Fluid Flow or Discharge
• When a fluid that
fills a pipe flows
through a pipe of
cross-sectional area
A with an average
velocity v, the flow
or discharge Q is
determined by:
Q Av
Equation of Continuity
• Suppose an
•
incompressible
(constant density) fluid
fills a pipe and flows
through it.
If the cross-sectional
area of the pipe is A1 at
one point and A2 at
another point, the flow
through A1 must equal
the flow through A2.
Q  A1  v1
Q  A2  v2
A1  v1  A 2  v 2
Equation of Continuity
• The equation of continuity in the form
A1  v1  A 2  v 2
applies only when the density of the fluid is
constant. If the density of the fluid is not
constant, the equation of continuity is
D1  A1  v1  D2  A 2  v 2
• mass flow rate (units kg/s):
m
 D  A v
t
• volume flow rate (units m3/s):
V
 A v
t
Equation of Continuity
• The equation of continuity shows that where the
•
cross-sectional area is large, the fluid speed is slow
and that where the cross-sectional area is small,
the fluid speed is large.
This explains why water shoots out of a hose faster
when you place your thumb across the opening,
reducing the cross-sectional area through which
the water can come out.
Viscosity and Viscous Flow
• Viscosity h of a fluid is a
measure of how difficult it is
to cause the fluid to flow.
• In an ideal fluid there is no
viscosity to hinder the fluid
layers as they slide past one
another.
• Within a pipe of uniform
cross-sectional area, every
layer of an ideal fluid moves
with the same velocity,
even the layer next to
the wall.
Viscosity and Viscous Flow
• When viscosity is present, the fluid layers
do not all have the same velocity.
– The fluid closest to the wall does not move at all,
while the fluid at the center of the pipe has the
greatest velocity.
– The fluid layer next to the wall surface does not
move because it is held tightly by intermolecular
forces.
– The intermolecular forces
are so strong that if a
solid surface moves,
the adjacent fluid layer
moves along with it and
remains at rest relative
to the moving surface.
Viscosity and Viscous Flow
• This is why a layer of dust lies on the
•
surface of fan blades even at high speeds.
The layer of air in contact with the fan
blade has no velocity relative to the fan
blade and does not blow off the dust.
Force F needed to move a layer of viscous
fluid with constant velocity:
–
–
–
–
h = viscosity
η A v
F 
A = area
d
v = velocity
d = distance from the immobile surface
Viscosity and Viscous Flow
• Viscosity of liquids and gases depend
on temperature.
– Usually, the viscosities of liquids decrease
as the temperature increases.
– The viscosities of gases increase as the
temperature increases.
• Viscous fluids have a high viscosity,
such as tar and molasses.
Forces Exerted By a Fluid
• If a fluid were
subjected to a
tangential force F, the
layers of the fluid
would slide past one
another without
friction.
• This means that a
fluid can sustain only
a perpendicular force,
and conversely, can
exert only a force
perpendicular to the
surface.
•If a fluid were subjected
to a tangential force F,
the layers of the fluid
would slide past one
another without friction.
•This means that a fluid
can sustain only a
perpendicular force, and
conversely, can exert
only a force
perpendicular to the
surface.
Forces Exerted By a Fluid
• Suppose that a
nonelastic fluid is
between 2 plates. If
the velocity v of the
upper plate is not
too large, the fluid
shears in the way
indicated. The
viscosity h is
related to the force
F required to
produce the velocity
v by:
vA η
F
d
Forces Exerted By a Fluid
• A = area of either plate
• d = distance between
•
•
•
plates
units for h: N·s/m2 or
kg/m·s or lb· s/ft2
1 poiseuille (Pl) =
1 N·s/m2 = 1 kg/m·s
1 poise (P) = 0.1 kg/m·s
Poiseuille’s Law
• The fluid flow through a cylindrical
pipe of length l and cross-sectional
radius r is given by:
4
π  r  (P1  P2 )
Q
8  η l
• P1 - P2 is the pressure difference
between the two ends of the pipe.
Work Done by a Piston
• Work done by a piston in forcing a
volume V of fluid into a cylinder
against an opposing pressure P is
given by: W = P·V
Bernoulli’s Principle
• If a fluid is incompressible - a change in pressure
does not cause a change in volume - the volume of
fluid entering per second must equal the volume
leaving per second.
V1 V2

