Warm-up For light of a given frequency, ice has an index of refraction of 1.31 and water has an index of refraction of Find the critical angle θ c for the ray of light at the ice/water interface. ice water θcθc

Answer 80º

Problems to skip/add to the HW Sets Problem SetSkipDon't skip 6311, 12,13 649, 12, , 13, 14, 1715, 18

Fluid Mechanics Buoyancy and Pressure

Fluids and Buoyant Force Fluids Have no definite shape Gases do not have definite volume They flow

Density Density = mass/volume (kg/m 3 ) ρ = m/v

Common Densities Water = 1000kg/m 3 Air = 1.29kg/m 3 Objects will float if their density is less than these

Pressure Force/Area P=F/A N/m 2 or Pa 1 N/m 2 = 1Pa P air = 101,300 Pa

Relationship between ρ, F g, and P 1m 3 of water has a weight of 9800N (1000kg/m 3 x 1m 3 x 9.8) Use 9.8 for all pressure problems 1.0m 9800N The pressure on the bottom equals the weight of the cube P = F/A = 9800N/1.0m 2 = 9800Pa

Relationship between ρ, F g, and P If we stack three cubes atop one another we get the following pressure P = 9800x 3/1.0m 2 = 29,400Pa.

Relationship between ρ, F g, and P Thus the pressure at a depth in fluid in Pa is numerically equivalent to the weight of a column of fluid whose cross sectional area is 1m 2. To find the pressure at any depth in a column of fluid of any given cross sectional area, the weight per unit volume is multiplied by the depth!

Relationship between ρ, F g, and P Works for gases also! The average sea level density of air at 25°C and 1 atm is 1.29kg/m 3. The density decreases rapidly with altitude. The total mass of a column of air extending from the top of the atmosphere down to seal level whose cross sectional area is 1m 2 is 10, 330kg and its weight is N. Thus the atmospheric pressure is 101,300 Pa or 101.3kPa.

Fluid Pressure P = ρgh

Pressure Varies with Depth Remember that air pressure is 1.01x10 5 Pa nn_4U nn_4U

Pressure Increasing with Depth The column of water above you builds as you go further and further down. P T = P + P o P T is the absolute pressure P o is the atmospheric or original pressure P is the fluid pressure (pressure from depth)

Gauge Pressure P g = P T - P o

Example I On a particular day, the atmospheric pressure was 101,000 Pa. A) What was the pressure at a depth of 94.0m below the surface of the sea? B) What was the gauge pressure at this depth? The density of seawater is 1025kg/m 3

Answer 1.05x10 6 Pa, 9.49x10 5 Pa

Example II The fluid in a u-tube is glycerin, whose density is 1260kg/m 3. Atmospheric pressure is 1.01x10 5 Pa. A) What is the pressure P in the container? B) What is the gauge pressure in the container?

gas P gas PxPx PxPx P 14 14cm taller PoPo

Answer 1.03x10 5 Pa, 2.0x10 3 Pa

Problems with Pressure Diving and ascending too fast The Bends eok

Pascal’s Principle

P 1 = F 1 /A 1 and P 2 = F 2 /A 2, but P 2 = P 1 Therefore P increase = F 1 /A 1 = F 2 /A 2

Example IV The pressure produced by the force F 1 on a frictionless piston with a cross sectional area A 1 is transmitted through the perfect fluid. Find the resulting upward force F 2 on the piston with area A 2. (F 1 = 75N and A 1 = 10cm 2, A 2 = 80cm 2 )

Answer 600N

Off Topic but Cool eqjWuDg eqjWuDg

Buoyancy Keeps objects afloat Archimedes’s Principle The fluid that is displaced by an object that is submerged is equal to the volume of the object. Any object completely or partially submerged in a fluid experiences an upward buoyant force equal in magnitude of the weight of the fluid displaced by the object.

