Chapter 13
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•Motion is steady: velocity, density,
pressure at a given point don’t change
•Moves without turbulence
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Steady/Unsteady flow
•
•
•
•
In steady flow the velocity of particles is constant with time
Unsteady flow occurs when the velocity at a point changes with
time. Turbulence is extreme unsteady flow where the velocity
vector at a point changes quickly with time e.g. water rapids or
waterfall.
When the flow is steady, streamlines are used to represent the
direction of flow.
Steady flow is sometimes called streamline flow
• Streamlines never cross. A set of streamlines can
define a tube of flow, the borders of which the fluid
does not cross
Turbulent Flow
Turbulent flow is an extreme kind of unsteady flow and occurs
when there are sharp obstacles or bends in the path of a fastmoving fluid.
In turbulent flow, the velocity at a point changes erratically
from moment to moment, both in magnitude and direction.
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Viscous/Non viscous flow
•
•
•
•
A viscous fluid such as honey does not flow readily, it has a large viscosity.
Water has a low viscosity and flows easily.
A viscous flow requires energy dissipation.
Zero viscosity – requires no energy. with no dissipation of energy. Some
liquids can be taken to have zero viscosity e.g. water.
• An incompressible, non viscous fluid is said to be an ideal fluid
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13. 2
Stream flow
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The volume flow rate, Q is defined as the volume of fluid that flows
past an imaginary (or real) interface.
A v represents the volume of fluid per second that passes through
the tube and is called the volume flow rate Q
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• If the fluid is incompressible, the density remains constant
throughout
Q  Av
A1v1  A2v2
• Av represents the volume of fluid per second that passes through
the tube and is called the volume flow rate Q
V  Ax  Avt
This just means that the amount of fluid moving in any “section of pipe”
must remain constant.
•If the area is reduced the fluid must speed up!
• Speed is high where the pipe is narrow and speed is
low where the pipe has a large diameter
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The product of the area and the fluid speed at all points along a pipe is constant
for an incompressible fluid.
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Q: Have you ever used your thumb to control the water flowing from the end of a
hose?
A: When the end of a hose is partially closed off, thus reducing its cross-sectional
area, the fluid velocity increases.
This kind of fluid behavior is described by the equation of continuity.
Example:
Oil is flowing at a speed of 1.22 m/s through a pipeline with a radius
of 0.305 m. How much oil flows in 1 day?
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example
example
• A river flows in a channel that is 40. m wide and 2.2 m deep with a speed of
4.5 m/s.
• The river enters a gorge that is 3.7 m wide with a speed of 6.0 m/s.
• How deep is the water in the gorge?
The area is width times depth.
A1 = w1d1
Use the continuity equation.
v1A1 = v2A2
v1w1d1 = v2w2d2
Solve for the unknown d2.
d2 = v1w1d1 / v2w2 =(4.5 m/s)(40. m)(2.2 m) / (3.7m)(6.0 m/s) = 18 m
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example
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example
If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30%
of total discharge and
pipe 3 diameter 60mm. What are the values of discharge and mean velocity in
each pipe?
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example
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example
example
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Example:
The volume rate of flow in an artery supplying the brain is 3.6x10-6 m3/s. If the
radius of the artery is 5.2 mm,
A) determine the average blood speed.
B) Find the average blood speed if a constriction reduces the radius of
the artery by a factor of 3 (without reducing the flow rate).
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Example:
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Example
if the cross-section area, A, is 1.2 x 10-3m2 and the discharge, Q is 24 l / s ,
then the mean velocity,
Example
if the area A 1 = 10 x10-3 m2 and A 2 = 3 x10-3 m 2 and the upstream mean
velocity,
1 = 2.1 m / s , then the downstream mean velocity
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Example, decrease in area of stream of water coming
from tap.
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•
•
Example:
The figure shows 3 straight pipes
through which water flows. The
figure gives the speed of the water
in each pipe.
Rank them according to the volume
of water that passes through the
cross-sectional area per minute,
greatest first.
6
same
Example:
•
•
•
•
Water flows smoothly through the
pipe shown in the figure, descending
in the process.
Rank the four numbered sections of
pipe according to
(a) the volume flow rate Q through
them
(b) the flow speed v through them
(c) the water pressure p within them,
greatest first.
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a)Same
b) 1,2,3,4 c ) 4,3,2,1
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1)
2)
3)
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Pressure
Pressure P is the magnitude of the force acting perpendicular to
a surface divided by the area A over which the force acts.
The greater the force, the greater the pressure is. The greater the
area, the smaller the force is.
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example
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Example
The mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm
deep. a) Find the weight of the water in the mattress.
The volume density of water at the normal condition (0oC and 1 atm) is
1000kg/m3. So the total mass of the water in the mattress is
m 
W
VM  1000  2.00  2.00  0.300  1.20 103 kg
Therefore the weight of the water in the
mattress is
W  mg  1.20 103  9.8  1.18 10 4 N
b) Find the pressure exerted by the water on the floor when the bed rests
in its normal position, assuming the entire lower surface of the mattress
makes contact with the floor.
Since the surface area of the
mattress is 4.00 m2, the
pressure exerted on the floor
is
F mg 1.18 10 4
3




