fluids physics

1. Chapter 11
FLUIDS

2. 11.1MassDensity
DEFINITION OF MASS DENSITY
The mass density of a substance is the mass of a
substance divided by its volume:
V
m


SI Unit of Mass Density: kg/m3
A convenient way to compare densities is to use the concept of
specific gravity
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 4°𝐶
=
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
1000 𝑘𝑔/𝑚3

3. 11.1MassDensity

4. 11.1MassDensity
Example 1 Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about
5.2x10-3 m3 of blood.
(a) Find the blood’s weight and (b) express it as a
percentage of the body weight.
   kg
5
.
5
m
kg
1060
m
10
2
.
5 3
3
3



 

V
m
   N
54
s
m
80
.
9
kg
5
.
5 2


 mg
W
%
8
.
7
%
100
N
690
N
54
Percentage 



5. 11.2 Pressure
A
F
P 
SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal

6. 11.2 Pressure
Example 2 The Force on a Swimmer
Suppose the pressure acting on the back
of a swimmer’s hand is 1.2x105 Pa. The
surface area of the back of the hand is
8.4x10-3m2.
(a)Determine the magnitude of the force
that acts on it.
(b) Discuss the direction of the force.
A
F
P 
  
N
10
0
.
1
m
10
4
.
8
m
N
10
2
.
1
3
2
3
2
5





 
PA
F
Since the water pushes perpendicularly against the back of the hand, the
force is directed downward in the drawing.

7. 11.2 Pressure
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere

8. 11.3PressureandDepthina StaticFluid
0
1
2 



 mg
A
P
A
P
Fy
mg
A
P
A
P 
 1
2

V
m
but 
g
V
A
P
A
P 

 1
2
g
Ah
A
P
A
P 

 1
2
gh
P
P 

 1
2
The equation indicates that if the pressure p1 is known at higher level,
the larger pressure p2 at deeper level can be calculated by adding gh


9. 11.3PressureandDepthina StaticFluid
Conceptual Example 3 The Hoover Dam
Lake Mead is the largest wholly artificial
reservoir in the United States. The water
in the reservoir backs up behind the dam
for a considerable distance (120 miles).
Suppose that all the water in Lake Mead
were removed except a relatively narrow
vertical column.
Would the Hoover Same still be needed
to contain the water, or could a much less
massive structure do the job?

10. 11.3PressureandDepthina StaticFluid
Example 4 The Swimming Hole
Points A and B are located a distance of 5.50 m beneath the surface
of the water. Find the pressure at each of these two locations.
gh
P
P 

 1
2
     
Pa
10
55
.
1
m
50
.
5
s
m
80
.
9
m
kg
10
00
.
1
Pa
10
01
.
1
5
2
3
3
pressure
c
atmospheri
5
2







 

 

P

11. 11.4PressureGauges
gh
P
P 

 1
2
gh
Patm 

 
  
mm
760
m
760
.
0
s
m
80
.
9
m
kg
10
13.6
Pa
10
01
.
1
2
3
3
5






g
P
h atm


12. 11.4PressureGauges
A
B P
P
P 

2
gh
P
PA 

 1
gh
P
P atm 







pressure
gauge
2
absolute pressure is the actual value of p2
Check your understanding page 316

13. 11.5Pascal’sPrinciple
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.

14. 11.5Pascal’sPrinciple
 
m
0
1
2 g
P
P 


1
1
2
2
A
F
A
F










1
2
1
2
A
A
F
F

15. 11.5Pascal’sPrinciple
Example 7 A Car Lift
The input piston has a radius of 0.0120 m
and 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 input
piston has a negligible weight and the bottom
surfaces of the piston and plunger are at
the same level. What is the required input
force?









1
2
1
2
A
A
F
F
   
 
N
131
m
150
.
0
m
0120
.
0
N
20500 2
2
2 



F

16. 11.6Archimedes’Principle
ARCHIMEDES’ PRINCIPLE
Any fluid applies a buoyant force to an object that is partially
or completely immersed in it; the magnitude of the buoyant
force equals the weight of the fluid that the object displaces:
 
fluid
displaced
of
Weight
fluid
force
buoyant
of
Magnitude
W
FB 

17. 11.6Archimedes’Principle
 A
P
P
A
P
A
P
FB 1
2
1
2 



gh
P
P 

 1
2
ghA
FB 

hA
V 
 g
V
FB
fluid
displaced
of
mass



18. 11.6Archimedes’Principle
If the object is floating then the
magnitude of the buoyant force
is equal to the magnitude of its
weight.
Do example 9 , 10 and 11

19. 11.6Archimedes’Principle
Example 9 A Swimming Raft
The raft is made of solid square
pinewood. Determine whether
the raft floats in water and if
so, how much of the raft is beneath
the surface.

