tag:blogger.com,1999:blog-89449315431650461432024-03-23T22:45:04.172-07:00External combustion - Internal cooling engineReflections on the motive power on heat of the environment and on machines fitted to develop that power
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.comBlogger36125tag:blogger.com,1999:blog-8944931543165046143.post-44977626569191556692023-09-13T22:48:00.003-07:002023-09-13T22:48:51.711-07:00Once upon a time<p> This blog is the embodiment of the evolution of my idea to turn the heat of nature into useful energy. The average temperature of air on earth (probably also of water, which I think is the more suitable source of energy) is 290 K, so down to 0 K we have a good chance of creating the temperature difference between the hot and cold part of the engine, that we need to converting heat into mechanical energy.</p><p> I published my ideas immediately the day they took shape, so the whole work is not well systematized, but you can see how I wandered into the unknown.</p><p>When I realized that I had nothing more to contribute, I decided to "forget" about this work - Theorizing is exhausting, one must put an end to it in, order to save oneself.</p><p>You can write to me by email:</p><p>megagreenenergy@gmail.com</p><p>Svetozar the Cold</p>Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-36971962594109689702017-04-07T23:40:00.001-07:002017-04-07T23:41:51.524-07:00off-topic "internally cooled engine " - To your attention : My theory "Perfect Living Creature"<div style="background-color: white; color: #1d2129; font-family: Helvetica, Arial, sans-serif; font-size: 14px; margin-bottom: 6px;">
Already drew first touches of my theory:</div>
<div style="background-color: white; color: #1d2129; font-family: Helvetica, Arial, sans-serif; font-size: 14px; margin-bottom: 6px;">
<br />
"Perfect Living Creature"<br />
(author : Svetozar the Soul )</div>
<div style="background-color: white; color: #1d2129; font-family: Helvetica, Arial, sans-serif; font-size: 14px; margin-bottom: 6px; margin-top: 6px;">
You can see on :<span class="text_exposed_show" style="display: inline; font-family: inherit;"><br /><span style="font-family: inherit;">www. perfect-living-creature.blogspot.bg</span></span></div>
<div style="background-color: white; color: #1d2129; font-family: Helvetica, Arial, sans-serif; font-size: 14px; margin-bottom: 6px; margin-top: 6px;">
<span class="text_exposed_show" style="display: inline; font-family: inherit;">or at : </span><span style="font-family: inherit;">"Perfect Living Creature- theory" (Facebook page)</span></div>
<div style="background-color: white; color: #1d2129; font-family: Helvetica, Arial, sans-serif; font-size: 14px; margin-bottom: 6px; margin-top: 6px;">
<br /></div>
<div style="background-color: white; margin-bottom: 6px; margin-top: 6px;">
<span style="color: #1d2129;">There is still much work on the theory. I hope for a few months to set out my views on this topic.</span></div>
<div style="background-color: white; margin-bottom: 6px; margin-top: 6px;">
<span style="color: #1d2129;">As for begining I want to outline the philosophy. </span></div>
<div style="background-color: white; margin-bottom: 6px; margin-top: 6px;">
<span style="color: #1d2129;">Let us go from this point of view</span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-12804336251621330972016-07-11T01:22:00.002-07:002016-07-11T01:22:41.959-07:00one turbine - one compressor (physics)<div class="MsoNormal">
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<span style="font-size: 14.0pt; line-height: 115%;">Diagram 1 - </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">one</span><span style="font-size: 14.0pt; line-height: 115%;"> turbine, </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">one</span><span style="font-size: 14.0pt; line-height: 115%;"> working substance, </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">one</span><span style="font-size: 14.0pt; line-height: 115%;"> compressor, </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">a </span><span style="font-size: 14.0pt; line-height: 115%;">refrigerant - an option which I will
discuss in this post.<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;"> The
power of the turbine will be superior to the power of the compressor if the
system for redistributing heat returns </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">h</span><span style="font-size: 14.0pt; line-height: 115%;">eat in the evaporator - </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">I</span><span style="font-size: 14.0pt; line-height: 115%;"> will try to prove it by dividing the
turbine (1) of two identical turbines and the same two turbines them turn</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">
on reverse</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%;"> </span><span style="font-size: 14.0pt; line-height: 115%;"> to become compressors - chart 2<o:p></o:p></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9M-zkSVs1gU3dDcWaXiz9t41WZaOjBSNOXXFUq6tRoLWZS7bC1emFIjtJP0FQrJW4zsHGZLbfLxoGfbi4ZOKLhAAC0qCAqNYbdhG6NreBhzkoO9qeJro6sHMUiUjeFSrArrXlBXzZl1s/s1600/2+%25D0%25B4%25D0%25B0.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="292" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9M-zkSVs1gU3dDcWaXiz9t41WZaOjBSNOXXFUq6tRoLWZS7bC1emFIjtJP0FQrJW4zsHGZLbfLxoGfbi4ZOKLhAAC0qCAqNYbdhG6NreBhzkoO9qeJro6sHMUiUjeFSrArrXlBXzZl1s/s640/2+%25D0%25B4%25D0%25B0.png" width="640" /></a></div>
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<span style="font-size: 14.0pt; line-height: 115%;">If the valve
</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">7a </span><span style="font-size: 14.0pt; line-height: 115%;">is closed and </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">valve
</span><span style="font-size: 14.0pt; line-height: 115%;">7b open will </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> work </span><span style="font-size: 14.0pt; line-height: 115%;">only turbine 1b.
Compressor 2b expands and compresses refrigerant R as</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> at</span><span style="font-size: 14.0pt; line-height: 115%;"> enlargement takes heat from the
working substance A</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;"> after the turbine in a heat
exchanger 4b to a temperature equal or
less</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> to</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%;"> </span><span style="font-size: 14.0pt; line-height: 115%;"> the boiling point of the working substance A
so that it liquefies. The heat that we take from the working substance A is
transmitted to the liquid working substance in a heat exchanger 5 where compressor</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> 2b</span><span style="font-size: 14.0pt; line-height: 115%;"> compresses the refrigerant R. As I
said above compressor 2b </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> is a</span><span style="font-size: 14.0pt; line-height: 115%;"> reverse turbine 1b so that</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> by</span><span style="font-size: 14.0pt; line-height: 115%;"> conservation of energy should: If
power</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> of</span><span style="font-size: 14.0pt; line-height: 115%;"> turbine apply it to the compressor working substance A before the
turbine and then the compressor must have the same parameters - temperature,
volume and pressure .<o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">aut</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">b = </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">in</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">b<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">The same
forces - no change in the parameters of the working substance - no effective
unit. There is no way to apply a small force </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">to </span><span style="font-size: 14.0pt; line-height: 115%;">the compressor</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">
so</span><span style="font-size: 14.0pt; line-height: 115%;"> to us remain useful
energy because we can not take away heat and give it to another body so that
the unit can not hold amounts of heat.<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">We begin to
open valve 7a. </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">T</span><span style="font-size: 14.0pt; line-height: 115%;">urbine</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> 1a</span><span style="font-size: 14.0pt; line-height: 115%;"> starts. Th</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">e same </span><span style="font-size: 14.0pt; line-height: 115%;">one reverse turbine</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">
is a</span><span style="font-size: 14.0pt; line-height: 115%;"> compressor 2a,
which expands and compresses refrigerant R so that </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">at </span><span style="font-size: 14.0pt; line-height: 115%;">enlargement </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">in</span><span style="font-size: 14.0pt; line-height: 115%;"> heat exchanger</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">
4a</span><span style="font-size: 14.0pt; line-height: 115%;"> removes heat from the
working substance </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">A </span><span style="font-size: 14.0pt; line-height: 115%;">after turbine 1a to liquefaction. The heat which is removed
from the working substance in a heat exchanger 4a return it to the evaporator
3. Pump refers working substance</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;"> A at a temperature
equal or lower than the boiling point of
the heat exchanger</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> 4a</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%;"> </span><span style="font-size: 14.0pt; line-height: 115%;"> to the heat exchanger 5.<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">To compare
the forces of tubes 1a and reverse turbine - compressor 2a </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">:</span><span style="font-size: 14.0pt; line-height: 115%;"> Turbine 1a works to a temperature
difference T2 / T</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">bp</span><span style="font-size: 14.0pt; line-height: 115%;"> and produces power Wout. By the laws of thermodynamics -
compressor 2a overcomes temperature difference T2 / </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Tbp</span><span style="font-size: 14.0pt; line-height: 115%;"> so that </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">it</span><span style="font-size: 14.0pt; line-height: 115%;"> would need force</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> Wina</span><span style="font-size: 14.0pt; line-height: 115%;"> equal to the force produced by the
turbine Wout</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">a<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">Wout</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> <span lang="EN-US">a</span></span><span style="font-size: 14.0pt; line-height: 115%;"> = </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">in</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> <span lang="EN-US">a</span></span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%;"> </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"><o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">How does the
inclusion of a turbine</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> 1a and</span><span style="font-size: 14.0pt; line-height: 115%;"> compressor 2a
of the balance of power </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Wo</span><span style="font-size: 14.0pt; line-height: 115%;">ut</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">b and </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">inb?? Compressor 2b must give warmth that refrigerant </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">R</span><span style="font-size: 14.0pt; line-height: 115%;"> is accepted at expansion in heat
exchanger 4b</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> of</span><span style="font-size: 14.0pt; line-height: 115%;"> working substance A</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> on</span><span style="font-size: 14.0pt; line-height: 115%;"> a large amount of working substance in a heat
exchanger 5</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">,</span><span style="font-size: 14.0pt; line-height: 115%;"> because working substance after heat exchanger 4a is collected by the
working substance of the heat exchanger 4b. This violates equality</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> Wo</span><span style="font-size: 14.0pt; line-height: 115%;">utb = </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">inb<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">Because the
compressor 2b to overcome a small temperature difference therefore:<o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">in</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">b <</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Wo</span><span style="font-size: 14.0pt; line-height: 115%;">ut</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">b<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;"> Net
power to the entire unit is:<o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;"> = Wout</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">a +</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%;"> </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Wo</span><span style="font-size: 14.0pt; line-height: 115%;">utb - Win</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">a</span><span style="font-size: 14.0pt; line-height: 115%;"> - </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">inb<o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">C</span><span style="font-size: 14.0pt; line-height: 115%;">onsidering that </span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">Wout</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> a</span><span style="font-size: 14.0pt; line-height: 115%;"> = Win</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> <span lang="EN-US">a</span></span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%;"> </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"><o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;"> = </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Wo</span><span style="font-size: 14.0pt; line-height: 115%;">ut</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">b – </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Win
b<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">So such</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> a</span><span style="font-size: 14.0pt; line-height: 115%;"> unit will have a beneficial </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">force</span><span style="font-size: 14.0pt; line-height: 115%;">.<o:p></o:p></span></div>
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<span style="font-size: 14.0pt; line-height: 115%;">As unite heat
exchangers 4a and 4b, </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">u</span><span style="font-size: 14.0pt; line-height: 115%;">nite turbine 1a and 1b and compressors 2a and 2b proceed to an
efficient engine where Wout</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> </span><span style="font-size: 14.0pt; line-height: 115%;">> </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">W</span><span style="font-size: 14.0pt; line-height: 115%;">in -</span><span style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> <span lang="EN-US">f</span></span><span style="font-size: 14.0pt; line-height: 115%;">igure 1a</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"><o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">In this line of thinking exchanger 5 may be
unnecessary - chart 3 (it is derived from the chart 1 depending on the setting
of valves 7)<o:p></o:p></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHGf-rFDs3ehfneBmCEpmjyhuDu22FDEUFfXVstYOpmO3Kb7TwxhULrwnOi9tChzgGOoNpMp0IDJ2zpEf-OVjHxfNx5qxzz1_Q5Kf1Mpt0n77A6sod_PUBh8-Q0da4-LvR2K9B35MZY_k/s1600/3+da.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="292" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHGf-rFDs3ehfneBmCEpmjyhuDu22FDEUFfXVstYOpmO3Kb7TwxhULrwnOi9tChzgGOoNpMp0IDJ2zpEf-OVjHxfNx5qxzz1_Q5Kf1Mpt0n77A6sod_PUBh8-Q0da4-LvR2K9B35MZY_k/s640/3+da.png" width="640" /></a></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">All waste heat return it to the evaporator 3. This
will lead to higher temperatures T2 and a small amount of circulation of the
working substance<o:p></o:p></span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-63306252341162489612016-07-07T13:11:00.000-07:002016-07-07T22:40:02.727-07:00n number of turbines engine on endothermic chemical processes<div class="MsoNormal">
<span style="font-size: 14.0pt; line-height: 115%;">If the
