Download Fluid Mechanics

Chapter 10
Density
 Recall that the density of
an object is its mass per
unit volume (SI unit is
kg/m3)
 The specific gravity of a
substance is its density
expressed in g/cm3
m

V
Pressure in Fluids
 Fluids exert a pressure in all directions
F Force
P 
A Area
SI Unit for Pressure is Pa
1 Pa= 1 N/m2
1 atm= 101.3 kPa=760 mm-Hg
 A fluid at rest exerts pressure perpendicular to any
surface it contacts
 The pressure at equal depths within a uniform fluid is
the same
P  gh  (density )( g )( depth)
Pressure in Fluids
 Gauge Pressure is a measure of the pressure over and
above the atmospheric pressure
 i.e. the pressure measured by a tire gauge is gauge
pressure. If the tire gauge registers 220 kPa then the
absolute pressure is 321 kPa because you have to add the
atmosphere pressure (101 kPa)
 If you want the absolute pressure at some depth in a
fluid then you have to add atmosphere pressure
P  Po  gh
Pressure in Fluids
 Pascal’s Principle: Pressure applied to a fluid in a
closed container is transmitted equally to every point
of the fluid and to the walls of the container
F1 F2

A1 A2
Buoyancy
 Buoyant force is the force acting
on an object that is immersed in a
fluid
 Archimedes Principle: The
buoyant force on a body immersed
in a fluid is equal to the weight of
the fluid displaced by the object
 Since the buoyant force acts
opposite of gravity, an object seems
to weigh less in a fluid
 Apparent Weight= Fg-FB
Fb = Buoyant Force
Fg = Gravity
Sinking vs Floating
 Think back to free body diagrams
 If the net external force acting on an object is zero then
it will be in equilibrium
FB
If Fb=Fg then the object
will be in equilibrium
and will FLOAT!
Fg
Cubes floating in a fluid
70% Submerged
20% Submerged
100% Submerged
Density determines depth of
submersion
 This equation gives the
percent of the object’s
volume that is submerged
 Vf is the volume of fluid
displaced
 Vo is the total volume of the
object
 ρo is the density of the
object
 ρf is the density of the fluid
o

Vo  f
Vf
Summary of Floating
For an object to float
FB= Fg
ρo ≤ ρf
If ρo = ρf then the
Object will be
Completely submerged
But not sinking.
If ρo is less than ρf
Then the amount
submerged
can be found with
Vf
Vo

o
f
Continuity Equation
•Continuity tells us that whatever the volume of fluid in
a pipe passing a particular point per second, the same
volume must pass every other point in a second.
•If the cross-sectional area decreases, then velocity
increases
Continuity Equation
A1v1  A2 v2
The quantity Av is the volume rate of flow
Bernoulli’s Principle
 The pressure in a fluid decreases as the fluid’s velocity
increases.
 Fluids in motion have kinetic energy, potential energy
and pressure
How do planes fly?
Bernoulli’s Equation
The kinetic energy of a fluid element is:
The potential energy of a fluid element is:
PE  mgh  ( V ) gh
Bernoulli’s Equation
 This equation is essentially a statement of
conservation of energy in a fluid. Notice that volume
is missing. This is because this equation is for
energy per unit volume.
Bernoulli' s Equation
1 2
1 2
P1  v1  gh1  P2  v2  gh2
2
2
Sample Problem p. 306 #40
 What is the lift (in newtons) due to Bernoulli’s
principle on a wing of area 80 m2. If the air passes over
the top and bottom surfaces at speeds of 350 m/s and
290 m/s, respectively.
 Let’s make point 1 the top of the wing and point 2 the
bottom of the wing
 The height difference between the top of the wing and
the bottom is negligible
1 2
1 2
P1  v1  gh1  P2  v2  gh2
2
2
Sample Problem p.306 #40
 The net force on the wing is a result of the difference
in pressure between the top and the bottom. P1 is
exerted downward, P2 is exerted upward
 If we know the difference in pressure we can use that
to find the force
1 2
1 2
P1  v1  P2  v2
2
2
1 2 1 2 1
P2  P1  v1  v2  (1.29kg / m3 )(340 2  290 2 )  20318 Pa upward
2
2
2
Sample Problem p.306 #40
 P2-P1=20318 Pa
F  ( P2  P1 ) A  (20318Pa)(80m2 )  1.63x106 N upward
P.306 #43
 Water at a pressure of 3.8 atm at street level flows into
an office building at a speed of 0.60 m/s through a
pipe 5.0 cm in diameter. The pipes taper down to 2.6
cm in diameter by the top floor, 20 m above street
level. Calculate the flow velocity and the pressure in
such a pipe on the top floor. Ignore viscosity. Pressures
are gauge pressures.
Find the flow velocity at the top
Continuity Equation
A1v1  A2 v2
 A1 is area of first pipe= πr2 = 1.96x10-3 m2
 A2 is area of second pipe= πr2 = 5.31x10-4 m2
 V1= 0.6 m/s
A1v1 (1.96 x103 m 2 )(0.6m / s)
m
v2 

 2.2
4
2
A2
5.31x10 m
s
Find pressure at the top
1 2
1 2
P1  v1  gh1  P2  v2  gh2
2
2
1 2 1 2
P2  P1  v1  v2  gh2
2
2
N
N
N
P2  3.84 x10 Pa  180 2  2420 2  196200 2
m
m
m
5
P2= 1.86 x 105 Pa= 1.8 atm
Sample Problem p.305 #37
 What gauge pressure in the water mains is necessary if
a fire hose is to spray water to a height of 12.0 m?
 Let’s make point 1 as a place in the water main where the
water is not moving and the height is 0
 Point 2 is the top of the spray, so v=0 , P= atmospheric
pressure, height = 12m
1 2
1 2
P1  v1  gh1  P2  v2  gh2
2
2
Sample Problem p.305 #37
P1  P 2  gh  Patm  (1000kg/m )(9.81m/s )(12m)
3
2
Remember that Gauge Pressure is the pressure above atmospheric
pressure. So to get gauge pressure, we need to subtract atmospheric
Pressure from absolute pressure.
Gauge Pressure  P1 P atm  1.2 x105 Pa