notes_13.pdf mechanical engineering Fluid

Fluid

1. CHAPTER 13
FLUIDS
• Density
! Bulk modulus
! Compressibility
• Pressure in a fluid
! Hydraulic lift
! Hydrostatic paradox
• Measurement of pressure
! Manometers and barometers
• Buoyancy and Archimedes Principle
! Upthrust
! Apparent weight
• Fluids in motion
! Continuity
! Bernoulli’s equation
To begin with ... some important definitions ...
DENSITY:

Mass
Volume
, i.e.,

ρ=
m
V
Dimension: ⇒

[M]
[L]
3
Units:

⇒ kg⋅m
−3
PRESSURE:

Force
Area
, i.e.,

P=
F
A
Dimension: ⇒

[M]
[L][T]
2
Units: ⇒ N⋅m
−2
⇒ Pascals (Pa)
FLUIDS
Liquids
Gases

2. Question 13.1: What, approximately, is the mass of air
in this room if the density of air is

1.29 kg⋅m
−3
?
Mass = Volume × Density
Volume of room ≈15 m×10 m×3 m

∴M≈450 m
3
×1.29kg⋅m
−3
=580 kg
(over half-a-ton!)

3. Question 13.2: How does the mass of a medium sized
apples compare with the mass of air in a typical
refrigerator? the density of cold air is

~1.3 kg/m
3
.
The weight of a
medium apple is
~1 N, so the mass
of a medium apple
is

~0.1 kg.
A typical refrigerator has a capacity of ~18 ft
3
.

∴18 ft
3
⇒18×(0.305 m)
3
=0.51 m
3
.
But 1 m
3
of air has a mass of

1.3 kg, so the mass of air in
the refrigerator is

0.51×1.3 kg=0.66 kg,
i.e., approximately the mass of 6 apples!
We don’t notice the weight of air
because we are immersed in air ... you
wouldn’t notice the weight of a bag of
water if it was handed to you
underwater would you?

4. definitions (continued) ...
BULK MODULUS:

B=
ΔP
ΔV
V
( )
Dimension:

⇒
[M]
[L][T]
2
Units: ⇒N⋅m
−2
Compressibility ⇒B
−1
Gases are easily compressed (B ⇒ very small).
Liquids and solids much less compressible.
V
ΔP
ΔV
ΔP
ΔP
ΔP
Atmospheric pressure:

P
o
=1×10
5
Pa
∴ Force on the ceiling from floor of room above is

pressure × area≈1×10
5
×(15 m×10 m)
≈1.2×10
7
N.
Why doesn’t it collapse under that weight ... ?
Because pressure operates equally in all directions!
When air molecules “bounce” off
the walls they produce an impulse:

FΔt=Δp⇒pressure
Since the molecules are traveling
with equal speeds in all directions
... the pressure is the same !

!
F Why?

5. Pressure at a depth in a fluid ...
Imagine a cylindrical body of the fluid with its top face at
the surface of the fluid. At equilibrium there is no net
force acting on the surfaces of the cylinder.

∴F
y∑=0 at the lower face,
i.e.,

P×A=P
!
×A+mg.
But the mass of fluid in the cylinder is

m=ρV=ρAh.

∴P=P
!
+ρgh.
What’s the pressure in water at a depth of 10m, say?

ρgh=1×10
3
kg/m
3
×9.81 m/s
2
×10 m≈10
5
Pa.

P
!
=P
at
≈1.01×10
5
Pa.

∴P≈2P
at
.

P
!
P

w=mg
h
Area=A

P
!
Question 13.3: A balloon has a radius of 10 cm. By how
much does the radius change if the balloon is pushed
down to a depth of 10 m in a large tank of water? (The
bulk modulus of air is 2×10
5
N⋅m
−2
.)

6. The pressure difference is

ΔP=hρg
But

B=
ΔP
ΔV
V
( )
.

∴
ΔV
V
=
ΔP
B
=
hρg
B
.
For an air-filled balloon at a
depth of 10m, we have

ΔV
V
=
10 m×1×10
3
kg/m
3
×9.81 m/s
2
2×10
5
N/m
2
≈0.5 (50%).
Assuming the balloon is spherical,

V=
4
3
πr
3
so ΔV=4πr
2
×Δr.

