Fluids _ Thermo

Document Sample

```					     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.
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.
Atmospheric pressure

 Atmospheric pressure is normally about 100,000
Pascals.
 Differences in atmospheric pressure cause winds
to blow
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    2. The depth of Lake
the atmospheric           Dam is 600 ft.
pressure to the           What is the hydrostatic
hydrostatic pressure      pressure at the base of
 Patm + ρgh = Pabs         the dam?
What is the absolute
pressure at the base of
the dam?
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     Scuba divers use a
neutrally buoyant, so a     buoyancy control
shark must swim             system to maintain
continuously or it will     neutral buoyancy
sink deeper                 (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
#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?
Buoyant Force Labs

1. Determine the density   2. Balloon Race:
of water by using the      Determine the
buoyant force.             buoyant force on your
Equipment:                  balloon with the
Beakers                  balloon, masses & a
String                   balance
Pulleys                  Without using the
Weights/Masses           balloon, design an
(NO BALANCES!)           when released, your
balloon will hit the
ceiling LAST.
Moving Fluids

 When a fluid flows,         The volume per unit
mass is conserved            time of a liquid flowing
 Provided there are no        in a pipe is constant
inlets or outlets in a       throughout the pipe
stream of flowing fluid,    We can say this
the same mass per unit       because liquids are
time must flow               generally not
everywhere in the            compressible, so mass
stream                       conservation is also
volume conservation
for a liquid
Fluid Flow Continuity

 V = Avt                        Comparing two points
   V: volume of fluid (m3)      in a pipe:
   A: cross sectional areas    A1v1 = A2v2
at a point in the pipe
 A1, A2: cross sectional
(m2)
areas at points 1 and 2
   v: the speed of fluid
flow at a point in the      v1, v2: speeds of fluid
pipe (m/s)                   flow at points 1 and 2
   t: time (s)
Practice 4 & 5

4. A pipe of diameter 6.0   5. The water in a canal
cm has fluid flowing        flows 0.10 m/s where
through it at 1.6 m/s.      the canal is 12 meters
How fast is the fluid       deep and 10 meters
flowing in an area of       across. If the depth of
the pipe in which the       the canal is reduced
diameter is 3.0 cm?         to 6.5 m at an area
How much water per          where the canal
second flows through        narrows to 5.0 m, how
the pipe?                   fast will the water be
moving through the
narrower region?
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        6. Knowing what you
 P = pressure (Pa)                know about
Bernoulli’s
 ρ = density of fluid (kg/m3)
principle, design an
 g = grav. accel. constant
airplane wing that
(9.81 m/s2)
you think will keep
 h = height above lowest
an airplane aloft.
point                            Draw a cross
 v = speed of fluid flow at a     section of the wing.
point in the pipe (m/s)
Thermodynamics

 Thermodynamics is the      Total Energy:
study of heat and
thermal energy             E = U + K + Eint
 Thermal properties         U   = potential
(heat & temperature)         energy
are based on the            K = kinetic energy
motion of individual
molecules, so               Eint= internal or
thermodynamics               thermal energy
overlaps with chemistry
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.
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
#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?
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)
8

8. Determine the number of moles of an ideal gas
that occupy 10.0 m3 at atmospheric pressure
and 25°C.
Ideal Gas Equation

9. Suppose a near
vacuum contains
25,000 molecules of
helium in one cubic
 P: pressure (Pa)           meter at 0°C. What is
 V: volume (m3)             the pressure?
 N: number of molecules
 kB: Boltzmann’s
constant
   1.38 x 10-23J/K
 T: temperature (K)
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   11. Suppose nitrogen
kinetic energy and        and oxygen are in a
average speed of          sample of gas at
oxygen molecules in a     100°C:
gas sample at 0C°?        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     If the boundary is
controls how the          “closed to mass,” that
environment affects       means mass can’t get
the system (for our       in or out
purposes, the system     If the boundary is
will almost always be     “closed to energy,”
an ideal gas)             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 –
how stupid is that?)
 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

 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   13. How much work does
J of heat energy from      the environment do on
the environment and        a system if its internal
does 100 J of work on      energy changes from
the environment.           40,000 J to 45,000 J
What is its change in      without the addition of
internal energy?           heat?
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
Isothermal Process
(Constant Temperature)
Isobaric Process
(Constant Pressure)
Isometric Process
(Constant Volume)
(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     15. What is the change
done by a gas that         in volume of a cylinder
expands from 0.020 m3      operating at
to 0.80 m3 at constant     atmospheric pressure if
atmospheric pressure.      its thermal energy
How much work is done        decreases by 230 J
by the environment         when 120 J of heat are
when the gas expands       removed from it?
this much?
Work (Isobaric)
Work is Path Dependent
16 & 17

16. One mole of a gas        17. One mole of a gas
goes from state A (200       goes from state A (200
kPa and 0.5 m3) to state     kPa and 0.5 m3) to state
B (150 kPa and 1.5 m3).      B (150 kPa and 1.5 m3).
What is the change in      a. Draw this process
temperature of the gas         assuming the smoothest
during this process?           possible transition
(straight line)
b. Estimate the work done
by the gas
c. Estimate the work done
by the environment
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
19

 Calculate the heat necessary to change the
temperature of one mole of an ideal gas from
600 K to 500 K
a.   At constant volume
b.   At constant pressure (assume 1 atm)
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
Heat Transfer
Heat Engines
Carnot Cycle
Work and Heat Engines

 QH = W + QC
 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
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?
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
dryer
21

 A coal-fired stream plant is operating with 33%
thermodynamic efficiency. If this is a 120 MW
plant, at what rate is heat energy used?
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   23. For #22, how much
efficiency of a heat       work is produced
engine operating           when 15 kJ of waste
between the                heat is generated?
temperature of 60 and
1500°C.
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

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 23 posted: 9/17/2012 language: simple pages: 65