capacity of the solution is</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> too</span><span style="font-size: 14.0pt; line-height: 115%;"> little to cool the working substance to a temperature
below its boiling point, will have to reduce the amount of waste heat. Let that
be a </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">n</span><span style="font-size: 14.0pt; line-height: 115%;"> of the number of working substances unit</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> as</span><span style="font-size: 14.0pt; line-height: 115%;"> I drew </span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">on</span><span style="font-size: 14.0pt; line-height: 115%;"> diagram 1.</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"><o:p></o:p></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxeOz3VZ1HJMqLDOGBG4DOc-_XVKN_a6crmTJtIvY_vxS9C2NXkHr0ScwaaFpEd6GTzoaEl9bGr-wIi8Uo_pSUpV-63MRxyBLxPayPGk3g9uVai6DgRmh705zA4tNIlgiKGm-6r1qSsBo/s1600/n+turbini+1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="292" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxeOz3VZ1HJMqLDOGBG4DOc-_XVKN_a6crmTJtIvY_vxS9C2NXkHr0ScwaaFpEd6GTzoaEl9bGr-wIi8Uo_pSUpV-63MRxyBLxPayPGk3g9uVai6DgRmh705zA4tNIlgiKGm-6r1qSsBo/s640/n+turbini+1.png" width="640" /></a></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">For me it is not known what are the possibilities of
endothermic solutions to cool, but at the expense of this issue of reducing
waste heat I am debate in physical
methods for creating cold part. So in short:<o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Several working substances - a, b ... n each with a
lower boiling point. Solvent </span><span style="font-size: 14.0pt; line-height: 115%;">α</span><span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;"> heated working substance a above its boiling point;
due to the heat exchange between a and b a liquefies and b boiling ... and so to working
substance n .<o:p></o:p></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Solvent </span><span style="font-size: 18.6667px; line-height: 21.4667px;">α</span><span lang="EN-US" style="font-size: 14pt; line-height: 115%;"> heat exchange with each of the evaporators on
working substances. The latter working substance n should have a boiling point
lower than the boiling point of the solvent </span><span style="font-size: 14pt; line-height: 115%;">α</span><span style="font-size: 14pt; line-height: 115%;"> <span lang="EN-US">and higher than the boiling point of
the solution</span></span><span lang="EN-US" style="font-size: 14pt; line-height: 115%;"> </span><span style="font-size: 14pt; line-height: 115%;">αβ</span><span style="font-size: 14pt; line-height: 115%;"> <span lang="EN-US">. So working substance n liquefies solvent </span></span><span style="font-size: 14pt; line-height: 115%;">α</span><span style="font-size: 14pt; line-height: 115%;"> <span lang="EN-US">, and the solution</span></span><span lang="EN-US" style="font-size: 14pt; line-height: 115%;"> </span><span style="font-size: 14pt; line-height: 115%;">αβ</span><span style="font-size: 14pt; line-height: 115%;"> <span lang="EN-US"> closes the cycle of working substance n.</span></span></div>
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">Probably every one variant of the physical method I've
drawn may be converted to chemical - with a one working substance, with n of
number of working substances and two working substances with a common cold part,
so small capacity (possibly) of cooling on solution hope it is not a problem.<o:p></o:p></span><br />
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<span lang="EN-US" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-US;">08.07.2016</span><br />
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<span lang="EN-US" style="font-size: 18.6667px; line-height: 21.4667px;">For а less waste heat would be appropriate to have another heat exchange between the solvent <span style="font-size: 18.6667px; line-height: 21.4667px;">α</span> and the last working substance n as in chart 2</span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-28404451354046528112016-07-05T11:37:00.000-07:002016-07-05T11:37:15.495-07:00one turbine engine on solution<div class="MsoNormal">
It is good<span lang="EN-US">
one</span> to give a break from work. "Vacationing makes champions" -
are increasingly convinced of the rightness of this maxim <span lang="EN-US">l</span>o<span lang="EN-US">l<o:p></o:p></span></div>
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So did I - I gave <span lang="EN-US">myself </span> three weeks
vacation on "an internal cooling<span lang="EN-US"> engine</span>" and <span lang="EN-US"> after </span>returned saw that the
engine using the endothermic solution for obtaining low temperatures of the
cold part is not necessarily to be on two working substances and two turbines. The
possibility - second working substance with low boiling point<span lang="EN-US"> as</span> doing work to create the liquid
solvent is not bad, but the possibility - solution to create liquid solvent is
even better because we eliminate one turbine<span lang="EN-US"> and</span> unit becomes small and compact. <span lang="EN-US"><o:p></o:p></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMYteeh7d43OBXHmabk773kw9pygo52DgMLAIw5_wyQkPR0PXV1Mmx3JRu03GXfvfkKuxo-DeHwj_fehvfBL25j70G4QpyTc-nHXBxrHvP-HMK6QomfgEYV_T69kM6R7G_agGJdLcaAcE/s1600/raztvor+1+turbina+en.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="292" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMYteeh7d43OBXHmabk773kw9pygo52DgMLAIw5_wyQkPR0PXV1Mmx3JRu03GXfvfkKuxo-DeHwj_fehvfBL25j70G4QpyTc-nHXBxrHvP-HMK6QomfgEYV_T69kM6R7G_agGJdLcaAcE/s640/raztvor+1+turbina+en.png" width="640" /></a></div>
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Here's patent application which handed today (Chart 1)</div>
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<span lang="EN-US" style="font-family: Calibri, sans-serif; line-height: 115%;"><span style="font-size: large;"> αβ - </span></span><span style="font-family: Calibri, sans-serif;"><span style="line-height: 18.4px;">endothermic solution</span></span></div>
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<o:p> </o:p><span style="font-family: Calibri, sans-serif; line-height: 115%;"><span style="font-size: large;">α - </span></span><span style="font-family: Calibri, sans-serif;"><span style="line-height: 18.4px;">solvent</span></span></div>
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<span lang="EN-US" style="font-family: Calibri, sans-serif; line-height: 115%;"><span style="font-size: large;"> β - </span> </span><span style="font-family: Calibri, sans-serif;"><span style="line-height: 18.4px;">solute</span></span></div>
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I would like to ask if anyone had guessed this opportunity as diagram1, аnd he has filed an application before me (not too hard to guess, and I think that at least 4 teams of very good
specialists in various places in the world working on internal cooling engine) - Please write to me at:</div>
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megagreenenergy@gmail.com</div>
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I will withdraw <span style="font-family: Calibri, sans-serif; font-size: 11pt; line-height: 115%;">this
application to </span>save my costs on this application</div>
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Many thanks</div>
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Svetozar the Cold</div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-60203917505557774372016-05-18T19:56:00.000-07:002016-05-19T06:29:02.043-07:00Wanted solvent and solute<div class="MsoNormal">
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<span style="font-size: 22pt; line-height: 115%;">Reflections
on the motive power on heat of the environment and on machines fitted to
develop that power<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">I imagined
that the issue goes beyond physics and moves in the field of chemistry, and
will have to leave ... but it is not easy to me stop thinking about it. These are
ten years till now - think themselves come, whether I want or do not want. So
since I announced that I stop, I saw development for the physical framework of
the process of making the cold part of the unit with endothermic chemical
processes. Even a "devil" whispered to me: "Do not be silly!
Patented it!" The "devil"
managed to tempt me ,and I handed application with the patent office. Here's
how things went with endothermic processes:<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Previous post
ended like this:<o:p></o:p></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgfKlP9TEYzyCz96XJ6HifMrMNS7SEXlVZwPHy9Fb4nD_3ssq2MD9njWluzd-BB0xtAfMEbzZJovHMC0lDBN1V0dZv0OBf6ZDcN3wV5gipwPVTg3ST_llTWO_KRI-hO8wsx_GYCLtT_VEk/s1600/8002.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="344" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgfKlP9TEYzyCz96XJ6HifMrMNS7SEXlVZwPHy9Fb4nD_3ssq2MD9njWluzd-BB0xtAfMEbzZJovHMC0lDBN1V0dZv0OBf6ZDcN3wV5gipwPVTg3ST_llTWO_KRI-hO8wsx_GYCLtT_VEk/s640/8002.png" width="640" /></a></div>
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<span lang="EN-US" style="line-height: 115%;">I thought,
Wait a minute! You should be warm and working substance in the environment. I was thinking about chemical processes and have neglected the physical basis for them and engine. But physics should outline the basics to open bigger opportunities for chemistry. So
things came to diagram 4<o:p></o:p></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhF2qed6jDS24NLCMvThyeSl_s3ZlkmBx9hnXJ8JWQ18TtRXkMU3jJEjoP8LsByNZ8DjxA-YzTiXKn2yqm2V9o-OURhgIVVEQA-CuTdBdNEiEZ6k7QyTBNw-_AqsLio4oGtmCh56nxb7C4/s1600/8003.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="344" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhF2qed6jDS24NLCMvThyeSl_s3ZlkmBx9hnXJ8JWQ18TtRXkMU3jJEjoP8LsByNZ8DjxA-YzTiXKn2yqm2V9o-OURhgIVVEQA-CuTdBdNEiEZ6k7QyTBNw-_AqsLio4oGtmCh56nxb7C4/s640/8003.png" width="640" /></a></div>
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<span lang="EN-US" style="line-height: 115%;">Then they
seized me thinks - Will I get cold in the cold part after heat up the working
substance in the environment, and then again with the exothermic reaction
between <span lang="EN-US" style="font-family: "calibri" , "sans-serif"; font-size: 14.0pt; line-height: 115%;">α</span><span lang="EN-US" style="font-family: "calibri" , "sans-serif"; font-size: 18.0pt; line-height: 115%;"> </span><span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">and</span><span lang="EN-US" style="font-family: "calibri" , "sans-serif"; font-size: 18.0pt; line-height: 115%;"> </span><span lang="EN-US" style="font-family: "calibri" , "sans-serif"; font-size: 14.0pt; line-height: 115%;">β </span>?</span><span lang="EN-US"> </span><span lang="EN-US" style="line-height: 115%;">Instead of turning to energy equations I saw technical solution:
We will use the mixer – diagram 5<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Control
valves 13 and 14, and so the temperature of the working substance we can change
from the temperature of the heat
exchanger 1 (when the valve 14 is closed) to ambient temperature (when the
valve 13 is closed).<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Then I saw
the flaw in this chart - waste a cold. Unit will become more efficient if we use
cold working substance and compound out of the cold part to cool the <span lang="EN-US" style="font-family: "calibri" , "sans-serif"; font-size: 14.0pt; line-height: 115%;">α</span> and <span lang="EN-US" style="font-family: "calibri" , "sans-serif"; font-size: 14.0pt; line-height: 115%;">β </span> before entering the heat exchanger 1. So the device evolved as chart 6<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Evolution
underwent and unit using an endothermic solution for low temperatures in the cold
part diagram 7<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">For example -
solvent α is ammonia (240K; boiling
point). At a temperature of 290K (17C) ammonia gases have a pressure of 8 MPa.
Solute αβ to be a mystical salts AxBy.
Hypothetically - AxBy was dissolved in ammonia by this chemical process is
endothermic.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Working substance
γ let's fluoromethyl CH3F (R-41; 195K bp).<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">In the heat
exchanger 1 of the drawing 7 ammonia boils and the solution was separated
into ammonia and AxBy. Ammonia gases perform work in the turbine 5, where they
enter the heat exchanger 2. In heat exchanger 2 ammonia liquefies due to heat
exchange with the liquid working substance CH3F. A pump (7) takes the liquid
ammonia in the heat exchanger 3.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Salts AxBy
separated from the solution pump 8 (probably screw) them ending up at 4 heat
exchanger to cool them before you take them in a heat exchanger 3 diagram 7.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;"> Working substance CH3F is pre-cooled and
liquefied to a temperature between its boiling point and its freezing point. In
heat exchanger 2, it is heated to a temperature of 240K. The pressure is increased
and it is boiling. Gases CH3F perform work in the turbine 6. From there enter
the heat exchanger 3 where liquefy due to low temperature created by the
dissolution of salts AxBy in the solvent ammonia. Pump 7 takes liquified working
substance CH3F first in heat exchanger 4 to cool the solute AxBy and then in
heat exchanger 2 to receive heat from the gases ammonia and so the cycle of
working substance is repeated.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Ammonia
boiling in the heat exchanger 1, perform work in the turbine 5, salts are
dissolved in ammonia in the heat exchanger 3, solution heat exchange first with
AxBy in a heat exchanger 4, then (possibly) with ammonia and fluoromethyl in a
heat exchanger 2 and out of the heat insulating part where is heat exchanger 1 to heat from the surrounding
environment and the process to begin again. In this renewable process would
receive mechanical energy from both turbines 5 and 6 at the expense of the heat
of the environment:<o:p></o:p></span></div>
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<span style="font-size: large;"><span style="font-family: "calibri" , sans-serif; line-height: 115%;">W = W<sub>1</sub> + W<sub>2</sub> = Q<sub> in</sub></span><sub><span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;"> </span></sub></span>Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-9760955635973582572016-04-30T06:54:00.000-07:002016-07-07T22:25:34.045-07:00An endothermic chemical reaction for closing cycles <span style="font-family: "calibri" , sans-serif; font-size: 18pt; line-height: 115%;">External
combustion- internal cooling engine</span><br />
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<span lang="EN-US"><span lang="EN-US" style="line-height: 115%;">In current engines we use exothermic chemical
reactions (chemical process that releases heat </span><span style="line-height: 115%;"> <span lang="EN-US">) to heat the hot part, so that the heat of the environment to use
for a cold part - Chart 1.</span></span></span></div>
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<tr><td class="tr-caption" style="text-align: center;">https://commons.wikimedia.org/wiki/File%3ARankine_cycle_layout.png </td></tr>
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<span lang="EN-US"><span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">I would venture to suggest that to using endothermic
chemical reaction (</span><span lang="EN-US"><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">chemical reaction in which the system absorbs energy from its surroundings</span><span style="line-height: 115%;">) to cool the cold part, so
that the heat of the environment to use to heat the hot part - Chart 2. <o:p></o:p></span></span></span></span></div>
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<span lang="EN-US" style="line-height: 115%;">Let working substance (</span><span lang="EN-US" style="line-height: 115%;">γ</span><span lang="EN-US" style="line-height: 115%;">)
of the engine has a boiling point lower than the temperature of the environment
- for example, </span><span lang="EN-US" style="line-height: 115%;">γ</span><span lang="EN-US" style="line-height: 115%;"> is ammonia ( 240K bp). At ambient
temperature 290K gases ammonia have 8MPa
pressure and driven turbine (piston). To close the cycle will use an
endothermic reaction between two substances </span><span lang="EN-US" style="line-height: 115%;">α</span><span lang="EN-US" style="line-height: 115%;"> and
</span><span lang="EN-US" style="line-height: 115%;">β </span><span lang="EN-US" style="line-height: 115%;">to the heat exchange with the gases ammonia, which
will liquefy the low temperature caused by a chemical reaction. Let α and β are
nitrogen and oxygen, such as by reacting with each other to give Nitrous oxide ( <span style="font-size: large;">N</span>2<span style="font-size: large;">O</span> ).
