Transcript Chapter 10 - TeacherWeb

PHYSICS OF FLUIDS
Fluids
Includes liquids and gases
Liquid has no fixed shape but nearly fixed volume
Gas has neither fixed shape or volume
Both can flow
Density
Mass per unit volume
r = m/V
(Greek letter “rho”)
m = rV
Example: Density of Mercury is 13.6 x 103
kg/m3 What is the mass of one liter?
M = rV = 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/liter
1 gram /cubic centimeter (cc)
1 cubic meter = 1000 liters
1 cubic centimeter = 1 milliliter
1 cubic meter = 1,000,000 cc
Table of Densities
Substance
Density kg/m3 x103
aluminum
Iron
Lead
Gold
Water
Sea water
Alcohol, ethyl
Mercury
Ice
2.7
7.8
11.3
19.3
1.0
1.025
0.79
13.6
0.917
Pressure in Fluids
Force per unit area
Pressure = P = F/A
Unit N/m2= Pascal(Pa)
Exerted in all directions
Force due to pressure is perpendicular to
surface in a fluid at rest
Pressure Varies with Depth as
rgh
Let depth be h. Assume incompressible.
Force acting on area is mg = rVg = rAhg
P = F/A = rgh
Pressure at equal depths is the same
If external pressure is also present it must be
added
DP = rg Dh
h
A
Example: Pressure at Bottom of a
Lake
• What
is the pressure at the bottom of a
20.0 meter deep lake?
P = rgh = 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
atmospheric
Absolute (total) pressure is gauge +
atmospheric P = PA + PG
What 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 straw
How 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?
rgH = 101,300 Pa; H = 101,300/(9.8
N/kg x 1000 Kg/m3) =10.3 m
Pascal’s Principle
Probably not tested
Pressure applied to a confined fluid increases
pressure throughout by the same amount
Pout = Pin
Fout/Aout = Fin/Ain
Fout/Fin = Aout/Ain
Multiplies force by ratio of areas
Principle 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 rgh
The (greater) pressure
P2 = P1 + rgh
How would this look if P1 was greater than P2 ?
Diagram courtesy Sensorsmag.com
Buoyancy
Submerged or partly submerged object
experiences an upward force called
buoyancy
Pressure in fluid increases with depth
Studied by Archimedes over 2000 years
ago.
Buoyant Force on Cylinder
h = h 2 – h1
FB = F2 – F1 = P2A – P1A
= rFgA(h2 – h1)
= rFgAh
= rFgV = mFg
h1
F1
A
h2
h
Buoyant Force equals
weight of fluid displaced
F2
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.
Weighing Submerged object
Sfy = 0
T +B - W = 0
T=W–B
T = W – rFVg
T is apparent weight
W’
Diagram courtesy Caduceus MCAT
Review
Example: King’s Crown
Given crown mass 14.7 kg but weighed
under water only 13.4 kg. Is it gold?
W’ = W – FB
W = rogV
W - W’ = FB = rF gV
W/(W-W’) = rogV / rF gV = ro / rF
ro / rH2O = W/(W-W’) = 14.7/(14.7 –
13.4) = 14.7/1.3 = 11.3 LEAD
Another way to solve: isolate ro = W/gV and then get V from
FB /rfg
Floating Objects
Objects float if density less
than that of fluid.
FB = W at equilibrium
rFVdisp g = ro Vo g
Vdisp / Vo = ro /rF
Example: If an object’s density is 80% of
the density of the surrounding fluid, 80% of
it will be submerged
Vo
Vdisp
Example: Floating Log
15 % of a log floats above the surface
of the ocean. What is the density of
the wood?
Vdisp / Vo = ro /rF
ro = Vdisp / Vo x rF = 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
rair Vg = (rHe V + 60 kg) g
V = 60 kg/(rair – rHe) = 60 kg/(1.29 – 0.18kg/m3)
= 54 m3
Fluid Flow
Equation of Continuity
Volume rate of flow is constant for
incompressible fluids (not turbulent)
A1v1 = A2 v2
v is velocity
Laminar vs. Turbulent Flow
Erratic,
contains
eddies
Fluid follows smooth path
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
A2
A1
l2
Bernoulli’s Equation
Where velocity of fluid is high, pressure is
low; where velocity is low, pressure is high
Consequence of energy conservation
P + ½ rv2 + rgy = constant for all
points in the flow of a fluid
P + ½ rv2 = 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 rv2 = constant
Speed of Water Flowing Through
Hole in Bucket
P1 = P2 + ½ rv2 + rgz
P1 = P2 since both
open
to air
½ rv22 + rgz = 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 = v1pr12/pr22 =1.2 m/s
2) Use Bernoulli’s Eq. to find pressure
P1 + ½ rv12 + rgy1 = P2 + ½ rv22 +
rgy2
P2 = P1 +rg(y1 – y2) + 1/2r(v12 –v22)
=(3.0 x 105 N/m2) + (1.0 x 103 kg/m3)x
(9.8 m/s2)(-5.0m) + ½ (1.0 x 103
2 – (1.18m/s)2] =
kg/m3)[(0.50
m/s)
5
2
= 2.5 x 10 N/m
No Change in Height
P1 + ½ rv12 = P2 + ½ rv22
Where speed is high, pressure is low
Where speed is low, pressure is high
Why Curveballs Curve
Courtesy Boston University Physics Dept. web site
How an Airfoil Provides Lift
Where is the pressure greater, less?
Courtesy The Aviation Group
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 Wind
Sails 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 works
Explains how carburetor works
Giancoli Website