Chapter 10 Fluids - Conroe High School
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Transcript Chapter 10 Fluids - Conroe High School
Chapter 10
Fluids
Phases
Solid
Liquid
Gas
Fluids
Plasma
Density
p “rho”
p = m/v
SI Units
Sometimes g/ cm3
1 kg/m3 = .001 g/cm3
Ex: Al p= 2.7 g/cm3
=2700 kg/m3
Kg/m3
specific gravity – the ratio of the density
of that substance to the density of
water at 4°C
SG
No Units
The SG of any substance will be equal
numerically to its density in g/cm3 or
10-3 times its density in kg/m3
continued…
SG
Pb=
11.3
Alcohol = .79
Al= 2.7
SG will tell you if substance floats or
not
>1 sink
<1 Float
Atmospheric Pressure
Pressure due to the Atmosphere,
changes with depth
Earth’s Atmosphere is complicated
P for air changes
No distinct top surface to measure h
Atmospheric Pressure is
1.013 x 105 Pa
14.7 psi
10 -3
This is another unit: Atm
1 atm = 1.013 x 105 N/m2
(Pa) = 101.3 kPa
another unit is the bar
1 bar= 1 x 105 N/m2
1 bar= 100 kPa
Gauges measure pressure. They
measure over and above
atmospheric pressure. To get
absolute pressure, one must add
atmospheric pressure to gauge
pressure.
P= Patm + PG
Example
car tire gauge reads 220 kPa,
Absolute pressure within the
tire is 220 kPa + 101 kPa
=321 kPa
33 psi + 14.7 psi = 47.7 psi
Pressure
P= f/A
Force/ Area
Force applied
to area
SI unit is N/m2
Pascal, Pa
1 Pa = 1 N/m2
PSI?
Feet
psi
Fluids exert a pressure equal in all
directions
*overhead picture*
car tire, swimming pool
In fluids, force always acts
to
the surface
as depth increases within a fluid,
so does pressure
Formulas
P= f/A
f= mg
=ma
m= pv
m=pAh
P=pA hg
A
P= pgh
Pressure
is directly
proportional to density of
liquid and to depth within
liquid
This is just for the liquidNOT any external force on
the liquid
Example Prob.
Blaise Pascal, French 1623 - 1662
States that pressure applied to a
confined fluid increases the pressure
throughout by the same amount
Pascal's Principle carries with it
hydraulics (pg. 280)
Changing the Area changes the force
For this to be true, the fluid must not
compress (effectively they don’t)
Pin = Pout
Input
Output
Fout Fin
Aout Ain
Fin= Fout(Ain)
Aout
A small force can be used to exert
a larger force by making the area
of one piston larger than the area
of another
Small input area, Large output area
greatly multiplies the input force
F= 200
in
A= 100
F = 2000
out
A= 1000
Fout
Fin
MA
Mechanical advantage
If area is 20x greater then output
force will be 20x greater
Homework!
Buoyancy
3
types of buoyancy:
+ rise
- sink
neutral equilibrium
All objects appear to weigh less
when submerged in a fluid
Why a buoyant force?
pg. 283
F1
F2> F1
F2
F2 is greater b/c
there is more
pressure at the
lower depth
This is how we derive the formula…
FB = F2 – F1
FB = Apf gh2-Apf gh1
= Apfg(h2- h1)
Apfgh
(Ah = Vol)
pfgV
Also… pV = mass
mfg=fB
Recall
F= AP
p= m/v
w= mg
Vol= Ah
so….
Fbuoyant = pfgV= mfg
This is Archimedes Formula
(Principle)
In English=
The buoyant force on a
body immersed in a fluid
is equal to the weight of
the fluid displaced by the
object
Lets Test out Archimedes Idea
and see if he was right…
65.95
.646
Find m of object ____g
weight ___N
3
3
7
7 x 10-6
Find V of Object ____cm
_____m
58.5 g
.573
Find m of submerged ____
weight ____N
Find buoyant force on object
1.
2.
3.
4.
fB = weight (air) – weight (submerged)
= _______
.646 N - _______
.573
N
.073
fB= _______N
Now check w/ formula
fB= pfgV
7 x 10 - 6
3
= 1 x 103 kg/m3(9.8m/s2) __________m
.069
= ______
What about fB = Mfg
% error = 5.7%
Why
do steel ships float?
How can fish suspend
themselves in H20
How do submarines
work?
Restatement :
fB=
Wf = pf Vg
The volume of an object can also
be found by:
Vf=
m
pf
=
w
gp
Archimedes Principle applies to both
submerged and floating objects
fB =
weight of object
objects
pfVdispg = poVog
Vdisp = po
Vo
pf
--for floating
--g cancels
Shortcut– sometimes applied (if p is
known)
Density of floating object = fraction
of object that density of fluid
floating in is submerged.
wood Log
900 kg/m3
water
1000 kg/m3
9/10 of log is
underwater
= .9
90 % of log is
submerged
fluid dynamics (hydrodynamics if water)
If flow is smooth- streamline or Laminar
if flow is erratic, currents present –
turbulent
Viscosity- internal resistance within fluid
Syrup (high viscosity)
Water (low viscosity)
Flux- term used to describe
Volume of fluid passing through a
given area each second
Velocity
D= Vt
Area
Rate of Flow - assume ideal fluid,
frictionless, laminar
flow
rate R
units= m3/s (volume/ time)
m/s x m2
(m3/s)
R= VA
Velocity x Area
Equation of continuity states that
rate of flow remains constant.
Velocity and area will change
(inversely), but rate (m3/s) stays
the same.
Area 1
Velocity 1
Area 2
Velocity 2
because R(m3/t) stays the same…
A1V1= A2V2
Large Velocity – Small Area
V1= A2V2
A1
Fast
Small Velocity- Large Area
River Ex.
Slow
Narrow
Wide
Blood Flow Example pg. 288
Rate of flow of blood in human
body stays same –Big Aorta to
Small Capillaries
Follow along w/ the book to see
how the sample problem is solved.
We can also find Power of a
moving fluid
Power = PR
Power = pressure x flow rate
N/m2 x m3/s
Nm/s
J/s
Power = work / time
Where the velocity of a fluid is
high, the pressure is low, where
the velocity is low, the pressure is
high.
1
2
V slow
V high
P high
P low
continued…
This
makes sense because if
the pressure at 2 were high
it would slow down the
fluid in 1, because the fluid
has sped up, this
corresponds to a lower
pressure
Want proof?!
Car’s Convertible Top -- jeep
Tarp in back of truck
carburetor
airplane wing
chimneys/ outhouses
perfume atomizer
Ventilation in burrows
hanging ping pong balls
ping pong ball = funnel
Here’s the equation
P1 + ½ pV12 + pgy1 = P2 + 1/2pV22 + pgy2
When solving for P2 the formula looks like
this:
P2 = P1 + pg(y1-y2)+ ½ p(v21-v22)
P= pressure
p= density of flowing fluid
g= gravity
y= height
V= velocity
**Overhead Picture**
Bernoulli’s
Equation is
really an expression of the
law of energy conservation.
The formula is derived from
work energy principle - pg. 290
Go to overhead for equation…
h
A
Vel