chapter 14: fluids - university of florida€¦ · fluids take the form of their container. when...

Chapter 14: FluidsLecture 3011/9/2009

FluidsGoals for this Lecture:

Study the behavior of fluids.

Static fluids:Pressure exerted by a static fluidMethods of measuring pressurePascal’s principleArchimedes’ principle, buoyancy

Real versus ideal Fluids in motion:Fluids Equation of continuityBernoulli’s equation

What is a Fluid?A substance that can flow

Gases (easy to compress)

Liquids (difficult to compress)

Solids ? (gel = toothpaste, glass)?

Basic properties:Fluids take the form of their container.

When shear stress is applied they flow.

Fluids can only exert a force perpendicular to their surface

The molecules in fluids are in constant random motion (Brownian motion).

The field of physics that describes the fluid motion is called “Fluid Mechanics”

Fluid Properties: DensityDensity (sometimes referred to as “mass density”) is the mass of a given material per unit volume.

Unit: ! = !m/!V = [kg]/[m3]

The density can be different at each point of the fluid.

An object is uniform (or homogenous) if the density is constant at any point of the body: ! = !m/!V⇒ mtot/Vtot

A fluid is incompressible if the density does not change as a result of an applied pressure.

! = !m/!V

Examples of Fluid Densities

Fluid Density [kg/m3]

Interstellar space 10-20

Laboratory vacuum 10-17

Air (1 atm 20° C) 1.21

Ice 917

Water 998

Seawater 1024

Earth 5500

Neutron star 10+18

Average person

Fluid Properties: PressurePressure is defined as force (perpendicular to a given surface) per unit area

Pressure is not just what a quarterback feels when Carlos Dunlap is closing in on him

Unit: p = !F/!A = [N]/[m2] ! [Pa] (Pa = Pascal)

Other popular units: 1 atm = 1 bar = 1.01x105 Pa = 760 torr = 14.7 lb/in2

Pressure is a scalar: it has the the same value in all directions!

p = !F/!A

Blaise Pascal

Origin of PressureWhy is pressure defined as force per unit area? Look at the microscopic picture:

There are many atoms bouncing off a surface in a given amount of time. Each one exerts a tiny bit of force on the surface

If the surface area is doubled, the number of collisions doubles

This is also why the pressureat a given point in the fluid is the same in all directions, i.e. a scalar,not a vector

Examples of Fluid Pressures

Location Pressure [Pa]

Sea level atm. pressure 1.013x10+5 = (29.92 in Hg)

Low pressure storm system 0.999x10+5 = (29.5 in Hg)

Hurricane 0.982x10+5 = (29 in Hg)

Bottom of the ocean 1x10+8

Center of the Earth 4x10+11

Center of the Sun 2x10+16

Pressure in Fluids at RestConsider a fluid with density ! in static equilibrium (i.e. not accelerating)

The portion of the fluid that is outlined as several forces acting on it

Downward force from gravity: Fg

Downward force from the pressure of the fluid above it: F1

Upward force from the pressure of the fluid below it: F2

The fluid is not accelerating so Fnet = 0

The fluid is not accelerating so Fnet = 0

-Fg - F1 + F2 = 0

F2 - F1 = mg

Recall that pressure is force per unit area: p = F/A ⇒ F = pA

Newton’s law on this portion of fuid gives us:

F2 - F1 = mg ⇒ p2A - p1A = mg

but, m = !V = !(hA) = !A(y1-y2)

p2A - p1A = !A(y1-y2) ⇒ p2 - p1 = !(y1-y2)g

Let the pressure at the surface (y=0) be p0

p = p0 + !gh (h = depth below the surface)

Pressure in Fluids at Rest

Area = A

F1

F2

Let the pressure at the surface (y=0) be p0

p is the absolute pressure and is given by: p = p0 + !gh (h = depth below the surface)

p - p0 is known as the “gauge pressure” and is given by pg = !gh

Pressure in Fluids at Rest

Area = A

F1

F2

You want to lift a 10 kg bowling ball with a vacuum cleaner hose attachment with a 1” radius. If the (less than excellent) vacuum cleaner can only provide a vacuum pressure of 50 kPa, would you be able to lift the ball?

Vacuum Pickup

Consider the forces on the ballFnet = F2 - F1 -Fg = pair A - pvac A - mg " 0 = A(pair - pvac) - mg " 0 ?A = #r2 = #(1 in)2 = #(0.0254 m)2 = 2x10-3 m2 pvac = 50 kPa, pair = 101.3 kPaA(pair - pvac) = (2x10-3 m2)(51.3 kPa) = 104 N

mg = (10 kg)(9.81 m/s2) = 98.1 N

⇒ A(pair - pvac) - mg " 0 ;)

Vacuum Pickup

Fg = mg

F1 = pvacA

F2 = pairA