Transcript AP Physics - Greensburg Salem

AP Physics
FLUIDS & THERMODYNAMICS
Fluids
 Fluids are substances that can flow, such as
liquids and gases, and even some solids

We’ll just talk about the liquids & gases
 Review of Density (remember this from chem?)
 ρ = m/V
 ρ = density
 m = mass (kg)
 V = volume (m3)
 density units: kg/m3
Pressure
 P = F/A
 P = pressure (Pa)
 F = force (N)
 A = area (m2)
 Units for pressure: Pascals
 1 Pa = 1 N/m2
 Pressure is always applied as a normal force
on a surface. Fluid pressure is exerted in all
directions and is perpendicular to every
surface at every location.
Atmospheric pressure
 Atmospheric pressure is normally about 100,000
Pascals.
 Differences in atmospheric pressure cause winds
to blow
Pressure Practice 1
 Calculate the net force on an airplane window
if cabin pressure is 90% of the pressure at sea
level and the external pressure is only 50 % of
the pressure at sea level. Assume the window is
0.43 m tall and 0.30 m wide.
Pressure Practice 1 solution
Atmospheric pressure = 100000 Pa
Pcabin = 90000 Pa
Poutside = 50000 Pa
P = F/A
net P = 90000Pa – 50000Pa = 40000Pa
A = lw
A = 0.3m · 0.43m = 0.129m2
F = AP
F = 0.129m2 · 40000Pa = 5160N
Pressure of a Liquid
 The pressure of a liquid is sometimes called
gauge pressure
 If the liquid is water, it is called hydrostatic
pressure
 P = ρgh
 P = pressure (Pa)
 ρ = density (kg/m3)
 g = 9.81 m/s2
 h = height of liquid column (m)
Absolute Pressure
 Absolute pressure is
obtained by adding
the atmospheric
pressure to the
hydrostatic pressure
 Patm + ρgh = Pabs
2. The depth of Lake
Mead at the Hoover
Dam is 180 m.
What is the hydrostatic
pressure at the base of
the dam?
What is the absolute
pressure at the base of
the dam?
Absolute Pressure (practice 2 – solution)
ρfresh water = 1000 kg / m3
P = ρgh
P = 1000 kg / m3 · 9.8m/s2 · 180 m = 1764000 Pa
Pabs = Patm + ρgh
Pabs = 100000Pa + 1764000Pa = 1864000 Pa
Buoyancy Force
 Floating is a type of equilibrium: An upward
force counteracts the force of gravity for
floating objects
 The upward force is called the buoyant force
 Archimedes’ Principle: a body immersed in a
fluid is buoyed up by a force that is equal to the
weight of the fluid it displaces
Calculating Buoyant Force
 Fbuoy = ρVg
Fbuoy: buoyant force exerted on a submerged or
partially submerged object
 V: volume of displaced fluid
 ρ: density of displaced fluid

 When an object floats, the upward buoyant
force equals the downward pull of gravity
 The buoyant force can float very heavy
objects, and acts upon objects in the fluid
whether they are floating, submerged, or even
resting on the bottom
Buoyant force on submerged objects
 A shark’s body is not
neutrally buoyant, so a
shark must swim
continuously or it will
sink deeper
 Scuba divers use a
buoyancy control
system to maintain
neutral buoyancy
(equilibrium)
 If the diver wants to
rise, he inflates his vest,
which increases his
volume, or the water
he displaces, and he
accelerates upward
Buoyant Force on Floating Objects
 If the object floats on the surface, we know that
Fbuoy = Fg! The volume of displaced water
equals the volume of the submerged portion of
the object
Practice 3
 Assume a wooden raft has 80.0 % of the density
of water. The dimensions of the raft are 6.0 m
long by 3.0 m wide by 0.10 m tall. How much of
the raft rises above the level of the water when
it floats?
Practice 3 solution
Buoyant Force Labs
1. Determine the density of water by using the
buoyant force.
Equipment:
Beakers
String
Pulleys
Weights/Masses
Graduated cylinder
(NO BALANCES!)
Moving Fluids
 When a fluid flows,
mass is conserved
 Provided there are no
inlets or outlets in a
stream of flowing fluid,
the same mass per unit
time must flow
everywhere in the
stream
 The volume per unit
time of a liquid flowing
in a pipe is constant
throughout the pipe
 We can say this
because liquids are
generally not
compressible, so mass
conservation is also
volume conservation
for a liquid
Fluid Flow Continuity
 V = Avt




