Unit 14
Fluid
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Density
Look at the two blocks below.
Both blocks are the same size (have the same volume).
However, one is Lead and the other is copper.
An internal look at these blocks reveals that there are many more atoms in the lead
block than there are in the copper block.
As a result, we say that the lead block is more dense than the copper block.
The formula for calculating density is as follows:
m

V
14-1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Density Calculations – WS 63 #2
The typical air in a room has a density of 1.29 kg/ m3.
Suppose your classroom has the dimensions of 3.5 m by 4 m by 2.5 m.
Calculate the volume of your classroom.
What is the mass of the air in your classroom?
V  lwh
m

V
14-2
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Continuity Equation – WS 65 #3
Consider the piping system below.
The black circles represent atoms of a fluid moving through the pipe.
The area on the left (A1) is twice as large as the area on the right (A2).
However, the amount of fluid per unit time flowing through this pipe is the same.
As a result, the speed of the fluid flowing through the right pipe segment (V2) must be twice
that as the speed on the left (V1).
The mathematical relationship used to demonstrate this fact is known as the continuity
equation.
A1v1  A2v2
A certain pipe has a radius of 0.25 m at point A (left) and a radius of 0.12 m at point B
(right).
If the fluid in the pipe is flowing at 5.2 m/s at point A, then how fast is it flowing at point B?
14-3
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
WS 65 #4
A piping system has a velocity of 14.8 m/s and a radius of 0.10 m at point
C. The radius at points A, B, and D are 0.50 m, 0.30 m, and 0.50 m
respectively.
Calculate the velocities of the fluid flowing through the pipes at points B
and D.
What do you think the velocity of the fluid at A will be? Why
Calculate the fluid velocity at A in order to prove or disprove your
prediction.
A
B
C
D
A1v1  A2v2
14-4
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Pressure and Bernoulli’s Principle – WS 66
Pressure – a force per unit area.
F
P
A
The unit of pressure is the Pascal (Pa) and is equal to a N/m2.
Bernoulli’s Principle – if the velocity of a fluid is high, then the pressure is low.
Conversely, if the velocity of a fluid is low, then the pressure is high.
When the pressure at a certain depth in a liquid is desired, we can use the following
formula:
P  yg
14-5
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Effects of Surface Area on Pressure
A man and a woman have made a footprint impression in the concrete by standing
on one foot.
They both weigh the same amount.
Which person, man or woman, exerted the most pressure on the ground?
The following problem is similar to WS 66 #1.
A man and a woman both weigh 750.0 N.
The woman’s shoe has a surface area that is 0.5 m2, and the area of the man’s shoe
is 0.8m2.
What are the pressures exerted on the ground by each of the people?
Woman
Man
14-6
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Pressure
A trunk full of valuable
treasure falls over as shown.
In which instance did the
trunk exert the most pressure
on the ground?
Why?
14-7
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Pressure
Which of the identical objects shown
below will exert the most pressure on
the ground below it?
Why?
14-8
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Pressure and Density in Fluids
The pressure experienced at any depth in fluid is constant everywhere at
that same depth.
The formula for calculating that pressure is given below.
WS 66 #3 - What is the pressure at the bottom of a swimming pool that is
2.25 m deep?
The density () of water is 1000.0 kg/m3.
P  yg
14-9
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Pressure at Various Depths
Suppose you had a coffee can full of water.
If there were three holes in the can, then from which hole would the water flow the
farthest? Why?
If we were to measure the fluid pressure at the top of the fluid, then we would see
that it would be “low.”
As we move the gauge deeper, we would observe and increase in the pressure.
However, if we were to move the gauge from left to right at the same depth, we
would see that there would be no change in pressure.
Pressure is constant at any given depth in a fluid that is opened to the atmosphere.
The same behavior is observed in oddly shaped containers.
14-10
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Bernoulli’s Principle
Bernoulli’s Principle – if the velocity of a fluid is high, then the pressure is
low. Conversely, if the velocity of a fluid is low, then the pressure is high.
The speed of the fluid coming into the pipe from the left is slow; therefore,
the pressure is high.
The speed of the fluid leaving the pipe is fast; therefore, the pressure is
low.
14-11
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Bernoulli’s Equation – WS 66
Water enters a house through the basement at a
speed of 0.60 m/s through a pipe that is 4.2 cm in
diameter.
When the water enters the house, it is under a
pressure of 303.9 kPa.
The water is pumped up to a height (y2) of 5.2 m
and out a faucet that is 2.7 cm in diameter.
Use the continuity equation to determine how fast
the water is going when it leaves the faucet.
Use Bernoulli’s equation to determine the pressure
of the water when it leaves the faucet.