This chemical reaction is related to the withdrawal of heat.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Another option (why not essential?) Is </span><span lang="EN-US" style="line-height: 115%;">α</span><span lang="EN-US" style="line-height: 115%;">
and </span><span lang="EN-US" style="line-height: 115%;">β</span><span lang="EN-US" style="line-height: 115%;"> are solvent and solute (salts), in
mixing that takes heat (endothermic solution). Guess - When solutions can probably achieve the
best option - to crystallize by the heat of the environment? So we can repeat the process again with the
same substances?<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Naturally – The unit and processes must be heat insulated
from the environment.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">PS I am a supporter on the physical method with use of mechanical
power (refrigerator) to obtain the cold in cold part.<span style="font-size: 18pt;"><o:p></o:p></span></span></div>
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<span style="font-size: large;">2 may 2016</span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">How nice it would be if we have any substances that are interconnected in endothermic reaction and the resulting chemical compound is unstable at ambient temperature! For example we use the chemical reaction between these substances to create a cold area with very low temperatures (to accept lower than 170K), and then after put out the resultant compound from thermal insulated cold part of the unit to the environment warm up in which disintegrate? The law of Lavoisier - Laplace collapse will be accompanied by heat, which we can use to heat the hot part of the unit / other unit. Or a catalyst to help the separation / reaction of the substances?</span></span></div>
<div style="font-family: calibri, sans-serif; line-height: 115%;">
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">So we can use the same amount of substances repeatedly to provide engine work</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> - a renewable process so that the resulting mechanical energy of the unit will be 100% renewable sources.</span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">I will try to convince chemists that is worth working on. </span></span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="font-size: large; line-height: 18.4px;">3 may 2016</span></span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Yes, the idea of cold part created by a chemical reaction is not new to me, but these days more and more solidified the idea that if there are appropriate substances, such unit will be useful in many cases even more than the aggregate with cold part created by physical methods. Probably chemical method will allow us smaller in size engines with more power . In the chemical method is likely to achieve greater temperature difference between the hot and cold part with less components - compressor dropped for example.</span></span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Of course the comparison between the physical and chemical unit is possible</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> olny in renewable chemical processes.</span></div>
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<span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">4 may 2016</span><br />
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<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">Wanted! Wanted!</span></div>
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<span lang="EN-US" style="line-height: 115%;">The attention of chemists:<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Seek solutions and substances that have properties
such substances </span><span lang="EN-US" style="line-height: 115%;">α </span><span lang="EN-US" style="line-height: 115%;">and </span><span lang="EN-US" style="line-height: 115%;">β</span><span lang="EN-US" style="line-height: 115%;">
of Chart 3:<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">1. To connect each other in an endothermic reaction
(1 on diagram3)<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">2. The resulting compound / solution is unstable at
ambient temperature (2 on diagram3)<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">3. With some intervention (heating, catalyst spark
etc.) unsustainable compound between them to break, assuming that this process
is accompanied by heat (3)<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">4. Divide substances and cooled them to ambient
temperature, so that we can fulfill all these four points again<o:p></o:p></span></div>
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<span lang="EN-US"><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Many thanks</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> </span></span></div>
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<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">Svetozar the Cold</span></div>
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<span lang="EN-US" style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; font-size: large;">5 may 2016</span></span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Looking at chart 3 one might think that the work of such a unit is impossible. At first glance, here is nonsense - two substances are connected and disconnected by chemical reactions, and these two chemical process with the same quantities of materials give us useful mechanical energy ?? This is contrary to the laws of nature - This is so that we can create energy ?? - Impossible!</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">This is at first glance. Actually we have two heat exchange with the environment -</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Position 2 - the compound is heated (assuming energy) from the environment during its transition from cold to warm part</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;"> Position 4 - substances cool - give out energy to the environment in their transition from warm to cold part</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">These two heat transfer (position 2, 4) form a useful mechanical energy that can be drawn from such an engine, and it is equal to the difference between the amount received heat from the environment during the passage of the compound from the cold to the warm part - <span style="font-size: large;">Q</span>in , and the amount of given heat the passage of substances from the warm to the cold part - <span style="font-size: large;">Q</span>out</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;"> <span style="font-size: large;">W</span>turbine = </span></span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in - (- </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out) = </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in +</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out</span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">In fact - the sum of the amounts of heat exchanged between the substances and the environment.</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Perhaps and you ask yourself - Why to use a chemical method once and it useful energy is formed by the heat exchange between a substance and environment as in physical method ?</span></span><br />
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">It should be chemical (assuming that we have renewable chemical processes) take advantage, because the warm part in this method can have a temperature higher than ambient temperature (while in physical strictly confined to the ambient temperature). It will shape a large temperature difference between the hot and cold side of the unit, respectively greater useful power.</span></span></div>
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<span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">7 may 2016</span></div>
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Oоps, the formula :</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-size: large;">W</span>turbine = </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in - (- </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out) = </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in +</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out </span><br />
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> for useful mechanical energy expressed by the quantities of heat which substances exchange with the environment is wrong.</span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Although I know that when it comes to converting heat into work before I have to express any statement on the matter have to think twice and still make mistakes to express an opinion without I have thought many times - I beg your pardon!</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">This formula will make / destroy energy</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Come on, let's remakes:</span></span><br />
<span style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"></span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">By the law</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> conservation of energy: the sum of the mechanical energy produced by the converter on heat into mechanical energy </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">W</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">turbine</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> and </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out</span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;"> must be equal to amount of heat given from the environment on the compound / solution - </span></span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">. </span><br />
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">The mechanical energy from the turbine will turn into heat and to be not create / destroy energy should have the following equality:</span><br />
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<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-size: large;">W</span>turbine + </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> = </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in </span><br />
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<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">so that:</span><br />
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<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-size: large;">W</span>turbine = </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">in - </span><span style="font-family: "calibri" , sans-serif; font-size: large; line-height: 18.4px;">Q</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">out</span></span><br />
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><br /></span></span>
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">I will use the case that I am in mine blog to present an analogy on external combustion - internal cooling engine . Because of frequent disputes if possible: the conversion of heat on the environment into mechanical energy, which will inevitably turn into heat and this circle of energy is repeated , want to give an analogy that I think is relevant (It applies to aggregate filled in the physical method, but probably appropriate in aggregate performed by chemical methods with renewable chemical processes):</span></span><br />
<span style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"></span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Will compare unit with a dam and power plant. The thermal insulation of the engine is the sluice that barred the river. If we have a river / environment with a suitable temperature;</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> and a turbine, to obtain mechanical energy than we should to dam the river / to insulated unit. The river will fill dam / should need more investment - to cool the cold part by external force ..... in both cases nature will do the rest. </span><br />
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">External combustion - internal cooling engine - This definition is probably not appropriate for the engine as this on chart 3, but the definition suggest to refine when we find a renewable chemical process.</span></span><br />
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<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">P.S. </span></span><br />
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> </span></span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Because I feel that I have nothing more to add on the subject </span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">external combustion - internal cooling engine,</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> I will concentrate on one of my "old love" and in the next few years (maybe soon) will present a theory that is in another area other than physics - </span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">To make a announcement :</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">If you see Svetozar the... (I have not yet decided what) to advertise a theory - read! I hope that will be interesting and will not waste time in vain.</span></span><br />
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Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-71949830368010847712016-03-10T08:00:00.006-08:002016-03-29T11:21:24.324-07:00Rankin cycle § Zero cycle on pistons<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;"> External combustion- internal cooling engine with two working substances on pistons§cylinders<o:p></o:p></span></span></div>
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<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;"> (Rankin cycle and zero cycle on pistons)</span><o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">I will present some reflections on the use of two
working substances (with different boiling points) who work in thermally
isolated environment by pistons. I will discuss several phases of performing a work of the substances and phase of application of force on one of them, and next
week will try to connect them in a "analog" type who will represent
external combustion- internal cooling engine of two working substances filled
with pistons .<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Let us have two Dewar containers with two working
substances in liquid state. One with a high boiling point will call it Alpha, and the other with a low boiling point will call it Beta substance. The
containers are connected to the cylinders in which the pistons move.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">All processes of course developed in thermal
insulated environment.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Let in one Dewar have some amount of liquid substance
Alpha with a temperature higher than its boiling point -T1 on diagram 1.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgh09gypBwZ-XGZXfNNlOC5VUXGy0ZhMk-mFPzEhmfXZjgs2rMm5buBfXCA4Q3FNfvyGUuM5mAKKAH2O6glppHW5rYaPbn8v5YBCpCdY62bkXvhe4f00gB4N4b2oElU1lZdtHaKTYH5N88/s1600/7000.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="384" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgh09gypBwZ-XGZXfNNlOC5VUXGy0ZhMk-mFPzEhmfXZjgs2rMm5buBfXCA4Q3FNfvyGUuM5mAKKAH2O6glppHW5rYaPbn8v5YBCpCdY62bkXvhe4f00gB4N4b2oElU1lZdtHaKTYH5N88/s640/7000.png" width="640" /></a></div>
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<span lang="EN-US" style="line-height: 115%;"> The container is
connected to the cylinder and piston position A on diagram1. We put weight N kg.
the piston for opening the valve substance will expand (evaporate) and pushed
the piston - respectively the weight of a distance -position B. Let
equilibrium between the pressure in the container and the weight on the piston
is in such an increase in volume, wherein the substance cools down to a
temperature T2 = (T1 - Tbp) / 2,where Tbp is a boiling point of the substance.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">In another Dewar we have some amount of liquid
substance Beta at temperature T1 which is the initial temperature of Alpha.
Container is also connected to the cylinder/piston - chart 2.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEitFFOl4K07P0k9VrzN587IsMOVvvdhu8fNpKpESFsC8zDDzBOEOUJ-mtm3i_JAxpa6inRjrLLSdZduswewcWQV-S7wM20XgHEkK5a9gD_N4iKnxlXFoOGflMk1MIs4XHJZHy9ARSAsk_g/s1600/7001.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="384" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEitFFOl4K07P0k9VrzN587IsMOVvvdhu8fNpKpESFsC8zDDzBOEOUJ-mtm3i_JAxpa6inRjrLLSdZduswewcWQV-S7wM20XgHEkK5a9gD_N4iKnxlXFoOGflMk1MIs4XHJZHy9ARSAsk_g/s640/7001.png" width="640" /></a></div>
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<span lang="EN-US" style="line-height: 115%;"> We take gas from
cylinders of the substance Alpha and put them in a heat exchanger to a
container of the substance Beta. On the piston put weight equal to ½ of weight N where Alpha is in equilibrium at T2. Let the amount of the substance Beta be
such that upon opening of the valve together with the gases of Alpha gravity
move the same distance, and the system goes into equilibrium at a temperature
T3 equal to the boiling point of the Alpha - position B in diagram 2. In its
equilibrium position volume on Beta has been extended so that the temperature of
the two substances (beta has a low boiling point and heat exchange between
them) is equal to the boiling point of the Alpha - gases Alpha liquefies at
equilibrium of the gas pressure of the Beta and weight equal to to ½ of gravity
N.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Now I want to discuss the question - What is the smallest
weight that if we put on the piston to return it to the starting position - to
return to the starting position parameters of volume, pressure and temperature
of the substances in these processes of charts 1 and 2? By low conservation of
energy this will be another added weight Nkg for Alpha, and 1 / 2Nkg for Beta - chart 1a for Alpha, </span></div>
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<span lang="EN-US" style="line-height: 115%;"><span style="line-height: 18.4px;"> and charts 2a and 2b for Beta.</span></span></div>
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<span lang="EN-US" style="line-height: 115%;"> As work has made the substance, so the force
applied to it to perform the same work on it, and the substance returns to its initial values of temperature, volume and pressure .<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;"> Let pistons of the two containers with different
substances are connected to the "scale" - diagram 3a,</span></div>
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<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
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<span lang="EN-US" style="line-height: 115%;"> or better of the
crankshaft in the opposite direction of movement - diagram 3c.</span></div>
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<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"> In the
condition of opening the valve Alpha will be in equilibrium with the weight Nkg. and Beta by weight 1 / 2Nkg i.e. the piston of Alpha acting force twice larger than
the force on the piston Beta. Alpha substance has power precisely so as to
return the substance beta to its initial state after opening the valves (as I
follow the logic of the previous charts 1a, 2a and 2b) - diagrams 3b; 3d</span></div>
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<span lang="EN-US" style="line-height: 115%;"> 3d.</span></div>
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<span style="line-height: 115%;"> The
