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Thermodynamics - statement of the obvious?

Thermodynamics - statement of the obvious? - Physics Forum

Thermodynamics - statement of the obvious? - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 02-12-2006, 12:31 PM
tadchem
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Default Thermodynamics - statement of the obvious?




"Doune60" <[Only registered users see links. ].uk> wrote in message
news:1139652976.147568.73850@g47g2000cwa.googlegro ups.com...

What may not be as obvious to some is that the Laws of Thermodynamics are
formulated for a *single* system, whether at equilibrium or not, which may
or may not be exchanging heat with its environment.

This is *explicitly* accounted for in the Law of Heat Transfer (sometimes
called Newton's Law of Cooling)
[Only registered users see links. ]
[click on the Heat and Thermodynamics bubble, then the Heat Transfer bubble]

Q/t = kA(Thot - Tcold)/d, where

Q is the heat transferred in time t from a temperature reservoir at the
temperature Thot to one at the temperature Tcold through an intervening
region of area A and thickness d *with thermal conductivity k*.

If the intervening region was not thermally conductive, then Q/t would be
zero.

This fact is just as much a part of the science of thermodynamics as the
well-known 'Laws', which basically set up the rules for what I think of as
'energy accounting' - the physicist's equivalent of double-entry
bookkeeping.

You are trying to re-invent the wheel.


Tom Davidson
Richmond, VA


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  #2  
Old 02-12-2006, 11:43 PM
tadchem
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Default Thermodynamics - statement of the obvious?


Doune wrote:

What you said before was "an available route." That is not the same
thing as "the most efficient route".

I believe you will find that when heat has multiple routes for transfer
from one reservoir to another, the analogy to electrical current (I =
E/R) will hold. The temperature difference (Thot - Tcold) becomes
analogous to the voltage E, the heat flow Q/t becomes analogous to the
current flow I, and the thermal conductance ka/d becomes analogous to
the electrical conductance 1/R.

IOW, the total rate of flow of heat from reservoir A to reservoir B
through multiple channels will be the sum of the heat flow through the
individual channels, each of which will separately obey the Law of Heat
Transfer

Q/t = kA(Thot - Tcold)/d

Formulated for parallel flows:

Qtot/t = (Thot - Tcold) * Sum{kA/d} [for all heat flow channels]

recognizing that k, A, and d will be different for each heat flow
channel.

The ease with which heat can flow through one particular channel will
not prevent the heat from simultaneously exploiting *all* available
channels.

Tom Davidson
Richmond, VA

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  #3  
Old 02-13-2006, 12:20 PM
Doune
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Default Thermodynamics - statement of the obvious?

Dear Tom

(please accept my apologies if this post appears out of sequence - I'm
using Google and I'm not too sure how this appears on other
newsreaders)


That's right, I modified it for clarity, the second is an expansion on
the first, but it means the same thing.

I agree with the rest of the post except for:

The basic premise I am putting forward is that it will *always* exploit
the most efficient (i.e. easiest) route first. This happens with *all*
instances, e.g. the paths of a lightning strike, the direction of a
river flow or even the slightest variations in A,d and k in any given
material condsidered in the Law of Heat Transfer .

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  #4  
Old 02-13-2006, 01:00 PM
N:dlzc D:aol T:com \(dlzc\)
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Default Thermodynamics - statement of the obvious?

Dear Doune:

"Doune" <[Only registered users see links. ].uk> wrote in message
news:1139833235.894169.163630@g47g2000cwa.googlegr oups.com...
....

"It" isn't sentient. "It" doesn't know "first", "second" or
"last". Tom said it right, *all* paths are invoked.


Lightning follows a "local minima" path, and frequently branches
(and coalesces) to/from all available local paths. Your
lightning strike isn't a particularly good example when
evaluating the First Law of thermodynamics.

David A. Smith


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  #5  
Old 02-13-2006, 05:35 PM
Don1
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Default Thermodynamics - statement of the obvious?

Doune wrote:

There probably aren't enough demons to open the doors of all of the
available paths at the proper times anyway.

There probably aren't enough demons to open the doors of all of the
available paths at the proper times anyway.

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  #6  
Old 02-13-2006, 09:45 PM
tadchem
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Default Thermodynamics - statement of the obvious?


Doune wrote:

<snip>


All routes are used simultaneously, to an extent that is inversely
proportional to the resistance. *That* much is in the equation.