∴
ΔV
V
=
3Δr
r
≈0.5,
so, for an initial radius r=0.1 m,

Δr≈
0.5r
3
=0.017m(1.7 cm).
h
Δr
r
What’s the difference in the air pressure from the
ceiling to the floor in this room ... ?
The room height is ~3m so the pressure difference is:

ΔP=hρg=3×1.29×9.81≈38Pa

∴
ΔP
P
!
≈
38
1×10
5
≈3.8×10
−4
(0.038%)
i.e., negligible.
Pascal’s principle ...
If additional pressure ( ΔP) is
applied, it is transmitted
through the whole fluid:

∴P
1
=P
!
+h
1
ρg+ΔP
and

P
2
=P
!
+h
2
ρg+ΔP.
Blaise Pascal (1623-1662)

h
1

P
1

h
2

P
2

P
!
+ΔP

7. Hydraulic lift:
If a force

F
1
is applied to the left hand piston, the
additional pressure,

P
1
=
F
1
A
1
, is transmitted through the
whole fluid. Therefore, on the surface of the right hand
piston,

P
2
=
F
2
A
2
=P
1
.

∴
F
1
F
2
=
A
1
A
2
, i.e.,

F
2
=
A
2
A
1
⎛
⎝
⎜
⎞
⎠
⎟ F
1
.
Wow ... the force is amplified!!
Mechanical advantage
h

F
1
Area

A
1
Area

A
2

F
2
same pressure
Get a larger force OUT than you put IN? Too good to be
true?
No, not really, because, to do work (like lift something
heavy) the force

F
2
is applied through a distance

Δx
2
.
But by conservation of energy

F
2
Δx
2
=F
1
Δx
1
⇒Δx
1
=
A
2
A
1
Δx
2
.
So, although

F
1
<F
2
, it is applied through a greater
distance

Δx
1
>Δx
2
.
Examples:
• lifts
• dentist’s chair
• hydraulic brake systems

F
1
Area

A
1
Area

A
2

F
2

Δx
1

Δx
2

8. Here’s something surprising ...
no matter the shape of a vessel,
the pressure depends only on the
vertical depth. It is known as the
hydrostatic paradox.

P
!
+ρgh
h

P
!
P
!
P
!
P
!
We can use these ideas to measure pressure:
P

y
1

y
2
h

P
!
h

y
2

y
1
P=0

P
!
P
!
Manometer Barometer

P+ρgy
1
=P
o
+ρgy
2
0+ρgy
2
=P
o
+ρgy
1
This is absolute pressure
i.e.,

P−P
!
=ρgh. i.e.,

P
!
=ρgh.
This is the Gauge pressure Atmospheric pressure

9. Barometer

P
!
=ρgh, i.e.,

h=
P
!
ρg

P
!
≈1.01×10
5
Pa
Using water:

ρ=1×10
3
kg⋅m
−3

∴h=
1.01×10
5
1×10
3
×9.81
~10m.
Using mercury:

ρ=13.6×10
3
kg⋅m
−3

∴h=
1.01×10
5
13.6×10
3
×9.81
~0.76m
‘Standard pressure’ is 760mm of Hg.
h
P=0

P
!
P
!
DISCUSSION PROBLEM 13.1:
The drawing shows two
pumps, #1 and #2 to be used
for pumping water from a very
deep well (~30 m deep) to
ground level. Pump #1 is
submerged in the water at the
bottom of the well; the other
pump, #2, is located at ground
level. Which pump, if either, can be used to pump water
to ground level?
A: Both pumps #1 and #2.
B: Pump #1.
C: Pump #2.
D: Neither pump #1 nor pump #2.
#1
#2

10. Buoyancy and the concept of upthrust
Archimedes Principle: when an object is partially or
wholly immersed in a fluid, the fluid exerts an upward
force ... upthrust ... (or buoyant force, B) on the object,
which is equal to the weight of fluid displaced.
Submerged: w > B
Weight of object

w=mg=ρ
s
V
s
g
Weight of liquid displaced

=ρ
L
V
s
g
If

ρ
s
>ρ
L
the object will sink.

w=mg
B

V
s
If the object is floating then ... w=B.
Weight of object is

w=mg=ρ
s
V
s
g. If

V
L
is the
volume submerged, then the weight of liquid displaced is

ρ
L
V
L
g. But according to Archimedes principle, this is
equal to the upthrust (B).