temperature of Alpha in the container and cylinder (liquids and gases) in equilibrium position </span><span style="line-height: 18.4px;">by default</span><span style="line-height: 115%;"> (position b) T2 = (T1 -Tbp) / 2, and the
temperature of Beta in position b (its initial state) is T1, respectively
liquid Alpha which heat exchange with Beta also has a temperature T1.</span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">If I remove the
liquid Alpha from heat exchanger , and in its place put gas Alpha from cylinder will return
to the starting position at which gases Alfa at temperature T2 and liquid Beta at temperature T1 perform work as push the piston respectively gravity 1 / 2Nkg to
their equilibrium position as I start - diagrams 4a; 4b; 4c .<o:p></o:p></span></div>
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<span style="line-height: 115%;">To return to the starting position the container
with liquid Alpha substance must be heated liquid alpha in the container of
temperature T2 to temperature T1. This heat has turned into mechanical energy.</span></div>
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<span lang="EN-US" style="line-height: 115%;">4c start (end) position</span></div>
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<span lang="EN-US" style="line-height: 115%;">Discussed above processes and actions with both
substances Alpha and Beta them harnessed in one unit to perform work on behalf
of a heat source. Naturally as with any patterns external combustion -
internal cooling engine heat source can be the environment - Alpha substance must must be a
boiling point lower than ambient temperature. These few several phases of
action I arrange them in a station that end (or initial) phase performs some work (raising the weight 1 / 2 Nkg ,of gases on substance Alpha, and substance Beta ,where the Nkg it is the strength of the alpha) on account of the heat source. I summarized :</span></div>
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<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;"> Based on pre-set temperatures, quantities and volumes of two working
substances can receive mechanical force as one substance - Alpha gets heat from source and works in Rankin cycle, and other Beta participate in the closing
cycle on Alpha, and in start / end point its parameters remain unchanged ( Zero cycle).</span></span><span style="line-height: 115%;"><o:p></o:p></span></div>
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<span style="line-height: 115%;"><br /></span></div>
<span style="font-size: large;">To be continued</span><br />
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<span lang="EN-US" style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif;">P.S. Right now I would like to propose for discussion a more interesting situation - Gases Alpha and Beta liquid heat exchange and perform work, so in their equilibrium position with weight 1 / 2n on the piston in the container temperature is close to freezing point of Alfa - diagram 5.</span></span><br />
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<span lang="EN-US" style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif;"><br /></span></span>
<span lang="EN-US" style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif;"> Let's Alpha be ammonia ( 240K bp, 196K mp) and Beta is nitrogen (77K bp). Let quantities Alpha and Beta are such that in the equilibrium position - position B from a temperature of 300K and 270K of nitrogen and ammonia gases temperature decreased to 200K - close to the freezing point of the ammonia,</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">due to the increase a volume on Beta</span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;"> . Now when I open the valve will have two forces - a piston which rises weight 1 / 2N kg, and another piston - over ammonia gases, due to contraction of temperature close to freezing point rise weight Xkg. Total work done from position A to position B will proportional on 1/ 2N kg + Xkg. What weight can raise (to N eventually) I do not know. I would prefer to check it empirically :)</span></span><br />
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<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">11.03.2016</span></span><br />
<span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;"><span style="line-height: 18.4px;">Another more close to our ideas example of a puzzle with the force of contraction - diagram 5a</span></span></span><br />
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<span style="font-size: large;">To be continued</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">12.03.2016</span><br />
Here are the reviewed processes in unit - two working substances on pistons - diagram 6<br />
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<br />
indications:<br />
1 - double-acting piston<br />
2 - evaporator<br />
3 - compression container<br />
4 - valve<br />
5 - reducer valve<br />
6 - heat exchanger with the heat source<br />
7 - pump<br />
8 - heat exchanger<br />
Good external combustion - internal cooling engine must have a good thermal insulation. In the case shall the pistons and cylinders must to be of materials with low thermal conductivity.<br />
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Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-17413788732250768652016-02-16T07:41:00.000-08:002016-03-29T11:28:17.064-07:00Theory on the topic: "Two working substances with a common cold part "<div class="MsoNormal">
<span style="line-height: 18.4px;">I will try to theorize on topic:</span><span style="line-height: 115%;"> "Two working substances with different </span><span style="line-height: 18.4px;">boiling point</span><span style="line-height: 115%;"> with common cold part"<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 115%;">I'll start
with a simple understanding of the process of converting heat into mechanical
energy. (And conversion of mechanical energy into heat)<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 115%;">Let us have
a particle substance in the cylinder piston. Let the particle be at a
temperature T<sub>1 </sub> higher than the
boiling point of the substance diagram 1.</span><br />
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<span style="line-height: 115%;"><br /></span></div>
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<span style="line-height: 115%;"> Let the other fraction of the same
substance at a temperature equal to the boiling point of the substance T</span><sub><span lang="EN-US" style="line-height: 115%;">bp</span></sub><span lang="EN-US" style="line-height: 115%;"> </span><span style="line-height: 115%;"> stands from other parts of the p</span><span lang="EN-US" style="line-height: 115%;">iston</span><span style="line-height: 115%;"> (</span><span lang="EN-US" style="line-height: 115%;">A</span><span style="line-height: 115%;">). Assume an ideal option where the
piston will mass and heat isolation - no heat loss. When a</span><span lang="EN-US" style="line-height: 115%;">
hot</span><span style="line-height: 115%;"> particle hits the piston that in turn will hit the particle on
the other side (B) and so the two particles will catch your energies, your
temperature accordingly (</span><span lang="EN-US" style="line-height: 115%;">C</span><span style="line-height: 115%;">). If the temperature - both particles will have a temperature</span><span lang="EN-US" style="line-height: 115%;">(</span><span style="line-height: 115%;"> T</span><sub><span lang="EN-US" style="line-height: 115%;">1</span></sub><span lang="EN-US" style="line-height: 115%;"> </span><span style="line-height: 115%;">– T</span><sub><span lang="EN-US" style="line-height: 115%;">bp</span></sub><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;"> )</span><span style="line-height: 115%;">/ 2. So I can make the following conclusion:<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 115%;">In Idel
version maximum work that can be done given amount of wor</span><span lang="EN-US" style="line-height: 115%;">king
substance</span><span lang="EN-US" style="line-height: 115%;"> </span><span style="line-height: 115%;"> in a converter </span><span lang="EN-US" style="line-height: 115%;"> on </span><span style="line-height: 115%;">heat into mechanical energy (whether in the cylinder piston
or turbine) is equal to the equivalent of the </span><span style="line-height: 18.4px;">arithmetical average</span><span style="line-height: 115%;"> its
temperature and its boiling point<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 18.4px;">W= cm(T1 - Tbp)/2</span><br />
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<span style="line-height: 115%;">If </span><span lang="EN-US" style="line-height: 115%;"> I </span><span style="line-height: 115%;">put a transformer </span><span lang="EN-US" style="line-height: 115%;"> of </span><span style="line-height: 115%;">heat into mechanical energy diagram 2, 2a (turbine </span><span lang="EN-US" style="line-height: 115%;">on
</span><span style="line-height: 115%;">diagram) between the two
tanks with a liquid substance, one has a temperature above the boiling point T</span><sub><span lang="EN-US" style="line-height: 115%;">1 </span></sub><span lang="EN-US" style="line-height: 115%;">,</span><span style="line-height: 115%;"> and the other has a temperature
equal to the boiling point of a substance T</span><sub><span lang="EN-US" style="line-height: 115%;">bp</span></sub><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">,</span><span style="line-height: 115%;">this turbine will perform work
equivalent</span><span lang="EN-US" style="line-height: 115%;"> to</span><span style="line-height: 115%;"> (T</span><sub><span lang="EN-US" style="line-height: 115%;">1</span></sub><span lang="EN-US" style="line-height: 115%;"> </span><span style="line-height: 115%;">– T</span><sub><span lang="EN-US" style="line-height: 115%;">bp</span></sub><span style="line-height: 115%;">) / 2. If work done on the substance
that comes out of the turbine</span><span style="line-height: 115%;"> <span lang="EN-US">with </span></span><span style="line-height: 115%;">a compressor and reducing valve</span><span lang="EN-US" style="line-height: 115%;">,</span><span style="line-height: 115%;">equal work which is carried turbine
then after valve parameters</span><span lang="EN-US" style="line-height: 115%;"> of</span><span style="line-height: 115%;"> substance must return to the starting position - the
work is performed, the work done on it and temperature, volume and pressure at
the beginning of the process and at the end of the processes are the same.<o:p></o:p></span></div>
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<span style="line-height: 115%;">So when we
use work to close the cycle of a working substance can not it remain useful
energy - the mechanical force we have received, so we have to use on the
working substance to close the cycle .<o:p></o:p></span></div>
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<span style="line-height: 115%;">This cycle
will call it - ZERO cycle.<o:p></o:p></span></div>
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<span style="line-height: 115%;">Let us have
two substances with different boiling point chart 3. The first working substance</span><span lang="EN-US" style="line-height: 115%;">-</span><span style="line-height: 115%;"> A has a higher boiling point than second</span><span lang="EN-US" style="line-height: 115%;"> -
</span><span style="line-height: 115%;"> B</span><span lang="EN-US" style="line-height: 115%;">. </span><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">F</span><span style="line-height: 115%;">or each substance have a two tanks, one being at a
temperature higher than their boiling point </span><span lang="EN-US" style="line-height: 115%;">– T<sub>1</sub>
, </span><span style="line-height: 115%;">the other at a
temperature equal to its boiling point</span><span lang="EN-US" style="line-height: 115%;"> - T<sub>bp</sub></span><span style="line-height: 115%;">. The temperature of the substance B
in the hot reservoir is equal to (T1a - Tbpa) / 2. Connect turbines and
compressors to tanks. Both substances have zero </span><span lang="EN-US" style="line-height: 115%;">cycle </span><span style="line-height: 115%;"> - </span><span lang="EN-US" style="line-height: 115%;">as </span><span style="line-height: 115%;">the work carried out, </span><span lang="EN-US" style="line-height: 115%;"> so </span><span style="line-height: 115%;">the work we perform on them</span><span style="line-height: 115%;"> <span lang="EN-US">working substances</span></span><span style="line-height: 115%;">. Will unite hot and cold tanks of both
substances (hot</span><span lang="EN-US" style="line-height: 115%;"> to</span><span style="line-height: 115%;"> hot tank 6a;6b </span><span lang="EN-US" style="line-height: 115%;">,</span><span style="line-height: 115%;">cold with cold tank5a;5b) to heat exchange with each other.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 115%;">Let the
amount of circulating substances are such that the cold p</span><span lang="EN-US" style="line-height: 115%;">art</span><span style="line-height: 115%;"> due to heat exchange the temperature
of the two substances is equal to the boiling point of the substance at a high
temperature - the first substance is liquefied. So after the heat exchange
between the two substances first liquefies. This defeats the compressor 4 and
the reducing valve 7a. </span><span lang="EN-US" style="line-height: 115%;">C</span><span style="line-height: 115%;">omp</span><span lang="EN-US" style="line-height: 115%;">r</span><span style="line-height: 115%;">es</span><span lang="EN-US" style="line-height: 115%;">sor 4 and </span><span style="line-height: 115%;">valve</span><span style="line-height: 115%;"> </span><span style="line-height: 115%;"> </span><span style="line-height: 115%;"> </span><span style="line-height: 115%;">became pump diagram 4.</span><span style="line-height: 115%;"> </span></div>
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<span style="line-height: 115%;"> So the zero cycle of the first
substance(ammonia on diagram 4) becomes familiar Rankine Cycle </span><span style="line-height: 18.4px;">which</span><span style="line-height: 115%;"> we use in external combustion engine.</span></div>
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<span style="line-height: 115%;">Due to heat
exchange conducted between the two substances in </span><span lang="EN-US" style="line-height: 115%;"> their </span><span style="line-height: 115%;">the cold and warm tanks</span><span lang="EN-US" style="line-height: 115%;">,</span><span style="line-height: 115%;"> compressor 3 will operate at high temperatures </span><span lang="EN-US" style="line-height: 115%;"> than </span><span style="line-height: 115%;">turbine 2, but at the same temperature range, which preserves
the balance of carried to the attached force</span><span lang="EN-US" style="line-height: 115%;">s</span><span lang="EN-US" style="line-height: 115%;"> </span><span style="line-height: 115%;">(</span><span lang="EN-US" style="line-height: 115%;">W</span><span style="line-height: 115%;">in / </span><span lang="EN-US" style="line-height: 115%;">W</span><span style="line-height: 115%;">out) operation for the second work</span><span lang="EN-US" style="line-height: 115%;">ing
substance </span><span style="line-height: 115%;">.<o:p></o:p></span></div>
<br />
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<span style="line-height: 115%;">By combining
hot and cold tanks of two Zero</span><span lang="EN-US" style="line-height: 115%;"> cycle,</span><span style="line-height: 115%;"> this cycle with high boiling point is converted</span><span lang="EN-US" style="line-height: 115%;">
in</span><span style="line-height: 115%;"> to Rankin cycle.<span style="font-size: 18pt;"><o:p></o:p></span></span><br />
<span style="line-height: 115%;"><br /></span>
<span style="line-height: 115%;">18.02.2016</span><br />
<span style="line-height: 18.4px;">For me remains unclear whether we can reach a temperature of the first working substance at outlet lower than (Tenv + Tbp)/2 (the average of ambient temperature and boiling point).</span><span style="line-height: 18.4px;">Whether we can set at practice the following temperatures as chart 5:</span><br />
<span style="line-height: 18.4px;">200K of the heat exchanger 5 (close to the melting point of ammonia)</span><br />
<span style="line-height: 18.4px;"> 210K heat exchanger 6</span><br />
<span style="line-height: 18.4px;">and keep them?</span><br />
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<span style="line-height: 115%;"><br /></span>
<span style="line-height: 18.4px;">As lower temperatures achieved (as long as they do not extend below the melting point :)) in the cold part - the higher the useful power will have the unit. Though 10 degrees difference do our work, I mean that after an initial "investment" - setting the cold part, will not be necessary "to spend a penny more."</span><br />
<span style="line-height: 115%;"><br /></span>
<span style="line-height: 115%;">20.02.2016</span><br />
<span style="line-height: 115%;"><br /></span>
<span style="line-height: 18.4px;">QUANTITIES OF HEAT AND WORK</span><br />
<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;"> Rankine cycle -diagram 6; Zero cycle -diagram 6a; Unit with two zero cycle with a common cold part- diagram 6b</span><br />
<span style="line-height: 18.4px;">Accept that a working substance at a temperature equal to its boiling point no pressure in the evaporator and can not perform work, so that will denote the amount of heat that has this substance 0Q.</span><br />
<span style="line-height: 18.4px;">Will denote the amount of heat that gets working substance from the heater with 1Q.</span><br />
<span style="line-height: 18.4px;">In the simplified version of the energy conversion (charts 1 and 2)</span><br />
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<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;"> Rankine cycle -diagrama 6</span><br />
<span style="line-height: 18.4px;">After heater 1Q</span><br />
<span style="line-height: 18.4px;">After turbine 1 / 2Q. As a result of work done by the turbine 1 / 2Q turns into mechanical energy and then have a working turbine substance with 1 / 2Q heat</span><br />
<span style="line-height: 18.4px;">After the heat exchanger - 0Q. Cooled heat exchanger working substance to its boiling point and after it has heat 0Q</span><br />
<span style="line-height: 18.4px;">Heater heated working substance, giving it heat 1Q</span><br />
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<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;">Zero cycle -diagram 6a</span><br />
<span style="line-height: 18.4px;">After heater 1Q</span><br />
<span style="line-height: 18.4px;">After turbine 1/2 Q . 1/2 Q heat has become mechanical energy</span><br />
<span style="line-height: 18.4px;"> After the compressor - 1Q. Compressor performs work equal to 1 / 2Q to liquefy the working substance in which mechanical energy is converted into heat, so after reducing valve total amount of heat which has the working substance is 1 / 2Q +1 / 2Q = 1Q</span><br />
<span style="line-height: 18.4px;"> So working substance with 1Q heat energy enters the heater and after him have the same heat. Zero cycle - 0Q heat has turned into mechanical energy</span><br />
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<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;">Two zero cycle with a common cold part - chart 6b</span><br />
<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;">The first working substance</span><br />
<span style="line-height: 18.4px;">After heater 1Q</span><br />
<span style="line-height: 18.4px;"> After 1 turbine 1/2Q</span><br />
