From: henley@eng.auburn.edu (James Paul Henley)
Subject: Re: Thermodynamics FAQ
Organization: Auburn University Engineering
Date: Fri, 31 Jul 1992 14:27:10 GMT

In article <1992Jul30.211613.149924@zeus.calpoly.edu> skroger@zeus.calpoly.edu (Seth L. Kroger) writes:

Sorry Seth, but the slight of hand trick won't work.  You made the most
common mistake in thermo - confusing the signs on an energy balance.
If the heat flow is out, then the value is negative relative to the earth,
and positive relative to the surroundings.  By putting a negative sign in
front of the outgoing heat transfer terms, you are making then heat transfer
relative to the surroundings, not the earth.

>
>
>A thermodynamic equation for entropy is dS = dQ/T + dP_s where dS is
>the change in entropy, dQ is heat transfer, T is temperature, and dP_s
>is the entropy production.  By the 2nd Law, dP_s >= 0[1, p.  206].
>
>Figure~1 is a diagram of the Sun and Earth.
>
>                                                   dQ_2  dQ_3
>                                                    ^     ^
>                                                    |     |
>                                                    |     |
>       /--------\                                   |     |
>      /          \                                   /---\
>     /            \                                 /  |  \
>     |            |                   dQ_1          |  |  |
>     |            |               ----------->      |  |  |
>     \            /                                 \  |  /
>      \          /                            T_d--> \---/ <--T_n
>       \--------/
>
>
>         Sun                                         Earth
>
>
>               Figure 1:  A simplified Sun-Earth system
>
>
>We receive heat from the sun (dQ_1) on our daylight side and radiate
>heat back out into space on both our day and night sides (dQ_2, dQ_3).
>Overall the Earth's energy stays constant, so dQ_1 = dQ_2 + dQ_3.  T_d
>and T_n are the average temperatures of the day and night sides,
>respectively.  It should be obvious that T_d is greater than T_n.
>
>Neglecting entropy production for the moment, the change of entropy
>for the Earth is:
>                             dQ_1   dQ_2   dQ_3
>                      dS =   ---- - ---- - ----
>                             T_d    T_d    T_n

  Only if dQ_1 is relative to the earth and dQ_2 and dQ_3 are relative to
  the surroundings.
  If dQ_1, dQ_2, and dQ_3 are all relative to your system then you can't
  use a positive sign on one and negative signs on the others.
  The equation is:

                               dQ_1   dQ_2   dQ_3
     dS = dS_1 + dS_2 + dS_3 = ---- + ---- + ---- 
                               T_d    T_d    T_n

  Where relative to the earth:  dQ_1 > 0, dQ_2 < 0 and dQ_3 < 0

  You can't just wave your hands and make a sign change.

  Whoops!

  Was this a peer reviewed article?

>Substituting dQ_2 + dQ_3 for dQ_1 (since they're equal):

  Should be dQ_1 = -(dQ_2 + dQ_3)  since both dQ_2, and dQ_3 are negative

  At steady state:   dQ_1 + dQ_2 + dQ_3 = 0     dQ_2 < 0,  dQ_3 < 0

>
>                         dQ_2 + dQ_3   dQ_2   dQ_3
>                    dS = ----------- - ---- - ----
>                             T_d       T_d    T_n
>
                         -dQ_2 - dQ_3   dQ_2   dQ_3
should be:           dS = ----------- + ---- + ----
                              T_d       T_d    T_n

>                         dQ_2   dQ_3   dQ_2   dQ_3
>                    dS = ---- + ---- - ---- - ----
>                         T_d    T_d    T_d    T_n
>
>                                dQ_3   dQ_3
>                           dS = ---- - ----
>                                T_d    T_n

Here is where you made your mistake, you have defined dQ_3 to be Positive
Positive means that it is heat flow TO SURROUNDINGS, and the sign is relative
to the surroundings.
If it is relative to the earth, then the value of dQ_3 will be negative, but
you will have a positive sign in front of it in the enthalpy balance.

                                   dQ_3   dQ_3
            should be:      dS = - ---- + ----
                                   T_d    T_n

where dQ_3 < 0

>
>Since T_n is smaller than T_d, dQ_3/T_n > dQ_3/T_d (like 1/2 is more
>than 1/3), therefore:
>
>                                dS < 0
>

Is -1/2 more than -1/3?  You left out the explicit negative sign that
you added to the original energy balance.  

Hmm...Maybe I should start counting off more for sign errors if they can
cause this much trouble.  Better yet, I'll give this to my Thermo students
as an example of how embarrassing sign mistakes can be. 

>The entropy of the Earth can actually decrease!  How is this possible?
>dQ_2 and dQ_3 are essentially transferring entropy away from the Earth
>to the universe at large.  The dS_universe is still positive, even
>though the entropy of a small part of the (Earth) has been reduced.
>
>[1] Reynolds, William C., and Perkins, Henry C., _Engineering_
>    _Thermodynamics_, 2nd ed., McGraw-Hill, Inc., 1977.
>
>=====================================================================
>>Interpretation:  To decrease system entropy, you have to have a *net* heat
>>loss.
>
>Sorry James, but no net heat loss is required if you get rid of heat at a
>lower temperature than you recieved it at.
>
>>James P. Henley Jr.
>
>|======================================================================|
>| Seth Kroger                       "If God made us in His image we    |
>| skroger@pan.calpoly.edu               have certianly returned the    |
>| Cal Poly, San Luis Obispo             compliment."  -Voltaire        |
>                                             ^---spelling error corrected

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

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