∴ρ
s
V
s
g=ρ
L
V
L
g⇒V
L
=
ρ
s
ρ
L
V
s
.
Example: what volume of an iceberg is submerged?

ρ
s
=0.92×10
3
kg⋅m
−3
.

ρ
L
=1.03×10
3
kg⋅m
−3
.

∴
V
L
V
s
=
ρ
s
ρ
L
=
0.92×10
3
1.03×10
3
=0.89,
i.e., 89% of an iceberg is submerged!
What ... you don’t believe me ...

w=mg
B

V
s

V
L

11. Question 13.4: On Earth, an ice cube floats in a glass of
water with 90% of its volume below the level of the
water. If we poured ourselves a glass of water on the
Moon, where the acceleration due to gravity is about

1
6
th
of that on Earth, and dropped in an ice cube, how
much of the ice cube would be below the level of the
water?

12. When an ice cube floats, the
weight of the ice cube (

=ρ
s
V
s
g)
equals the weight of the water
displaced (

=ρ
L
V
L
g), which is
proportional to

V
L
, the volume of
the ice cube below the surface.

∴ρ
L
V
L
g=ρ
s
V
s
g,
i.e.,

V
L
=
ρ
s
ρ
L
V
s
,
which is independent of g. So, the volume submerged
would remain the same!
Since both the weight of an object, which is floating, and
the weight of a fluid it displaces are proportional to the
local value of g, the submerged volume does not depend
on g.
Question 13.5: A piece of copper of mass 0.50 kg is
suspended from a spring scale. If it is fully submerged in
water, what is the reading (in N) on the spring scale?
(The density of copper is

9.0×10
3
kg⋅m
−3
.)

13. Identify the forces acting on the block.
At equilibrium

F
y∑=T+B+(−w)=0.


∴T=w−B=ρ
s
V
s
g−ρ
L
V
s
g=ρ
s
V
s
g1−
ρ
L
ρ
s
⎛
⎝
⎜
⎞
⎠
⎟ .
True weight

=mg. Upthrust
But

ρ
L
ρ
s
=
1×10
3
kg/m
3
9.0×10
3
kg/m
3
=0.111.

∴T=0.5×9.81×(1−0.111)=4.36 N (0.444kg).
In air instead of water ...

ρ
air
ρ
s
=
1.29 kg/m
3
9.0×10
3
kg/m
3
=0.143×10
−4
.
So, the mass of the block would be

0.9998mg less than
in vacuum (0.50 kg).
T

w=mg=V
s
ρ
s
g B
T(apparent weight)
Question 13.6: A beaker containing water is placed on a
balance; its combined mass reading on the balance is
1.200 kg. In (a) below, a copper block is hanging freely
from a spring scale, which has a mass reading of 0.20 kg.
If the copper block is totally immersed in the water, as
shown in (b), what are the readings on the balance and
the spring scales?
(The density of copper is

9.0×10
3
kg⋅m
−3
.)
1.200kg
??
0.20kg
??
(a) (b)

14. (a) Initially, the spring scale registers the weight of the
copper block and the balance registers the weight of
(water + beaker). (b) When the block is lowered into
the water, the water exerts an upward buoyant force (the
upthrust) on the copper block. Then

T
2
+B=mg, i.e.,

T
2
=mg−B,
where B is the upthrust, which is equal to the weight of
water displaced, i.e.,

B=ρ
w
V
s
g. The submerged
volume is

V
s
=
m
s
ρ
s
=
0.20 kg
9.0×10
3
kg/m
3
=2.22×10
−5
m
3
1.200kg
0.20kg
(a)(a)