[Only registered users see links. ]


[Only registered users see links. ]


http://volcanoes.usgs.gov/Imgs/Jpg/K...-054_large.jpg

Tom Davidson
Richmond, VA

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  #7  
Old 02-13-2006, 11:59 PM
Doune
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Default Thermodynamics - statement of the obvious?

Hi Tom



So, you agree then?



You're introducing empirical evidence? The image in the link you give
shows a series of clearly defined paths upon which electricity is
discharged - which proves the point exactly.



I refer you to the above comment on the image of a lightning strike for
the same reasons.




I think this image show quite nicely that if the area A, the distance d
or the thermal conductivity k (this assumes that that there is some
variation in the material in the image), then we get heat permeating at
different rates, hence the obvious variation in temperature.



I take it that the point of your post to add weight to my argument?

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  #8  
Old 02-14-2006, 01:29 AM
N:dlzc D:aol T:com \(dlzc\)
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Default Thermodynamics - statement of the obvious?

Dear Doune:

"Doune" <[Only registered users see links. ].uk> wrote in message
news:1139836779.047961.71030@g47g2000cwa.googlegro ups.com...

It clearly isn't. *All* processes are invoked, only the
assignment of "quantities" makes one path "primary" and the rest
less so.


This is entirely "too late". All paths evolve locally, so all
paths are "investigated". Once you assign quantities, you can
claim "most efficient".


No.


There are a host with varying degrees of success. All of them
fail in specifics, if not in general.


The first law says energy is conserved. When you have an event
in an uncontained space, that emits EM radiation even into the
vaccum of space (witness various lightning event in the rings of
Saturn), that disspates its energy as heat and sound, how do you
go about drawing a "control volume" around the Universe? It can
be done, for very short time durations. It is just not a good
example.

David A. Smith


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  #9  
Old 02-14-2006, 01:14 PM
N:dlzc D:aol T:com \(dlzc\)
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Default Thermodynamics - statement of the obvious?

Dear Doune
"Doune" <[Only registered users see links. ].uk> wrote in message
news:1139903867.930443.248180@f14g2000cwb.googlegr oups.com...

No. Either English is not your first language, or you are simply
pulling chains.

When you say "It follows any given process
because it is the most energetically efficient", this implies
that:
- it doesn't follow the other processes, and
- it does choose the most efficient without actually sending any
energy that way (indicating prescience and/or intelligence).


Your argument implies that energy has intelligence. Energy
follows all available paths. Period.

How much energy following a certain path makes it primary. It
doesn't mean there aren't secondary, tertiary, quaternary, or
"lower" order paths also. Heat transfer is commonly accomplished
by forced convection. It is also transferred in the same device
by natural convection, conduction, and direct radiation.


"Primary"? Surely this is an anthropomorpic, nay, an
*accounting*, term?


No. It does not "restate what you've said". The outcomes are
plethora. "Most efficient" is an anthropomorpic term.


I tire of arguing with a stump. Goodbye.
<plonk>

David A. Smith


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  #10  
Old 02-14-2006, 08:03 PM
tadchem
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Default Thermodynamics - statement of the obvious?


Doune wrote:

The data reinforces quite unambiguously that the flow of energy will
always exploit *all* the available paths *simultaneously*. The
evidence that the *extent* to which the energy flow exploits *any
given* channel will vary directly with the conductance (or, in the
words of the semi-literate layman, the 'efficiency') of that particular
channel.

[In thermodynamics the word 'efficiency' has an explicit definition in
terms of reservoir temperatures which applies to cyclic heat engines
and *not* to 'one-shot' processes such as we have been discussing.]

There is NO evidence that there is any temporal priority in the
selection of one route over another - i.e. no route gets used *first*.

In view of the empirical data, your above statement (to wit: "The basic
premise I am putting forward is that it will *always* exploit the most
efficient (i.e. easiest) route first.") should be rephrased as "it will
*always* exploit every available route simultaneously."

If you cannot see the difference in meaning between your original
statement and my paraphrase of it, then you need to refine your
language skills before attempting to expand your activity in physics
beyond that of an undergraduate student.

The ability to express oneself clearly, concisely, and unambiguously
using the commonly defined terms of your specialty field as they are
commonly accepted is critical to the use of the scientific method.
Without such clear communication there can be no peer review, and
without peer review, there is only bloviation.

Tom Davidson
Richmond, VA

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