Volumetric flow rate: t
t
1 = entering
2 = leaving
Let v represent the speed with which a liquid moves
in a cylindrical pipe, so that during the time t the
liquid moves a distance equal to v·t (where v is
velocity.
Bernoulli’s Principle
• The volume of liquid passing a cross-sectional area
A is given by
V  Avt
• Volumetric flow rate Q:
V
Q  Av
t
• Because the liquid is incompressible, the
•
volumetric flow rate is the same entering and
leaving the system.
Volumetric flow rate (system): A  v  A  v
1 1
2
2
Bernoulli’s Principle
•
2
0
.
5

D

v
 h  D  g  P has the same
The quantity
value at every point in an incompressible fluid
moving in streamline (non-turbulent) flow.
• Bernoulli’s equation:
0.5  D  v12  h1  D  g  P1  0.5  D  v 22  h2  D  g  P2
Bernoulli’s Principle
• If the fluid is not moving, then both speeds are zero. The
fluid is static. If the height at the top of the column is h1 is
defined as zero, and h2 is the depth, then Bernoulli’s
equation reduces to the equation for pressure as a function of
depth:
P1  P2  D  g  h2
• If the fluid is flowing through a horizontal pipe with a
constriction, as shown in the figure on the next slide, there is
no change in height and the gravitational potential energy
does not change. Bernoulli’s equation reduces to:
P1 
2
0.5  D  v1
 P2  0.5  D  v 2
2
1
1
2
2
P1  h1  D  g  D  v1  P2  h 2  D  g  D  v 2
2
2
Work per
unit volume
done by the
fluid
Potential
energy
per unit
volume
Kinetic
energy
per unit
volume
Points 1 and 2
must be on the
same streamline
Bernoulli’s Principle
• The flow rate Q in the tube has
to be constant, therefore, the
fluid has to move faster through
the constriction to maintain the
constant flow rate Q.
• The velocity at point a is greater
than at either the meter
entrance or the meter exit.
• The pressure in a fluid is related
to the speed of flow, therefore
the pressure in the fluid is less
at point a and greater at the
meter entrance, as illustrated by
the liquid levels in the U-tube
manometer.
• The pressure difference is equal
to:
P  hDg
Bernoulli’s Principle
• Bernoulli's principle describes the
relationship between pressure and
velocity in a fluid and describes the
conservation of energy as it applies to
fluids.
• Bernoulli’s principle also explains why
a roof blows off of a house in violent
winds.
– Wind creates a low pressure region above
the peak of the roof, creating a pressure
difference inside and outside the house
which results in the loss of the roof.
Bernoulli's Equation and Lift
• The shape of a wing forces
•
•
air to travel faster over the
curved upper surface than
it does over the flatter
lower surface.
According to Bernoulli’s
equation, the pressure
above the wing is lower
(faster moving air), while
the pressure below the
wing is higher (slower
moving air).
The wing is lifted upward
due to the higher pressure
on the bottom of the wing.
Bernoulli's Equation and Lift
• Air flows over the top of an airplane wing of area A
•
with speed vt, and past the underside of the wing
(also of area A) with speed vu.
the magnitude FL of the upward lift force on the
wing will be:

