Chapter 11

Fluids

Density and Specific Gravity

The density ρ of an object is its mass per unit volume:

The SI unit for density is kg/m3. Density is also sometimes given in g/cm3; to convert g/cm3 to kg/m3, multiply by 1000.

Water at 4°C has a density of 1 g/cm3 = 1000 kg/m3.

The specific gravity of a substance is the ratio of its density to that of water.

(10-1)

Pressure in FluidsPressure is defined as the force per unit area.

Pressure is a scalar; the units of pressure in the SI system are pascals:

1 Pa = 1 N/m2

Pressure is the same in every direction in a fluid at a given depth; if it were not, the fluid would flow.

A

FP

Atmospheric Pressure and Gauge Pressure

•At sea level the atmospheric pressure is about

• this is called one atmosphere (atm).

• 1 atm = 14.7psi = 760 torr = 760mm Hg

•Another unit of pressure is the bar:

•Standard atmospheric pressure is just over 1 bar.•This pressure does not crush us, as our cells maintain an internal pressure that balances it.

Pressure at a Depth P = F = mg = ρVg = ρAhg = ρhg A A A A

330 m

Find the pressure on the bottom of the submarine due to the waterabove it.

Find the pressure on the bottom of the submarine due to the waterAnd air above it.

Pa150,416,3P101,300PaPa850,314,3P

Pa850,314,3)s/8.9)(m330)(kg/m1025(

T

T

2 2

PmhgP

10-4 Atmospheric Pressure and Gauge Pressure

Most pressure gauges measure the pressure above the atmospheric pressure – this is called the gauge pressure.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

gaugeatmabsolute PPP

Archimedes’s Principle

Any object completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the fluid displaced by the object.

Buoyant Force

• The upward force is called the buoyant force

• The physical cause of the buoyant force is the pressure difference between the top and the bottom of the object

11.6 Archimedes’ Principle

APPAPAPFB 1212

ghPP 12

ghAFB

hAV

gVFB

fluiddisplaced

of mass

Buoyant Force, cont.

• The magnitude of the buoyant force always equals the weight of the displaced fluid

• For a floating object, the buoyancy force must equal the weight of the object.

B = W• Some objects may float high or low

fluidobjectfluid wgVB

Example: A balloon having a volume of 1.5 cubic meters is filled with ethyl alcohol and is tethered to the bottom of a swimming pool. Calculate the tension in the cord tethering it to the bottom of the swimming pool.

Fy = B – T – W = 0

Therefore, T = B – W

T = waterVg – alcoholVg

T = (water – alcohol )Vg

T = (1000kg/m3 – 806kg/m3) (1.5m3)(9.8m/s2)

T = 2851.8 N

B

T W

FBD

11.6 Archimedes’ Principle

Example 9 A Swimming Raft

The raft is made of solid squarepinewood. Determine whetherthe raft floats in water and ifso, how much of the raft is beneaththe surface.

11.6 Archimedes’ Principle

N 47000

sm80.9m8.4mkg1000 233

max

gVVgF waterwaterB

m 8.4m 30.0m 0.4m 0.4 raftV

11.6 Archimedes’ Principle

N 47000N 26000

sm80.9m8.4mkg550 233

gVgmW raftpineraftraft

The raft floats!

11.6 Archimedes’ Principle

gVwaterwaterN 26000

Braft FW

If the raft is floating:

23 sm80.9m 0.4m 0.4mkg1000N 26000 h

m 17.0sm80.9m 0.4m 0.4mkg1000

N 2600023

h

11.5 Pascal’s Principle

PASCAL’S PRINCIPLE

Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls.

11.5 Pascal’s Principle

m 012 gPP

1

1

2

2

A

F

A

F

1

212 A

AFF

11.5 Pascal’s Principle

Example 7 A Car Lift

The input piston has a radius of 0.0120 mand the output plunger has a radius of 0.150 m.

The combined weight of the car and the plunger is 20500 N. Suppose that the inputpiston has a negligible weight and the bottomsurfaces of the piston and plunger are atthe same level. What is the required inputforce?

11.5 Pascal’s Principle

N 131m 150.0

m 0120.0N 20500 2

2

1

F

2

121 A

AFF

11.8 The Equation of Continuity

The mass of fluid per second that flows through a tube is called the mass flow rate.

11.8 The Equation of Continuity

2222 vAt

m

111

1 vAt

m

distance

tvAVm

11.8 The Equation of Continuity

222111 vAvA

EQUATION OF CONTINUITY

The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow.

SI Unit of Mass Flow Rate: kg/s

11.8 The Equation of Continuity

Incompressible fluid:

2211 vAvA

Volume flow rate Q:

AvQ

11.8 The Equation of Continuity

Example 12 A Garden Hose

A garden hose has an unobstructed openingwith a cross sectional area of 2.85x10-4m2. It fills a bucket with a volume of 8.00x10-3m3

in 30 seconds.

Find the speed of the water that leaves the hosethrough (a) the unobstructed opening and (b) an obstructedopening with half as much area.

11.8 The Equation of Continuity

AvQ

sm936.0

m102.85

s 30.0m1000.824-

33

A

Qv

(a)

(b) 2211 vAvA

sm87.1sm936.0212

12 v

A

Av

Bernoulli’s Equation

• Relates pressure to fluid speed and elevation

• Bernoulli’s equation is a consequence of Conservation of Energy applied to an ideal fluid

• Assumes the fluid is incompressible and nonviscous, and flows in a nonturbulent, steady-state manner

Bernoulli’s Equation, cont.

• States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline

constant gyvP 2

2

1

Example: Water is contained in a tank. The water level is 5 meters above a hole in the tank where water exits through a hole to the atmosphere as shown below. The diameter of the hole is 0.01 meters, and the diameter of the tank is 3 meters. The tank is open to atmosphere. Calculate the velocity of the water exiting the tank.

eeettt gyvPgyvP

gyvP

22

2

2

1

2

1

2

1

:Therefore

constant

•The pressure at the top of the tank is equal to the pressure at the exit since they are both open to atmosphere. Therefore Pt and Pe cancel out.•Since the tank diameter is so much larger than the exit hole, the velocity of the water level drop can be approximated as zero.•The exit height is taken as zero, therefore ye = 0

The equation above reduces to:et

et

vgy

Therefore

vgy

2

:2

1 2

Where yt=hve= 9.9m/s

How Airplanes Wings Work