PHYSICS OF FLUIDS

Fluids

Includes liquids and gasesLiquid has no fixed shape but nearly fixed volumeGas has neither fixed shape or volumeBoth can flow

Density Mass per unit volume

= m/V (Greek letter “rho”)

m = V Example: Density of Mercury is 13.6 x

103 kg/m3 What is the mass of one liter?

M = V = 13.6 x 103 kg/m3 x 10-3 m3 = 13.6 kg

Specific Gravity is ratio of its density to that of water

(1.00 x 103 kg/m3 = 1.00 g/cm3)

Density of water

1000 Kg per m3

1 Kg/liter1 gram /cubic centimeter (cc)

1 cubic meter = 1000 liters1 cubic centimeter = 1 milliliter1 cubic meter = 1,000,000 cc

Table of DensitiesSubstance Density kg/m3

x103

aluminum 2.7

Iron 7.8

Lead 11.3

Gold 19.3

Water 1.0

Sea water 1.025

Alcohol, ethyl 0.79

Mercury 13.6

Ice 0.917

Pressure in FluidsForce per unit areaPressure = P = F/AUnit N/m2= Pascal(Pa)Exerted in all directionsForce due to pressure is perpendicular to surface in a fluid at rest

Pressure Varies with Depth as

Let depth be h. Assume incompressible.Force acting on area is mg = VgAhgP = F/A = ghPressure at equal depths is the sameIf external pressure is also present it must be added

P = g h

gh

Ah

Example: Pressure at Bottom of a Lake

• What is the pressure at the bottom of a 20.0 meter deep lake?

P = gh = 1.0 x 103 kg/m3 x 9.8 m/s2 x 20m =

1.96 x 105 N/m2 due to the water

What about the atmosphere pressing down on the lake?

Atmospheric, Gauge and Absolute Pressure

Average sea level pressure is 1 atm = 1.013 x 105 N/m2 (14.7 lbs/sq inch)Pressure gauges read pressure above atmosphericAbsolute (total) pressure is gauge + atmospheric P = PA + PGWhat is total pressure at bottom of lake?

Add 1.01 x 105 N/m3 to 1.96 x 105

Example: Water in a Straw

Finger holds water in strawHow does pressure above water

compare with atmospheric? (hint:atmospheric pushes up from below)

Pressure less because it plus weight of water must balance atmospheric

What is the tallest column of water that could be trapped like this?

gH = 101,300 Pa; H = 101,300/(9.8 N/kg x 1000 Kg/m3) =10.3 m

Pascal’s Principle

Probably not testedPressure applied to a confined fluid increases pressure throughout by the same amountPout = Pin

Fout/Aout = Fin/Ain

Fout/Fin = Aout/Ain

Multiplies force by ratio of areasPrinciple of hydraulic jack and lift

Diagram courtesy Caduceus MCAT Review

Pascal’s Principle in Action

Measuring Pressure

Open tube manometer simulation

               

        

The pressure difference is gh

The (greater) pressure

P2 = P1 + gh

Diagram courtesy Sensorsmag.com

How would this look if P1 was greater than P2 ?

BuoyancySubmerged or partly submerged object experiences an upward force called buoyancyPressure in fluid increases with depthStudied by Archimedes over 2000 years ago.

Buoyant Force on Cylinder

FB = F2 – F1 = P2A – P1A

= FgA(h2 – h1)

= FgAh

= FgV = mFg

Buoyant Force equalsweight of fluid displaced

F2

F1h2

h1

A

h = h2 – h1

h

Archimedes Principle

Buoyant force on a body immersed (or partly immersed) in fluid equals weight of fluid displaced.Argument in general: consider immersed body in equilibrium of any shape with same density as fluid. FB up must equal weight down. Replacing body by one with different density does not alter configuration of fluid so conclusion would not change.

fy = 0

T +B - W = 0

T = W – B

T = W – FVg

T is apparent weight W’

Diagram courtesy Caduceus MCAT Review

Weighing Submerged object

Example: King’s CrownGiven crown mass 14.7 kg but weighed under water only 13.4 kg. Is it gold?W’ = W – FB W = ogV

W - W’ = FB = F gV

W/(W-W’) = ogV / F gV = o / F

o / H2O = W/(W-W’) = 14.7/(14.7 – 13.4) = 14.7/1.3 = 11.3 LEAD

Another way to solve: isolate o = W/gV and then get V from FB /fg

Floating Objects

                                                                   

Objects float if density less than that of fluid.

