Chapter 10 Fluids

Fluids A fluid is a gas or a liquid. A gas expands to fill any container A liquid (at fixed pressure and temperature), has a fixed volume, but deforms to the shape of its container.

The atoms in a liquid are closely packed while those in a gas are separated by much larger distances. Gas have a density ~ 1/1000 x liquid density

Density and Pressure  The density of a substance of uniform composition is defined as its mass per unit volume:  Units are kg/m 3 (SI) or g/cm 3 (cgs)  1 g/cm 3 = 1000 kg/m 3

Density, cont.  The densities of most liquids and solids vary slightly with changes in temperature and pressure  Densities of gases vary greatly with changes in temperature and pressure

 Density = Mass/Volume  = M/V  = M/V SI unit: [kg/m 3 ] SI unit: [kg/m 3 ]  Densities of some common things (kg/m 3 ) Water 1000 Water 1000 ice 917(floats on water) ice 917(floats on water) blood 1060(sinks in water) blood 1060(sinks in water) lead 11,300 lead 11,300 Copper 8890 Copper 8890 Mercury 13,600 Mercury 13,600 Aluminum 2700 Aluminum 2700 Wood 550 Wood 550 air 1.29 air 1.29 Helium 0.18 Helium 0.18 Density

 The specific gravity of a substance is the ratio of its density to the density of water at 4° C The density of water at 4° C is 1000 kg/m 3 The density of water at 4° C is 1000 kg/m 3  Specific gravity is a unitless ratio

Pressure Pressure P is the amount of force F per unit area A: Pressure is the outward force per unit area that the fluid exerts on its container. A2A2 F1F1 A1A1 F2F2 By the Action-Reaction principle, Pressure is the inward force per unit area that the container exerts on the fluid.

 The force exerted by a fluid on a submerged object at any point if perpendicular to the surface of the object

A woman’s high heels sink into the soft ground, but the larger shoes of the much bigger man do not. Pressure = force/area

The pressure exerted on the piston extends uniformly throughout the fluid, causing it to push outward with equal force per unit area on the walls and bottom of the cylinder.

 The spring is calibrated by a known force  The force the fluid exerts on the piston is then measured

You are walking out on a frozen lake and you begin to hear the ice cracking beneath you. What is your best strategy for getting off the ice safely? 1) stand absolutely still and don’t move a muscle 2) jump up and down to lessen your contact time with the ice 3) try to leap in one bound to the bank of the lake 4) shuffle your feet (without lifting them) to move towards shore 5) lie down flat on the ice and crawl toward shore ConcepTest 10.3 On a Frozen Lake

You are walking out on a frozen lake and you begin to hear the ice cracking beneath you. What is your best strategy for getting off the ice safely? 1) stand absolutely still and don’t move a muscle 2) jump up and down to lessen your contact time with the ice 3) try to leap in one bound to the bank of the lake 4) shuffle your feet (without lifting them) to move towards shore 5) lie down flat on the ice and crawl toward shore As long as you are on the ice, your weight is pushing down. What is important is not the net force on the ice, but the force exerted on a given small area of ice (i.e., the pressure!). By lying down flat, you distribute your weight over the widest possible area, thus reducing the force per unit area. ConcepTest 10.3 On a Frozen Lake

Atmospheric Pressure Atmospheric pressure comes from the weight of the column of air above us. At sea level, atmospheric pressure is: P at = 1.01  10 5 N/m 2 = 1.01  10 5 Pa 1 Pascal= 1 N/m 2 = 1.01  10 5 Pa 1 Pascal= 1 N/m 2 = 14.7 lb/in 2 (psi) = 14.7 lb/in 2 (psi) = 1 bar (tire pressure gauges in Europe read 1, 2,..bar) = 1 bar (tire pressure gauges in Europe read 1, 2,..bar) Hurricane Rita 2005: P = 882 millibar = bar F=Mg F=PA

Pressure examples 1. Estimate the force of the atmosphere on the top of your head. A = (10cm)(15cm)=0.015m 2 A = (10cm)(15cm)=0.015m 2 F=PA = [1.01  10 5 N/m 2 ][0.015 m 2 ] = 1.5 kN F=PA = [1.01  10 5 N/m 2 ][0.015 m 2 ] = 1.5 kN A = (4in)(6in)=24 in 2 A = (4in)(6in)=24 in 2 F=PA = [15 lb/in 2 ][24in 2 ] = 360 lb. F=PA = [15 lb/in 2 ][24in 2 ] = 360 lb. 2. Is atmospheric pressure on top of a mountain greater or less than at sea level? Less. At higher altitude, there is less mass above. Less. At higher altitude, there is less mass above.

