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Re: Need help with a new solar engine

 
 
amdx
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Posts: n/a
 
      10-09-2010, 07:18 PM

"Morris Dovey" <(E-Mail Removed)> wrote in message
news:i7u6km$47s$(E-Mail Removed)...
> On 9/28/2010 2:23 AM, Morris Dovey wrote:
>
>> Sorry, I'm just too tired to make another drawing tonight.

>
> Another day, another sketch. I've done a bit of polishing and moved the
> explanation to a web page with a sketch to illustrate. It's at
>
> http://www.iedu.com/DeSoto/Misc/Lesson1.html
>
> The explanation is mostly conjecture, since I've never been able to see
> or measure what's going on inside.
>
> Comments, corrections, and suggestions will be welcome.
>
> --
> Morris Dovey
> http://www.iedu.com/DeSoto/
> PGP Key ID EBB1E70E
>

Hi Morris,
I looked over the page above and have a couple of questions.
Do you have any ballpark idea of the frequency it will operate at, or you
want it to operate at?
I ask realizing it all depends on length, diameter, fluid density (
everything),
the video on your website has one that runs at about 1-1/2 hertz,
Do you expect it to operate as fast as 60 hz?
How fast can the cold head cool? (somewhat rhetorical)
The math is way over me, I have a physicist buddy I have twice tried to get
interested,
so far no go, I may have to put a working model in front of him and say how
long would
I have to make this tube to ........
If I do plan a build, I would use electric to heat the hot head for design
work.
How many watts would you suggest for a 2" Dia. 2 ft. hot head tube.
Or maybe a better question I can get 1500 watts from the outlet,
what size hot head tube would you start with?
MikeK


 
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daestrom
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      10-13-2010, 11:19 PM
On 10/10/2010 3:37 PM, BobG wrote:
> Lets assume that the hot side and the cool side heat and cool above
> and below the avg temperature. Lets use ice water and boiling water.
> (212 deg F + 32 deg F)/2= 122 deg F. So the experiment is to get a
> cylinder full of 122 deg air, plunge it into the tank of boiling
> water, and time how fast the heat rises. It seems to me that the
> 'sweet spot' is when the temp gets 'half way' to the hot temp. The
> time to cool to half way to the cold temp should be the same. Since
> PV=nRT, when the T goes up, the P goes up. So we started with 15 psi
> in the cylinder and the T rise in Rankine is (167+453)/(122+453)=1.08,
> so the P rise is 15*1.08=16.2. That pressure on a piston of a certain
> area will puch with a certain force. So far I've only used algebra! No
> calculus needed!


Quite right.

Now, the next step is instead of plunging the whole thing in hot then
cold water, just make the air inside move from a 'hot' area to a 'cold'
area. In early Stirling engines, this was done with a 'displacer' a
sort of slug positioned in the cylinder that would shuttle back and
forth as the crankshaft turns. It would 'displace' the air in the hot
end, forcing it around to the cold end. Then as the air contracts and
sucked the piston inward, the turning crank would move the displacer and
it would 'displace' the air in the cold end, forcing it back to the hot end.

Now you get into heat-transfer across the tube and into/out of the air.
If the engine is running really fast, you don't have a lot of time to
cool/heat the air. That means less temperature rise/fall and lower
Carnot efficiency (also less pressure difference between
expansion/compression stroke). Remember, Carnot efficiency is based on
the hottest and coldest temperature that the working fluid reaches, not
the temperature of the flame or ultimate heat sink.


It's my understanding that the various 'acoustic' designs are actually
using the inertia of the air itself as it is thrust up/in and expands
out/down to keep moving in the same direction for a few moments. This
'piles up' some extra air in the hot/cold end for a moment as the piston
goes through dead center. With that tiny 'pressure wave' of air moving
back and forth, it acts something like a displacer. A tiny extra bit of
air end up in the section being heated so that you get more push when
expanding it than it took to compress it.

Trying to figure out the frequency of oscillation would be quite a
calculation. The acoustic speed of the pressure wave (i.e. the speed of
sound in air at those temperatures/pressures) is probably a key part.
If you run too slow, the pressure wave would simply reflect off the
cylinder end and carry the extra bit of air away from where you want it.
Too fast and you don't have enough time to transfer any significant
heat into/out of the air.

daestrom

 
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Josepi
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      10-14-2010, 02:43 AM
Apparently from people constructing these engines, the mass of the damping
material at one end is critical.


"daestrom" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
Quite right.

Now, the next step is instead of plunging the whole thing in hot then
cold water, just make the air inside move from a 'hot' area to a 'cold'
area. In early Stirling engines, this was done with a 'displacer' a
sort of slug positioned in the cylinder that would shuttle back and
forth as the crankshaft turns. It would 'displace' the air in the hot
end, forcing it around to the cold end. Then as the air contracts and
sucked the piston inward, the turning crank would move the displacer and
it would 'displace' the air in the cold end, forcing it back to the hot end.

Now you get into heat-transfer across the tube and into/out of the air.
If the engine is running really fast, you don't have a lot of time to
cool/heat the air. That means less temperature rise/fall and lower
Carnot efficiency (also less pressure difference between
expansion/compression stroke). Remember, Carnot efficiency is based on
the hottest and coldest temperature that the working fluid reaches, not
the temperature of the flame or ultimate heat sink.


