Chapter 10

DensityRecall that the density

of an object is its mass per unit volume (SI unit is kg/m3)

The specific gravity of a substance is its density expressed in g/cm3

V

m

Pressure in FluidsFluids exert a pressure in all directions

A fluid at rest exerts pressure perpendicular to any surface it contacts

The pressure at equal depths within a uniform fluid is the same

Area

Force

A

FP

))()(( depthgdensityghP

SI Unit for Pressure is Pa1 Pa= 1 N/m2

1 atm= 101.3 kPa=760 mm-Hg

Pressure in FluidsGauge Pressure is a measure of the pressure

over and above the atmospheric pressurei.e. the pressure measured by a tire gauge is

gauge pressure. If the tire gauge registers 220 kPa then the absolute pressure is 321 kPa because you have to add the atmosphere pressure (101 kPa)

If you want the absolute pressure at some depth in a fluid then you have to add atmosphere pressure ghPP o

Pressure in FluidsPascal’s Principle: Pressure applied to a fluid

in a closed container is transmitted equally to every point of the fluid and to the walls of the container

2

2

1

1

A

F

A

F

BuoyancyBuoyant force is the force

acting on an object that is immersed in a fluid

Archimedes Principle: The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the objectSince the buoyant force acts

opposite of gravity, an object seems to weigh less in a fluid

Apparent Weight= Fg-FB

Fb = Buoyant Force

Fg = Gravity

Sinking vs FloatingThink back to free body diagrams

If the net external force acting on an object is zero then it will be in equilibrium

If Fb=Fg then the object will be in equilibrium and will FLOAT!

Fg

FB

Cubes floating in a fluid

70% Submerged 100% Submerged20% Submerged

Density determines depth of submersionThis equation gives the

percent of the object’s volume that is submergedVf is the volume of fluid

displacedVo is the total volume of

the objectρo is the density of the

objectρf is the density of the

fluid

f

o

o

f

V

V

Summary of FloatingFor an object to float

FB= Fgρo ≤ ρf

If ρo = ρf then the Object will be

Completely submergedBut not sinking.

If ρo is less than ρf Then the amount

submerged can be found with

f

o

o

f

V

V

Continuity Equation•Continuity tells us that whatever the volume of fluid in a pipe passing a particular point per second, the same volume must pass every other point in a second.

•If the cross-sectional area decreases, then velocity increases

2211

Equation Continuity

vAvA

The quantity Av is the volume rate of flow

Bernoulli’s PrincipleThe pressure in a fluid decreases as the

fluid’s velocity increases.

Fluids in motion have kinetic energy, potential energy and pressure

How do planes fly?

Bernoulli’s EquationThe kinetic energy of a fluid element is:

The potential energy of a fluid element is:

ghVmghPE )(

Bernoulli’s EquationThis equation is essentially a statement of

conservation of energy in a fluid. Notice that volume is missing. This is because this equation is for energy per unit volume.

22221

211 2

1

2

1

Equation sBernoulli'

ghvPghvP

Sample Problem p. 306 #40What is the lift (in newtons) due to

Bernoulli’s principle on a wing of area 80 m2. If the air passes over the top and bottom surfaces at speeds of 350 m/s and 290 m/s, respectively.Let’s make point 1 the top of the wing and

point 2 the bottom of the wingThe height difference between the top of the

wing and the bottom is negligible

22221

211 2

1

2

1ghvPghvP

Sample Problem p.306 #40The net force on the wing is a result of the

difference in pressure between the top and the bottom. P1 is exerted downward, P2 is exerted upward

If we know the difference in pressure we can use that to find the force 2

22211 2

1

2

1vPvP

upward Pa 20318)290340)(/29.1(2

1

2

1

2

1 22322

2112 mkgvvPP

Sample Problem p.306 #40P2-P1=20318 Pa

upward 1063.1)80)(20318()( 6212 NxmPaAPPF

P.306 #43Water at a pressure of 3.8 atm at street level

flows into an office building at a speed of 0.60 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6 cm in diameter by the top floor, 20 m above street level. Calculate the flow velocity and the pressure in such a pipe on the top floor. Ignore viscosity. Pressures are gauge pressures.

Find the flow velocity at the top

A1 is area of first pipe= πr2 = 1.96x10-3 m2

A2 is area of second pipe= πr2 = 5.31x10-4 m2 V1= 0.6 m/s

2211

Equation Continuity

vAvA

s

m

mx

smmx

A

vAv 2.2

1031.5

)/6.0)(1096.1(24

23

2

112

Find pressure at the top

22221

211 2

1

2

1ghvPghvP

222

2112 2

1

2

1ghvvPP

2225

2 19620024201801084.3m

N

m

N

m

NPaxP

P2= 1.86 x 105 Pa= 1.8 atm

Sample Problem p.305 #37What gauge pressure in the water mains is

necessary if a fire hose is to spray water to a height of 12.0 m?Let’s make point 1 as a place in the water main

where the water is not moving and the height is 0

Point 2 is the top of the spray, so v=0 , P= atmospheric pressure, height = 12m

22221

211 2

1

2

1ghvPghvP

Sample Problem p.305 #37

)12)(m/s81.9)((1000kg/m P 23atm21 mghPP

Pax 5atm1 102.1PP Pressure Gauge

Remember that Gauge Pressure is the pressure above atmosphericpressure. So to get gauge pressure, we need to subtract atmosphericPressure from absolute pressure.