<span style="line-height: 18.4px;"> After heat exchange performed in a heat exchanger 5 0Q</span><br />
<span style="line-height: 18.4px;">1 / 2Q after heat exchange with the second working substance in a heat exchanger 6</span><br />
<span style="line-height: 18.4px;">1Q after the heater is increased heat energiya- 1 / 2Q +1 / 2 Q = 1Q</span><br />
<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;">The second working substance</span><br />
<span style="line-height: 18.4px;"> 1Q after heat exchange in heat exchanger 5 with the first working substance </span><br />
<span style="line-height: 18.4px;"> Compressor performs work such as mechanical energy equal to 1 / 2Q becomes heat -1Q + 1/2Q = </span><br />
<span style="line-height: 18.4px;"> = 3 / 2Q after compressor</span><br />
<span style="line-height: 18.4px;">1Q after the heat exchange in heat exchanger 6</span><br />
<span style="line-height: 18.4px;">1 / 2Q after turbine 2</span><br />
<span style="line-height: 18.4px;">1Q after heatexchanger 5</span><br />
<span style="line-height: 18.4px;"><br /></span>
<span style="line-height: 18.4px;">1 / 2Q thermal energy is converted into mechanical energy of the unit with two working substances with a common cold part compared to Rankine cycle</span><br />
<span style="line-height: 115%;"></span>
<span style="line-height: 115%;"><br /></span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-62332754568593851982016-02-01T03:55:00.001-08:002016-07-07T22:30:41.338-07:00Two working substances with a common cold part as to start the internal cooling<div class="MsoNormal">
<span style="line-height: 115%;">Reflecting
on the method for converting heat into mechanical energy that the cold part I
create by doing work I saw an untapped so far from me the opportunity for
effective work - It would be better to replace</span><span style="line-height: 115%;"> </span><span style="line-height: 115%;">expansion valve </span><span lang="EN-US" style="line-height: 115%;">on
</span><span style="line-height: 115%;">"refrigerator"with
a turbine</span><span style="line-height: 115%;"> </span><span style="line-height: 115%;"> (or piston / cylinder; generally speaking
converter heat into mechanical energy)</span><span style="line-height: 115%;"> <span lang="EN-US">diagram (1a)( 1)</span></span><span style="line-height: 115%;">.</span></div>
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<div class="MsoNormal">
<span style="line-height: 115%;"> On both sides of the expansion
valve system for redistributing heat ("refrigerator") there is a
temperature difference (differential pressure) which is a prerequisite turbine
to perform work. By doing this work converter</span><span lang="EN-US" style="line-height: 115%;"> of </span><span style="line-height: 115%;">heat into mechanical energy (as well
as all others who work in the method) achieved two important goals for us:<o:p></o:p></span></div>
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<span style="line-height: 115%;">- Mechanical
energy<o:p></o:p></span></div>
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<span style="line-height: 115%;">- Cold<o:p></o:p></span></div>
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<span style="line-height: 115%;">The
mechanical energy is our goal, and the cold part we need in the process of
internal cooling in the absence of such a natural.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 115%;">Let the
operating cycle of the n-th working substance creating cold part by using
cooler. Replace the expansion valve with a turbine. We now have a further
beneficial </span><span lang="EN-US" style="line-height: 115%;">force</span><span style="line-height: 115%;"> that is a result of the conversion of heat into mechanical energy.<o:p></o:p></span></div>
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<span style="line-height: 115%;">Replacing
expansion valve</span><span lang="EN-US" style="line-height: 115%;"> with turbine</span><span style="line-height: 115%;"> convert</span><span lang="EN-US" style="line-height: 115%;">s refrigerant</span><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">in</span><span style="line-height: 115%;"> working substance. Now in the cold part (the last n-th
cycle of the working substance with the lowest boiling point</span><span lang="EN-US" style="line-height: 115%;">)</span><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">h</span><span style="line-height: 115%;">ave two working substances with a
common cold part - the cold</span><span style="line-height: 115%;"> </span><span style="line-height: 115%;">part of the system for redistributing heat.<o:p></o:p></span></div>
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<span style="line-height: 115%;">In the cold
part the working substance liquefies
because of the low temperature generated in the expansion of refrigerant due to
the compressor. In the warm part of the </span><span lang="EN-US" style="line-height: 115%;">working</span><span style="line-height: 115%;"> cycle of the refrigerant already
liquid working substance is heated with the same amount of heat which is taken
away from him at liquefaction. Thus, after the heat exchanger in the hot p</span><span lang="EN-US" style="line-height: 115%;">art</span><span style="line-height: 115%;"> will have a liquid working substance
having a temperature ½ of the difference
between the temperature on entry into the turbine and its boiling point<o:p></o:p></span></div>
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<span style="line-height: 115%;"> Working
cycle </span><span lang="EN-US" style="line-height: 115%;">on </span><span style="line-height: 115%;">refrigerant becomes neutral - no change in temperatures,</span> <span style="line-height: 115%;">considering that the same amount of
heat is removed and transferred from refrigerant gas</span><span lang="EN-US" style="line-height: 115%;"> at</span><span style="line-height: 115%;"> working substance</span><span lang="EN-US" style="line-height: 115%;">.</span><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">T</span><span style="line-height: 115%;">herefore the work of the compressor and turbine are the
same - </span><span lang="EN-US" style="line-height: 115%;"> as </span><span style="line-height: 115%;">the mechanical power is converted
into heat in the course of operation of the compressor on refrigerant, </span><span lang="EN-US" style="line-height: 115%;"> so </span><span style="line-height: 115%;">the</span><span lang="EN-US" style="line-height: 115%;"> same</span><span style="line-height: 115%;"> heat turbine is turned into mechanical power. Cycle</span><span lang="EN-US" style="line-height: 115%;">
on</span><span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">refrigerant
</span><span style="line-height: 115%;">agent is a neutral
shade.</span><span lang="EN-US" style="line-height: 115%;"><o:p></o:p></span></div>
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<span style="line-height: 115%;">Energy
balance </span><span lang="EN-US" style="line-height: 115%;"> on </span><span style="line-height: 115%;">n-th working substance + refrigerant
(provided that the turbines convert 50% of the heat into mechanical energy)
will be:</span><span lang="EN-US" style="line-height: 115%;"><o:p></o:p></span></div>
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<br /></div>
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<span style="line-height: 115%;">A zero cycle
– </span><span lang="EN-US" style="line-height: 115%;">the </span><span style="line-height: 115%;"> "refrigerator"<o:p></o:p></span></div>
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<span style="line-height: 115%;">A cycle in
which 50% of the heat is converted into mechanical energy - the working
substance<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="line-height: 115%;">The outcome
of the two cycles is a liquid </span><span lang="EN-US" style="line-height: 115%;"> working </span><span style="line-height: 115%;">substance with </span><span lang="EN-US" style="line-height: 115%;"> a </span><span style="line-height: 115%;">temperature 1/2 of the difference between its temperature
before entering the turbine and then the heat exchanger in the hot part of the
"fridge"</span><span lang="EN-US" style="line-height: 115%;">.<o:p></o:p></span></div>
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<span style="line-height: 115%;">Thus, in the
last cycle (n) working substance decelerated for some amount of heat</span><span lang="EN-US" style="line-height: 115%;">
as</span><span style="line-height: 115%;"> has become a
mechanical energy, whereby cooled</span><span lang="EN-US" style="line-height: 115%;"> the previous(n-1)</span><span style="line-height: 115%;"> to allow the unit to operate.</span><span lang="EN-US" style="line-height: 115%;"><o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Now I can not go back heat into the evaporator - diagram 2<span style="font-size: 18pt;"><o:p></o:p></span></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfjWV_aHTqGCplU_L3ZUwQMjCcX8ke6I5XMw0Z7-Swvv7znAJgfV8BCULwvv3GoIbva7OPsoZ-VFSLyWwTXGMVxqJJJSAj2m_71KqvZDDFZlLz3_dqJu4jwOhDPaIlINCQEoyq6wCd4W4/s1600/diagram+2004.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfjWV_aHTqGCplU_L3ZUwQMjCcX8ke6I5XMw0Z7-Swvv7znAJgfV8BCULwvv3GoIbva7OPsoZ-VFSLyWwTXGMVxqJJJSAj2m_71KqvZDDFZlLz3_dqJu4jwOhDPaIlINCQEoyq6wCd4W4/s640/diagram+2004.png" width="640" /></a></div>
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<span style="line-height: 18.4px;">Variant of еngine with two working substances (ammonia and ethylene in the case) - chart 3</span><br />
<span style="line-height: 18.4px;"><span lang="EN-US" style="line-height: 115%;"></span></span><br />
<span style="line-height: 18.4px;">In optimal load of the turbine and compressor at best we can achieve a temperature difference between input and output =<span style="font-size: large;"> (Tenv - Tbp) / 2</span></span><br />
<span style="line-height: 18.4px;">where : Tenv- temperature on environment; </span><br />
<span style="line-height: 18.4px;"> Tbp - boiling point of the first working substance</span><br />
<span style="line-height: 18.4px;">The power will be expressed (ideally):</span><br />
<span style="line-height: 18.4px;"><span style="line-height: 18.4px;"></span></span><br />
<span style="line-height: 18.4px;"><span style="font-size: large;">P = cm (Tenv - Tbp) / 2</span></span><br />
<span style="line-height: 18.4px;"><span style="font-size: large;"><br /></span></span>
<span style="line-height: 18.4px;">where: <span style="font-size: large;">c</span> - specific heat capacity of the first working substance</span><br />
<span style="line-height: 18.4px;"> <span style="font-size: large;">m </span>- mass of first working substance circulating for a given time</span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhahyphenhyphent0n9sQo6e2CvsrKaUeTL7lvT0Mpr7PfMOLKugLWSJe0GANDtIiW5lU0OSVKjwxI4WIiY70nUH-RFKlmP2OOEt0ZcKp_1DnJd9GqGM40bSLV9SjxKWRU0KEEYq1t4hwzntJsA-J2F4/s1600/3003.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="384" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhahyphenhyphent0n9sQo6e2CvsrKaUeTL7lvT0Mpr7PfMOLKugLWSJe0GANDtIiW5lU0OSVKjwxI4WIiY70nUH-RFKlmP2OOEt0ZcKp_1DnJd9GqGM40bSLV9SjxKWRU0KEEYq1t4hwzntJsA-J2F4/s640/3003.png" width="640" /></a></div>
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<span style="line-height: 18.4px;"><br /></span>
<span lang="EN-US" style="line-height: 18.4px;">Note again: As with all engine variants of the external combustion - internal cooling method cold/s part create it beforehand using external force.</span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-16517153071590047072015-11-11T02:37:00.000-08:002016-01-05T16:18:55.994-08:00External combustion engine VS External combustion - internal cooling engineCharts 1 and 2 depict external combustion engine, such as a free cooler imaginary (as in the familiar environment external combustion engine) closes the cycle. Accept that the 20% waste heat engines produced 100kW power for a given capacity of the converters of heat into mechanical energy (turbines, pistons in cylinders).<br />
<br />
chart 1<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhS4U63qit7fWegzC5vdiZWENp5uWP2ROBSHyXzOKP_3brxP_F3yfgl-8VBuci9fb4Zst3vY3sM1AhsQO_Vyr9qsyNvzubZFO6JkebU7GzIk6DMAzHCZbvI6Aic05_HZ_1CVmOJBaSV3ak/s1600/1100.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhS4U63qit7fWegzC5vdiZWENp5uWP2ROBSHyXzOKP_3brxP_F3yfgl-8VBuci9fb4Zst3vY3sM1AhsQO_Vyr9qsyNvzubZFO6JkebU7GzIk6DMAzHCZbvI6Aic05_HZ_1CVmOJBaSV3ak/s640/1100.png" width="640" /></a></div>
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chart 2<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrbOh73E6y3A42YYopHzrVvrllNHiP37EJzatdg8qABa-DThy8FzaxPewc3KN2QHdO0kQdZrCxgGXrUy_Y2u9iAotWGjjIqPTol1naaAELY1Z4MCvuIFXtMqqzKzBX4VXAhc54hIPAzHM/s1600/1102.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgrbOh73E6y3A42YYopHzrVvrllNHiP37EJzatdg8qABa-DThy8FzaxPewc3KN2QHdO0kQdZrCxgGXrUy_Y2u9iAotWGjjIqPTol1naaAELY1Z4MCvuIFXtMqqzKzBX4VXAhc54hIPAzHM/s640/1102.png" width="640" /></a></div>
<br />
chart 3<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqbj1fxqdB8kxWFG8Rbnm2zyZMIxHSwWB6hyiSsCp5KSTFidDattnvWRd8cNNDwWc3BA9_kI7mG5V7XtooLLqIYoB12-GzfypuaKq8re0AByjwn8u9tMd_jje9t30t64JVXsqHrG9DHpM/s1600/1101.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqbj1fxqdB8kxWFG8Rbnm2zyZMIxHSwWB6hyiSsCp5KSTFidDattnvWRd8cNNDwWc3BA9_kI7mG5V7XtooLLqIYoB12-GzfypuaKq8re0AByjwn8u9tMd_jje9t30t64JVXsqHrG9DHpM/s640/1101.png" width="640" /></a></div>
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chart 4<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMOENF15jbtK-bD7mm2lzWda2d9S7azP4IMtFBOO67qFml2RidUh8cKpeG4lZnZCxv8KisJ7X71ev53u7MDrxYlh2USAam-0wfVq2J3yyZZOGYIIAAL3NO_GDpvrb8Do_Re7nBOJZrOSA/s1600/1103.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMOENF15jbtK-bD7mm2lzWda2d9S7azP4IMtFBOO67qFml2RidUh8cKpeG4lZnZCxv8KisJ7X71ev53u7MDrxYlh2USAam-0wfVq2J3yyZZOGYIIAAL3NO_GDpvrb8Do_Re7nBOJZrOSA/s640/1103.png" width="640" /></a></div>
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Charts 3 and 4 am figuring engines of the same capacity of the converters of heat into mechanical energy as chart 1 and 2, but here I have no imaginary "free" cooler. Close the cycle by using force to close the cycle - cooling gas as I use cooler - External combustion - internal cooling engines. Waste heat (20%) cooler return it in the evaporators of the unit. Accept that besides waste heat in evaporators returns heat equal to the force I used to close the cycle + 20% of heat . So aggregates of diagrams 3 and 4 for the same capacity of the turbines will produce 40% less mechanical energy than units with imaginary "free cooling". If a cooler is free - 100kW, and if cooling is "paid" - 60 kW of power.<br />
External combustion 100kW<br />
External combustion - internal cooling - 60kW<br />
for the same capacity of the turbine (piston / cylinder)<br />
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chart 5<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgw1HwdqkZpu6B5T74hTXEJ1W7s83TDL2mSUZDJwio6dfDygNmolrVjHYWzBrY5vQ78Ld-gGRKQvBtkpagSvUiGtXRH8mkVZ7PVmchbrGw96pc0yjO3nOj4RvU4KkyYTT_MbM3h0qAWecs/s1600/1104.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgw1HwdqkZpu6B5T74hTXEJ1W7s83TDL2mSUZDJwio6dfDygNmolrVjHYWzBrY5vQ78Ld-gGRKQvBtkpagSvUiGtXRH8mkVZ7PVmchbrGw96pc0yjO3nOj4RvU4KkyYTT_MbM3h0qAWecs/s640/1104.png" width="640" /></a></div>
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chart 6<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhLNXblZAXJUwTcu2W_Z9wUZf4r98HxWSbRbgA-mzxkeupOk0C-OKC-dmPZFuq_3QdE5m2_lU6iL4NEPMDXrQSdazoQL8O9k-Na_rN1AqAHIOTbtr9L2FQorjZUcSjYmNwN3ooGXjFaFjE/s1600/1105.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhLNXblZAXJUwTcu2W_Z9wUZf4r98HxWSbRbgA-mzxkeupOk0C-OKC-dmPZFuq_3QdE5m2_lU6iL4NEPMDXrQSdazoQL8O9k-Na_rN1AqAHIOTbtr9L2FQorjZUcSjYmNwN3ooGXjFaFjE/s640/1105.png" width="640" /></a></div>
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On chart 5 am drawn external combustion engine with real free cooler - environment. On chart 6 - engine by method - external combustion - internal cooling, with the same parameters.<br />
Assume - on 20% waste heat - 100kW power of external combustion engine<br />
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<span style="font-size: large;">Verdict:</span><br />
<br />
1.On 20% waste heat - 100kW useful power of external combustion engine<br />
On 20% waste heat - 60kW useful power of external combustion - internal cooling engine for one and the same capacity of the turbines.<br />
<br />
2.For the same capacity of the converters of heat into mechanical energy engine - external combustion - internal cooling gives 40% less mechanical power, but uses 40% less heat - no waste heat.<br />
<br />
3. Another small deficiency on engine by external combustion method - necessarily need heater to heat the hot part, so that the engine to have a cold part. On the engine by external combustion - internal cooling method the cold part we create it, so that we can use any heat, including of heater.<br />
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chart7<br />
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chart 8<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhEx6sFZ2vKAJCt-5cN-B6137_BPcNkNhvbh3Zj7OD-pjM9PJ5jjZAarv-kpU7UijJNCc1nxsh4bxMEihBnGy1oH9Fl_reZkT_XeZU9J8NXMSkXSgTkV9b5hEujm_MJ0TTVbZpkGlJtiiU/s1600/1107.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhEx6sFZ2vKAJCt-5cN-B6137_BPcNkNhvbh3Zj7OD-pjM9PJ5jjZAarv-kpU7UijJNCc1nxsh4bxMEihBnGy1oH9Fl_reZkT_XeZU9J8NXMSkXSgTkV9b5hEujm_MJ0TTVbZpkGlJtiiU/s640/1107.png" width="640" /></a></div>
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<span lang="EN-US" style="line-height: 115%;">Let's
take one external combustion engine as
the chart 2 filled with the same three working substances
(ammonia, R41, R14) and go to Planet X, which has an atmosphere with a
temperature of 130K - chart 7. Now, for such a unit will have free cooling,
as we have in mind that the last working substance (R14) has a boiling point of
145K, and the atmosphere of Planet X on which the temperature is 130K. Light a
burner and heat ammonia to 290K. I accept that for a 20% waste heat engine will
gives 100kW mechanical energy.<o:p></o:p></span></div>
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<span lang="EN-US" style="line-height: 115%;">Redesign
the External combustion engine to an External
combustion- internal cooling by removing heat exchanger which cools the last
working substance to liquefy in the atmosphere of the planet X - chart 8. Set in its
place cooler loaded with nitrogen. Waste heat set it back into evaporators. As
I said above, now I lose 40% of the power output of the unit, but also
decreased 40% fuel in the burner.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">On Planet X unit working with these substances can only work with heater, whether
external combustion engine or an external combustion - internal cooling engine.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">On
Earth we do not need a heater - Sun heated the atmosphere at 290K and hence
heat ammonia. But on Earth mandatory for these working substances unit must be performed by External combustion - Internal cooling method, because no natural cold part.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;">Conclusions:</span><o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;">Engine
- external combustion</span><o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">1
With external combustion engine ever we need a heater.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">2
In these units have free cooling <o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">3.
