Physics 1501: Lecture 31 Today ’ s Agenda

Physics 1501: Lecture 31, Pg 1

Physics 1501: Lecture 31Physics 1501: Lecture 31TodayToday’’s Agendas Agenda

Homework #10

(due Friday Nov. 18) Midterm 2: Nov. 16

Students with schedule issues … Honor’s student … Topics: Fluids

Pascal’s Principle (hydraulic lifts etc.)Archimedes’ PrincipleFluid dynamicsBernouilli’s equationExample of applications


FluidsFluids

A

n

Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface.

Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as:

What parameters do we use to describe fluids?Density

Pressure

units :

kg/m3 = 10-3 g/cm3

units : 1 N/m2 = 1 Pa (Pascal)


FluidsFluids

Bulk Modulus

LIQUID: incompressible (density almost constant) GAS: compressible (density depends a lot on pressure)

Bulk modulus (Pa=N/m2)

H2O SteelGas (STP)Pb


When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure:

incompressible fluid

For an incompressible fluid, the density is the same everywhere, but the pressure is NOT!

Pressure vs. DepthIncompressible Fluids (liquids)


PascalPascal’’s Principles Principle

So far we have discovered (using Newton’s Laws):Pressure depends on depth: p = gy

Pascal’s Principle addresses how a change in pressure is transmitted through a fluid.

Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel.

Pascal’s Principle explains the working of hydraulic liftsi.e. the application of a small force at one place can result

in the creation of a large force in another.Does this “hydraulic lever” violate conservation of energy?

» Certainly hope not.. Let’s calculate.



Consider the system shown:A downward force F1 is applied to

the piston of area A1.This force is transmitted through the

liquid to create an upward force F2.Pascal’s Principle says that

increased pressure from F1 (F1/A1) is transmitted throughout the liquid.

F2 > F1 : Have we violated conservation of energy??



Consider F1 moving through a distance d1.

How large is the volume of the liquid displaced?

Therefore the work done by F1 equals the work done by F2

We have NOT obtained “something for nothing”.

This volume determines the displacement of the large piston.


A1 A10

A2 A10

A1 A20

M

M

MdC

dB

dA

A) dA = (1/2)dB B) dA = dB C) dA = 2dB

» If A10 = 2A20, compare dA and dC.

A) dA = (1/2)dC B) dA = dC C) dA = 2dC

Lecture 31, Lecture 31, ACT 1ACT 1HydraulicsHydraulics

Consider the systems shown to the right.In each case, a block of mass M is placed

on the piston of the large cylinder, resulting in a difference dI in the liquid levels.

» If A2 = 2A1, compare dA and dB.


At what depth is the water pressure two atmospheres? It is one atmosphere at the surface. What is the pressure at the bottom of the deepest oceanic trench (about 104 meters)?

P2 = P1 + gd

2.02105 Pa = 1.01105 Pa

+ 103 kg/m3*9.8m/s2*d

d = 10.3 m

P2 = 1.01105 Pa + 103 kg/m3*9.8m/s2*104 m

= 9.81107 Pa = 971 Atm

Solution:

d is the depth.

The pressure increases

one atmosphere for

every 10m.

This assumes that water

is incompressible.

For d = 104 m:

Example Problems (1)Example Problems (1)


Two masses rest on two pistons in a U-shaped tube of water, as shown. If M1 = 3 kg and M2 = 4.5 kg, what is

the height difference, h? The area of piston 1 is A1 =

200 cm2, and the area of piston 2 is A2 = 100 cm2.

Example Problems (2)Example Problems (2)

Solution:

M1

M2h

The pressure difference due to the two masses must be balanced by the water pressure due to h.

Note that g does not appear in the answer.

M1g/A1 + gh = M2g/A2

h = (M2/A2-M1/A1)/

= 0.3 m


Using Fluids to Measure PressureUsing Fluids to Measure Pressure

Barometer

Use Barometer to measure Absolute Pressure

Top of tube evacuated (p=0)

Bottom of tube submerged into pool of mercury open to atmosphere (p=p0)

Pressure dependence on depth:


Using Fluids to Measure PressureUsing Fluids to Measure Pressure

1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H20

p0

h

Manometerp1

Use Manometer to measure Gauge Pressure

Measure pressure of volume (p1) relative to atmospheric

pressure (gauge pressure)

The height difference (h) measures the gauge pressure


ArchimedesArchimedes’’ Principle Principle

Suppose we weigh an object in air (1) and in water (2).How do these weights compare?

W2?W1

W1 < W2 W1 = W2 W1 > W2

Why?

» Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.


The buoyant force is equal to the difference in the pressures times the area.

W2?W1

Archimedes:

The buoyant force is equal to the weight of the liquid displaced.

The buoyant force determines whether an object will sink or float. How does this work?

ArchimedesArchimedes’’ Principle Principle


Sink or Float?Sink or Float?

The buoyant force is equal to the weight of the liquid that is displaced.

If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink.

F mgB

y

We can calculate how much of a floating object will be submerged in the liquid:

Object is in equilibrium


The Tip of the IcebergThe Tip of the Iceberg

What fraction of an iceberg is submerged?