Pressure  Example

 If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium  All points at the same depth must be at the same pressure Otherwise, the fluid would not be in equilibrium Otherwise, the fluid would not be in equilibrium The fluid would flow from the higher pressure region to the lower pressure region The fluid would flow from the higher pressure region to the lower pressure region

 Examine the darker region, assumed to be a fluid It has a cross- sectional area A It has a cross- sectional area A Extends to a depth h below the surface Extends to a depth h below the surface  Three external forces act on the region

 P = P o + ρgh  P o is normal atmospheric pressure x 10 5 Pa = 14.7 lb/in 2 = 1 atm x 10 5 Pa = 14.7 lb/in 2 = 1 atm  The pressure does not depend upon the shape of the container

Pressure in a Fluid Pressure in a fluid depends only on the depth h below the surface. P = P at +  gh  = density of fluid P = P at +  gh  = density of fluid Weight/Area of fluid Weight/Area of atmosphere above fluid IF the density of the fluid is constant and it has atmospheric pressure (P at ) at its surface. Mass of fluid above depth h is (density)(volume) =  hA Force of gravity on fluid above depth h: W=  ghA

Pressure under water To what depth in water must you dive to double the pressure exerted on your body? P = P at +  gh  gh = P at, h= P at /  g Start to feel strong pressure at 3m

Pressure variation in fluid The variation in pressure at two different depths is given by: P 2 = P 1 +  gh

Pressure and Depth Barometer: a way to measure atmospheric pressure p 2 = p 1 +  gh p atm =  gh Measure h, determine p atm example--Mercury  = 13,600 kg/m 3 p atm = 1.05 x 10 5 Pa  h = m = 757 mm = 29.80” (for 1 atm) h p 2 =p atm p 1 =0

 Absolute vs. Gauge Pressure  The pressure P is called the absolute pressure Remember, P = P o +  gh Remember, P = P o +  gh  P – P o =  gh is the gauge pressure

 One end of the U- shaped tube is open to the atmosphere  The other end is connected to the pressure to be measured  Pressure at B is P o +ρgh

 One atmosphere of pressure is defined as the pressure equivalent to a column of mercury exactly 0.76 m tall at 0 o C where g = m/s 2  One atmosphere (1 atm) = 76.0 cm of mercury (760mm = 1 torr) 76.0 cm of mercury (760mm = 1 torr) x 10 5 Pa x 10 5 Pa 14.7 lb/in lb/in 2

 Example:

 A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. First recognized by Blaise Pascal, a French scientist (1623 – 1662) First recognized by Blaise Pascal, a French scientist (1623 – 1662)

 The hydraulic press is an important application of Pascal’s Principle  Also used in hydraulic brakes, forklifts, car lifts, etc.

A small force F 1 applied to a piston with a small area produces a much larger force F 2 on the larger piston. This allows a hydraulic jack to lift heavy objects.

Pascal’s Principle, Force  A external pressure P applied to any area of a fluid is transmitted unchanged to all points in or on the fluid.  This is just an application of the Action-Reaction principle.  Hydraulic Lift A Force F 1 is applied to area A 1, displacing the fluid by a distance d 1. The pressure increase in the fluid is P=F 1 /A 1. The Pressure F 1 /A 1 creates a force on the car F 2 = A 2 (F 1 /A 1 ) = F 1 (A 2 /A 1 ). A small force acting on a small area creates a big force acting over a large area!

Archimedes’ Principle: The buoyant force acting on an object fully or partially submerged in a fluid is equal to the weight of the fluid displaced by the object.

The weight of a column of water is proportional to the volume of the column. The volume V is equal to the area A times the height h. Equilibrium…

 The upward force is called the buoyant force  The physical cause of the buoyant force is the pressure difference between the top and the bottom of the object

 The magnitude of the buoyant force always equals the weight of the displaced fluid  The buoyant force is the same for a totally submerged object of any size, shape, or density

 The buoyant force is exerted by the fluid  Whether an object sinks or floats depends on the relationship between the buoyant force and the weight

 The upward buoyant force is B=ρ fluid gV obj  The downward gravitational force is w=mg=ρ obj gV obj  The net force is B-w=(ρ fluid -ρ obj )gV obj

 The object is less dense than the fluid  The object experiences a net upward force

 The object is more dense than the fluid  The net force is downward  The object accelerates downward Question: How do steel ships float if steel is roughly 6 times more dense than water?

 The object is in static equilibrium  The upward buoyant force is balanced by the downward force of gravity  Volume of the fluid displaced corresponds to the volume of the object beneath the fluid level

 The forces balance 

Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up causing the water to spill. 2. Go down. 3. Stay the same. Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice When ice melts it will turn into water of same volume

 Example 9.9  A raft is constructed of wood having a density of 6.00 x 10 2 kg/m 3. Its surface area is 5.70m 2, and volume is 0.60m 3. When the raft is placed in fresh water, what depth h is the bottom of the raft submerged?

