Fluids Dr. Robert MacKay Clark College. Mass Density, Mass Density,

FluidsFluids

Dr. Robert MacKay

Clark College

Mass Density, Mass Density,

mV

or

m V

Weight Density, DWeight Density, D

D wV

or

w DV

VolumeVolume

Some DensitiesSome DensitiesDensity g/cm3 kg/m3 lb/ft3Air .0012 1.2 .075water 1.00 1000 62.4steel 7.8 7800 487Benzene 0.88 880 55gold 19.3 19300 1204ice 0.92 920 58lead 11.4 11400 705mercury 13.6 13600 850

Specific GravitySpecific Gravity

SGx x

H2O

Dx

DH2O

PressurePressure

Pressure =ForceArea

P =FA

F = P A

Quick Quiz 14.1

Suppose you are standing directly behind someone who steps back and accidentally stomps on your foot with the heel of one shoe. Would you be better off if that person were

(a) a large professional basketball player wearing sneakers

(b) a petite woman wearing spike-heeled shoes?

Answer: (a). Because the basketball player’s weight is distributed over the larger surface area of the shoe, the pressure (F / A) that he applies is relatively small. The woman’s lesser weight is distributed over the very small cross-sectional area of the spiked heel, so the pressure is high.

Quick Quiz 14.1

In about 1657 Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemispheres. Two teams of eight horses each could pull the hemispheres apart only on some trials, and then "with greatest difficulty," with the resulting sound likened to a cannon firing (Fig. P14.62). (a) Show that the force F required to pull the evacuated hemispheres apart is R2(P0 – P), where R is the radius of the hemispheres and P is the pressure inside the hemispheres, which is much less than P0. (b) Determine the force if P = 0.100P0 and R = 0.300 m.

Figure P14.62

Pressure with increasing depthm=V & w=mg

BarometerBarometerPatms=1.013x105 N/m2

=1.013x105 Pa =1.013 bar = 1 Atmosphere =1013 mb = 760 mm Hg = 29.92 in Hg = 760 Torr =14.7 lb/in2

manometermanometer

Pabs=Patm+Pgauge

Pgauge = gh

Quick Quiz 14.2

The pressure at the bottom of a filled glass of water (ρ = 1 000 kg/m3) is P. The water is poured out and the glass is filled with ethyl alcohol (ρ = 806 kg/m3). The pressure at the bottom of the glass is

(a) smaller than P

(b) equal to P

(c) larger than P

(d) indeterminate

Answer: (a). Because both fluids have the same depth, the one with the smaller density (alcohol) will exert the smaller pressure.

Quick Quiz 14.2

Archimedes’ PrincipleArchimedes’ Principle

Active Figure 1409

Archimedes’ PrincipleArchimedes’ Principle

p fg(h2 h1)

FB pA fg(h2 h1)A fVg

FB mfg

The Buoyant force acting on an object submerged in a fluid equals the weight of fluid displaced by the object.

BuoyancyBuoyancy

BuoyancyBuoyancy

400 g

500g

?

obj= ?

V= m/= 100 cm3

f= 1.00 g/cm3

m= 100 g

BuoyancyBuoyancy

400 g

500g

?

obj= 5 g/cm3

V= = m/= 100 cm3

f= 1.00 g/cm3

m= 100 g

BuoyancyBuoyancy

400 g

500g

?

obj=8.0 g /cm3

V= ?

f= ?

m= 100 g

BuoyancyBuoyancy

400 g

500g

?

obj=8.0 g /cm3

V= 500g/8.0 g /cm3

V=62.5 cm3

f= 100g/62.5 cm3

=1.6 g /cm3

m= 100 g

BuoyancyBuoyancy

A 20,000,000 tonn shipFloats in salt water. (1 tonn=1000kg)

How much water does this shipDisplace?

Active Figure 1410

For a physics experiment, you drop three objects of equal mass into a swimming pool. One object is a piece of pine, the second object is a hunk of copper and the third object is a hunk of lead. The relationship between the magnitudes of the buoyant forces on these three objects will be

a) Fcopper > Fpine > Flead, b) Fpine > Fcopper > Flead, c) Flead > Fcopper > Fpine or d) Fcopper > Flead > Fpine.

(end of section 14.4)QUICK QUIZ 14.2QUICK QUIZ 14.2

(b).(b). From Archimedes From Archimedes’’ principle, the magnitude of the buoyant force will be equal to principle, the magnitude of the buoyant force will be equal to the weight of the water displaced. From Table 14.1, lead and copper are more dense the weight of the water displaced. From Table 14.1, lead and copper are more dense than water and will therefore sink, while pine is less dense than water and will than water and will therefore sink, while pine is less dense than water and will therefore float. The buoyant force for the pine must equal the weight, therefore float. The buoyant force for the pine must equal the weight, mgmg, of the pine , of the pine since these two forces balance. For completely submerged objects, the buoyant force since these two forces balance. For completely submerged objects, the buoyant force will be equal to the weight of the water displaced, will be equal to the weight of the water displaced, mmwwgg, and will be less for the denser , and will be less for the denser

lead, because of its smaller volume, than for the copper. In addition, the mass of the lead, because of its smaller volume, than for the copper. In addition, the mass of the water displaced will be less than the mass of the equal volume of metal displacing it, water displaced will be less than the mass of the equal volume of metal displacing it, so that so that mmww < < mm. Therefore, the buoyant force on each metal is less than the buoyant . Therefore, the buoyant force on each metal is less than the buoyant

force on the pine.force on the pine.