We have waste heat<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;">Engine
- External combustion- internal cooling</span><o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">1 Heater is not mandatory<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">2
For the same capacity of converters of heat into mechanical energy
(turbines; pistons / cylinders) has a lower power than external combustion
engine<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">3
No waste heat<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;">Summary
of advantages and disadvantages of the method for converting heat into
mechanical energy - External combustion - internal cooling</span><o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">When
using External combustion engine we pay for heating and cooling is free.</span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">When
using External combustion - internal
cooling engine is not required to pay for heating, but must pay for cooling.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"><br /></span></div>
<span lang="EN-US" style="font-family: Calibri, sans-serif; line-height: 115%;"> <span style="font-size: large;">The price we play for
cooling </span></span><span style="font-family: Calibri, sans-serif; font-size: large;"><span style="line-height: 18.4px;">with internal cooling method</span></span><span style="font-family: Calibri, sans-serif; font-size: large; line-height: 115%;"> is - less power for a given capacity of the converters of heat into
mechanical energy (turbines; pistons/cylinders )</span><br />
<span style="font-family: Calibri, sans-serif; font-size: large; line-height: 115%;"><br /></span>
<span style="font-family: Calibri, sans-serif; font-size: large; line-height: 115%;"><br /></span>
<span style="font-family: Calibri, sans-serif; font-size: large;"><span style="line-height: 115%;"> 06 </span><span style="line-height: 27.6px;">January</span><span style="line-height: 115%;"> 2016</span></span><br />
<span style="font-family: Calibri, sans-serif; font-size: large; line-height: 115%;"><br /></span>
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;">I made some mistakes in the calculations for external combustion - internal cooling. True comparison between the two engines in the same capacity of the converters of heat into mechanical energy should look like:</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;">At 100 kW useful power and 20 kW waste heat</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><br /></span></span>
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><span style="font-size: large;">1</span>. <span style="font-size: large;">External combustion engine</span>:</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><span style="font-size: large;">120</span> kW power of the heater</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><span style="font-size: large;">100</span> kW useful power</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"> <span style="font-size: large;">20</span> kW waste heat</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><br /></span></span>
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><span style="font-size: large;">2</span>. <span style="font-size: large;">External combustion - internal cooling engine</span>:</span></span><br />
<br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><span style="font-size: large;">80</span> kW power from the heater </span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><span style="font-size: large;">80</span> kW useful force</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"> <span style="font-size: large;">0</span> kW waste heat </span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"> 20 kW from the gross mechanical power is converted into heat (so the net power becomes: 100-20 = 80 kW)</span></span><br />
<span style="font-family: Calibri, sans-serif; line-height: 27.6px;">+ 20 kW waste heat - these two amounts of heat</span><span style="font-family: Calibri, sans-serif; line-height: 27.6px;"> returned to the evaporator(s). This requires to reduce the power output of the heater wiht 40 kW : 120 - (20 + 20) = 80 kW </span><br />
<span style="font-family: Calibri, sans-serif; line-height: 27.6px;"><br /></span>
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;">It would be good to think about things in depth before presenting them to readers. I beg your pardon!</span></span><br />
<br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;">:) :) For my next invention I </span><span style="line-height: 27.6px;">intend</span></span><span style="font-family: Calibri, sans-serif; line-height: 27.6px;"> to never wrong :) :)</span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><br /></span></span>
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;">06 Jan 2016 </span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;">Svetozar the Cold</span></span><br />
<span style="font-family: Calibri, sans-serif;"><span style="line-height: 27.6px;"><br /></span></span>Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-46694095412864103852015-11-05T04:33:00.000-08:002015-11-08T00:35:20.227-08:00Transferring heat from the cold to the warm part<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">Assuming that:</span><br />
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;"> 1. For the
liquefaction of a given quantity of gas by performing work on it, the amount of
heat which will have received liquid substance is equal to the amount of heat
the on gas plus heat equivalent of the work done on it </span><br />
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">2.The amount of heat which can be converted into mechanical energy of a liquid substance having a temperature higher than its boiling point is equivalent to a difference in temperature to its boiling point</span><br />
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">3.</span> The <span style="font-family: "calibri" , sans-serif; line-height: 115%;">converter of heat into
mechanical energy on the last cycle</span> of the working substance in the "external
combustion - internal cooling by
more than one working substance" converted at less than 50% of the amount of heat<br />
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">Then we will need to transfer
waste heat from last cycle to previous cycles, to enable the latter substance to cool the
foregoing, and so the unit to work - diagram 1</span><br />
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;"><br /></span>
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;">diagram 1</span><br />
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;"><br /></span>
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7jVl-G3MQl4oYVJEJTc7MgoEo1_KRxgGGNeYVx8xix9hDeNOOBLO6vCBWq0ZDGvhuaUVBgy5idrOSSkWZhFhKSX-7CCGnrbJRqmAKTpkFZG2BwGN8p9jlqxDvI94IeLHHAgvb9oZfXPk/s1600/1017.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7jVl-G3MQl4oYVJEJTc7MgoEo1_KRxgGGNeYVx8xix9hDeNOOBLO6vCBWq0ZDGvhuaUVBgy5idrOSSkWZhFhKSX-7CCGnrbJRqmAKTpkFZG2BwGN8p9jlqxDvI94IeLHHAgvb9oZfXPk/s640/1017.png" width="640" /></a></div>
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;"><br /></span>
<span lang="EN-US"><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">Оn this diagram </span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">I chose option whereby heat of the last cycle is attributed to the liquefied gases the last working substance and on the evaporator of the previous working substance, </span></span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">but it can distribute heat to all evaporators of the unit.</span></span><br />
<span lang="EN-US" style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif;">Comparing the amount of heat and temperature differences of useful and opposite force - the unit would be effective. </span></span><br />
<span lang="EN-US" style="line-height: 18.4px;"><span style="font-family: "calibri" , sans-serif;">Because of the heat that will bring in the </span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">evaporators on</span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> preceding substances will allow to reduce the </span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">flow of pumps -the</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> amount of circulation of the </span><span style="font-family: "calibri" , sans-serif;"><span style="line-height: 18.4px;">relevant</span></span><span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"> working substance.</span><br />
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><br /></span>
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;">diagram 2</span><br />
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><br /></span>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgnGe_NOWCYnZmKsnP8_la2bE69hLTyEfLmsZrt-WguHchZDndgaE1DWIpw_lgpnDIpnf-gtIflzbLP6oyfK5WqGN_K1RKbkAoq1Aprnw9_1AxBDXKZw9eX2Jxtt2RvXT2hOho3-436y1A/s1600/1022.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgnGe_NOWCYnZmKsnP8_la2bE69hLTyEfLmsZrt-WguHchZDndgaE1DWIpw_lgpnDIpnf-gtIflzbLP6oyfK5WqGN_K1RKbkAoq1Aprnw9_1AxBDXKZw9eX2Jxtt2RvXT2hOho3-436y1A/s640/1022.png" width="640" /></a></div>
<span style="font-family: "calibri" , sans-serif; line-height: 18.4px;"><br /></span>
<br />
In method of heat exchange between liquid and gaseous working substance is not necessarily the last evaporator (n on diagram 2) to be at low temperature. We can have an effective unit in which gases from all evaporators enters the cooler, which returns heat in the evaporator n . His temperature will not be the lowest in the unit, but the useful forces are bigger than the opposite and this option.<br />
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 16pt;"><o:p></o:p></span></div>
<span lang="EN-US" style="font-family: "calibri" , sans-serif; line-height: 115%;"><br /></span>Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-78025952873460487612015-11-02T13:47:00.001-08:002015-11-03T18:19:12.920-08:00Option to transfer waste heat of the last cycle to previous cycles (puzzle)<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">On
unit filled with more than one working substance can transfer part of the waste
heat of the last cycle to the previous working cycles of substances with a higher
boiling point. In each evaporator place warm heat exchanger on cooler.
Adjustable valves are adjusted so that the working substance of the cooler to
give heat on the working substance of the respective evaporator. By valves can
choose to transfer heat to one or more evaporators (in the case of diagram 1 waste heat from the last cycle transfer to the previous
cycle, but I can configure the system to transfer heat to all evaporators which
has a unit)</span><span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">.<o:p></o:p></span><br />
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span style="line-height: 24.5333px;">diagram 1</span></div>
<div class="MsoNormal">
<span style="line-height: 24.5333px;"><br /></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgH7jjJiWZZ4jDpd0sjCofy89Fbg4qGfowgLCTnvKxp7bwyQmefWbfkWnJmWip6IpoNG5L2AK9nAOknnPu8CA1dRnaADJMSWi0r9rbQIxfRslV1KsjthHL09N_8iweWZFeGHVkeor3m5j4/s1600/1013.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgH7jjJiWZZ4jDpd0sjCofy89Fbg4qGfowgLCTnvKxp7bwyQmefWbfkWnJmWip6IpoNG5L2AK9nAOknnPu8CA1dRnaADJMSWi0r9rbQIxfRslV1KsjthHL09N_8iweWZFeGHVkeor3m5j4/s640/1013.png" width="640" /></a></div>
<div class="MsoNormal">
<span style="line-height: 24.5333px;"><br /></span></div>
<div class="MsoNormal">
<span style="line-height: 24.5333px;"><span style="font-size: large;">Wout / Win = ?</span></span><br />
<span style="line-height: 24.5333px;"><span style="font-size: large;"><br /></span></span>
<span style="line-height: 24.5333px;">diagram 2</span><br />
<span style="line-height: 24.5333px;"><br /></span>
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjOHFpw_LGhw89IqIiUlBpoGDoe63I1TV_NopYmbtD6IBK5eqKG7ICzzrGamYkLQ9Rf0ih8Y40cLjTW5n4ejVYzfJt6dja9HrU6Jx1Z-RDOUBBdzaObjRPxikyS7I-nJ3tMWMhIpUVHEI0/s1600/1015.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjOHFpw_LGhw89IqIiUlBpoGDoe63I1TV_NopYmbtD6IBK5eqKG7ICzzrGamYkLQ9Rf0ih8Y40cLjTW5n4ejVYzfJt6dja9HrU6Jx1Z-RDOUBBdzaObjRPxikyS7I-nJ3tMWMhIpUVHEI0/s640/1015.jpg" width="640" /></a></div>
<span style="line-height: 24.5333px;"><br /></span>
<div class="separator" style="clear: both; text-align: center;">
</div>
<span style="line-height: 24.5333px;"><br /></span>
<span style="line-height: 24.5333px;">Water cooled propylene glycol, and carbon dioxide cooled all (we are used to associate it with increasing temperature, but I intend to entrust it the task to cool) as transfer waste heat from the last working substance to the previous two.</span><br />
<span style="font-size: large;">Wout / Win = ?</span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-70497595723999755742015-10-17T06:08:00.000-07:002015-11-05T23:55:10.043-08:00thermodynamic - mechanical puzzlesSome thermodynamic - mechanical puzzles<br />
1. drawing<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiUn8xA5e-nuQIVaocGes6qywx8oXEY1H7qIXvWGtalpYGmj9yhJ5QFdcaf1aDcctVursGd4vswPliDIu4OPA2w0Yqe9cz2Ck6ndQcwwAjoFg8ArySs9udgw6-3tcBQLBKYH7r5XFc9Dq4/s1600/1001a.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiUn8xA5e-nuQIVaocGes6qywx8oXEY1H7qIXvWGtalpYGmj9yhJ5QFdcaf1aDcctVursGd4vswPliDIu4OPA2w0Yqe9cz2Ck6ndQcwwAjoFg8ArySs9udgw6-3tcBQLBKYH7r5XFc9Dq4/s640/1001a.png" width="640" /></a></div>
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<br />
2.drawing<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdjOkR6zW289W2sCFrT-eqz0obYbv8qQOKhAR_UEgid4ufHFP5agToGkYCdHG8hDcmI49Xsdn9nVGki3yeq9WtULeIekQdlFl9x31Q4rGawZ51qS00mvSoEaBHZT2gNHIFPJeSFLjW5zk/s1600/1003a.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdjOkR6zW289W2sCFrT-eqz0obYbv8qQOKhAR_UEgid4ufHFP5agToGkYCdHG8hDcmI49Xsdn9nVGki3yeq9WtULeIekQdlFl9x31Q4rGawZ51qS00mvSoEaBHZT2gNHIFPJeSFLjW5zk/s640/1003a.png" width="640" /></a></div>
<br />
<br />
<br />
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</div>
<br />
drawing 3<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhe_lcTt1zx0VEMbcwe7_y2T1CcXmud01fX0t7JBty0xKkLoxEdTimUpzFnbWuS8osOsMbIsu6QGbOzY9X6H_9GKpa2BjiYuudobGJrf-Z4ewMg_ZysHh8dWlPMgOUBLd7qmzomGRNpMLQ/s1600/1004.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhe_lcTt1zx0VEMbcwe7_y2T1CcXmud01fX0t7JBty0xKkLoxEdTimUpzFnbWuS8osOsMbIsu6QGbOzY9X6H_9GKpa2BjiYuudobGJrf-Z4ewMg_ZysHh8dWlPMgOUBLd7qmzomGRNpMLQ/s640/1004.png" width="640" /></a></div>
<br />
<br />
drawing 4<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVtWQ4hbMfVb9CCIo1mGZhEoE4dQGzTVjuNhYtUbFCH3zKZBl8cVdNmMjcmkvqbA1FwCjPqJlrRWWhYwBGTUVKDGhJ9YhVz3_jRzwurezt6hUKFEcibUv6APQdDo9IEf1HMG4gwOp7x0k/s1600/1005.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVtWQ4hbMfVb9CCIo1mGZhEoE4dQGzTVjuNhYtUbFCH3zKZBl8cVdNmMjcmkvqbA1FwCjPqJlrRWWhYwBGTUVKDGhJ9YhVz3_jRzwurezt6hUKFEcibUv6APQdDo9IEf1HMG4gwOp7x0k/s640/1005.png" width="640" /></a></div>
<br />
<br />
<br />
What power we will need to rotate the crank?<br />
It will change the temperature of the working substance?<br />
<br />
drawing 5<br />
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Win/Wout = ?<br />
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drawing 6<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilce6wA07eqgNEAUoT4KfqJcVO5V96Qy0lCQPiJ9OgftRDSj_xSjJFn7Fk0uoFt6BBd44B6QPfSa9V4K7hz4KAANEKih0MIGn_CAtGG4vJhwEFGzfhpHb04KxxydiCUlQyqqON_uvdO38/s1600/1009a.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilce6wA07eqgNEAUoT4KfqJcVO5V96Qy0lCQPiJ9OgftRDSj_xSjJFn7Fk0uoFt6BBd44B6QPfSa9V4K7hz4KAANEKih0MIGn_CAtGG4vJhwEFGzfhpHb04KxxydiCUlQyqqON_uvdO38/s640/1009a.png" width="640" /></a></div>