V: volume of fluid (m3)
A: cross sectional areas
at a point in the pipe
(m2)
v: the speed of fluid
flow at a point in the
pipe (m/s)
t: time (s)
 Comparing two points
in a pipe:
 A1v1 = A2v2
 A1, A2: cross sectional
areas at points 1 and 2
 v1, v2: speeds of fluid
flow at points 1 and 2
Practice 4 & 5
4. A pipe of diameter 6.0
cm has fluid flowing
through it at 1.6 m/s.
How fast is the fluid
flowing in an area of
the pipe in which the
diameter is 3.0 cm?
How much water per
second flows through
the pipe?
5. The water in a canal
flows 0.10 m/s where
the canal is 12 meters
deep and 10 meters
across. If the depth of
the canal is reduced
to 6.5 m at an area
where the canal
narrows to 5.0 m, how
fast will the water be
moving through the
narrower region?
Practice 4 solution
Practice 5 solution
Bernoulli’s Theorem
 The sum of the pressure, the potential energy
per unit volume, and kinetic energy per unit
volume at any one location in the fluid is equal
to the sum of the pressure, the potential energy
per unit volume, and the kinetic energy per unit
volume at any other location in the fluid for a
non-viscous incompressible fluid in streamline
flow
 All other considerations being equal, when fluid
moves faster, pressure drops
Bernoulli’s Theorem
P + ρgh + ½ ρv2 = constant
P = pressure (Pa)
 ρ = density of fluid (kg/m3)
 g = grav. accel. constant
(9.81 m/s2)
 h = height above lowest
point
 v = speed of fluid flow at a
point in the pipe (m/s)

6. Knowing what you
know about
Bernoulli’s
principle, design an
airplane wing that
you think will keep
an airplane aloft.
Draw a cross
section of the wing.
practice 6 solution
Thermodynamics
 Thermodynamics is the
study of heat and
thermal energy
 Thermal properties
(heat & temperature)
are based on the
motion of individual
molecules, so
thermodynamics
overlaps with chemistry
 Total Energy:
 E = U + K + Eint
U
= potential
energy
 K = kinetic energy
 Eint= internal or
thermal energy
Total Energy
 Potential and kinetic energies are specifically for
“big” objects, and represent mechanical
energy
 Thermal energy is related to the kinetic energy
of the molecules of a substance
Temperature & Heat
 Temperature is a measure of the average
kinetic energy of the molecules of a substance.
(like how fast the molecules are moving) The
unit is °C or K. Temperature is NOT heat!
 Heat is the internal energy that is transferred
between bodies in contact. The unit is Joules (J)
or sometimes calories (cal)
 A difference in temperature will cause heat
energy to be exchanged between bodies in
contact. When two bodies are the same temp,
they are in thermal equilibrium and no heat is
transferred.
Laws of Thermodynamics
 O: When two systems are sitting in equilibrium with a third system,
they are also in thermal equilibrium with each other.
 1: The first law states that when heat is added to a system, some
of that energy stays in the system and some leaves the system.
The energy that leaves does work on the area around it. Energy
that stays in the system creates an increase in the internal energy
of the system.
 2: It is impossible to have a cyclic (repeating) process that
converts heat completely into work. Also, heat never flows
spontaneously from a colder body to a hotter spontaneously
from a colder body to a hotter body.
 3: It is not possible to reach a temperature of absolute zero
(about -273 C). Since temperature is a measure of molecular
movement, there can be no temperature lower than absolute
zero.
Ideal Gas Law
 P: initial & final pressure (any unit)
 V: initial & final volume (any unit)
 T: initial & final temperature (K)
 T in Kelvins = T in °C + 273
Practice 7
7. Suppose an ideal gas occupies 4.0 L at 23°C
and 2.3 atm. What will be the volume of the
gas if the temperature is lowered to 0°C and the
pressure is increased to 3.1 atm?
Practice 7 solution
Ideal Gas Equation
 If you don’t remember this from chem, you
shouldn’t have passed!
 P: pressure (Pa)
 V: volume (m3)
 n: number of moles
 R: gas law constant 8.31 J/(mol K)
 T: temp (K)
Practice 8
8. Determine the number of moles of an ideal gas
that occupy 10.0 m3 at atmospheric pressure
and 25°C.
Practice 8 solution
Ideal Gas Equation
 P: pressure (Pa)
 V: volume (m3)
 N: number of molecules
 kB: Boltzmann’s
constant