T
1

mg
??
??
(b)

mg

T
2 B

∴T
2
=m−ρ
w
V
s
( )g

=0.20 kg−1×10
3
kg/m
3
×2.22×10
−5
m
3
( )
g

=0.178g,
i.e., the reading on the spring scale is

0.178 kg.
The upthrust B is the force on the water on the block; by
Newton’s 3rd law the block must exert an equal and
opposite force on the water. Consequently, the reading
on the balance will increase by 0.178 kg, i.e., it will read
1.378 kg when the bock is suberged.
So, when you dip a teabag into your cup, the weight of
the teabag is reduced, but the weight of the cup (and
contents) is increased!
Image from Paul Hewitt’s
Conceptual Physics website:
http://www.arborsci.com/ConceptualPhysics/

15. Question 13.7: A barge, loaded with steel cannisters, is
floating in a lock. If the cargo is thrown into the water,
what happens to the level of water at the side of the
lock? Does it rise, stay the same, or fall?
With cargo on the barge, as in (a), the weight of water
displaced is equal to the weight of the barge plus the
cargo, i.e.,

w
w
(a)=w
b
+w
s
=w
b
+NV
s
ρ
s
g,
where

w
b
is the weight of the barge, N is the number of
blocks,

V
s
is the volume of each block and

ρ
s
their
density. When the blocks are thrown into the water, as
in (b), the weight of water displaced is equal to the
weight of the barge plus the weight of the water
displaced by the blocks, i.e.,

w
w
(b)=w
b
+NV
s
ρ
w
g.
Since

ρ
w
<ρ
s
,

w
w
(b)<w
w
(a), i.e., less water is
displaced in (b) than in (a). Since the volume of the lock
is the same, the depth of water is less in (b) than in (a),
i.e., the water level falls.
(a) (b)

16. Question 13.8: A block of balsa wood with a rock tied to
it floats in water. When the rock is on top, exactly one-
half of the wood is submerged below the water line. If
the block is turned over so that the rock is now
underneath and submerged, what can you say about the
amount of the block below the water line? Is it less than
one-half of the block, one-half of the block, more than
one-half of the block.
By Archimedes principle, the rock plus wood displaces
its combined weight of water whether the rock is on top
or underneath. So, the weight and volume of water
displaced is the same in both cases. If the volume of the
block is

V
b
and the volume of the rock is

V
r
, when the
rock is on top the volume of water displaced is

0.5V
b
. If
the fraction of the wood submerged when the rock is
underneath is

xV
b
, then, since the volume of water
displaced with the rock underneath is the same as with
the rock on top,

0.5V
b
=xV
b
+V
r
,
i.e., x<0.5, so less than one-half of the wood is
submerged.

17. Fluids in motion; mass continuity
We assume the fluid is incompressible, i.e., a liquid, so
there is no change in density from 1→2. Then, the mass
and volume must be conserved from 1→2 as the liquid
flows down the pipe,
i.e.,

V
1
=V
2
.
Then in time Δt, we have

A
1
v
1
Δt=A
2
v
2
Δt,
i.e.,

A
1
v
1
=A
2
v
2
⇒ constant.
This is called the continuity equation for an
incompressible liquid. The conserved quantity ...

Area × velocity
... is called the volume flow rate

dV
dt
(m
3
⋅s
−1
)
⎛
⎝
⎜
⎞
⎠
⎟ .

v
1
Δt

v
2
Δt

Area A
1
Area A
2

v
1

v
2
If the density changes (from

ρ
1
⇒ρ
2
) then, since mass is
conserved we have ...

m
1
=m
2
i.e.,

A
1
v
1
Δt
( )
ρ
1
=A
2
v
2
Δt
( )
ρ
2

∴A
1
v
1
ρ
1
=A
2
v
2
ρ
2
This is the mass continuity equation.

v
1
Δt

v
2
Δt

Area A
1
Area A
2

v
1

v
2

18. The continuity equation in everyday life ...
[1] The garden hose:
If you squeeze the end of a
garden hose the area is
reduced and so the water
velocity increases, since

