From: btd@IASTATE.EDU (Benjamin T Dehner)
Subject: Re: Thermodynamics FAQ
Reply-To: btd@IASTATE.EDU (Benjamin T Dehner)
Organization: Iowa State University
Date: Fri, 31 Jul 1992 17:11:59 GMT

In article <henley.920731092710@wilbur.eng.auburn.edu>, henley@eng.auburn.edu
(James Paul Henley) writes:
> In article <1992Jul30.211613.149924@zeus.calpoly.edu> skroger@zeus.calpoly.edu
(Seth L. Kroger) writes:
> 
> 
> >
> >
> >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

	Seth explicitly declared dQ_3 > 0, then put in signs into the equations
as appropriate.  Here you are allowing the sign of dQ_3 to be arbitrary, and
keeping careful track of the signs within the equations.

> 
> >                         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

	Which, from Seth's sign convention, gives dS < 0, since T_d > T_n > 0
and dQ_3 > 0.

> 
> 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
> 

	So, by Jame's sign convention, dQ_3 = -|dQ_3|, where | | denotes
absolute value, to make the signs unambiguous.  This gives

				    |dQ_3|   |dQ_3|
                               dS = ------ - ------
                                      T_d      T_n

	which still give dS < 0 , since now all quantities are explicitly
positive, and T_d > T_n > 0 still holds.

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

Yes, but -(-1/2) is more than -(-1/3).  By leaving in the '-' sign, you left
the sign of dQ arbitrary.  (It turns out to be negative.)  Seth defined dQ_3 to
be positive, then filled in the signs as appropriate, then rumbled through the
arithmetic.  You left the sign of dQ_3 arbitrary, then rumbled through the 
arithmetic.  However, when you fill in the sign of dQ_3 as the final step
(dQ_3 < 0 ) you arrive at the same result.  I did this explicitly by putting
in the absolute value functions.
 
> 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. 
 
	Don't you hate it when it happens at the blackboard? :)

> >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.
> >
> >|======================================================================|
> >| 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.
> 

'Teacher, leave those kids alone!' -- Pink Floyd
-----------------------------------------------------------------------------
Benjamin T. Dehner          Department of Physics and Astronomy 
btd@iastate.edu             Iowa State University 
                            Ames, IA 50011

Newsgroups: talk.origins,sci.skeptic