1.38 x 10-23J/K
 T: temperature (K)
9. Suppose a near
vacuum contains
25,000 molecules of
helium in one cubic
meter at 0°C. What is
the pressure?
Practice 9 solution
Kinetic Theory of Gases
1. Gases consist of a large number of molecules
that make elastic collisions with each other
and the walls of their container
2. Molecules are separated, on average, by
large distances and exert no forces on each
other except when they collide
3. There is no preferred position for a molecule in
the container, and no preferred direction for
the velocity
Average Kinetic Energy of a Gas
 Kave = 3/2 kBT
 Kave
= average kinetic energy (J)
 kB = Boltzmann’s constant (1.38 x 10-23J/K)
 T = Temperature (K)
 The molecules have a range of kinetic
energies, so we take the Kave
10 & 11
10. What is the average
kinetic energy and
average speed of
oxygen molecules in a
gas sample at 0C°?
11. Suppose nitrogen
and oxygen are in a
sample of gas at
100°C:
a) What is the ratio of
the average kinetic
energies for the two
molecules?
b) What is the ratio of
their average speeds?
Practice 10 solution
Practice 11 solution
Suppose nitrogen and oxygen are in a sample of gas at 100°C:
a) What is the ratio of the average kinetic energies for the two
molecules?
b) What is the ratio of their average speeds?
Thermodynamics
 The system boundary
controls how the
environment affects
the system (for our
purposes, the system
will almost always be
an ideal gas)
 If the boundary is
“closed to mass,” that
means mass can’t get
in or out
 If the boundary is
“closed to energy,”
that means energy
can’t get in or out
 What type of
boundary does the
earth have?
First Law of Thermodynamics
 The work done on a system + the heat
transferred to the system = the change in
internal energy of the system.
 ΔU = W + Q
ΔU = Eint = thermal energy (NOT potential energy –
crazy???)
 W = work done on the system (related to change in
volume)
 Q = heat added to the system (J) – driven by
temperature difference – Q flows from hot to cold

First Law of Thermodynamics
More about “U”
 U is the sum of the kinetic energies of all the
molecules in a system (or gas)
 U = NKave
 U = N(3/2 kBT)
 U = n(3/2 RT)

