Phys 250 Ch10 p1 Chapter 10: Fluids Fluids: substances which flow Liquids: take the shape of their container but have a definite volume Gases: take the shape and volume of their container Pressure in a fluid: force per area P = F/A Force = normal force, pressure exerts a force perpendicular to the surface. pressure of the bottom of a container on a liquid balances the pressure the liquid exerts on the container bottom Units for pressure: 1 N/m 2 = 1 Pa 1 Bar = 10 5 Pa ~ atmospheric pressure (14.7 psi)* 1 atmosphere = 1.01 E5 Pa 1 mm Hg= 1.33E2 Pa 1 torr= 1.33E2 Pa 1 lb/in 2 (psi)= 6.89 E3 Pa *atmospheric pressure varies from.970 bar to bar

Phys 250 Ch10 p2 Most pressure gages detect pressure differences between the measured pressure and a reference pressure. absolute pressure: the actual pressure exerted by the fluid. gauge pressure: the difference between the pressure being measured and atmospheric pressure. P = P gauge + P atm Some important aspects of pressure in a fluid The forces a fluid at rest exerts on the walls of its container (and visa versa) always perpendicular to the walls. An external pressure exerted on a fluid is transmitted uniformly throughout the volume of the fluid. The pressure on a small surface in a fluid is the same regardless of the orientation if the surface.

Phys 250 Ch10 p3 Density

Phys 250 Ch10 p4 Pressure and Depth A fluid supports itself against its weight with pressure. The fluid also must support itself against external pressure P = F/A = P external + weight of fluid w = mg =  VgV = Ah P = P external +  gh   P =  g  h A P externa l P h

Phys 250 Ch10 p5 Example: A tank is filled with water to a depth of 1.5 m. What is the pressure at the bottom of the tank due to the water alone? Example: How high above an IV insertion point into the patient’s arm must the saline bag be hung if the density of the saline solution is 1E3 kg/m3 and the gauge pressure inside the patient's vein is 2.4E3 Pa?

Phys 250 Ch10 p6 Example: An application of pressure in a fluid is the hydraulic press. The smaller piston is 3 cm in diameter, and the larger piston is 24 cm in diameter. How much mass could be lifted by a 50 kg woman putting all her weight on the smaller piston? p F 1 = PA 1 F 2 = P A 2 Pascal’s Principle: The pressure applied at one point in an enclosed fluid is transmitted to every part of the fluid and to the walls of the container.

Phys 250 Ch10 p7 Buoyant force: pressure balances gravity for a fluid to support itself. F net = w =  Vg F net =  fluid Vg Archimedes’ principle: Buoyant force = weight of fluid displaced F buoyant = V  g Example: An object of density  is submerged in a liquid with density  0. What is the effective weight of the object in terms of the densities and the original weight of the object.

Phys 250 Ch10 p8 Example: Icebergs are made of freshwater (density of 0.92 E3 kg/m 3 at 0ºC). Ocean water, largely because of dissolved salt, has a density of 1.03E3 kg/m 3 at 0ºC. What fraction of an iceberg lies below the surface?

Phys 250 Ch10 p9 Surface Tension: attraction of molecules in liquid for each other result in imbalance in the net force for charges near the surface. Surface Tension is a force per unit length. Example: lifting a ring of circumference C out of a liquid surface Surface Tensions  = F/2C F F  F  h Capillary Action

Phys 250 Ch10 p10 Fluid Flow with approximations: incompressible fluid no viscosity (friction) laminar flow (a.k.a. streamline flow) in contrast with turbulent flow Rate of flow: volume per time AA vt  V = vt A If no fluid is added/lost, flow rate must be the same throughout R in = v 1 A 1 R out = v 2 A 2 R in = R out v 1 A 1 = v 2 A 2

Phys 250 Ch10 p11 Example: A horizontal pipe of 25 cm 2 cross section carries water at a velocity of 3.0 m/s. The pipe feeds into a smaller pipe with a cross section of only 15 cm 2. What is the velocity of the water in the smaller pipe?

Phys 250 Ch10 p12 Bernoulli’s Equation: flow with changing heights and pressure Work-Energy Theorem + incompressible fluid A2A2 h1h1 h2h2 A1A1 p 1, v 1 p 2, v 2

Phys 250 Ch10 p13 Example: Determine the pressure change that occurs in the previous example skip sections 10.7, 10.8

Phys 250 Ch10 p14 Applications of Bernoulli’s Equation Liquid at rest: p 2  p 1 =  g(h 1 - h 2 ) No pressure difference, one part “at rest”: Torricelli’s theorem typically atmospheric pressure for both v h A boat strikes an underwater rock and opens a pencil-sized crack 7mm wide and 150mm long in its hull 65 cm below the waterline. The crew takes 5.00 minutes to locate the crack and plug it up. How much water entered the boat during those 5 minutes? Conceptual Question 10.11