15fluid.pdf

2/27/2012

1

Fluids and Elasticity

Chapter 15

Density ( )

= mass/volume

Rho ( ) Greek letter for density

Units - kg/m3

Specific Gravity = Density of substance

Density of water (4oC)

Unitless ratio

Ex: Lead has a sp. Gravity of 11.3 (11.3 times

denser than water

Ex: 1

Estimate the mass of air in this classroom

Pressure

Force per unit area

P = F/A

Unit - N/m2 (Pascal)

The larger the area, the less the pressure

Shoeshoes

Elephant feet

Bed of nails

Fluid Pressure

A fluid exerts the same pressure in all directions

at a given depth

P = Po + gh

The atmosphere is a fluid

Po often 1 atm (101.3 kPa)

2/27/2012

2

Pressure: Example 1

A water storage tank is 30 m above the water faucet

in a house. Calculate the pressure at the faucet:

We will neglect the atmospheric pressure since it is

the same at the tank and at the surface

P = gh = (1000 kg/m3)(9.8 m/s2)(30 m)

P = 29,000 kgm2/m3s2 = 29,000 kg m/s2m2

P = 29,000 N/m2

Pressure: Example 2 The Kraken can live at a depth of 200 m. Calculate

the pressure the creature can withstand (neglect

atmospheric pressure)

Pressure: Example 2

P = gh = (1000 kg/m3)(g)(200 m)

P = 1.96 X 106 N/m2

Atmospheric Pressure

1 atm = 1.013 X 105 N/m2 = 101.3 kPa

1 bar = 1 X 105 N/m2 (used by meteorologists)

Gauge pressure

P = Patm + PG

Absolute pressure atmospheric pressure Gauge pressure

We usually measure gauge pressure

Ex: A tire gauge reads 220 kPa, what is the

absolute pressure?

P = Patm + PG

P = 101.3 kPa + 220 kPa = 321 ka

Straw Example

You can pick up soda in a

straw using your finger. Why

doesnt the soda fall out?

2/27/2012

3

Another Straw Example

What pushes soda up a straw when you drink

through it?

Torrecillis Work

P = Patm + PG

P = Patm + gh

What is the highest column of water

that the atmosphere can support?

P = Patm + gh

0 = 1.013 X 105 N/m2 + (1000kg/m3)(9.8m/s2)(h)

h = 10.3 m

No vacuum pump can pump more than ~30 feet

Try the same calculation with mercury

P = Patm + gh

0 = 1.013 X 105 N/m2 + (13,600kg/m3)(9.8m/s2)(h)

h = 0.760 m (760 mm)

1 atm = 760 mm Hg (760 torr)

Can an astronaut attach suction cups to the boots of

his spacesuit to help him climb around the space

shuttle while in space?

2/27/2012

4

Pascals Principle

Pressure applied to a confined fluid increases the

pressure the same throughout

Pin = Pout

Fin = Fout

Ain Aout

for pistons:

F = g(A1 + A2)d2

Pascals Principle: Ex 1

A hydraulic lift can produce 200 lb of force. How

heavy a car can be lifted if the area of the lift is

20 times larger that the input of the Force?

Fin = Fout

Ain Aout

Fout = Fin Aout = (200 lb) (20) = 4000 lbs

Ain 1

Pascals Principle: Ex 1

A hydraulic lift has the car rest ona 25 cm pipe.

The lift the car, compressed air pushes on a 6.0

cm pipe.

a. Calculate the force needed to lift a 1300 kg car.

b. Calculate how much the air pressure force must

be increased to lift the car 2.0 m.

Buoyancy

Buoyancy

The lift provided by water

Objects weight less in water than out

Caused by pressure differential between top and

bottom of an object.

Fbouyant = gV

Derivation of the Buoyancy Formula

Fb = F2 F1

P = F/A

F = PA

F = ghA

Fb = gh2A gh1A

Fb = gA(h2 - h1)

Fb = gV

Archimedes Principle

The bouyant force on an

object equals the weight

of fluid displaced by the

object

w = weight of an object in

water (or any liquid)

w = mg - Fb

2/27/2012

5

Buoyancy: Example 1

A 7000-kg ancient statue lies at the bottom of the

sea. Its volume is 3.0 m3. How much force is

needed to lift it?