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Wout > Win ?<br />
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drawing 7<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOAk82UpdKNjeF5qF598JUkw_cI32IpML25bMlLniwDIdCF8xebyOIzLGatNM47atXQmbnL3ALjspFv5z46HEBR_mWBdv1IuIcUUcOozXYzfFO1KEcNQ7td3kv1aWXbyqkzZ3ylVZleko/s1600/1010.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOAk82UpdKNjeF5qF598JUkw_cI32IpML25bMlLniwDIdCF8xebyOIzLGatNM47atXQmbnL3ALjspFv5z46HEBR_mWBdv1IuIcUUcOozXYzfFO1KEcNQ7td3kv1aWXbyqkzZ3ylVZleko/s640/1010.png" width="640" /></a></div>
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Wout > Win ?<br />
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drawing 8<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5biD1HA9NpGAPog97qprtrlVGDLYfqIYwfGVtgck4vJQ18JdWTPa2d4M9-aTWx0xSfqbH7ki5i3PF1qvVnuwkxNBqHYfywlYtVeCvho7Qk6K0FTBGsmX2q9MkhkoJedCky5zlBkO9Jd8/s1600/1011.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5biD1HA9NpGAPog97qprtrlVGDLYfqIYwfGVtgck4vJQ18JdWTPa2d4M9-aTWx0xSfqbH7ki5i3PF1qvVnuwkxNBqHYfywlYtVeCvho7Qk6K0FTBGsmX2q9MkhkoJedCky5zlBkO9Jd8/s640/1011.png" width="640" /></a></div>
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Wout / Win = ?<br />
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drawing 9<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjlfppb6bS_uMfwyccjxNa3WSqCsufGx-5UmP0B1R7MxSAsbmamisFKoisyapjojaYpv_xwV6UlnYmU2vcCfIZirIztrJm_eaHuG20NGGtPQlejcH8ufUdXZRlhxREtrvV_0EVGRrjkFig/s1600/1020.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjlfppb6bS_uMfwyccjxNa3WSqCsufGx-5UmP0B1R7MxSAsbmamisFKoisyapjojaYpv_xwV6UlnYmU2vcCfIZirIztrJm_eaHuG20NGGtPQlejcH8ufUdXZRlhxREtrvV_0EVGRrjkFig/s640/1020.png" width="640" /></a></div>
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Wout / Win = ?<br />
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drawing 10<br />
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Wout / W in = ?Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-73795234400086574902015-10-05T11:29:00.000-07:002015-10-05T11:29:05.511-07:00Points of view<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 18px;">To be more clearly</span><span style="font-size: 13.5pt;"> what happens with heat,
what about the cold, </span><span style="font-size: 18px;">whether we lose</span><span style="font-size: 13.5pt;"> (to be warm up ) the cold part will present
two perspectives that I hope to be </span><span style="font-size: 18px;">useful</span><span style="font-size: 13.5pt;"> to readers:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">1. Will present the method from the following perspective - look
at turbine as cooler.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;"> Turbines and cooler removes heat from a given substance, as
the turbine makes transforming the heat of the substance into mechanical
energy, and cooler removes heat from the substance as gives this heat to the
other body by heat exchange. So thus each turbine is a cooler for the previous
turbine - transforming waste heat into mechanical force in a closed cycle of
its working substance. So for aggregate with n of number of turbines will have
to "pay" for cooling only the last, and n-1 of number turbines will
have a "free" cooling. After each cycle of working substance the
amount of waste heat progressively reduces, as to cooler of last turbine
remains small amount of kilowatts of heat, because turbines are turned into
mechanical energy initial amount of heat.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">To be equal power between the turbine and the cooler (useful to
opposite) there must be equality:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">- Temperature difference that will have substance before and after
the turbine has performed work, and the temperature difference to which the
cooler will cool the substance and the temperature on the body which receive
this heat<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">- Kilowatts of heat which turbine converts into mechanical force
to be equal on kilowatts of heat that the cooler will take from the substance
and will transmitted to another body<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">But comparing:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">-Kilowatts of heat which the turbines are turned into mechanical
power and kilowatts of heat that the cooler has to take from the substance and
to transmitted it to another body<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">-The temperature difference between the hot and cold part of the
unit, and the temperature difference between the hot and cold part of the
cooler:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;"> Useful forces exceed the force that we will need to apply on
compressor of cooler to perform its task - to provide a precondition of the
last turbine to work, and hence the entire unit to work.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">2. </span><span style="font-size: 18px;">Another point of view</span><span style="font-size: 13.5pt;">: The turbine and compressor are not on the same axis
(drive shaft),and supply compressors and pumps with external force. For example
- supply compressor and pumps from an electric grid, and turbines operate
independently produceing energy to be transmitted to the grid.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">Let in a thermodynamic system designed by the "external
combustion - internal cooling" method a compressor of cooler powered
by external force creates prerequisite of a last turbine to operate. By doing
this work turbine creates a prerequisite of another turbine to work, and
for 3 turbines unit - the work of the second turbine creates the premise of the
first turbine to work.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">For example - Compressor on cooler with a capacity of 100 kW
.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">These 100kW power of the cooler can "take" 100 kW heat
of a substance and it to him back*. Assuming that the turbines
convert 50% of the amount of heat into mechanical energy , from </span><span style="font-size: 18px;">200 kW waste heat from the previous cycle, the turbine converts into mechanical energy 100kW, and 100 kW remain </span><span style="font-size: 13.5pt;"> a waste heat which the
cooler should "deal".<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">200kW of waste heat for the s</span><span style="font-size: 18px;">econd cycle</span><span style="font-size: 13.5pt;"> - so it goes:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">200kW of mechanical force + 200kW of waste heat = 400kW<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">For the first cycle:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">400kW mechanical work + 400kW waste heat = 800kW<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;"> (Here may momentarily confusion in the reader, because the
last cycle differs from the previous in that the waste heat is returned to the
cycle, but it will only mean a large amount of the circulating working
substance in it.)<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;"> Thus, for a unit with three working substances and three
turbines account is as follows:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">- From a grid will draw 100kW for compressor on cooler<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">+ The turbines will return to the grid 400 + 200 + 100 = 700 kW
power<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">To this we must add the energy that will take from electricall
grid to drive the pumps.<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<br /></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">*I accept that if:<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;">- As a result of work done from a turbine by turning some amount
of heat into mechanical energy of a given working substance, a working
substance cools with N° degrees<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;"> - Cooler to take the same amount of heat from the same
substance and transmit it to another body in which overcomes the same
temperature difference as it is created the turbine (N °)<o:p></o:p></span></div>
<div style="margin-bottom: .0001pt; margin: 0cm;">
<span style="font-size: 13.5pt;"> The mechanical force </span><span style="font-size: 18px;">which produces turbine</span><span style="font-size: 13.5pt;"> and the</span><span style="font-size: 18px;"> force which need to apply the compressor on cooler over working substance</span><span style="font-size: 13.5pt;"> are the
same.<o:p></o:p></span></div>
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Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-59003962392020066472015-06-25T02:59:00.000-07:002015-10-24T12:13:06.595-07:00Let's burn hydrocarbons<span style="font-size: large;">Continued from previous post</span><br />
<span style="font-size: large;"><br /></span>
I can add more operating cycles of substances with increasingly lower boiling points, such as waste heat of any previous heated the next to a temperature higher than its boiling point. On the last working substance will close his cycle with а cooler. So I could get somewhere around 0 Kelvin with helium or hydrogen - to where the materials can withstand . This will convert the heat from the combustion of hydrocarbons in the mechanical work in full - with each turbine (piston in the cylinder) the quantity of waste heat will decrease (I think we can make 50% of the heat into mechanical energy of each working substance).<br />
But what do we need to burn hydrocarbons, such as space and nature give us some average 290K?<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5FlIyXt9ggVNAPhPbj1d4rXPpv0RAhiU1cEq4elHilQxva1SUhSw_vz5j-6QL8WspkIoTm-IIbb7Zg-kMgKSvIB3XJtLzTXbYWBd2cAjrOSShtExA57pb1QDAi0QS7Uyp0Ae_IEmsuBw/s1600/CO2+e2.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5FlIyXt9ggVNAPhPbj1d4rXPpv0RAhiU1cEq4elHilQxva1SUhSw_vz5j-6QL8WspkIoTm-IIbb7Zg-kMgKSvIB3XJtLzTXbYWBd2cAjrOSShtExA57pb1QDAi0QS7Uyp0Ae_IEmsuBw/s640/CO2+e2.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">"Rankine cycle layout" by Wikipedia (user:andrew.ainsworth) user [[:User:Andrew.Ainsworth:User:{{{3}}}|{{{3}}}]]. Licensed under CC BY-SA 3.0 via Commons - https://commons.wikimedia.org/wiki/File:Rankine_cycle_layout.png#/media/File:Rankine_cycle_layout.png</td></tr>
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Some 100 degree difference between hot and cold part will probably be able to achieve. Well, the difference between hot and cold part using the heat from the combustion of hydrocarbons is very large (eg 790K ), respectively of the same capacity of the turbines will receive greater power. But hydrocarbons are not inexhaustible, and burning them is bad for the climate and nature.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhusikv8tcB_J3TuggCZbJjvJ1mlGbX5v6Ps7-RZXY73CQUBS12V_NJnBHMZb_ZZ_BE_Y5Qa_fFwvq-YMIZ0RNxO3om2VLw57weiO3noTGWUmbpayT7AdTSXNsOmT8TJxq-lWqahgW-TvM/s1600/CO2+f2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhusikv8tcB_J3TuggCZbJjvJ1mlGbX5v6Ps7-RZXY73CQUBS12V_NJnBHMZb_ZZ_BE_Y5Qa_fFwvq-YMIZ0RNxO3om2VLw57weiO3noTGWUmbpayT7AdTSXNsOmT8TJxq-lWqahgW-TvM/s640/CO2+f2.png" width="640" /></a></div>
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<br />Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-35202473102705387172015-06-23T22:36:00.000-07:002015-10-24T12:09:08.202-07:00Let's burn hydrocarbons<span style="font-size: large;">Continued from previous post</span><br />
<br />
I will try to become more mechanical energy from our valuable heat produced by the combustion of hydrocarbons. Still hydrocarbons cost money and combustion affects the climate so we must be frugal. Will use the waste heat from the ammonia cycle to warmed R-41 to a temperature higher than its boiling point. A turbine (maybe piston in the cylinder) will add to turn half on the amount of heat seething R-41 into mechanical energy - chart 4.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7sWbE6LkBk5CijN7uJJLGQ_XBIK5gSl6C3JH2Zjj5Uz68K90KZW4wWNJjpC3s5OLSuNMpPfYnG0YzRE1MJMl2fQhgr7xccXajgrpyzZuj1hSMgLhKrOirJ7WuCH-qVrTdXIWqGVUG5ng/s1600/CO2+c2.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="406" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7sWbE6LkBk5CijN7uJJLGQ_XBIK5gSl6C3JH2Zjj5Uz68K90KZW4wWNJjpC3s5OLSuNMpPfYnG0YzRE1MJMl2fQhgr7xccXajgrpyzZuj1hSMgLhKrOirJ7WuCH-qVrTdXIWqGVUG5ng/s640/CO2+c2.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">"Rankine cycle layout" by Wikipedia (user:andrew.ainsworth) user [[:User:Andrew.Ainsworth:User:{{{3}}}|{{{3}}}]]. Licensed under CC BY-SA 3.0 via Commons - https://commons.wikimedia.org/wiki/File:Rankine_cycle_layout.png#/media/File:Rankine_cycle_layout.png</td></tr>
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It is worth recalling that the low temperatures of the working substances must be preset before launching the operation of the unit.<br />
For the same capacity on turbine 3/4 of waste heat on diethyl ether which releases into the environment will become a mechanical force.<br />
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To be continuedSvetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-43528343458791626142015-06-22T03:08:00.000-07:002015-10-24T12:06:00.061-07:00Let's burn hydrocarbons<span style="font-size: large;">Continued from previous post</span><br />
<br />
It will become mechanical work from waste heat of the water cycle by using ammonia to close the cycle of diethyl ether. - chart 3.<br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhTN5SwLYVH-2Kg3ZuI54ZwCSye4QfBxFwBZ6VWXhgB9C-Uv16WOx-UqBYkHiU8gL_faRKsGaSqNKHzMl8vneDRV2rMG_d1EWEORuBRW8WVswXx6aRDmIfZBoiGREM5tNUZtCNwMnAABY/s1600/CO2+b3.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="404" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhTN5SwLYVH-2Kg3ZuI54ZwCSye4QfBxFwBZ6VWXhgB9C-Uv16WOx-UqBYkHiU8gL_faRKsGaSqNKHzMl8vneDRV2rMG_d1EWEORuBRW8WVswXx6aRDmIfZBoiGREM5tNUZtCNwMnAABY/s640/CO2+b3.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">"Rankine cycle layout" by Wikipedia (user:andrew.ainsworth) user [[:User:Andrew.Ainsworth:User:{{{3}}}|{{{3}}}]]. Licensed under CC BY-SA 3.0 via Commons - https://commons.wikimedia.org/wiki/File:Rankine_cycle_layout.png#/media/File:Rankine_cycle_layout.png</td></tr>
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Because ammonia is liquid at temperatures lower than nature can offer us, I would have to close the cycle of ammonia as the cooling ammonia vapor to liquefaction.Compressor on cooler will take the useful power. As mechanical energy producing cycles of ammonia, so will spending to close its cycle. The unit of this figure and unit of the chart 2 will be equally useful.<br />
<br />
It is worth recalling that the low temperatures of the working substances must be preset before launching the operation of the unit.<br />
<br />
But the topic to the burning of hydrocarbons did not think to end up here.<br />
<br />
<span style="font-size: 26.6666660308838px; line-height: 30.6666641235352px;">To be continued</span>Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-41040067768311283532015-06-21T09:52:00.000-07:002015-06-23T22:37:57.288-07:00Let's burn hydrocarbons<div class="MsoNormal">
<span lang="EN-US" style="font-size: 22.0pt; line-height: 115%;">Mechanical
work by burning hydrocarbons<o:p></o:p></span><br />
<span lang="EN-US" style="font-size: 22.0pt; line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">I
will discuss a few ways to get more of heat from the combustion of hydrocarbons.