A1v1  A2v2
1
1
2
P1   v1   gy1  P2   v2 2   gy2
2
2
14-12
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
WS 64 #5
At point A on the pipe to the left, the water’s speed is 4.8 m/s pressure is
52.0 kPa.
The water drops down 14.8 m to point B where the pipe’s cross sectional
area is twice that at point A.
Calculate the velocity of the water at point B.
Calculate the pressure at point B.
A
1
1
2
2
P1  v1  gy1  P2  v2  gy 2
2
2
B
14-13
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A Tie Race
Consider the two ball’s on the track below.
If they race and tie, then what can you tell me about the speed of the orange ball
when compared to the yellow ball?
Why?
The orange ball traveled further; therefore, it had to go faster in order to reach the
end at the same time as the yellow ball.
A similar effect helps to partially explain how a wing produces lift enabling an
airplane to fly.
14-14
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Bernoulli’s Principle & the Wing
Bernoulli’s Principle – if the velocity of a fluid is high, then the pressure is low. Conversely,
if the velocity of a fluid is low, then the pressure is high.
The distance across the top of the wing is farther; therefore, the top molecule must go faster
in order to reach the rear wing at the same time as the bottom molecule.
The air above the wing moves faster; therefore, the downward pressure acting on the top of
the wing is less than the upward pressure acting on the bottom of the wing.
Recall, P = F/A: therefore, F = PA. This means that the upward force on the wing is greater
than the downward force on the wing.
This difference in force results in the generation of lift which enables a plane to fly.
FUP
FDOWN
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
14-15
Laminar Flow – WS 67
Consider the piping system below.
Depending on the fluid and piping material properties, the fluid may move
easily and smoothly through the reduction in the pipe diameter.
We can replace the molecules with flow lines that represent the paths of the
different layers of fluid.
In the case of Laminar flow, the flow lines would look like those shown
below.
14-16
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Turbulent Flow – WS 65
Consider the piping system below.
Depending on the fluid and piping material properties, the fluid may find it
hard to move and may move roughly through the reduction in the pipe
diameter.
We can replace the molecules with flow lines that represent the paths of the
different layers of fluid.
In the case of Turbulent flow, the flow lines would look like those shown
below.
14-17
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Pressure Difference – WS 66 #1
When the pressure in two chambers is uniform, no fluid flows from one chamber to
the other.
However, when there is a pressure difference, fluid moves from one chamber to the
other.
In a fluid system, work is done when a pressure Difference causes liquids to move.
Notice how the pressure is constant everywhere after the pressure has equalized after the
valve is opened.
14-18
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Open Fluid System
An open fluid system is one in
which the fluid is not retained and is
not recirculated.
When you flush a toilet, the fluid
drains into a septic tank or a municipal
sewage system.
The toilet retention tank is then
refilled from a well or a municipal
water source.
The water from the septic tank does
not return to the municipal water
source.
As a result, a toilet is an open fluid
system.
14-19
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Closed Fluid System
An closed fluid system is one in which the
fluid is retained and is recirculated.
In isolated rest areas, the toilets use a
special oil instead of water.
When you flush the toilet, the fluid drains
into a septic tank.
The oil is clean because the solids and urine
automatically settle to the bottom.
The oil is able to flow over a baffle (left
side).
It is pumped from the left side of the tank
back up to the toilet retention tank.
This toilet is an example of a closed fluid
system because the fluid is retained and
recirculated.
14-20
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Work Done in Closed Fluid Systems
In a closed fluid system, the pressure throughout the system is constant as long as
there is no change in volume.
When the pump in a hydraulic cylinder is turned on, it moves a volume of fluid
from one side of the piston to the other.
When fluid is pumped from one side of the piston, a low pressure is established on
that side.
When fluid is pumped to the other side of the piston, a high pressure is established on that
side.
As a result, the pressure difference causes the piston to move in order to reestablish a
uniform pressure throughout the system.
The work done is calculated using the following equation.
V is the change in volume of the system.
The final system pressure is equal to the initial system pressure.
W  P  V
14-21
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Work Done in an Open Fluid System
One example of an open fluid system is shown below.
The fluid must be pumped from the reservoir to the storage tank.
Both the storage tank and the reservoir a under the influence of atmospheric
pressure.
In order to move the fluid up to the storage tank, the pump must be able to
overcome the pressure associated with the weight of the fluid moved.
This pressure may be calculated using the following equation where w is the
weight density of the fluid and h is the height to which the fluid is pumped.
P  w  h
The work done in an open fluid system may be calculated using the equation below
where V is the volume of fluid moved.
W  P  V
14-22
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
This presentation was brought to you by
Where we are committed to Excellence
In Mathematics And Science
Educational Services.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A
A
13-1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.