QUICK QUIZ 14.2 ANSWERQUICK QUIZ 14.2 ANSWER

BuoyancyBuoyancy

A cubicalblock of wood has30.0 cm sides. IfIts density is 600 kg/m3 how muchof it is submergedwhen floating in water?

BuoyancyBuoyancy

A cubicalblock of wood has30.0 cm sides. IfIts density is 600 kg/m3 how muchmass must be placed on its top to just barely submerge it?

M=?

Quick Quiz 14.6

A glass of water contains a single floating ice cube as in the figure below. When the ice melts, the water level

(a) goes up

(b) goes down

(c) remains the same

Answer: (c). The ice cube displaces a volume of water that has a weight equal to that of the ice cube. When the ice cube melts, it becomes a parcel of water with the same weight and exactly the volume that was displaced by the ice cube before.

Quick Quiz 14.6

A barge loaded with steel floats in a lock. If the steel is then thrown overboard. Does the water level in the lock a) go up b) go down c) stay at the same level?

A barge loaded with steel floats in a lock. If the steel is then thrown overboard. Does the water level in the lock a) go up b) go down c) stay at the same level?

Before H2O displaced =W(boat)+W(steel)

After H2O displaced =W(boat)+little more

Quick Quiz 14.5

An apple is held completely submerged just below the surface of a container of water. The apple is then moved to a deeper point in the water. Compared to the force needed to hold the apple just below the surface, the force needed to hold it at a deeper point is

(a) larger

(b) the same

(c) smaller

(d) impossible to determine

Answer: (b). For a totally submerged object, the buoyant force does not depend on the depth in an incompressible fluid.

Quick Quiz 14.5

ContinuityContinuity

Equation of Equation of ContinuityContinuity

A1x1 A2x2

A1x1

tA2

x2

tA1v1 A2v2

A1x1 A2x2

A1x1

tA2

x2

tA1v1 A2v2

Volume rate of flowVolume rate of flow

Q Volume

tA

xtAv

A r2 4

D2

Water flows out of a 1.0 inch hose at a speed of 2.0 ft/s. How long will it take this hose to fill a 50 gallon drum?

Quick Quiz 14.8

You tape two different sodas straws together end-to-end to make a longer straw with no leaks. The two straws have radii of 3 mm and 5 mm. You drink a soda through your combination straw. In which straw is the speed of the liquid the highest?

(a) whichever one is nearest your mouth

(b) the one of radius 3 mm

(c) the one of radius 5 mm

(d) Neither – the speed is the same in both straws.

Answer: (b). The liquid moves at the highest speed in the straw with the smaller cross sectional area.

Quick Quiz 14.8

ExampleExample

A 1.0 inch diameter hose is connected to a 0.25 in diameter nozzle. If the water shoots up in the air to a maximum height of 15.0 m, what is the flow rate (gallons/min) in the hose? (231 in3=1 gallon)

You would like to change the opening on the nozzle You would like to change the opening on the nozzle of a fire hose so that the water exiting the hose can of a fire hose so that the water exiting the hose can reach a height that is four times the present reach a height that is four times the present maximum height the water can reach. To do this, maximum height the water can reach. To do this, you should decrease the cross sectional area of the you should decrease the cross sectional area of the opening by a factor of opening by a factor of a) 16, a) 16, b) 8, b) 8, c) 4 c) 4 d) 2.d) 2.

(end of section 14.5)QUICK QUIZ 14.5QUICK QUIZ 14.5

(d). From the continuity equation, the velocity of the water exiting the hose is inversely proportional to the cross sectional area or v 1/A. However, the kinetic energy of the water that exits the hose will be equal to the potential energy of the water at its maximum height (when you point the hose straight up), or

So to increase the height by a factor of four, you must decrease the area by a factor of 2.

1 2 2 2 12

1, so that ( ) , or .

hmv mgh h v A

A

QUICK QUIZ 14.5 ANSWERQUICK QUIZ 14.5 ANSWER

Bernoulli’s PrincipleBernoulli’s Principle

Where the fluid speed is high the internal pressure is low. [based on the conservation of energy]


http://www.aerospaceweb.org/question/aerodynamics/


y1

y2

F2=P2A2

F1=P1A1

v1 A1

A2

v2∆m2

∆x2 =v2 ∆t

∆x1 =v1∆t

∆m1

∆m1 = ∆m2


P1 12v1

2 gh1 P2 12v2

2 gh2

V1 = 0.0

h2 = 0.0 , P2=P1

h1 = h

0.00.0


P1 12v1

2 gh1 P2 12v2

2 gh2

V1 = 0.0

h2 = 0.0 , P2=P1

h1 = h

0.00.0

gh 12v2

2

2gh v2

v 2gh


gh 12v2

2

2gh v2

v 2gh


P1 12v1

2 gh1 P2 12v2

2 gh2


P1 12v1

2 gh1 P2 12v2

2 gh2

(P1-P2)A=Lift

Surface TensionSurface Tension

L = F Water , T0.076 N/m 0°C0.072 N/m 20°C0.059 N/m 100 °C

Capillary ActionCapillary Action

2 r = r2 h gr

h

ViscosityViscosity

F Av

h

ViscosityViscosity

Viscosity (Pa s) Fluid0.0018 0°C water0.0010 20°C water0.0003 100 °C water0.0040 37°C blood

Poiseuille’s LawPoiseuille’s Law

Q r2p

8L

Stoke’s Law, terminal velocity Stoke’s Law, terminal velocity of a sphereof a sphere

Fluids Dr. Robert MacKay Clark College. Mass Density, Mass Density,

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