Taking into account that incineration is bad for the climate would be better to
be thrifty.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">In
ordinary water Rankin cycle waste heat is too much, given the fairly high
boiling point of water relative to the cold part – environment - diagram 1.<o:p></o:p></span></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcfIFk65MLMVEwRr4h_BGjyapaX5jLE4r9ScuP9UnxmSmkql14P_4DtmOf2Hzp2h20xEAFwQ9KNTaQkLq67QS6ymjiJ-66grG3kBYFkNzK1tHaD2U0Y3uTSTi1E4nImC6HS4DmyOe60z4/s1600/Rankine_cycle_layout.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="258" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcfIFk65MLMVEwRr4h_BGjyapaX5jLE4r9ScuP9UnxmSmkql14P_4DtmOf2Hzp2h20xEAFwQ9KNTaQkLq67QS6ymjiJ-66grG3kBYFkNzK1tHaD2U0Y3uTSTi1E4nImC6HS4DmyOe60z4/s400/Rankine_cycle_layout.png" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">"<a href="https://commons.wikimedia.org/wiki/File:Rankine_cycle_layout.png#/media/File:Rankine_cycle_layout.png">Rankine cycle layout</a>" by Wikipedia (user:andrew.ainsworth) user [[:User:Andrew.Ainsworth:User:{{{3}}}|{{{3}}}]]. Licensed under <a href="http://creativecommons.org/licenses/by-sa/3.0/" title="Creative Commons Attribution-Share Alike 3.0<p></p>">CC BY-SA 3.0</a> via <a href="//commons.wikimedia.org/wiki/">Wikimedia Commons</a>.</td></tr>
</tbody></table>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">I
will use steam to warm diethyl ether (308K b.p.) to be able to extract
mechanical energy from its heat - chart 2. </span><br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtBrMQ-nau5a_FLY7vv-owsiQ0Hxlpd6_rMVRUZNgz-JkLtsrnw69hYt3j46mrFX8emNUlG4nCa8zY7lap3culQ76JyXIkaR0Kiyld6s2TXVY0sow5duE35nz_9itqzENHqnKLKlklQk0/s1600/CO2+a.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="414" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtBrMQ-nau5a_FLY7vv-owsiQ0Hxlpd6_rMVRUZNgz-JkLtsrnw69hYt3j46mrFX8emNUlG4nCa8zY7lap3culQ76JyXIkaR0Kiyld6s2TXVY0sow5duE35nz_9itqzENHqnKLKlklQk0/s640/CO2+a.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">"Rankine cycle layout" by Wikipedia (user:andrew.ainsworth) user [[:User:Andrew.Ainsworth:User:{{{3}}}|{{{3}}}]]. Licensed under CC BY-SA 3.0 via Wikimedia Commons - https://commons.wikimedia.org/wiki/File:Rankine_cycle_layout.png#/media/File:Rankine_cycle_layout.png</td></tr>
</tbody></table>
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;"><br /></span></div>
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</div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">It will use waste heat from the
water cycle to produce mechanical energy. And we will throw a little bit of
valuable heat.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">I
intend also to further decreased waste heat, but diethyl ether exhausts in
general ability of nature to close the cycle and will need to use closed cycle as
a result of work done.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 16.0pt; line-height: 115%;">How
to become a more mechanical energy from
the heat from the combustion of hydrocarbons will discuss in the next post.<o:p></o:p></span></div>
<br />
<div class="MsoNormal">
<span lang="EN-US" style="font-size: 20.0pt; line-height: 115%;">To
be continued<o:p></o:p></span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-16617269711812456452015-06-13T22:30:00.000-07:002015-10-06T00:12:50.291-07:00Cold reality - heretical thoughts<div class="MsoNormal">
<span lang="EN-US" style="font-size: 20.0pt; line-height: 115%;">Reflections
on the motive power of heat on the environment and on machines fitted to develop that power<o:p></o:p></span></div>
<span lang="EN-US" style="font-size: 20.0pt; line-height: 115%;"><br /></span>
<br />
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;"> </span><span lang="EN-US" style="line-height: 115%;">I
will formulate conclusions which led me the work on the external combustion
engine which is driven by the heat of the environment.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">The
only way we can use the heat of the environment as a driving force is to create
a cold part. Cooling the cold part to
have the required temperature difference so hot part on the engine to be the
environment. To keep the cold part I need to heat isolate it from the
environment. If I do not do it due to the heat exchange environment will warm
the cold part. Because theoretically and practically can not turn all heat into
mechanical energy will always have waste heat which will heat up the cold part.
So I need to cool the cold part by cooler. As less amount of waste heat, so the
little force it would need for the compressor on cooler to maintain the low
temperature of the cold part. Small amount of waste heat will achieve by thermal
isolating the entire unit so that the
process of work working substance to cool down and the amount of waste heat to
decrease. Another obstacle for maintaining the low temperature of the cold part
will be the inevitable leakage of heat from the thermal insulation. Practically
we can not achieve 100% thermal insulation so that heat from the environment
will pass through the insulation and warms the cold part. This heat will call
it - harmful heat.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">Effective
unit will have when the sum of the quantities of waste heat and harmful heat at
any one time is less than the amount of heat which the turbines (the pistons in
the cylinders) are transformed into mechanical energy. Then I'll have enough
power to keep the cold part. The difference between forces that generate
the converters on heat into mechanical energy (turbines, pistons) and a force
necessary to maintain the cold part will give us useful mechanical work.</span><span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"><o:p></o:p></span></div>
<div class="MsoNormal">
<span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;">When we have plenty of heat the cold part is important for us to receive
mechanical power. We should strive </span><span lang="EN-US" style="line-height: 115%;">waste</span><span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"> heat and</span><span lang="EN-US" style="line-height: 115%;"> harmful</span><span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"> heat </span><span lang="EN-US" style="line-height: 115%;"> to be</span><span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"> in small amounts</span><span lang="EN-US" style="line-height: 115%;">, so</span><span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"> to spend a
small portion of the useful energy </span><span lang="EN-US" style="line-height: 115%;"> for</span><span style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"> the maintenance
of low temperature of the cold part.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">Several
decisions on how to convert heat into mechanical work, so the amount of waste
heat to be a small part of the total heat input from environment in the engine I described in my
blog. If we do well </span><span lang="EN-US" style="font-size: 16.0pt; line-height: 115%; mso-ansi-language: BG;"> </span><span lang="EN-US" style="line-height: 115%;">with harmful heat we can achieve perpetual motion .<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">Perpetual
motion would be possible with a thermal insulating external combustion engine.
The warmth of the environment will be the driving force and the temperature
difference between the hot and cold part we create and maintain due to perform
work. Perpetual motion of such an engine is provided by two pairs supporting
one another physical characteristics:<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">1.
The temperature difference is a prerequisite to perform work - By doing work we
can create a temperature difference.<o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">2.
The heat is converted into mechanical energy - mechanical energy is converted
into heat<o:p></o:p></span><br />
<span lang="EN-US" style="line-height: 115%;"><br /></span>
<span style="line-height: 18.4px;">By definition (Wikipedia) Perpetual motion is:</span><br />
<span style="line-height: 18.4px;">"Perpetual motion is motion that continues indefinitely without any external source of energy.This is impossible to ever achieve because of friction and other sources of energy loss"</span><br />
<span style="line-height: 18.4px;">Next follows a right conclusions:</span><br />
<span style="line-height: 18.4px;">"There is a scientific consensus that perpetual motion in an isolated system violates either the first law of thermodynamics, the second law of thermodynamics, or both" .... There is no dispute - true.</span><br />
<span style="line-height: 18.4px;"> BUT - each ONE isolated system can divide it into TWO </span><span style="line-height: 18.4px;">isolated systems</span><span style="line-height: 18.4px;"> as:</span><br />
<span style="line-height: 18.4px;">1. Heat isolate one part of it to the other</span><br />
<span style="line-height: 18.4px;">2. Remove some amount of heat from one to the other</span><br />
<span style="line-height: 18.4px;">So we can "avoid" the laws of thermodynamics which apply to an ONE isolated system - we divide an ONE isolated system completely legally on TWO </span><span style="line-height: 18.4px;">isolated systems</span><span style="line-height: 18.4px;">.</span><br />
<span style="line-height: 18.4px;">Unit that converts heat into mechanical energy is in one part, and converts the heat from the other side into mechanical energy.</span><br />
<span style="line-height: 18.4px;">In one part the heat ultimately turns into mechanical energy - in the other part the mechanical energy ultimately converted into heat.</span><br />
<span style="line-height: 18.4px;"><span lang="EN-US"></span></span><br />
<span style="line-height: 18.4px;"> Nor violate the laws of thermodynamics ... neither conversion of heat from one part to mechanical energy in the other part will stop.</span><br />
<br /></div>
<div class="MsoNormal">
<span lang="EN-US" style="line-height: 115%;">I
think that the cold reality will allow us perpetuum mobile.<span style="font-size: 16pt;"><o:p></o:p></span></span><br />
<span lang="EN-US" style="line-height: 115%;"><br /></span>
<span lang="EN-US" style="font-size: large; line-height: 18.4px;">But most important for us is to look at the method as an opportunity for renewable energy, one more opportunity to reduce the footprint we leave on the planet.</span></div>
Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-23484539485081618372015-06-12T04:51:00.000-07:002015-06-12T04:51:52.659-07:00Reduce waste heat Two working substances - ammonia (240bp) and R 41 (195bp) - diagram 5. Working substance on cooler - R14 (145bp). Prior we cooled the liquid R41 to its boiling point. After opening the valves and drive the pumps waste heat on ammonia cycle heated R41 to 238K. Ammonia liquefies and pumps it appear to be heated in the atmosphere to 290K. R41 works by the method of divided load and heat exchange between liquid and gaseous working substance. Cooler closes the cycle of R41 and pumps sent liquid R41 back to the heat exchange with the gases ammonia, as quantization of the two substances are such that ammonia liquefies. Too small amount of waste heat after the work on R41 - Effective unit.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6bS4UPD6A_CTiGAK_fFvtgGcjKoAVxdF7hMLWx9dcKqoUtpGsIuvZNDSEfJYOpWYxZjGRkKAcmYjo5akWVqthgQVaTvjhmCtINfozTcCOSlNXonicbwNXyQdLo1aHhKU_YeiCuFMN4X4/s1600/diagram+5.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6bS4UPD6A_CTiGAK_fFvtgGcjKoAVxdF7hMLWx9dcKqoUtpGsIuvZNDSEfJYOpWYxZjGRkKAcmYjo5akWVqthgQVaTvjhmCtINfozTcCOSlNXonicbwNXyQdLo1aHhKU_YeiCuFMN4X4/s640/diagram+5.png" width="640" /></a></div>
<br />
By the logic:<br />
- Environment heated first working substance to a temperature above its boiling point<br />
- all other working substances work in an heat isolated environment<br />
- each additional working substance with -low boiling closes cycle on previous<br />
- substances work by the method / or not by the method of separately loading<br />
- on the last working substance closing cycle as a result of work done<br />
We can fulfill unit with so much substances as nature allows us. After the working cycle of each substance waste heat remains in a small amount compared to the amount of heat which we turned into mechanical work.Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-84692817870933763282015-06-10T04:46:00.000-07:002015-06-10T04:46:06.703-07:00divided load and workMethod will be effective and if we apply "clean power" to close the cycle - diagram 4c.<br />
But if the compressor works for a cooler is more effectively.<br />
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<br />Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-51253055515330999682015-06-08T10:44:00.000-07:002015-06-08T10:44:16.721-07:00option for waste heatWill draw the waste heat from the evaporator 3 to simplify things - diagram 1 (2).<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi7SF7ABBv5N17UIAJUxvuhUdFTaX5M-l3mXi3dvLM_gw4EikehVWngyJ2k9vPaYr9-iLM4F9Oon5Z9_qRAeQ-UsAYdU5fO-tN9FYvYpvNQs22kG_LlIIbZkbjcgKvjOch_RLpCQurUXiM/s1600/diagram+1+%25282%2529.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="340" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi7SF7ABBv5N17UIAJUxvuhUdFTaX5M-l3mXi3dvLM_gw4EikehVWngyJ2k9vPaYr9-iLM4F9Oon5Z9_qRAeQ-UsAYdU5fO-tN9FYvYpvNQs22kG_LlIIbZkbjcgKvjOch_RLpCQurUXiM/s640/diagram+1+%25282%2529.png" width="640" /></a></div>
<br />
<br />
Liquefied gases collected with the liquid working substance coming out of the evaporator 3 and the heat exchange with the compressed refrigerant gas in the heat exchanger 25. One such option unit will be less effective than when part of the waste heat returns to the evaporator as a chart 1.<br />
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<br />
<br />
On the one hand the system for redistributing heat (cooler) must overcome a large temperature difference in this option. We also need to increase the circulation of liquid working substance in the cycle, which is another loss of useful power.Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-79615759565396389162015-06-08T05:33:00.000-07:002015-06-08T05:33:13.181-07:00Let's go back to the philosophy of external combustion engineLet's go back to the philosophy of external combustion engine - is required temperature difference between the hot and cold part to be able to convert heat into mechanical work. Now we heat the warm part to have a temperature difference. I suggest that the temperature difference to be a result of work done - temperature difference between the hot and cold part we can create it as a cooling cold part. And to get efficient engine has the power we need to apply to create the necessary temperature difference at any time should be less than the force that generates the working substance.<br />
Lets to think - how much is the heat which we can turn into mechanical work of a working substance with a given temperature? To be able to work working substance must have a temperature -high from its boiling point. All the heat of a working substance which can be converted into mechanical work it is equal to the difference between the temperature to which it heats the heat source up to its boiling point. Theoretically and practically we can make all this heat into mechanical work always remained waste heat.<br />
Let's perform work on the working substance as it cooled in the cold part to have a temperature difference and the engine running. To be an effective engine running on this principle should amount to heat is less than the amount of heat which has become into mechanical energy. This will achieve it by carrying out the process of converting heat into mechanical energy into heat - isolated environment - on the work process working substance cools. So like I can become a a large part of the heat into mechanical force balance between the useful power and strength that must apply in order to maintain the temperature difference will be in favor of the beneficial forces. I think there is a way beneficial forces are more powerful than the opposite and so we have an effective engine of which the temperature difference between hot and cold part is held as a result of work done.<br />
We need to create conditions for the majority of the amount of heat on the working substance to turn into mechanical energy so that the waste heat to be a small part of the total heat quantity.<br />
In my previous posts I have given some suggestions (It is better to begin consideration of the method with two working substances). I have a few more ideas that will present soon.<br />
One primary method is to use more than one working substance. Each working substance can convert think ½ of the amount of heat into mechanical work. So the power we need to use for running the unit at any time will be measuring less than power produced by the working substances in their work.<br />
Another method is to split the load and repeatedly perform work as the opportunity to make heat exchange between liquid and gaseous working substance in the course of his work. So I want to achieve a small amount of waste heat before closing the cycle of working substance. Less waste heat = good balance between useful and opposite force.<br />
To draw attention once again to something important - Every start we need to have set a low temperature of the cold part. This will use external power.Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0tag:blogger.com,1999:blog-8944931543165046143.post-65902288951410258412015-06-06T23:49:00.000-07:002015-06-06T23:49:20.845-07:00To load repeatedly and exchanging heat to be effective unit.<span lang="EN-US" style="font-size: 18.0pt; line-height: 115%;">If
we resolve on working substance to expand (evaporates) without conducting enough
work then to close the cycle we will need a large force to apply on it. At the
same time there is no way to spread the load of single evaporator- turbine to
decline in the course of work - to make the most of the opportunities of
working substance. I therefore suggest that the same amount of working substance
to perform multiple work as it passes from one evaporator to another with
increasingly lower temperatures. In this way I can load the given amount of
substance to work up and to take to cool and gaseous working substance due to
heat exchange. A "cooler" loaded with a substance with a low boiling
point(R41) at the end. Its task is to
cool the gas to liquefaction - Chart 4.</span><br />
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<span lang="EN-US" style="font-size: 18.0pt; line-height: 115%;"><br /></span>
<span lang="EN-US" style="font-size: 18.0pt; line-height: 115%;"><br /></span>
<span lang="EN-US" style="font-size: 18.0pt; line-height: 115%;"> When proceeding to closing the cycle
the gaseous working substance will have a temperature close to the boiling
point and the power necessary for compressor 5 will be smaller than the power
received from the plurality of converters of heat into mechanical energy 7.
Such a unit will be effective.</span>Svetozar the Coldhttp://www.blogger.com/profile/09682150619060923552noreply@blogger.com0