Download Gauge pressure as a function of depth d below the surface of a fluid

Fluids (continued) Last class:
Density: ρ =
m
V
 kg 
 3
m 
F
p
=
Pressure:
A
 N

 m 2 ≡ Pa 


5
=
1 atm 1.013x10
=
Pa 760
=
Torr 14.7 lb / in 2
Gauge pressure as a function of depth d below the surface of
a fluid:
p = ρgd
Absolute pressure as a function of depth d below the surface of
a fluid on earth:
p= po + ρgd (po = local atmospheric pressure)
Pascal’s principle: the change in pressure occurring in an
incompressible fluid, in a closed container, is transmitted
to every portion of the fluid and the walls of the container.
(allows for hydraulic lifts)
U-tubes (determining the density of an immiscible fluid)
Tube cross-sectional area A.
The pressure on the two
sides at the lowest dashed
line must be equal (or the
fluid would move).
H-h
H
h
p
p
pleft = p right
mug
mk g
po +
=
po +
A
A
mu = mk
Initially – fluid
of known density
ρk
Add column h of
fluid of unknown
density ρu
( ρ u > ρ k case)
ρ u Ah =
ρ k AH
H
ρu =
ρk
h
If ρ u < ρ k
The pressure on the two
sides at the lowest dashed
line must be equal (or the
fluid would move).
pleft = p right
h-H
H
h
p
p
mug
mk g
po +
=
po +
A
A
mu = mk
ρ u Ah =
ρ k AH
So again,
Initially – fluid
of known density
ρk
Add column h of
fluid of unknown
density ρu
H
ρu =
ρk
h
But now H < h.
Archimedes' principle
Consider the pressure on the top and lower faces of the cube of
water labeled C, somewhere in the tall stationary column of
water.
The pressure at the upper face, pt , is due
to the mass of the column of water above
C.
The pressure at the lower face, pl , is due to
the mass of the column above C plus the
mass of C itself.
With mf the mass of the fluid contained
in C, we can write that,
mf g
p=l p t +
A
pt
C
pl
area A
Or since, p = F
A
Fl Ft m f g
=
+
A A
A
F=l Ft + m f g
Fl is the force downwards due to C and the
column of water above it but since the water is
stationary there must be an equal and opposite
force, due to the surrounding water on the
lower face of C upwards.
Hence the net forces acting on C are
Fl − Ft = Ft + m f g − Ft = m f g
Ft
C
Fl mf g
So we’ve found that the water that surrounds C provides
a buoyant force Fb upwards given by,
Fb = m f g
This net force from the surrounding
water is actually independent of what
material occupies that volume.
Hence if we replace the cube of
water with a different material
having a different mass, m, Newton’s
2nd law gives,
Fb − mg =
ma
m f g − mg =
ma
C
Fb = mfg
mg
mf − m
a=
g
m
Period 2 class see slide at end
(ignoring hydrodynamic drag)
If the material has a smaller mass than
the equal volume of water that it displaced
(meaning that it has a lower density than
water, e.g. Styrofoam) then the acceleration
due to the buoyant force will be positive
(i.e. upwards).
Fb= mfg
If the material has a greater mass than
the equal volume of water that it displaced
(meaning that it has a higher density than
water, e.g. metal) then the acceleration
due to the buoyant force will be negative
(i.e. downwards), but smaller than its free fall
acceleration if only gravity were acting.
Styro
foam
mg
Another way to look at this is to to consider the apparent weight of
the object. Suppose the object is placed on a spring scale (in the
fluid). Then
FS + Fb − mg = ma = 0
So the scale reads,
=
FS mg − Fb
Thus its apparent weight is
the actual weight minus the
buoyant force.
C
FB
FS mg
The buoyant force depends (only)
on the weight of the fluid
displaced, i.e.
Fb = mf g
This is Archimedes' principle.
If the buoyant force exactly balances the weight of the object the
apparent weight FS = 0 and the object neither rises nor sinks.
The object is then said to be neutrally buoyant.
FS = mg − FB = 0
This is what permits a submarine to hover
at particular depth. A submarine has
internal ballast tanks that are designed to
be filled with seawater.
To dive a neutrally buoyant submarine
pumps water into these chambers making
the weight of the hull plus the water taken
on greater than the weight of the water
displaced by the hull (i.e. greater than the
buoyant force).
neutral
sinks
As the desired depth is approached
this water is blown out, making the
ship neutrally buoyant again.
To rise again, still more water is
blown out, making the submarine
have an apparent weight that is
negative (Fb > mg).