F mgB

y


Lecture 31, Lecture 31, ACT 2ACT 2BuoyancyBuoyancy

A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown.

If you turn the styrofoam+Pb upside down, what happens?

styrofoam

Pb

A) It sinks C)B) D)styrofoam

Pb


ACT 2-AACT 2-AMore Fun With BuoyancyMore Fun With Buoyancy

Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it.

Which cup weighs more?

Cup I Cup II

(A) Cup I (B) Cup II (C) the same (D) can’t tell


ACT 2-BACT 2-BEven More Fun With BuoyancyEven More Fun With Buoyancy

A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (oil < ball < water) is slowly added to the container until it just covers the ball.

Relative to the water level, the ball will: water

oil

(A) move up (B) move down (C) stay in same place


Fluids in MotionFluids in Motion

Up to now we have described fluids in terms of their static properties:

density pressure p

To describe fluid motion, we need something that can describe flow:

velocity v

There are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / ideal


Simplest situation: consider ideal fluid moving with steady flow - velocity at each point in the flow is constant in time

In this case, fluid moves on streamlines

streamline

Ideal FluidsIdeal Fluids

Fluid dynamics is very complicated in general (turbulence, vortices, etc.)

Consider the simplest case first: the Ideal Fluidno “viscosity” - no flow resistance (no internal friction) incompressible - density constant in space and time


Flow obeys continuity equation

volume flow rate Q = A·v is constant along flow tube.

follows from mass conservation if flow is incompressible.

streamline

A1v1 = A2v2

Ideal FluidsIdeal Fluids streamlines do not meet or cross

velocity vector is tangent to streamline

volume of fluid follows a tube of flow bounded by streamlines


Steady Flow of Ideal FluidsSteady Flow of Ideal Fluids(actually laminar flow of real fluid)(actually laminar flow of real fluid)


1) Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe?

Lecture 31 Lecture 31 Act 3Act 3ContinuityContinuity

A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house.

v1 v1/2

a) 2 v1 b) 4 v1 c) 1/2 v1 c) 1/4 v1


Recall the standard work-energy relationApply the principle to a section of flowing fluid with

volume V and mass m = V (here W is work done on fluid)

V

Conservation of Energy for Conservation of Energy for Ideal FluidIdeal Fluid

Bernoulli Equation


Lecture 31 Lecture 31 Act 4Act 4BernoulliBernoulli’’s Principles Principle

A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house.

2) What is the pressure in the 1/2” pipe relative to the 1” pipe?

a) smaller b) same c) larger

v1 v1/2


Some applicationsSome applications Lift for airplane wing

Enhance sport performance

More complex phenomena: ex. turbulence


More applicationsMore applications

Vortices: ex. Hurricanes

And much more …


Bernoulli says: high velocities go with low pressure

Airplane wing shape leads to lower pressure on top of wing faster flow lower pressure lift

» air moves downward at downstream edge wing moves up

Ideal Fluid: Bernoulli ApplicationsIdeal Fluid: Bernoulli Applications


Warning: the explanations in text books are generally over-simplified!

Curve ball (baseball), slice or topspin (golf) ball drags air around (viscosity) air speed near ball higher on bottom lower pressure force sideways acceleration or lift



Bernoulli says: high velocities go with low pressure

“Atomizer” moving air ‘sweeps’ air away from top of tube pressure is lowered inside the tube air pressure inside the jar drives liquid up into tube



In ideal fluids mechanical energy is conserved (Bernoulli) In real fluids, there is dissipation (or conversion to heat) of

mechanical energy due to viscosity (internal friction of fluid)

Real Fluids: ViscosityReal Fluids: Viscosity

Viscosity measures the force required to shear the fluid:

where F is the force required to move a fluid lamina (thin layer) of area A at the speed v when the fluid is in contact with a stationary surface a perpendicular distance y away.

area A


Viscosity arises from particle collisions in the fluid as particles in the top layer

diffuse downward they transfer some of their momentum to lower layers

Real Fluids: ViscosityReal Fluids: Viscosity

Viscosity (Pa-s)

oilair glycerinH2O

area A

lower layers get pulled along (F = p/t)


p+p Qr

L

pR

Because friction is involved, we know that mechanical energy is not being conserved - work is being done by the fluid.

Power is dissipated when viscous fluid flows: P = v·F = Q ·p the velocity of the fluid remains constant power goes into heating the fluid: increasing its entropy

Real Fluids: Viscous FlowReal Fluids: Viscous Flow

How fast can viscous fluid flow through a pipe? Poiseuille’s Law


1) Given that water is viscous, what is the ratio of the flow rates, Q1/Q1/2, in pipes of these sizes if the pressure drop per meter of pipe is the same in the two cases?

Consider again the 1 inch diameter pipe and the 1/2 inch diameter pipe.

a) 3/2 b) 2 c) 4

L/2 L/2

Lecture 31 Lecture 31 Act 5Act 5Viscous flowViscous flow

Physics 1501: Lecture 31 Today ’ s Agenda

Documents

Transcript of Physics 1501: Lecture 31 Today ’ s Agenda