What????? This means that F B = ρV fluid g (density of fluid)(volume of fluid)(9.8) Buoyant force = weight of the displaced fluid

Buoyancy Extended No matter what depth the cube is submerged, the difference in the total force pressing down on the cube and the total force pushing up on the cube will always equal the weight of the volume of water displaced by the cube!!!!

Further Extensions Archimedes’ Principle works no matter the shape of the object. It also works for hollow objects compared to full objects (hollow sphere of lead compared to a ball of lead with the same volume). Both volumes of lead would have the same buoyant force acting upon them

Example V A hollow sphere whose mass is 250kg has a radius of 1.0m. If it is submerged in salt water at a depth of 10m by a cable. What is the tension in the cable? The density of salt water is 1030kg/m 3

Answer 40,000N

Example VI The rectangular block below is placed in water and 2.0cm of the block remains above the water. What is the density of the block? 10.0cm 40.0cm 10.0cm

Fluids in Motion Fluid Statics – the study of the relationships between forces and fluids at rest. Fluid Dynamics – the study of the relationships between moving fluids and the forces that cause the motion

Fluid Dynamics When a fluid is in motion in a pipe there are friction forces at play between the individual particles of the fluid and between the fluid and pipe. If the viscosity is not too great, all particles follow similar streamlines.

Fluid Dynamics Streamlines form flow tubes which depict fluid motion. Particles in a streamline do not cross paths with other streamlines in laminar (smooth) motion

In the figure below we see that the streamlines get closer together at the middle due to the decrease in area. The fluid is moving faster at this point.

The streamlines above an airplane wing are closer indicating greater speed.

Fluids Dynamics Fluid Flow Laminar – Smooth and repeating motion Turbulent – flow becomes irregular Eddy currents

Ideal Fluids We will only deal with “ideal fluids” Nonviscous The density of the fluid is constant. (cannot be compressed) The fluid motion is steady (P, ρ, and v at any given point does not change). The flow is streamline and laminar.

So What????? TKP50 TKP50

Flow Rate Volume of fluid/unit time Rate = Area x speed of fluid R = Av A fluid is pumped into a pipe at a flow rate of 80cm 3 /s. If the diameter of the pipe is 1.4cm, what is the average speed of the fluid at this point?

Making Fluids Flow Faster Pump it faster (no duh!) Pinch the hose or make the pipe smaller

Continuity Equation

A 1 v 1 = A 2 v 2

Example VII The radius of a fluid-carrying pipe at a certain point is 2.0cm and the average speed of the fluid is 14cm/s. What is the average speed of the fluid at a point where the radius is only 1.3cm?

Answer 33cm/s

Warm-Up 4/15/14 Get out your notes and a calculator

Bernoulli’s Principle The pressure in a fluid decreases as the velocity increases This is how lift on an airplane wing works mfCf8 mfCf8

Bernoulli’s Equation

Based upon conservation of Energy Flow rate changes when a pipe changes area (thickness) and elevation. Pressure + PE/volume + KE/volume = constant Sum of pressures before = sum of pressures after

Bernoulli’s Equation P 1 + ½  1 v  gh 1 = P 2 + ½  2 v  gh 2 Uses Pipes moving water throughout a city Water flow into your house Works for all fluids (liquids and gases) moving below the speed of sound

Example VIII Water whose density is 1000kg/m 3 flows through a pipe with a flow rate of 0.80m 3 /s. The cross sectional area of the pipe is initially 0.25m 2 and the pressure is 5.2Pa. The cross sectional area of the pipe becomes 0.40m 2, what is the new pressure of the fluid.

Answer 3100Pa

Example IX Ethyl alcohol (ρ = 810kg/m 3 ) has a flow rate of 0.24m 3 /s through a pipe whose cross sectional area is 0.060m 2 and in which the pressure is 9.5x10 4 Pa. The centerline of the pipes rises 4.0m and the area is reduced to 0.020m 2. What is the fluid pressure in the elevated pipe?

Answer 11000Pa