From: henley@eng.auburn.edu (James Paul Henley)
Subject: Re: Thermodynamics FAQ
Organization: Auburn University Engineering
Date: Wed, 29 Jul 1992 22:08:04 GMT

In article <schinder.712422766@leprss.gsfc.nasa.gov> schinder@leprss.gsfc.nasa.gov writes:
>In <henley.920728071653@wilbur.eng.auburn.edu> henley@eng.auburn.edu (James Paul Henley) writes:
>
>
>>For any system that you want to apply a thermodynamic analysis to, you have
>>to use the appropriate means of studying the system - microscopic or
>>macroscopic.  When dealing with a very large system such as the earth, a
>>macroscopic analysis is the only one that makes sense.
>
>Not true.  You run the risk of making an error similar to Kelvin's

Point taken.  You can't ignore the microscopic considerations, but likewise,
you can't ignore the macroscopic considerations.  I will talk about them
independently.   Let's take a macroscopic view:

Take a sphere, the center of which is at the center of the earth, and the
radius is such that the surface of the sphere is above any measurable
concentrations of air.  Let that be your system.  Now do an energy balance
around this sphere.  You have the highest radiation at the point closest to
the sun, and as you move away from that point you have a decreasing radiation.
You can fairly well define the levels of radiation at each point on the sphere.
The side away from the earth has very little incoming radiation, and a lot of
outgoing radiation.  It should be fairly simple to determine a radiation
profile - I'm sure it has been done at NASA, right?  OK, this is only one
term in the energy balance.  Next comes the rate of energy lost due to
air molecules escaping into space.  Then the rate of energy lost due to
Rockets going into space, then the rate of energy gained by meteorites.
We can't forget about the tidal forces due to the Moon's gravity.  Of course
the moon has its own reflected radiation that is not in sync with the sun,
so the overall energy balance follows the cycles of the Sun and the Moon.
If you look at the magnitude of the energy transfer due to meteorites,
escaping air molecules, and rockets, and compare that with the other terms,
the order of magnitude is so small that we can consider it to be negligible.
If the rate of mass transfer is negligible, then we can eliminate change
in overall mass from the energy balance and call the mass of the earth
constant.  This is what is called a closed, diabatic system.

Are there any microscopic considerations that will alter this overall
energy balance?  If so, tell me what they are, and I will include them
in the energy balance.  The only one that I can think of would be
a short term change in cloud cover.  Enough to significantly alter the
energy balance because of reflected radiation.

>(about the time it would take for the earth to cool), and ignoring
>internal degrees of freedom (on the microscopic level) that are
>important for the question you are asking.  That's why it's always
>wise to keep the "little picture" in mind.  You can go from
>Boltzmann's equation to hydrodynamics, but you can't do the reverse.
>
>
>
>>Entropy is a measure of the disorder of the energy of a system.  Ordered
>>energy is available to do work, disordered energy is not. So, mathematically
>
>Missed this one the first time around.  I don't know exactly what you
>mean by "disordered energy", but if you mean heat energy (random
>kinetic energy of molecules), this is wrong as stated, as anyone who's
>played with a toy steam engine knows.  If you take a hot gas and let
>it expand, useful work can be done.  The second law says that not
>*all* of the heat energy can be extracted.  I don't know how you
>distinguish between "ordered" and "disordered" energy.

It's very simple.  You get out your steam tables, calculate the enthalpy
and entropy of the steam, determine what the enthalpy of the steam will
be at the same entropy but at the temperature that it is exhausted,
and delta H - difference in enthalpy, is how much work you can get out
of the steam.  In doing that, you are using the First Law to do an energy
balance, and the Second Law to do an entropy balance.  Elementary engireering
thermodynamics.  You are assuming an adiabatic, reversible expansion, so
to get the actual amount of work, you need to know how much friction was
involved and how much heat was lost to the surroundings.  The heat you
measure, and the amount of friction involved requires an entropy balance.
You measure the actual pressure of the exhausted steam, calculate the
entropy of the steam at that pressure,  and do an overall energy and
enthalpy balance to determine the amount of lost work. 

>Paul J. Schinder
>NASA Goddard Space Flight Center
>schinder@leprss.gsfc.nasa.gov


James P. Henley Jr.
Chemical Engineering Dept.
Auburn University

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