2
.
95

10
P A A
4.00
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• Relates pressure to fluid speed and elevation
• Bernoulli’s equation is a consequence of Work
Energy Relation applied to an ideal fluid
• Assumes the fluid is incompressible and
nonviscous, and flows in a nonturbulent,
steady-state manner
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P2 A
P1A
area A
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This work is negative because the force on the segment of fluid is to the left
and the displacement is to the right. Thus, the net work done on the
segment by these forces in the time interval
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• 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
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Bernoulli’s Principle
• In steady flow, the speed, pressure and elevation of an ideal fluid are
related. Two situations.
• (1) Horizontal pipe, changing Area A.
– Pressure drop in thin pipe why?
– Fluid speeds – acceleration – force required
• (2) Elevation change but same area A
– Lower fluid is under more pressure
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Atmospheric Pressure
• The atmosphere exerts a pressure on the surface of the
Earth and all objects at the surface
• Atmospheric pressure is generally taken to be
1.00 atm = 1.013 x 105 Pa = Patm
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13.4 static consequences of Bernoulli's equation
 Fluid At Rest In a Container :
•Pressure in a continuously distributed uniform static fluid varies only with
vertical distance and is independent of the shape of the container.
• The pressure is the same at all points on a given horizontal plane in a fluid.
Y a = y b = y c = yd
• For liquids, which are incompressible, we have:
pb + ρgh1 = patm + ρgh2
yA =yB=yC=yD
pb = patm + ρg (h2 - h1) = patm + ρgd
d
h2
h1
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Absolute Pressure and Gauge Pressure
The pressure P is called the absolute pressure
, because it is the actual value of the system’s pressure.
Remember, P = Patm + ρ g h
P – Patm = ρ g h
( so ρ g h is the gauge pressure)
Gauge pressure: Pressure expressed as the difference between the pressure of
the fluid and that of the surrounding atmosphere
Usual pressure gauges record gauge pressure. To calculate absolute pressure:
P = P atm + P gauge
As a fluid moves along if it changes speed or elevation then the pressure
changes and vice versa.
Bernoulli’s equation is really just conservation of energy for the fluid
• Bernoulli’s equation.
• P + ½ ρv2 + ρg h = constant
• Bernoulli’s equation shows that as the velocity of a fluid speeds up
it’s pressure goes down…..this idea used in airplane wings and
frisbees (difference in pressure leads to upward force we call Lift)
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PA>PB
PA=PC
PA – Patm = ρ g h
P  Patm  gh
PA = PB = PC = PD
• P atm is atmospheric pressure =1.013 x 105 Pa
• The pressure does not depend upon the shape of the
container
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Variation of Pressure with Depth
• If a fluid is at rest in a container, all
portions of the fluid must be in static equilibrium
• All points at the same depth must be at the same pressure
– Otherwise, the fluid would not be in
equilibrium
• Pressure changes with elevation
•The pressure gradient in the vertical direction is
negative
• The pressure decreases as we move upward
in a fluid at rest
• Pressure in a liquid does not change due to
the shape of the container
• The fluid would flow from the higher pressure
region to the lower pressure region
•the pressure at a point in a fluid depends only on density, gravity and
depth
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example
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the Manometer is a device to measure pressures. A common simple
manometer consists of a U shaped tube of glass filled with some liquid. Typically the
liquid is mercury because of its high density
Case 1
both ends of the tube are open to the atmosphere. Thus both points
A and B are at atmospheric pressure. The two points also have the
same vertical height.
Case 2
the top of the tube on the left has been closed. We imagine that there
is a sample of gas in the closed end of the tube.
The right side of the tube remains open to the atmosphere. The point
A, then, is at atmospheric pressure.
The point C is at the pressure of the gas in the closed end of the tube.
The point B has a pressure greater than atmospheric pressure due to
the weight of the column of liquid of height h.
The point C is at the same height as B, so it has the same pressure as B.
And we have already seen that this is equal to the pressure of the gas
in the closed end of the tube.
Thus, in this case the pressure of the gas that is trapped in the closed
end of the tube is greater than atmospheric pressure by the amount of
pressure exerted by the column of liquid of height h.
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Case 3
Perhaps the closed end of the tube contains a sample of gas as before, or
perhaps it contains a vacuum.
The point A is at atmospheric pressure.
The point C is at whatever pressure the gas in the closed end of the tube
has, or if the closed end contains a vacuum the pressure is zero.
Since the point B is at the same height as point A, it must be at
atmospheric pressure too. But the pressure at B is also the sum of the
pressure at C plus the pressure exerted by the weight of the column of
liquid of height h in the tube.
We conclude that pressure at C, then, is less than atmospheric pressure
by the amount of pressure exerted by the column of liquid of height h.
If the closed end of the tube contains a vacuum, then the pressure at
point C is zero, and atmospheric pressure is equal to the pressure exerted
by the weight of the column of liquid of height h. In this case, the