20. 11.6Archimedes’Principle
   
N
47000
s
m
80
.
9
m
8
.
4
m
kg
1000 2
3
3
max



 g
V
Vg
F water
water
B 

    m
8
.
4
m
30
.
0
m
0
.
4
m
0
.
4 

raft
V

21. 11.6Archimedes’Principle
   
N
47000
N
26000
s
m
80
.
9
m
8
.
4
m
kg
550 2
3
3




 g
V
g
m
W raft
pine
raft
raft 
The raft floats!

22. 11.6Archimedes’Principle
g
Vwater
water


N
26000
B
raft F
W 
If the raft is floating:
     
2
3
s
m
80
.
9
m
0
.
4
m
0
.
4
m
kg
1000
N
26000 h

     m
17
.
0
s
m
80
.
9
m
0
.
4
m
0
.
4
m
kg
1000
N
26000
2
3


h

23. 11.6Archimedes’Principle
Conceptual Example 10 How Much Water is Needed
to Float a Ship?
A ship floating in the ocean is a familiar sight. But is all
that water really necessary? Can an ocean vessel float
in the amount of water than a swimming pool contains?

24. 11.6Archimedes’Principle

25. 11.7FluidsinMotion
In steady flow the velocity of the fluid particles at any point is constant
as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a
point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity
of the fluid particles at a point change erratically in both magnitude and
direction.

26. 11.7FluidsinMotion
Fluid flow can be compressible or incompressible. Most liquids are
nearly incompressible.
Fluid flow can be viscous or nonviscous.
An incompressible, nonviscous fluid is called an ideal fluid.
When the flow is steady, streamlines are often used to represent
the trajectories of the fluid particles.

27. 11.7FluidsinMotion
Making streamlines with dye
and smoke.

28. 11.8TheEquationof Continuity
The mass of fluid per second that flows through a tube is called
the mass flow rate.

29. 11.8TheEquationof Continuity
2
2
2
2
v
A
t
m



 1
1
1
1
v
A
t
m





distance
t
v
A
V
m 


 

If a fluid enters one end of a pipe at a certain rate (e.g., 5 kilograms per
second), then fluid must also leave at the same rate, assuming that there
are no places between the entry and exit points to add or remove fluid.

30. 11.8TheEquationof Continuity
2
2
2
1
1
1 v
A
v
A 
 
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

31. 11.8TheEquationof Continuity
Incompressible fluid: 2
2
1
1 v
A
v
A 
Volume flow rate Q: Av
Q 

32. 11.8TheEquationof Continuity
Example 12 A Garden Hose
A garden hose has an unobstructed opening
with 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 hose
through (a) the unobstructed opening and (b) an obstructed
opening with half as much area.
(a) Av
Q 
    s
m
936
.
0
m
10
2.85
s
30.0
m
10
00
.
8
2
4
-
3
3






A
Q
v
(b) 2
2
1
1 v
A
v
A     s
m
87
.
1
s
m
936
.
0
2
1
2
1
2 

 v
A
A
v

33. 11.9Bernoulli’sEquation
The fluid accelerates toward the
lower pressure regions.
According to the pressure-depth
relationship, the pressure is lower
at higher levels, provided the area
of the pipe does not change.

34. 11.9Bernoulli’sEquation
   
2
2
2
2
1
1
2
1
2
1
nc mgy
mv
mgy
mv
W 



       V
P
P
As
P
s
F
s
F
W 1
2 





 
     
2
2
2
2
1
1
2
1
2
1
1
2 mgy
mv
mgy
mv
V
P
P 




     
2
2
2
2
1
1
2
1
2
1
1
2 gy
v
gy
v
P
P 


 




2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 



 BERNOULLI’S EQUATION
In steady flow of a nonviscous, incompressible fluid, the pressure, the
fluid speed, and the elevation at two points are related by:
2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 





35. Home work
Chapter 11 problem no 27, 30, 39, 60, 61 and 69

36. 11.10Applicationsof Bernoulli’sEquation
Conceptual Example 14 Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.