2
FL  0.5  D  A  v t  v u
• Ski jumpers use this same principle to help
2

themselves stay in the air longer during jumps.
• A boomerang with a curved surface will turn in the
direction of the curved face due to pressure
differences created by the different air velocities
over the two surfaces.
Control surfaces on a plane
• By extending the
slats, the wing area
A can be increased
to generate more lift
at low speeds for
take off and
landing.
Curveball Pitch
Torricelli’s Theorem
• If an opening exists in a
tank containing a liquid
at a distance h below the
top of the liquid, then the
velocity v of outflow from
the opening is:
v  2 g h
provided the liquid obeys
Bernoulli’s equation and
the top of the liquid may
be regarded as motionless
(v = 0 m/s).
Tidal Waves
• Tidal waves are the dissipation of energy in a viscous fluid over
an inclined plane; tidal waves have nothing to do with tides.
• The energy source is usually an under-sea earthquake (it could
also be an under-sea explosion or a meteor strike); the viscous
fluid is the ocean; the inclined plane is the ocean floor sloping
upward toward land.
• Earthquake:
– When the Earth moves up and down it also moves the ocean water
up and down. This generates a huge wave traveling outward in a
series of concentric rings.
– In deep water, most of the tidal wave (tsunami) remains hidden
beneath the surface. But as the tidal wave moves toward more
shallow water, its enormous energy is forced to the surface.
– In the open ocean, tidal waves are hundreds of miles wide and
travel at jetliner speeds. Near land they slow down to freeway
speeds.
– What makes a tidal wave so destructive is the speed and
tremendous volume of water delivered onto a coastal or island
environment as the tidal wave is forced by the inclining ocean floor
onto the land.
Bernoulli’s Principle and Syringes
• The force applied to the plunger is equal to the
•
•
•
pressure times the area of the plunger.
Viscous flow will occur within the barrel of the
syringe and only a little pressure difference is
needed to move the fluid through the barrel to
point 2, where the fluid will enter the
narrow needle.
The pressure applied
to the plunger is nearly
equal to the pressure
P2 at point 2.
The pressure at point 1, P1,
is also known as the gauge
pressure.
Bernoulli’s Principle and Syringes
• Apply Bernoulli’s principle,
0.5  D  v12  h1  D  g  P1  0.5  D  v 22  h2  D  g  P2
and if the needle is held horizontally,
P1 
2
0.5  D  v1
 P2  0.5  D  v 2
2
• Poiseuille’s Law may also be needed to
solve this type of problem.
P1  P2 
8  η l  Q
πr
4
Bernoulli’s Principle and Siphons
• A siphon is an inverted U-shaped pipe or tube that
can transfer water from a higher container to a lower
container by lifting the water upward from the
higher container and then lowering it into the lower
container. The water is simply seeking its level, just
as it would if you connected the two containers with
a pipe at their bottoms. In that case, the water in the
higher container would flow out of it and into the
lower container, propelled by the higher water
pressure at the bottom of the higher container. In
the case of a siphon, it's still the higher water
pressure in the higher container that causes the
water to flow toward the lower container, but in the
siphon the water must temporarily flow above the
water level in the higher container on its way to the
lower container.
Bernoulli’s Principle and Siphons
• Two means of initiating the
liquid flow (assume the liquid is
water):
– You can make a siphon using a rubber
hose and gravity is the key to getting it
to work. A siphon needs to have the
"dry" end of the hose lower than the
end that is stuck in the water (the
"wet" end). You can get the siphon
started by first filling the hose with
water. Once the hose is full, use your
thumb to plug the end of the hose that
will be removed from the water. Place
the “dry” end into the second
container and then remove your
thumb. Gravity does all the work from
there...
How does it work? Think of the water in terms of distinct “packets".
Since the dry end of the hose is lower than the wet end, there are
more water "packets" towards the dry end. As such, the column of
water being pulled downward by gravity is heavier than the column
of water at the wet end of the tube. Gravity pulls on one “packet" of
water on the dry end of the tube causing it to move down the tube.
As it moves, it creates a small vacuum behind itself. This vacuum
pulls the next “packet” forward (downward) as well. This suction is
strong enough to pull other “packets” up the tube (against gravity) at
the wet end. Once a given “packet” passes the highest point in the
tube, gravity pulls it downward and the
process continues. The siphon will work
as long as the vertical (up and down)
column of water outside the container is
larger than the vertical column inside the
container. If the two ends of the hose are
exactly the same height (the columns are
equal), the pull of gravity will be the same
on each side and the flow of water will stop.
If you then lower the free end, the flow of
water will begin once again.
Bernoulli’s Principle and Siphons
– Sucking on the lower end of the tube
causes a partial vacuum (a region of
space with a pressure that's less
than atmospheric pressure) at the
top of the siphon. The partial
vacuum results in a difference in
pressure between the bottom of the
tube and the top of the tube. With
greater fluid pressure at the top than
the bottom, the water is pushed up
into the tube and over to the lower
container. The same kind of partial
vacuum exists in a drinking straw
when you suck on it and is what
allows atmospheric pressure to push
the beverage up toward your mouth.
Bernoulli’s Principle and Siphons
Bernoulli’s Principle and Siphons
• To get water up to the top of the tube, need to
reduce pressure in the tube so that the
atmospheric pressure will do the work to push
the fluid up to the top of the tube where gravity
will do the work to pull the fluid out the end of
the hose.
• Apply Bernoulli’s equation
between points A and B to
determine the maximum height
h1 to which water can be
lifted above the water surface.
1
1
2
2
Patm  h  D  g   D  vt  PB  h1  D  g   D  vB
2
2
h  0  set top surface of water as 0  ; vt  0
PB  0 (want this to be as small as possible)
vB  vC  2  g  (d  h 2 )  to be explained soon 
Patm
Patm
Patm