since kB = R/NA
12 & 13
12. A system absorbs 200
J of heat energy from
the environment and
does 100 J of work on
the environment.
What is its change in
internal energy?
13. How much work does
the environment do on
a system if its internal
energy changes from
40,000 J to 45,000 J
without the addition of
heat?
Practice 12 solution
Practice 13 solution
Gas Process
 The thermodynamic state of a gas is defined by
pressure, volume, and temperature.
 A “gas process” describes how gas gets from
one state to another state
 Processes depend on the behavior of the
boundary and the environment more than they
depend on the behavior of the gas
Isobaric Process
(Constant Pressure)
Isometric Process
(Constant Volume)
Isothermal Process
(Constant Temperature)
Adiabatic Process
(Insulated)
Work
 Calculation of work done on a system (or by a
system) is an important part of thermodynamic
calculations
 Work depends upon volume change
 Work also depends upon the pressure at which
the volume change occurs
Work
Done BY a gas
Done ON a gas
14 & 15
14. Calculate the work
done by a gas that
expands from 0.020 m3
to 0.80 m3 at constant
atmospheric pressure.
How much work is done
by the environment
when the gas expands
this much?
15. What is the change
in volume of a cylinder
operating at
atmospheric pressure if
its thermal energy
decreases by 230 J
when 120 J of heat are
removed from it?
Practice 14 solution
Practice 15 solution
Work (Isobaric)
Work is Path Dependent
16 & 17
16. One mole of a gas
goes from state A (200
kPa and 0.5 m3) to state
B (150 kPa and 1.5 m3).
What is the change in
temperature of the gas
during this process?
17. One mole of a gas
goes from state A (200
kPa and 0.5 m3) to state
B (150 kPa and 1.5 m3).
a. Draw this process
assuming the smoothest
possible transition
(straight line)
b. Estimate the work done
by the gas
c. Estimate the work done
by the environment
Practice 16 solution
Practice 17 solution
Work Done by a Cycle
 When a gas undergoes a complete cycle, it
starts and ends in the same state. the gas is
identical before and after the cycle, so there is
no identifiable change in the gas.
 ΔU = 0 for a complete cycle
 The environment, however, has been changed
Work Done By Cycle
 Work done by the gas
is equal to the area
circumscribed by the
cycle
 Work done by the gas
is positive for clockwise
cycles, and negative
for counterclockwise
cycles. Work done by
the environment is
opposite that of the
gas
18
 Consider the cycle ABCDA, where
 State A: 200 kPa, 1.0 m3
 State B: 200 kPa, 1.5 m3
 State C: 100 kPa, 1.5 m3
 State D: 100 kPa, 1.0 m3
a. Sketch the cycle
b. Graphically estimate the work done by the
gas in one cycle
c. Estimate the work done by the environment in
one cycle
Practice 18 solution
 A: 200 kPa, 1.0 m3
C: 100 kPa, 1.5 m3
B: 200 kPa, 1.5 m3
D: 100 kPa, 1.0 m3
19
 Calculate the heat necessary to change the
temperature of one mole of an ideal gas from
600 K to 500 K
a.
b.
At constant volume
At constant pressure (assume 1 atm)
Practice 19 solution
 Calculate the heat necessary to change the temperature of one mole
of an ideal gas from 300 K to 500 K
a.
At constant volume
b.
At constant pressure (assume 1 atm)
 W = PDV so when volume is constant, W = 0
 Since V = nRT/P then W = nR(T2-T1) when P const
 W = (1 mol)(8.31 J/mol/K)(500-300)
 W = 1662 J
Second Law of Thermodynamics
 No process is possible whose sole result is the
complete conversion of heat from a hot
reservoir into mechanical work (Kelvin-Planck
statement)
 No process is possible whose sole result is the
transfer of heat from a cooler to a hotter body
(Clausius statement)
 Basically, heat can’t be completely converted
into useful energy
Heat Engines
 Heat engines can convert heat into useful work
 According to the 2nd Law of Thermodynamics,
Heat engines always produce some waste heat
 Efficiency can be used to tell how much heat is
needed to produce a given amount of work
Work and Heat Engines
 QH = W + Q C
 QH: Heat that is put into the system and comes from
the hot reservoir in the environment
 W: Work that is done by the system on the
environment
 QC: Waste heat that is dumped into the cold
reservoir in the environment
Heat Transfer
Heat Engines
Adiabatic vs. Isothermal Expansion
Carnot Cycle
 an ideal heat-engine cycle in which the
working substance goes through the four
successive operations of isothermal expansion
to a desired point, adiabatic expansion to a
desired point, isothermal compression, and
adiabatic compression back to its initial state.
Carnot Cycle
20
20. A piston absorbs 3600 J of heat and dumps
1500 J of heat during a complete cycle. How
much work does it do during the cycle?
Practice 20 solution
Efficiency of Heat Engine
 In general, efficiency is related to what fraction
of the energy put into a system is converted to
useful work
 In the case of a heat engine, the energy that is
put in is the heat that flows into the system from
the hot reservoir
 Only some of the heat that flows in is converted
to work. The rest is waste heat that is dumped
into the cold reservoir
Efficiency of Heat Engine
 Efficiency = W/QH = (QH – QC) / QH
 W: Work done by the engine on the environment
 QH: Heat absorbed from hot reservoir
 QC: Waste heat dumped into cold reservoir
 Efficiency is often given as percent efficiency
 HOW COULD YOU: find the efficiency of your
hair dryer?
21
 A coal-fired steam plant is operating with 33%
thermodynamic efficiency. If this is a 120 MW
plant, at what rate is heat energy used?
Practice 21 solution
 A coal-fired steam plant is operating with 33%
thermodynamic efficiency. If this is a 120 MW
plant, at what rate is heat energy used?
 Efficiency = W/QH
 0.33 = 120MW/QH
 QH = 363.6 MW
Carnot Engine Cycle
Efficiency of Carnot Cycle
 For a Carnot engine, the efficiency can be
calculated from the temperatures of the hot
and cold reservoirs.
 Carnot Efficiency = (TH – TC) / TH


TH: temperature of hot reservoir (K)
TC: temperature of cold reservoir (K)
22 & 23
22. Calculate the Carnot
efficiency of a heat
engine operating
between the
temperature of 60 and
1500°C.
23. For #22, how much
work is produced
when 15 kJ of waste
heat is generated?
Practice 22 solution
Practice 23 solution
 23. For #22, how much work is produced when





15 kJ of waste heat is generated?
15 kJ (Qc = the amount of heat wasted) is 19%
of QH (the amount of heat to begin with) since it
has 81% efficiency (see last slide)
Therefore…15kJ/.19 = QH = 78.95 kJ
QH = Q C + W
78.95 = 15 + W
W = 63.95 kJ
 Or use Efficiency = (QH – QC) / QH & solve for QH then W
Entropy
 Entropy is disorder, or randomness
 The entropy of the universe is increasing.
Ultimately, this will lead to what is affectionately
known as “Heat Death of the Universe.”
Entropy
 ΔS = Q/T
 ΔS: change in entropy (J/K)
 Q: heat going into the system (J)
 T: temperature (K)
 If change in entropy is positive, randomness or
disorder has increased
 Spontaneous changes involve an increase in
entropy
 Generally, entropy can go down only when
energy is put into the system