F2 /A = F1/A + rVg/A p2 = p1 - rg y
Pressure vs. Depth • For a uniform fluid in an open container pressure same at a given depthindependentof the container • Fluid level is the same everywhere in a connected container, assuming no surface forces
Pressure Measurements: Barometer • Invented by Torricelli • A long closed tube is filled with mercury and inverted in a dish of mercury • The closed end is nearly a vacuum • Measures atmospheric pressure as 1 atm = 0.760 m (of Hg)
(A)r1 < r2 (B)r1 = r2 (C)r1 > r2 Exercise Pressure • What happens with two fluids?? • Consider a U tube containing liquids of density r1 and r2 as shown: • At the red arrow the pressure must be the same on either side. r1 x = r2 (d1+ y) • Compare the densities of the liquids: dI r2 y r1
W2? W1 W1 > W2 W1 < W2 W1 = W2 Archimedes’ Principle: A Eureka Moment • Suppose we weigh an object in air (1) and in water (2). How do these weights compare? • Buoyant force is equal to the weight of the fluid displaced
W2? W1 W1 > W2 W1 < W2 W1 = W2 Archimedes’ Principle • Suppose we weigh an object in air (1) and in water (2). • How do these weights compare? • Why? Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.
y F mg B Sink or Float? • The buoyant force is equalto the weight of the liquid that is displaced. • If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. • We can calculate how much of a floating object will be submerged in the liquid: • Object is in equilibrium
piece of rock on top of ice Bar Trick What happens to the water level when the ice melts? Expt. 1 Expt. 2 B. It stays the same C. It drops A. It rises
Exercise V1 = V2 = V3 = V4 = V5 m1 < m2 < m3 < m4 < m5 What is the final position of each block?
Exercise V1 = V2 = V3 = V4 = V5 m1 < m2 < m3 < m4 < m5 What is the final position of each block? But this Not this
Pascal’s Principle • So far we have discovered (using Newton’s Laws): • Pressure depends on depth: Dp = r g Dy • Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel.
Pascal’s Principle in action: Hydraulics, a force amplifier • Consider the system shown: • A downward force F1 is applied to the piston of area A1. • This force is transmitted through the liquid to create an upward force F2. • Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid. • F2 > F1 with conservation of energy
M dA A1 A10 M dB A2 A10 Exercise • Consider the systems shown on right. • In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference di in the liquid levels. • If A2 = 2A1, how do dA and dB compare? V10 = V1= V2 dA A1= dB A2 dA A1= dB 2A1 dA = dB 2
Fluids in Motion • To describe fluid motion, we need something that describes flow: • Velocity v • There are different kinds of fluid flow of varying complexity • non-steady/ steady • compressible / incompressible • rotational / irrotational • viscous / ideal
Types of Fluid Flow • Laminar flow • Each particle of the fluid follows a smooth path • The paths of the different particles never cross each other • The path taken by the particles is called a streamline • Turbulent flow • An irregular flow characterized by small whirlpool like regions • Turbulent flow occurs when the particles go above some critical speed
Types of Fluid Flow • Laminar flow • Each particle of the fluid follows a smooth path • The paths of the different particles never cross each other • The path taken by the particles is called a streamline • Turbulent flow • An irregular flow characterized by small whirlpool like regions • Turbulent flow occurs when the particles go above some critical speed
A 2 A 1 v 1 v 2 Ideal Fluids • Streamlines do not meet or cross • Velocity vector is tangent to streamline • Volume of fluid follows a tube of flow bounded by streamlines • Streamline density is proportional to velocity • Flow obeys continuity equation Volume flow rate (m3/s) Q = A·vis constant along flow tube. Follows from mass conservation if flow is incompressible. Mass flow rate is just r Q (kg/s) A1v1 = A2v2
(A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1 Exercise Continuity • A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. v1 v1/2 • Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe?
v1 v1/2 Exercise Continuity (A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1 • For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast. • But if the water is moving 4 times as fast then it has 16times as much kinetic energy. • Something must be doing work on the water (the pressure drops at the neck) and we recast the work as P DV = (F/A) (A Dx) = F Dx
P1 P2 Conservation of Energy for Ideal Fluid W = (P1– P2 ) DV and W = ½ Dm v22 – ½ Dm v12 = ½ (r DV) v22 – ½ (r DV)v12 (P1– P2 ) = ½ r v22 – ½ r v12 P1+ ½ r v12= P2+ ½ r v22= const. and with height variations: Bernoulli Equation P1+ ½ r v12 + r g y1 = constant
Lecture 21 • Question to ponder: Does heavy water (D2O) ice sink or float? • Assignment • HW9, due Wednesday, Apr. 8th • Thursday: Read all of Chapter 16