A
1
v
1
=A
2
v
2
i.e.,

v
2
=
A
1
A
2
⎛
⎝
⎜
⎞
⎠
⎟ v
1
.
If you don’t restrict the pipe too much, the volume flow
rate remains constant, which means you will fill a bucket
in the same time whether the end of the hose is restricted
or not!
[2] Lanes at highway tolls:
~ See useful notes on web-site ~

v
2

v
1

A
1

A
2
Bernoulli’s equation
We state Bernoulli’s equation without proof. It relates
the pressure (P), elevation (y) and speed (v) of an
incompressible fluid in steady (i.e., non-turbulent) flow
down a pipe,

P
1
+ρgy
1
+
1
2
ρv
1
2
=P
2
+ρgy
2
+
1
2
ρv
2
2
,
i.e.,

P+ρgy+
1
2
ρv
2
= constant.

Note, if fluid is at rest

v
1
=v
2
=0.

∴P
2
−P
1
=ρg(y
2
−y
1
)⇒ΔP=ρgΔy,
a result we obtained earlier.
“potential energy” “kinetic energy”
Area

A
1
Area

A
2

v
1

v
2

y
2

y
1

19. h

y
2

y
1
x
Question 13.9: A large tank of water has an outlet a
distance h below the surface of the water. (a) What is
the speed of the water as it flows out of the hole? (b)
What is the distance x reached by the water flowing out
of the hole? (c) What value of h would cause the water
to reach a maximum value of x? You may assume that
the tank has a very large diameter so the level of the
water remains constant.
(a) We start with Bernoulli’s equation

P
1
+ρgy
1
+
1
2
ρv
1
2
=P
2
+ρgy
2
+
1
2
ρv
2
2
.
But,

P
2
=P
1
=P
!
since both the hole and the top of the
tank are at atmospheric pressure. Then

∴
1
2
ρv
1
2
=ρg(y
2
−y
1
)+
1
2
ρv
2
2
=ρgh+
1
2
ρv
2
2
.
However, if

A
2
>>A
1
, then

v
1
>>v
2
; in fact, we were
told that

v
2
=0. Then

1
2
ρv
1
2
=ρgh, i.e.,

v
1
=2gh.
(Does this seem familiar?)
h

y
2

y
1
x
1
2

A
2

A
1

20. (b) To find x we model the water leaving the hole as a
projectile. The time it takes for a volume element of
water to exit the hole and reach the ground is given by,

y
1
2
=v
yi
t−
1
2
gt
2
,
but

v
yi
=0, as the water emerges horizontally.

∴t=
2y
1
g
.
The “range” is

x=v
xi
t, but

v
xi
=v
1
, i.e., in projectile
motion it remains constant.

∴x=v
1
2y
1
g
=2hy
1
.
Also, substituting for h we find

x=2y
1
y
2
−y
1
2
( )
.
h

y
2

y
1
x
1
2

A
2

A
1
(c) To find the maximum value of x, we take

x=2y
1
y
2
−y
1
2
( )
and set

dx
dy
1
=0. (Note: we were told that the water level
remains constant, so

y
2
is constant.)

∴
dx
dy
1
=
1
2
(2)y
1
y
2
−y
1
2
( )
−
1
2
(y
2
−2y
1
)=0.
This equation is satisfied if

(y
2
−2y
1
)=0,
i.e., when

y
1
=
1
2
y
2
.
Thus, the hole should be halfway between the bottom of
the tank and the surface of the water.

21. Question 13.10: A large tank of water has an outlet a
distance d below the surface of the water. The outlet is
curved so that the exiting water is directed vertically
upward. Prove that the vertical height achieved by the
water, h, is equal to the depth of the outlet below the
surface of the water, d.
h d
Using Bernoulli’s equation we have

P
1
+ρgy
1
+
1
2
ρv
1
2
=P
2
+ρgy
2
+
1
2
ρv
2
2
.
But

P
1
=P
2
=P
!
. Also, if the initial speed of an element
of water is

v
1
, then

v
2
2
=v
1
2
=2(−g)h.
But

v
2
=0.

∴v
1
=2gh.
Substituting in Bernoulli’s equation we obtain

1
2
ρv
1
2
=ρg(y
2
−y
1
), i.e.,

1
2
ρ(2gh)=ρgd,
since

(y
2
−y
1
)=d, then
d=h.
h d
1
2