FB = W at equilibrium

FVdisp g = Vg

Vdisp / Vo = o /F

Vo

Vdisp

Example: If an object’s density is 80% of the density of the surrounding fluid, 80% of it will be submerged

Example: Floating Log

15 % of a log floats above the surface of the ocean. What is the density of the wood?

Vdisp / Vo = o /F

o = Vdisp / Vo x F = 0.85 x 1.025 x 103 kg/m3

= 0.87125 = 0.87 x 103 kg/m3

Example: Lifted by Balloon

What volume of helium is needed to lift a 60 Kg student?FB = (mHe + 60 kg)g

air Vg = (He V + 60 kg) g

V = 60 kg/(air – He) = 60 kg/(1.29 – 0.18kg/m3)

= 54 m3

Fluid Flow

Equation of ContinuityVolume rate of flow is constant for incompressible fluids (not turbulent)A1v1 = A2 v2

v is velocity

Laminar vs. Turbulent Flow

Fluid follows smooth path

Erratic, contains eddies

Courtesy MIT Media Laboratory

Example: Narrows in a River

A river narrows from 1000m wide to 100m wide with the depth staying constant. The river flows at 1.0 m/s when wide. How fast must it flow when narrow?

10 m/s

Example: Heating Duct

What must be the cross sectional area of a heating duct carrying air at 3.0 m/s to change the air in a 300 m3 room every 15 minutes?A1v1 = A2v2 = A2l2/t = V2/t

A1 = V2/ v1t = 300m3 /(3.0 m/s x 900s) = 0.11m2

A2A1

l2

Bernoulli’s EquationWhere velocity of fluid is high, pressure is low; where velocity is low, pressure is highConsequence of energy conservation

P + ½ v2 + gy = constant for all points in the flow of a fluidP + ½ v2 = constant if all on same level

Question

If A1 is six times A2 how will the pressure in the narrow section compare with than in the wide section?

Hint: P + ½ v2 = constant

Speed of Water Flowing Through Hole in Bucket

P1 = P2 + ½ v2 + gz

P1 = P2 since both open to air

½ v22 + gz = 0

v2 = (2gz)1/2

Torricelli’s Theorem

Speed and Pressure In Hot Water Heating System If water pumped at 0.50 m/s through

4.0 cm diameter pipe in basement under 3.0 atm pressure, what will be flow speed and pressure in 2.6 cm diameter pipe 5.0m above?

1) find flow speed using continuity A1v1 = A2 v2

v2 = v1A1/A2 = v1r12/r2

2 =

1.2 m/s

2) Use Bernoulli’s Eq. to find pressure

P1 + ½ v12 + gy1 = P2 + ½ v2

2 + gy2

P2 = P1 +g(y1 – y2) + 1/2(v12 –v2

2)

=(3.0 x 105 N/m2) + (1.0 x 103 kg/m3)x (9.8 m/s2)(-5.0m) + ½ (1.0 x 103

kg/m3)[(0.50 m/s)2 – (1.18m/s)2] = = 2.5 x 105 N/m2

No Change in Height

P1 + ½ v12 = P2 + ½ v2

2

Where speed is high, pressure is lowWhere speed is low, pressure is high

Why Curveballs Curve

Courtesy Boston University Physics Dept. web site

How an Airfoil Provides Lift

Courtesy The Aviation Group

Where is the pressure greater, less?

Crowding of streamlines indicates air speed is greater above wing than below

Courtesy http://www.monmouth.com/~jsd/how/htm/airfoils.html

Lift Illustrated

Courtesy NASA and TRW, Inc.

Sailing Against the WindSails are airfoils

Low pressure between sails helps drive boat forward

Courtesy Dave Culp Speed Sailing

Venturi Tube

Courtesy http://www.abdn.ac.uk/physics/streamb/fin13www/sld001.htm

Bernoulli’s Principle also

Helps explain why smoke rises up a chimney (air moving across top)Explains how air flows in underground burrows (speed of air flow across entrances is slightly different)Explains how perfume atomizer worksExplains how carburetor works

Giancoli Website