1
 h1  D  g   D  2  g  (d  h 2 )
2
1
 h1  D  g   D   2  g  (d  h 2 ) 
2
 h1  D  g  D  g  (d  h 2 )

2
Bernoulli’s Principle and Siphons
• To determine the speed of the liquid flow at the
bottom of the siphon, start with Bernoulli’s
equation:
0.5  D  v t 2  ht  D  g  Pt  0.5  D  v b 2  hb  D  g  Pb
• Atmospheric pressure is found at
the top of the liquid and at the
bottom of the siphon, therefore,
Pt and Pb are equal and cancel
out.
• Consider the velocity vt at the top
of the liquid to be 0 m/s.
Bernoulli’s Principle and Siphons
• Consider the lower end of the
siphon to be the point at which
the height is 0 m. From the figure,
the distance from the bottom of the
to the siphon to the upper level of
the liquid is d + h2.
(d  h2 )  D  g  0.5  D  v b
2
• The density cancels out:
(d  h2 )  g  0.5  v b
• Solve for vb:
2
v b  2  d  h2   g
Equation of Continuity Example
• What is the flow rate of water in a pipe whose
diameter is 10 cm when the water is moving
with a velocity of 0.322 m/s?
d  10 cm  0.1 m; r  0.05 m
A  π  r 2  π  0.05 m2  0.007854 m2
m
Q  A  v  0.007854 m  0.322
s
2
m3
Q  0.002529
s
Equation of Continuity Example
• If the diameter of Qleft  Qright ; A L  v L  A R  v R
2
2
the pipe to the
π  rL  v L  π  rR  v R
right is reduced
2
2
to 4 cm, what is rL  v L  rR  v R
m
2
the velocity of

0.05 m   0.322
2
rL v L
s
the fluid in the
vR 

2
r
R
right-hand side of
0.02 m2
the pipe?
m
v R  2.0125
s
Bernoulli’s Example
• The pressure P1= 53913.24 N/m2, whereas the velocity of the water v1
= 0.322 m/s. The diameter of the pipe at location 1 is 10 cm and it is at
ground level. If the diameter of the pipe at location 2 is 4 cm, and the
pipe is 5 m above the ground, find the pressure P2 of the water at
position 2.
– From the previous example, we know that the velocity of the water at
location 2 is 2.015 m/s.
Bernoulli’s Example



 
P1  h1  D  g   0.5  D  v12  P2  h 2  D  g  0.5  D  v 2 2
h1  0 m (ground level);D  1000



kg
m3
 
P1  0.5  D  v12  P2  h 2  D  g  0.5  D  v 2 2


 
P2  P1  0.5  D  v12  h 2  D  g  0.5  D  v 22


2

kg
m
kg
m

  

P2  2940 2  0.5  1000 3  0.322    5 m  1000 3  9.8 2 

s   
m
m 
m
s 

N
2

kg
m


  0.5  1000 3  2.015  


s
m




N
N
N
N
P2  53913 .24 2  51 .842 2  49000 2  2030 .1125 2
m
m
m
m
N
P2  2934 .97 2
m
Helpful Online Links
• Hyperphysics Fluids
• Work-Energy Applet (to determine the
power needed in the pump for the waterjet to pass over the wall)
• Gallery of Fluid Mechanics