Fb = gV

Fb = (1000 kg/m3)(9.8 m/s2)(3.0m3)

Fb = 2.94 X 104 kg-m/s2

Fb = 2.94 X 104 N mg

Fb

w = mg - Fb

w = (7000 kg)(9.8m/s2) - 2.94 X 104 N

w = 3.92 X 104 N

Say, isnt w just the sum of the forces?

Yep.

F = w mg

Fb

Buoyancy: Example 2

Archimedes tested a crown for the king. Out of

water, it masses 14.7 kg. In water, it massed 13.4

kg. Was the crown gold?

w = mcrg Fb

w = mcrg gVcr

(13.4 kg)(g) = (14.7 kg)(g) (1000 kg/m3)(g)(Vcr)

131 N = 144 N (9800 kg/ms2)(Vcr)

Vcr = 0.00133 m3

Now we can calculate the density of the crown:

= m/V = 14.7 kg/0.00133 m3

= 11,053 kg/m3

Golds density is about 19,000kg/m3. This is much

closer to lead.

Example 3

A cube of wood that is 10 cm on a side is held

underwater by tying a string to the cube and the

bottom on the container. The wood has a density

of 700 kg/m3.

a. Draw a free body diagram showing all the forces

on the block.

b. Calculate the force of bouyancy

c. Calculate the tension in the string.

Floating Objects that are less dense than water will float

Part of the object will be above the water line

A case of static equilibrium

F = 0

mg

Fb

2/27/2012

6

Floating: Example 1

A 1200 kg log is floating in water. What volume of

the log is under water?

F = 0

F = 0 = mg Fb

mg = Fb

mg = gVlog

Vlog = mg

g mg

Fb

Vlog = m (Hey, the gs cancel!)

Vlog = 1200 kg = 1.2 m3

1000 kg/m3

Floating: Example 1

A wooden raft has a density of 600 kg/m3, an area

of 5.7 m2, and a volume of 0.60 m3. How much

of the raft is below water in a freshwater lake?

Lets first calculate the mass of the raft:

= m/V

m = V = (600 kg/m3)(0.60 m3) = 360 kg

Now we can worry about the raft.

F = 0

F = 0 = mg Fb

mg = Fb

mg = gVsubmerged

mg = ghsubmergedA

mg = ghsubmergedA

m = hsubmergedA (Hey, the gs cancelled!)

hsubmerged = m/ A

hsubmerged = 360 kg = 0.063 m

(1000 kg/m3)(5.7 m2)

Floating: Example 3

Suppose a continent is floating on the mantle rock.

Estimate the height of the continent above the

mantle (assume the continent is 35 km thick).

2/27/2012

7

F = 0 = mg Fb

0 = mcg mangVc(submerged)

mcg = mangVc(submerged)

mc = manVc(submerged)

We dont know the mass of the continent

c = mc/Vc(total)

mc = cVc(total)



mc = cVc(total)

manVc(submerged) = cVc(total)

Vc(submerged) = c = (2800 kg/m3) = 0.85

Vc(total) man (3300 kg/m3)

This means that 85% of the continent is submerged, and only 15% is above:

(0.15)(35 km) = 5.25 km

Floating: Ex 4

A block is placed in water and 5.8 cm is

submerged. The same block is placed in an

unknown liquid and 4.6 cm is submerged.

Calculate the density of the unknown liquid.

Assume the same face of the block pointed

downward in both cases (A).

Fluid Flow

Laminar Flow Smooth, streamline flow

(laminar means in layers)

Turbulent Flow erratic flow with eddies

Viscosity Internal friction of a liquid

High viscosity = slow flow

Viscosity is NOT the same as density

Equation of Continuity

A1v1 = A2v2

A = Area of a pipe

v = velocity of the liquid

2/27/2012

8

Equation of Continuity

v1A1 = v2A2

Fluid will flow faster through a smaller opening

Placing your finger over a hose opening.

The term vA is the volume rate of flow

A = m2

v = m/s

vA = m3/s

Q = vA

Eqn. Of Continuity: Example 1

A garden hose has a radius of 1.00 cm and the

water flows at a speed of 0.80 m/s. What will be

the velocity if you place your finger over the hose

and narrow the radius to 0.10 cm?

A1 = r2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2

A2 = r2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2

A1v1 = A2v2

v2 = A1v1

A2

v2 = (3.14 X 10-4 m2)(0.80 m/s) = 80 m/s

(3.14 X 10-6 m2)


A water hose 1.00 cm in radius fills a 20.0-liter

bucket in one minute. What is the speed of water

in the hose?