When the ship gets to the surface, it
continues to rise until its total weight
equals the weight of the water it
displaces.
neutral
rises
mg
mg
This sets the depth at which an object floats, i.e. an object will
sink into a fluid to a level until the weight of the object equals
the weight of fluid displaced.
Example
Floating in the very salt rich (dense) waters of the Dead Sea keeps
about 1/3 of your body above the water line. What is the density
of the water there? Assume your density to be 1.0 g/cm3.
Let your mass be m,
FB – mg = 0
but
FB = mf g , mf = displaced water
so
mf g – mg = 0
mf = m
ρfVf = ρV
mg
FB
1/3
2/3
mf = m
ρfVf = ρV
ρf = unknown density of the seawater
ρ = 1.0 g/cm3 (your density)
Now the volume of fluid displaced is 2/3 of your volume so,
Vf = 2/3V then,
ρf 2/3V = ρV
ρf 2/3 = ρ
ρf = 3/2ρ
ρf = 1.5 g/cm3
mg
FB
1/3
2/3
HITT challenge question
Iron
The iron block is thrown off the float into the pool.
Does the water level in the pool
A) go up
B) go down
C) stay the same
Answer
B) The water level actually goes down
Fully submerged the block displaces a quantity of water equal to
its volume.
On the float, the block displaces a quantity of water equal to its mass.
That displaced water raises the level in the pool
Since iron has greater density than water, it displaces more water
when on the float. So the water level is higher with the block on the
float.
Fluid Dynamics
Types of fluid flow:
We can visualize the flow of a fluid by injecting jets of smoke or dye
into the moving fluid. There are two distinct regimes of flow:
laminar flow and turbulent flow. These are both evident in the
figure below.
laminar
flow
turbulent
flow
laminar
flow
turbulent
flow
For laminar flow the velocity of the fluid at each point is a constant,
i.e. at each point the speed and the direction of the fluid remains the
same over time. For turbulent flow this is not true.
For the situation above two photographs taken at different times
would show that the streamlines on the left and upper right would
be unchanged, while the smoke pattern on the lower right would
be quite different.
laminar
flow
turbulent
flow
For laminar flow a volume element of the fluid that begins on a
streamline remains on that streamline and streamlines do not cross
each other. In the following developments we assume an ideal fluid
for which:
1) The flow is laminar
2) The fluid is incompressible
3) The fluid has very low (negligible) viscosity
Equation of continuity
Consider the flow of an ideal fluid through a uniform pipe.
Cross-sectional
area = A
∆x
If the fluid is flowing with speed v then an element of the
fluid goes a distance ∆x in time ∆t so
∆x
v=
→ ∆x = v∆t
∆t
The volume of fluid crossing a point along the pipe, in
that time is then ∆V = A∆x = Av∆t
The volume flow rate RV , which is the volume of fluid passing a
point along the pipe in given time can then be written,
R
=
V
∆V
= Av
∆t
So the volume flow rate equals the cross-sectional area of the
pipe times the velocity of the fluid going through it.
Given the volume flow rate through a hose, RV , we can determine
e.g. the time taken to fill a pool of volume VP , since with t unknown
we can write,
VP = R V t
VP
t=
RV
Consider now a pipe that has a non-uniform cross-section,
necking down as shown below
A1
A2
v2
v1
The volume of fluid crossing a point along the tube on the left (in the
fatter section) in time ∆t must equal the volume of fluid crossing a
point on the right (in the thinner section) in the same time, so
∆V(left) = ∆V(right)
A1v1∆t = A2v2∆t
A1v1 = A2v2
A1v1 = A2v2
A1
A2
v2
v1
This is the equation of continuity. It says that in order to keep the
volume flow rate constant,
∆V
R=
= Av
= constant
V
∆t
the velocity of the fluid in the thinner section must be greater (in
proportion to the ratio of the cross-sectional areas), i.e.
A1
v2 =
v1
A2
Example
A stationary lawn sprinkler consists of a pipe feeding a dead ended
container with 12 identical holes of 0.13 cm diameter. Water in the
1.9 cm (inner) diameter pipe has a speed of 0.91 m/s. With what
speed does the water leave the holes?
The pipe has cross-sectional area,
Ap = π (dp / 2)
2
The holes have total cross-sectional area,
2
A
=
(12)
π
d
/
2
( h ) (12 holes × area/hole)
h
Eqn. of continuity,
dp )
Ap
(
1  1.9 cm 
=
=
vh =
vp
vp
0.91 m / s
2


