Ch. 5

1
Ch. 5 2nd Law of Thermodynamics

• Entropy and Potential Temperature
• The right hand sides of our equations for
  potential temperature and specific entropy are
  the same, so we have
• Leading to the final result
• Changes in entropy (for a reversible process) are
  directly proportional to changes in potential
  temperature.
• Adiabatic (reversible) processes are isentropic.

2
Ch. 5 2nd Law of Thermodynamics

• Entropy and Potential Temperature
• We could write a similar equation for the
  extensive variable, S, as
• What about an irreversible adiabatic process with
  ds gt 0?
• For any adiabatic process, d? 0 ? ds for an
  irreversible process.
• Entropy increase due to irreversible work
  (uncompensated heat), such as frictional
  dissipation.
• Isentropic equals adiabatic, but not always the
  other way.

3
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
  (following Bohren and Albrecht, 1998)
• On a flat earth (an approximation), the total
  mass per unit area (kg m-2) between altitudes z1
  and z2 is
• Assuming hydrostatic equilibrium we can
  substitute for the density, ?, to get
• Considering an atmospheric layer between 2
  constant pressure surfaces, the total mass of the
  layer is constant.

4
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Assume the layer is
• Isolated from its surroundings
• Not heated or cooled by direct radiation (solar
  or terrestrial)
• Not heated by conduction from surroundings or
  surface (if in contact)
• QUESTION What is the equilibrium temperature
  profile in this layer?
• Total enthalpy per unit area of layer is

5
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Aside from the influence of the gravitational
  field, the total enthalpy would be conserved.
• Because of gravitational field, the layer has
  potential energy, given by
• Recalling that the units on cpdT are (J
  kg-1K-1)(K), which J kg-1 m2 s-2, and the
  units on gz are (m s-2)(m) m2 s-2, also J kg-1,
  the enthalpy is an energy equivalent.

6
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Given the assumptions concerning isolation, the
  enthalpy and the potential energy are the only
  forms of energy the layer can have.
• We can add them up and realize that their sum
  must be constant, yielding,
• The quantity cpdT ( h) gz is called the dry
  static energy (see equation (4.31) in the text).

7
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• We can also calculate the total entropy of the
  layer as,
• Note that the additive constant has been left off
  and we are playing fast and loose by taking the
  natural log of something that has dimensions.
• The maximization of entropy is addressed by
  asking, what temperature and pressure profiles
  maximize the layer entropy subject to the
  constancy of the dry static energy?

8
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Since we are talking temperature and pressure, we
  convert our equations to pressure from height
  coordinates.
• The dry static energy becomes (using the
  hydrostatic equation)
• We have reversed the order of integration to get
  rid of the minus sign introduced by the
  hydrostatic equation, and the extra g has been
  included in the constant.

9
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• The equation for layer entropy becomes
• We can also write the relationship between
  entropy and potential temperature as

10
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• From the potential temperature equation we have
• Factoring out ?/T (p0/p)?, we have

11
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Simplifying a little further, we have
• Substituting this into our expression for ds/dz,
  we get
• We can multiply thru by T, take cpd inside the ()
  and put the last term in terms of a derivative,
  giving

12
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• We can put two equations of interest in
  p-coordinates as
• From the potential temperature equation we have
  an expression for the ratio of T/? (p/p0)0.286.
• At p 1000 mb, T/? 1.0.
• At p 200 mb (80 of the atmosphere), T/?
  0.631.
• If we restrict consideration of atmospheric
  layers to those a few hundred meters thick, a few
  10s of mb, we can use T/? ? 1.0.

13
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Then we have
• We can integrate this to get
• We had earlier that
• So we can write

14
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• We are almost there.
• We apply a change of variables p to x where
• This transforms the coordinates so that x 0 is
  in the middle of the layer, since ½ (p1 p2)
  defines the arithmetic midpoint of the layer, and
  dx dp.
• We can rewrite the last integral as

15
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Lets play with this a little.
• When p p1 (at the bottom of the layer), what is
  x?
• x gt 0 since p1 gt p2 (pressure decreasing with z).
• When p p2 (at the top of the layer), what is x?
• Here x lt 0, so the sign sense is reversed, but
  intervals are preserved.

16
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Now the limits of integration are symmetric about
  x, and ? the limit at p1 is given by
• So now we want to find the function T for which
  the entropy, S (divided by the constant cpd/g),
  is a maximum, subject to the constraint indicated
  by the previous equation, i.e.,

17
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• For simplicity we will consider only the set of
  linear potential temperature profiles, such that
• Here a and b are arbitrary.
• Putting this in our constraint equation yields
• Therefore a is a constant. Since S is related to
  ln?, once a is fixed, S can only depend on b.

18
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• The condition for S to be an extremum (maximum or
  minimum) is that its derivative vanish
• Looking at the derivative of the integral (taking
  the derivative operator inside the integral), we
  have
• Since d(lnu)/dx 1/u du/dx. Now we carry out
  the integral

19
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• We have to solve
• Solving this yields

20
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Since a is a constant, the only solution for this
  equation is if b 0.
• We determine whether this is a maximum or minimum
  by taking the 2nd derivative with respect to b.
  Doing this we get,
• Because the integrand is positive definite, the
  result is negative, indicating that this is a
  maximum.

21
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Of all linear potential temperature profiles, b
  0 indicates a constant ?, since ? a and a was
  determined to be a constant.
• Therefore, ? const maximizes the entropy of an
  isolated layer of the atmosphere in hydrostatic
  equilibrium.
• The dry adiabat delineates between stable and
  unstable air.
• Entropy maximization requires the equilibrium
  temperature of an isolated atmospheric layer
  subjected to mixing, but with no condensation or
  evaporation, to decrease with height at the dry
  adiabatic lapse rate.

22
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• This is not intuitive.
• We might expect the equilibrium profile to be
  isothermal.
• A solid isolated from the environment with an
  initial non-uniform temperature gradient, reaches
  a constant temperature by conduction eliminating
  gradients.
• In the atmosphere, energy transfer is dominated
  by convection rather than conduction.
• Thus, convective equilibrium is established,
  rather than conductive or radiative equilibrium.
• Fluids are inherently different from solids.

23
Ch. 5 2nd Law of Thermodynamics

• Maximization of Entropy in the Atmosphere
• Mass motion in solids is absent, but it dominates
  fluids.
• Suppose a parcel moves to another level within
  the isolated layer and then mixes with its
  surroundings.
• Temperature of parcels is not conserved during
  vertical motion.
• If a parcel moves up (cools) and mixes with air
  at that level, the layer is destabilized the
  net effect is cooling.
• Similarly for a downward moving (warming) parcel.
• Mixing increases the entropy of the layer even
  though it is stable. Well mixed layers have an
  adiabatic profile.