A1 = r2 = (3.14)(1 cm)2 = 3.14 cm2

Remember that Av is volume rate of flow.

A2v2=20.0 L 1 min 1000 cm3 = 333 cm3/s

1 min 60 s 1 L

A1v1 = A2v2

v1 = A2v2/A1

v1 = 333 cm3/s = 160 cm/s or 1.60 m/s

3.14 cm2

2/27/2012

9


A sink has an area of about 0.25 m2. The drain has

a diameter of 5 cm. If the sink drains at 0.03 m/s,

how fast is water flowing down the drain?

Ad = r2 = ( )(0.025 m)2 = 1.96 X 10-3 m3

Advd = Asvs

vd = Asvs/Ad=[(0.25 m2)(0.03 m/s)]/(1.96 X 10-3 m3)

vd = 3.82 m/s


The radius of the aorta is about 1.0 cm and blood

passes through it at a speed of 30 cm/s. A typical

capillary has a radius of about 4 X 10-4 cm and

blood flows through it at a rate of 5 X 10-4 m/s.

Estimate how many capillaries there are in the

human body.

Aava = NAcvc (N is the number of capillaries)

Aa = r2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2

Ac = r2 = (3.14)(4 X 10-6 cm)2 = 5.0 X 10-11 cm2

N = Aava/ Acvc

N = (3.14 X 10-4 m2)(0.30 m/s) = ~ 4 billion

(5.0 X 10-11 cm2)(5 X 10-4 m/s)


How large must a heating duct be to replenish the

air in a room 300 m3 every 15 minutes? Assume

air moves through the vent at 3.0 m/s.

Advd = Arvr

Arvr = volume rate of flow:

Arvr = 300 m3 1 min = 0.333 m3/s

15 min 60 s

Ad = Arvr/vd

Ad = 0.333 m3/s = 0.11 m2

3.0 m/s

Bernoullis Principle

The velocity and

pressure of a fluid are

inversely related.

2/27/2012

10

Why does a shower curtain sometimes attack a

person taking a shower?

What will happen to closed windows during a

tornado? Will they blow in or out?

Applications of Bernoullis Principle

1. Airplane wing


2. Prairie Dog Burrows

1. Air moves faster (lower pressure) at the top

2. Draws air through the burrow

3. The exact same thing happens with our chimneys


3. Spray Paint Flow of air (low pressure)


4. Dime in a cup

Bernoullis Equation

Pt + vt2 + gyt = Pb + vb

2 + gyb

(note: you often have to use the Eqn. Of Continuity

in these situations:)

A1v1 = A2v2

2/27/2012

11

Often useful when you have both a change in

height and area:

Pipe from a water

reservoir to a house Pipe from a house into

a sewer pipe

If there is no change in altitude, the equation

simplifies:


2 + gyb

Pt + vt2 = Pb + vb

2

Bernoullis Equation: Example 1

A water heater in the basement of a house pumps

water through a 4.0 cm pipe at 0.50 m/s and 3.0

atm. What will be the flow speed and pressure

through a 2.6 cm spigot on the second floor, 5.0

m above?

3.0 atm 1.013 X 105 N/m2 = 3.0 X 105 N/m2

1 atm

Flow speed:

Atvt = Abvb

vt = Abvb/At (Remember A = r2)

vt = r2bvb (Hey, the s cancel!)

r2t

vt = r2

bvb = (0.02 m)2(0.50m/s) = 1.2 m/s

r2t (0.013 m)

2

Now the pressure:


2 + gyb

Pt + (1000)(1.2)2 + (1000)(9.8)(5) = 3.0X105 +

(1000)(0.50)2 + (1000)(9.8)(0)

Pt = 2.5 X 105 N/m2

Bernoullis Equation: Example 2

A drunken redneck shoots a hole in the bottom of

an aboveground swimming pool. The hole is 1.5

m from the top of the tank. Calculate the speed

of the water as it comes out of the hole.

yt = 1.5 m

yb = 0 m

2/27/2012

12

The top of the pool is a much larger area than the

hole. We will assume that the vt = 0.


