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
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