Fluids _ Thermo

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Fluids _ Thermo Powered By Docstoc
					     AP Physics


 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

 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
 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

 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
  obtained by adding        Mead at the Hoover
  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

 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
    Graduated cylinder       apparatus so that
    (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
                                  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
 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
  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 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
 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
                  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. 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. 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
    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
                            b) What is the ratio of
                            their average speeds?

 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
  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   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)
Adiabatic Process

 Calculation of work done on a system (or by a
  system) is an important part of thermodynamic
 Work depends upon volume change
 Work also depends upon the pressure at which
  the volume change occurs

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
 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

 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

 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
 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
Adiabatic vs. Isothermal Expansion
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
   QC: Waste heat that is dumped into the cold
    reservoir in the environment

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
 YOUR TASK: find the efficiency of your hair

 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

 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.”

 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
 Generally, entropy can go down only when
  energy is put into the system

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