manometer can be used as a barometer to measure atmospheric
pressure.
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Pressure Measurements: The manometer
•
Manometers are devices in which one or more columns of a liquid are used to
determine the pressure difference between two points.
•
•
•
U-tube manometer
One end of the U-shaped tube is open to the atmosphere
The other end is connected to the pressure to be measured
P P
P  P  gy
P  P  gy
P  gy  P  gy
P  P  gy  gy
P  Patm  g ( y 2  y1 )
A
B
A
B
atm
A
A
1
2
1
atm
atm
2
2
1
P  Patm  gh
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• Pressure at B is P=Patm+ρgh
Pressure at A > Patm
Pressure in a continuous static fluid is
the same at any horizontal level so,
Pressure at A = Pressure at B
P A = P B = Patm+ ρgh
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Blood Pressure measurements by cannulation
Pressure at c =pressure at D
PC = PD
PC = PA + ρsgh
PB = Patm + ρ gh
PA + ρsgh = Patm + ρ gh
PA = Patm + ρ g h – ρsgh
Pblood =PA = Patm + ρ g h – ρsgh
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Blood Pressure
Sphygmometer
• Blood pressure is measured
with a special type of
manometer called a
sphygmomano-meter
• Pressure is measured in mm
of mercury
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Pressure with depth
Estimate the amount by which blood pressure changes in an
actuary in the foot P2 and in the aorta P1 when the person is lying
down and standing up
Take density of blood = 1060kg/m3
Lying down
P2  P1  gh  0 Pa
s tan ding
P2  P1  1060  9.8 1.35  1.4 10 4 Pa
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26.8 K P a
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example
A fluid of constant density ρ = 960 kg / m3 is flowing steadily through the above
tube. The diameters at the sections are d 1 =100 mm and d 2 = 80 mm . The gauge
pressure at 1 is p1 = 200 k N/ m2 ,and the velocity here is u 1 = 5 m /s . We want to
know the gauge pressure at section 2.
The tube is horizontal, with y1 = y2 so
Bernoulli gives us the following equation
for pressure at section 2:
P2 = 2 x10 5 + 960 { 52 – ( 7.8 )2} /2
= 182796.8 pa
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An example of the U-Tube manometer Using a u-tube manometer to
measure gauge pressure of fluid density ρ = 700 kg/m3, and the
manometric fluid is mercury, with a relative density of 13.6.
What is the gauge pressure if:
h1 = 0.4m and h2 = 0.9m?
pB = pC
pB = pA + ρgh1
pB = pAtmospheric + ρman gh2
We are measuring gauge pressure so patmospheric
=0
pA = ρman gh2 - ρgh1
a) pA = 13.6 x 103 x 9.81 x 0.9 - 700 x 9.81 x 0.4
= 117 327 N,
b) pA = 13.6 x 103 x 9.81 x (-0.1) - 700 x 9.81 x 0.4
= -16 088.4 N, The negative sign indicates that the pressure is
below atmospheric
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example
A MANOMETER (U-TUBE) is a variation of the Barometer in which the pressure of two
gases may be compared. Here the difference in the pressures of the gases on the arms
of the manometer is equal to the pressure 'exerted' by the column of fluid, ph = ρ g h.
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Example:
The horizontal constricted pipe illustrated in the Figure, known as a Venturi
tube, can be used to measure the flow speed of an incompressible fluid.
Determine the flow speed at point 2 if the pressure difference P1 & P2 is
known.
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example
Calculate the column height of water
corresponding to a pressure of 50 kPa
P=ρgh
h = P/ρg
= 50 X 103 / 10 3 X 9.8
= 5.1 m
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Example:
Fluid is flowing from left to right through the pipe. Points A and B are at the same
height, but the cross-sectional areas of the pipe differ. Points B and C are at different
heights, but the cross-sectional areas are the same.
Rank the pressures at the three points, from highest to lowest.
A) A and B (a tie), C
B) C, A and B (a tie)
C) B, C, A
D) C, B, A
E) A, B, C
E) PA > PB > PC
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Example:
A stream of water of water d = 0.10 m flows steadily from a tank of diameter D = 1.0
m as shown in the figure. What flow-rate is needed from the inlet to maintain a
constant water volume in the header tank depth? The depth of water at the outlet is
2.0 m . Can regard outlet as a free jet (note water level at
(1) is not going down).
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Pressure Measurements:
• A long closed tube is filled with mercury
and inverted in a dish of mercury
• Measures atmospheric pressure as ρ g h
•Since the closed end is at vacuum, it
does not exert any force.
Vacuum pressure = 0
Patm
Patm
  gh  h 
g
p=0
patm
Vacuum!
h
For mercury, h = 760 mm.
How high will water rise?
No more than h = patm/g = 10.3 m
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example
erusaem ew erehw si ro dnuorg eht no si eloh eht emussa( We can set
atm. So we have1
evah osla eW .)morf thgieh
h
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example
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example
P2 =p1 + ρ (v12 –v22 ) / 2
= 180X103 + 103X(22 – 182)
= 20X 103 pa
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example
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If a fluid is at rest in a container, all portions of
the fluid must be in static equilibrium
• If the height doesn’t change much,
Bernoulli becomes: y1 = y2
P1  12 v12  P2  12 v22
FL
• Where speed is higher, pressure is
lower.
• Speed is higher on the long surface
of the wing – creating a net force of
lift.
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example
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example
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Water flows through the pipe as shown at a
rate of .015 m3/s. If water enters the lower
end of the pipe at 3.0 m/s, what is the
pressure difference between the two ends?
1.5m
A2 = 20 cm2
Q  AV
Q
0.015


