13
Chapter 14: Fluids Lecture 30 11/9/2009

Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

  • Upload
    others

  • View
    2

  • Download
    0

Embed Size (px)

Citation preview

Page 1: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

Chapter 14: FluidsLecture 3011/9/2009

Page 2: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 3: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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”

Page 4: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 5: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 6: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 7: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 8: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 9: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 10: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 11: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 12: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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

Page 13: Chapter 14: Fluids - University of Florida€¦ · Fluids take the form of their container. When shear stress is applied they flow. Fluids can only exert a force perpendicular to

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