2 + gyb

Pt + gyt = Pb + vb2 + gyb

Also, both the top and the hole are open to the

atmosphere, so Pt = Pb

Pt + gyt = Pb + vb2 + gyb

gyt = vb2 + gyb

Set the bottom of the pool as yb = 0.

gyt = vb2 + gyb

gyt = vb2

vb2 = 2 gyt/

vb2 = 2gyt

vb2 = (2)(9.8m/s2)(1.5 m)

vb = 5.42 m/s

Example 3

A hydroelectric dam is 200 m above the power

plant. The inlet hose at the top has a diameter of

100 cm, and the outlet hose to the turbine has a

diameter of 50 cm.

a. Calculate the speed of the water into the turbine

(both are open to the atmosphere)

Elasticity: Hookes Law

Hookes Law usually

used with a spring

Can consider anything to be

like a spring

F = k L (F=kxspring)

k = proportionality (spring)

constant

Cant stretch things forever

Elastic region material will still bounce back

Plastic region material will not return to original

length (but has not broken)

F = k L

This is only linear

in the proportional

region

HOPE I DONT REACH MY

ELASTIC LIMIT!

2/27/2012

13

Elastic Region: Youngs

Modulus(E) Stress = Force = F

Area A

Strain = Change in length = L

Original length Lo

Y = stress

strain

Y = F/A or F = Y L

L/Lo A Lo

Youngs Modulus: Example 1

A 1.60 m long steel piano wire has a diameter of

0.20 cm. How great is the tension in the wire if it

stretches 0.30 cm when tightened?

A = r2 = (0.0010 m)2 = 3.1 X 10-6 m2

F = Y L

A Lo

F = AY L

Lo

F = AY L

Lo

F = (3.1 X 10-6m2)(200 X 1011N/m2)(0.0030 m)

1.60 m

F = 1200 N


A steel support rod of radius 9.5 mm and length 81

cm is stretched by a force of 6.2 X 104 N (about 7

tons).

a. Calculate the stress.

b. Calculate the elongation.

Area = r2 = (0.0095 m2

Stress = Force = 6.2 X 10-4m = 2.2 X 108 N/m2

Area 2.84 X 10-4 m2

F = Y L

A Lo

L = FL = (6.2 X 104 N)(0.81m)

YA (200 X 109 N/m2)(2.84 X 10-4 m2)

L = 8.84 X 10-4 m = 0.89 mm

2/27/2012

14


A 2.0 m long, 1.0 mm diameter wire is suspended.

Hanging a 4.5 kg mass stretches the wire length

by 1.00 mm.

a. Calculate Youngs modulus

b. Identify the material from the table.

The Three Types of Stress

Stretching Squeezing Horizontal

Other Modulus

Shear Modulus Used for shear stress

Bulk Modulus Used for even compression on all

sides (an object when submerged)

Fracture

Breaking Point

Uses

Tensile Strength Stretching

Compressive Strength under a load

Shear Stress Shearing

Safety Factor reciprocal that is multiplied by

the tensile strength

Ex: A safety factor of 3 means you will only use

1/3 of the maximum stress

Fracture: Example 1

A concrete column 5 m tall will have to support 1.2

X 105 N (compression). What area must it have

to have a safety factor of 6?

Max stress = (1/6)(compressive strength)

Max stress = (1/6)(20 X 106 N/m2)= 3.3X106 N/m2

Stress = F

A

2/27/2012

15

Stress = F

A

Area = F = (1.2 X 105 N) = 3.64 X 10-2 m

Stress 3.3 X 106N/m2

How much will the column compress under the

load?

F = E L

A Lo

L = FL = (1.2 X 105 N)(5 m)

EA (20 X 109 N/m2)(3.64 X 10-2 m2)

L = 8.3 X 10-4 m = 0.83 mm

Fracture: Example 2

Spider-mans webbing has

a tensile strength of 600

X 106 N/m2 and he

wishes to use a safety

factor of 3. What is the

diameter of the webbing

if the maximum force at

the bottom of a swing is

1500 N?

Maximum Stress

(1/3)(600 X 106 N/m2) = 200 X 106 N/m2

Stress = Force

Area

Area = Force = 1500 N = 7.5 X 10-6m2

Stress 200X106N/m2

Area = r2

r = (A/ )1/2 = 1.55 X 10-3 m or 1.55 mm

Diameter = 3.10 mm

Concrete

Concrete is much stronger under compression

than tension

Tensile Strength 2 X 106N/m2

Compressive Strength 20 X 106N/m2

Prestressed concrete rods or mesh are stretched

when the concrete is poured. Released after

concrete dries.

Now under compression

15fluid.pdf

Documents

nm2 pressure

pressure differential

nm2 atmospheric pressure

gauge pressure ex

air pressure force

pascals principle pressure

classroom pressure force

patm pg p