 7.5 m /s
V 2 A 20 
4
10
1
1
2
2



g



 g y
y
p
V
V
1
2
1
1
2
2
2
2
1
2
2


p1 p2 2  V 2 V 1  g y 2  y1
1
3
2
2
3
P 
10 7.5  3 10  9.8  1.5
2
p







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Example
Estimate the force exerted on your eardrum due to the water above
when you are swimming at the bottom of the pool with a depth 5.0
m.
We first need to find out the pressure difference that is being exerted on the
eardrum. Then estimate the area of the eardrum to find out the force
exerted on the eardrum.
Since the outward pressure in the middle of the eardrum is the same as
normal air pressure
P
 Patm  W gh  1000  9.8  5.0  4.9 10 4 Pa
Estimating the surface area of the eardrum at 1.0cm2=1.0x10-4 m2, we
obtain
F  P  P0 A  4.9 10 4 1.0 10 4  4.9 N
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Absolute and Relative Pressure
How can one measure pressure?
One can measure pressure using an open-tube
manometer, where one end is connected to the
system with unknown pressure P and the other open
h
to air with pressure P0.
The measured pressure of the system P  P  gh
0
is
This is called the absolute pressure, because it is
the actual value of the system’s pressure.
In many cases we measure pressure difference with
P  P0  gh
respect to atmospheric pressure due to changes in P0
depending on the environment. This is called gauge or
relative pressure.
P
P0
The common barometer which consists of a mercury column with one end closed at
vacuum and the other open to the atmosphere was invented by Evangelista
Torricelli.
Since the closed end is at vacuum, P0  gh  (13.595 103 kg / m3 )(9.80665m / s 2 )(0.7600m)
it does not exert any force. 1 atm
is
 1.013 105 Pa  1atm
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Q: Is it possible to stand on the roof of a
five-story (20 m) building and drink, using
a straw, from a glass on the ground?
Even if a person could completely remove
all of the air from the straw, the height to
which the outside air pressure pushes
the water up the straw would not be high
enough for him/her to drink the water, no
matter how hard he/she sucks
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Applications of Bernoulli's
Equation
The tarpaulin that covers the cargo is flat when the truck is stationary but bulges
outward when the truck is moving.
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•Air speeds up in the constricted space between the
car & truck creating a low-pressure area. Higher
pressure on the other outside pushes them together.
•Hold two sheets of paper together, as shown
here, and blow between them. No matter how
hard you blow, you cannot push them more
than a little bit apart!
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