37. 11.10Applicationsof Bernoulli’sEquation

38. 11.10Applicationsof Bernoulli’sEquation

39. 11.10Applicationsof Bernoulli’sEquation

40. 11.10Applicationsof Bernoulli’sEquation
Example 16 Efflux Speed
The tank is open to the atmosphere at
the top. Find and expression for the speed
of the liquid leaving the pipe at
the bottom.

41. 11.10Applicationsof Bernoulli’sEquation
2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 




atm
P
P
P 
 2
1
0
2 
v
h
y
y 
 1
2
gh
v 
 
2
1
2
1
gh
v 2
1 

42. Archimedes supposedly was asked to
determine whether a crown made for the king
consisted of pure gold. According to legend,
he solved this problem by weighing the crown
first in air and then in water as shown.
Suppose the scale read 7.84 N when the
crown was in air and 6.84 N when it was in
water. What should Archimedes have told the
king?

48. Mercury is poured into a U-tube as in The
left arm of the tube has cross-sectional area
A1 of 10.0 cm2, and the right arm has a
cross-sectional area A2 of 5.00 cm2. One
hundred grams of water are then poured
into the right arm as in Figure P14.18b. (a)
Determine the length of the water column
in the right arm of the U-tube. (b) Given
that the density of mercury is 13.6 g/cm3,
what distance h does the mercury rise in
the left arm?

51. v
h
y
A
B
C
O
l A siphon is used to drain water from a
tank (beside). The siphon has a
uniform diameter. Assume steady flow
without friction, and h=1.00 m. You
want to find the speed v of the outflow
at the end of the siphon, and the
maximum possible height y above the
water surface.
Example
Fluid dynamics
l Use the 5 step method
ç Draw a diagram that includes all the relevant quantities for this
problem. What quantities do you need to find v and ymax ?

52. Example: Solution
Fluid dynamics
lDraw a diagram that includes all the relevant quantities for this problem.
What quantities do you need to find v and ymax ?
ç Need P and v values at points O, A, B, C to find v and ymax
ç At O: P0=Patm and v0=0
ç At A: PA and vA
ç At B: PB=Patm and v0=v
ç At C: PC and vC
ç For ymax set PC=0
v
h
y
A
B
C
O

53. Example: Solution
Fluid dynamics
What concepts and equations will you use to solve this problem?
◦ We have fluid in motion: fluid dynamics
◦ Fluid is water: incompressible fluid
◦ We therefore use Bernouilli’s equation
◦ Also continuity equation

54. Example: Solution
Fluid dynamics
lSolve for v and ymax in term of symbols.
ç Let us first find v=vB
ç We use the points O and B
where : P0=Patm=1 atm and v0=0 and y0=0
where: PB=Patm=1 atm and vB=v and yB=-h
ç Solving for v
v
h
y
A
B
C
O

55. Example: Solution
Fluid dynamics
lSolve for v and ymax in term of symbols.
ç Incompressible fluid: Av =constant
ç A is the same throughout the pipe  vA= vB= vC = v
ç To get ymax , use the points C and B (could also use A)
where: PB=Patm=1 atm and vB=v and yB=-h
set : PC=0 (cannot be negative) and vC=v and yC= ymax
ç Solving for ymax
v
h
y
A
B
C
O

56. Example: Solution
Fluid dynamics
lSolve for v and ymax in term of numbers.
ç h = 1.00 m and use g=10 m/s2
ç Patm=1 atm = 1.013  105 Pa (1 Pa = 1 N/m2 )
ç density of water water = 1.00 g/cm3 = 1000 kg/m2

57. Example: Solution
Fluid dynamics
l Verify the units, and verify if your values are plausible.
ç [v] = L/T and [ymax] = L so units are OK
ç v of a few m/s and ymax of a few meters seem OK
◦ Not too big, not too small
l Note on approximation
ç Same as saying
PA= PO =Patm or vA=0
ç i.e. neglecting the flow in
the pipe at point A
v
h
y
A
B
C
O