Ah
12  0.13 cm 
12 ( d h )
2
Ahvh = Apvp
vh= 16.2 m/s
2
A1
A2
v2
v1
For the ideal fluid the volume flow rate, R V (the volume of fluid passing
a point in unit time) is related to the mass flow rate, R m (the mass of
fluid passing a point in unit time) via,
R m = ρR V
The textbook uses the continuity equation and conservation of energy
to derive Bernoulli’s equation.
The Bernoulli Equation
P1 + 12 ρv12 + ρgy1= P2 + 12 ρv 22 + ρgy 2
For the flow of an ideal fluid this relates the absolute pressure and
flow velocity at one elevation to the absolute pressure and flow
velocity (in a possibly different diameter pipe) at a different elevation.
Examples
A1
P1
P2
A2
v2 = 15 m/s
v1
Water flows through the pipe and out into the atmosphere at 15 m/s.
The diameter of the right side is 3 cm and the left side is 5 cm.
a) What volume of water flows into the atmosphere in 10 minutes?
2
 0.03 m 
∆V =R V ∆t =A 2 v 2 ∆t =π 
 (15 m / s)(10 min)(60 s / min)
 2 
∆V =
6.4 m 3
b) What is the speed v1?
A1
P1
P2
A2
v2 = 15 m/s
v1
By the continuity equation
A1v1 = A2v2
2
 d2 
π(d 2 / 2)
A2
=
=
v1 =
v2
v2   v2
2
π(d1 / 2)
A1
 d1 
2
 3
v1 =
=
 (15 m / s) 5.4 m / s
5
2
b) What is the pressure P1?
A1
P1
P2
A2
v2 = 15 m/s
v1
By Bernoulli’s equation (for no change in elevation)
P1 + 12 ρv12 = P2 + 12 ρv 22
Since the right side flows into the atmosphere the pressure
P2 = 1 atm. Then
P1 = P2 + 12 ρv 22 − 12 ρv12
P1 = P2 + 12 ρ(v 22 − v12 )
P1 =
1.01x105 Pa + 12 (1000 kg / m 3 )[(15 m / s) 2 − (5.4 m / s) 2 )
P1 = 1.01x105 Pa + 9.8x10 4 Pa = 1.99x105 Pa
P1
1 atm
1.99x10
=
Pa
1.97 atm
5
1.01x10 Pa
5
The pressure in the larger pipe is greater than that in the smaller
pipe.
This is absolute pressure. If we were asked for gauge pressure,
Pgauge = Pabsolute – Patm= 1.97 atm – 1 atm = 0.97 atm
Example
The water intake at a hydroelectric power station has a 1.0 m
diameter and the inflow rate is 0.40 m/s. The inlet to the turbine at
the generator building has a smaller diameter and the flow rate
there is 9.5 m/s. What is the pressure difference between the
inlet to the turbine and the intake?
This is a straightforward application of Bernoulli’s equation
P1 + 12 ρv12 + ρgy1= P2 + 12 ρv 22 + ρgy 2
P2 − P1 =12 ρv12 − 12 ρv 22 + ρgy1 − ρgy 2
∆P =
1
2
ρ(v12 − v 22 ) + ρg(y1 − y 2 )
Let 2 be at the bottom (y2 = 0) then,
∆P
1
2
(1000 kg / m 3 )[(0.40m / s) 2 − (9.5 m / s) 2 ]
+ (1000 kg / m 3 )(9.8m / s2 )(180 m)
∆P =−4.5x10 4 Pa + 1.764x106 Pa
 1 atm

Pa 
17 atm
∆P 1.72x10
=

5
 1.01x10 Pa 
6
mf − m
a=
g
m
(ignoring hydrodynamic drag)
Period 2 class: The slides for this section are
correct. If the mass of the object, m, is less
than the mass of the fluid it displaces the
acceleration should be upwards and so the
acceleration should be positive (which is
what this expression gives).
Fb= mfg
The acceleration only comes out negative
(i.e. downwards) if m > mf again what
should be expected.
Styro
foam
mg