Concept Question Which weighs more: 1. A large bathtub filled to the brim with water. 2. A large bathtub filled to the brim with water with a battle-ship floating in it. 3. They will weigh the same. Tub of water Tub of water + ship Overflowed water CORRECT Weight of ship = Buoyant force = Weight of displaced water

 Streamline flow Every particle that passes a particular point moves exactly along the smooth path followed by particles that passed the point earlier Every particle that passes a particular point moves exactly along the smooth path followed by particles that passed the point earlier Also called laminar flow Also called laminar flow  Streamline is the path Different streamlines cannot cross each other Different streamlines cannot cross each other The streamline at any point coincides with the direction of fluid velocity at that point The streamline at any point coincides with the direction of fluid velocity at that point

Streamline flow shown around an auto in a wind tunnel

 The flow becomes irregular exceeds a certain velocity exceeds a certain velocity any condition that causes abrupt changes in velocity any condition that causes abrupt changes in velocity  Eddy currents are a characteristic of turbulent flow

 The rotating blade (dark area) forms a vortex in heated air The wick of the burner is at the bottom The wick of the burner is at the bottom  Turbulent air flow occurs on both sides of the blade

 Viscosity is the degree of internal friction in the fluid Measure of a fluid's ability to resist gradual deformation by shear or tensile stresses  The internal friction is associated with the resistance between two adjacent layers of the fluid moving relative to each other

 Viscous Liquid!

 The fluid is nonviscous There is no internal friction between adjacent layers There is no internal friction between adjacent layers  The fluid is incompressible Its density is constant Its density is constant  The fluid motion is steady Its velocity, density, and pressure do not change in time Its velocity, density, and pressure do not change in time  The fluid moves without turbulence No eddy currents are present No eddy currents are present The elements have zero angular velocity about its center The elements have zero angular velocity about its center

Equation of Continuity  Mass is conserved as the fluid flows. If a certain mass of fluid enters a pipe at one end at a certain rate, the same mass exits at the same rate at the other end of the tube (if nothing gets lost in between through holes, for instance). Mass flow rate at position 1 = Mass flow rate at position 2 Mass flow rate at position 1 = Mass flow rate at position 2   1 A 1 v 1 =  2 A 2 v 2   1 A 1 v 1 =  2 A 2 v 2  A v = constant along a tube that has a single entry and a single exit point for fluid flow.

 What goes in comes out!  If density is constant: A 1 v 1 = A 2 v 2 A 1 v 1 = A 2 v 2  The product of the cross- sectional area of a pipe and the fluid speed is a constant Speed is high where the pipe is narrow and speed is low where the pipe has a large diameter Speed is high where the pipe is narrow and speed is low where the pipe has a large diameter  Av is called the flow rate

 The equation is a consequence of conservation of mass and a steady flow  A v = constant This is equivalent to the fact that the volume of fluid that enters one end of the tube in a given time interval equals the volume of fluid leaving the tube in the same interval This is equivalent to the fact that the volume of fluid that enters one end of the tube in a given time interval equals the volume of fluid leaving the tube in the same interval Assumes the fluid is incompressible and there are no leaksAssumes the fluid is incompressible and there are no leaks

Bernoulli’s Equation  Work-Energy Theorem : W nc = change of total mechanical energy applied to fluid flow : Difference in pressure => net force is not zero => fluid accelerates Pressure is due to collisional forces which is a nonconservative force: W nc = (P 2 -P 1 ) V W nc = (P 2 -P 1 ) V Consider a fluid moving from height h 1 to h 2. Its total mechanical energy is given by the sum of kinetic and potential energy. Thus, W nc = E tot,1 –E tot,2 = ½ m v 1 2 +m g h 1 –( ½ m v 2 2 +m g h 2 ) W nc = E tot,1 –E tot,2 = ½ m v 1 2 +m g h 1 –( ½ m v 2 2 +m g h 2 )

 States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline

 Shows fluid flowing through a horizontal constricted pipe  Speed changes as diameter changes  Can be used to measure the speed of the fluid flow  Swiftly moving fluids exert less pressure than do slowly moving fluids

Objects Moving Through a Fluid  Many common phenomena can be explained by Bernoulli’s equation At least partially At least partially  In general, an object moving through a fluid is acted upon by a net upward force as the result of any effect that causes the fluid to change its direction as it flows past the object

 The air speed above the wing is greater than the speed below  The air pressure above the wing is less than the air pressure below  There is a net upward force Called lift Called lift  Other factors are also involved Designed to produce lift Designed to produce lift Racecars designed to produce faster airflow on the bottom Racecars designed to produce faster airflow on the bottom  High velocity implies low pressure, IN the fluid

 Example:  A jet of water squirts out horizontally from a hole near the bottom of the tank with a velocity of 1.33 m/s. What is the height of the water level in the tank?