Download - PHYSICS OF FLUIDS
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PHYSICS OF FLUIDS
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Fluids
Includes liquids and gasesLiquid has no fixed shape but nearly fixed volumeGas has neither fixed shape or volumeBoth can flow
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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)
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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
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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
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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
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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
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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?
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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
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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
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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
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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
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Pascal’s Principle in Action
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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 ?
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BuoyancySubmerged or partly submerged object experiences an upward force called buoyancyPressure in fluid increases with depthStudied by Archimedes over 2000 years ago.
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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
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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.
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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
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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
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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
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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
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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
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Fluid Flow
Equation of ContinuityVolume rate of flow is constant for incompressible fluids (not turbulent)A1v1 = A2 v2
v is velocity
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Laminar vs. Turbulent Flow
Fluid follows smooth path
Erratic, contains eddies
Courtesy MIT Media Laboratory
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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
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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
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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
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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
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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
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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
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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
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No Change in Height
P1 + ½ v12 = P2 + ½ v2
2
Where speed is high, pressure is lowWhere speed is low, pressure is high
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Why Curveballs Curve
Courtesy Boston University Physics Dept. web site
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How an Airfoil Provides Lift
Courtesy The Aviation Group
Where is the pressure greater, less?
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Crowding of streamlines indicates air speed is greater above wing than below
Courtesy http://www.monmouth.com/~jsd/how/htm/airfoils.html
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Lift Illustrated
Courtesy NASA and TRW, Inc.
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Sailing Against the WindSails are airfoils
Low pressure between sails helps drive boat forward
Courtesy Dave Culp Speed Sailing
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Venturi Tube
Courtesy http://www.abdn.ac.uk/physics/streamb/fin13www/sld001.htm
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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
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Giancoli Website