It's my understanding that the various 'acoustic' designs are actually
using the inertia of the air itself as it is thrust up/in and expands
out/down to keep moving in the same direction for a few moments. This
'piles up' some extra air in the hot/cold end for a moment as the piston
goes through dead center. With that tiny 'pressure wave' of air moving
back and forth, it acts something like a displacer. A tiny extra bit of
air end up in the section being heated so that you get more push when
expanding it than it took to compress it.

Trying to figure out the frequency of oscillation would be quite a
calculation. The acoustic speed of the pressure wave (i.e. the speed of
sound in air at those temperatures/pressures) is probably a key part.
If you run too slow, the pressure wave would simply reflect off the
cylinder end and carry the extra bit of air away from where you want it.
Too fast and you don't have enough time to transfer any significant
heat into/out of the air.

daestrom


 
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vaughn
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      10-14-2010, 01:08 PM

"Proteus" <(E-Mail Removed)> wrote in message
news:0568748f-c5a3-4809-b729-(E-Mail Removed)...
>I AM PROTEUS


....and you are history...


 
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daestrom
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      10-14-2010, 03:40 PM
On 10/14/2010 9:20 AM, Morris Dovey wrote:
> On 10/13/2010 6:19 PM, daestrom wrote:
>> On 10/10/2010 3:37 PM, BobG wrote:
>>> Lets assume that the hot side and the cool side heat and cool above
>>> and below the avg temperature. Lets use ice water and boiling water.
>>> (212 deg F + 32 deg F)/2= 122 deg F. So the experiment is to get a
>>> cylinder full of 122 deg air, plunge it into the tank of boiling
>>> water, and time how fast the heat rises. It seems to me that the
>>> 'sweet spot' is when the temp gets 'half way' to the hot temp. The
>>> time to cool to half way to the cold temp should be the same. Since
>>> PV=nRT, when the T goes up, the P goes up. So we started with 15 psi
>>> in the cylinder and the T rise in Rankine is (167+453)/(122+453)=1.08,
>>> so the P rise is 15*1.08=16.2. That pressure on a piston of a certain
>>> area will puch with a certain force. So far I've only used algebra! No
>>> calculus needed!

>>
>> Quite right.
>>
>> Now, the next step is instead of plunging the whole thing in hot then
>> cold water, just make the air inside move from a 'hot' area to a 'cold'
>> area. In early Stirling engines, this was done with a 'displacer' a
>> sort of slug positioned in the cylinder that would shuttle back and
>> forth as the crankshaft turns. It would 'displace' the air in the hot
>> end, forcing it around to the cold end. Then as the air contracts and
>> sucked the piston inward, the turning crank would move the displacer and
>> it would 'displace' the air in the cold end, forcing it back to the hot
>> end.

>
> Ok - I'm with you this far...
>
> In my toy Stirling, the power piston and displacer pistons are pinned to
> a flywheel in a crankshaft arrangement, with the displacer connection at
> a 90° remove from the power connection.


Right. The idea is to get the air moved to the opposite end and start
it cooling/heating while the power piston is mid-stroke.

>
>> Now you get into heat-transfer across the tube and into/out of the air.
>> If the engine is running really fast, you don't have a lot of time to
>> cool/heat the air. That means less temperature rise/fall and lower
>> Carnot efficiency (also less pressure difference between
>> expansion/compression stroke).

>
> Now you've lost me. It seems to me that the speed of the engine is
> determined by the rate at which heat is transferred into the hot head
> and out of the cold head - not the other way around.
>


What I was trying to say is that heat transfer into/out of the air is a
process that takes time. For a given engine speed, the air is only at
the hot or cold end for a fixed amount of time. In that short time you
have to get as much heat into /out-of the air as you can.

But there's a limit as to just how fast you can transfer heat. So if
the air is in the hot end for a short time, you can only get so much
heat into it and that limits how hot the air can get.

So it's a sort of feedback mechanism. If the engine develops a lot of
torque and accelerates, it spins faster. But a faster engine cannot
heat/cool the air as far and that reduces the amount of torque that can
be developed. So the engine can't overspeed and if it slows down the
opposite happens and you develop more torque.

Yes, if you make the hot end hotter, then there is a larger temperature
difference between the casing and the air, so heat transfers from the
casing into the air faster. And that means the ultimate temperature
that the air reaches is higher.

It all works together.

>> Remember, Carnot efficiency is based on
>> the hottest and coldest temperature that the working fluid reaches, not
>> the temperature of the flame or ultimate heat sink.

>
> Yes.
>
>> It's my understanding that the various 'acoustic' designs are actually
>> using the inertia of the air itself as it is thrust up/in and expands
>> out/down to keep moving in the same direction for a few moments. This
>> 'piles up' some extra air in the hot/cold end for a moment as the piston
>> goes through dead center. With that tiny 'pressure wave' of air moving
>> back and forth, it acts something like a displacer. A tiny extra bit of
>> air end up in the section being heated so that you get more push when
>> expanding it than it took to compress it.

>
> The acoustic designs may take advantage of inertia as you describe, but
> I'm having difficulty seeing that as a requirement for operation. It
> seems to me that the pressure wave would be the overriding element.


Well, perhaps I've mixed terms. Any pressure wave in a fluid travels at
'sonic velocity' so that's where the term 'acoustic' comes into play.

>
>> Trying to figure out the frequency of oscillation would be quite a
>> calculation. The acoustic speed of the pressure wave (i.e. the speed of
>> sound in air at those temperatures/pressures) is probably a key part. If
>> you run too slow, the pressure wave would simply reflect off the
>> cylinder end and carry the extra bit of air away from where you want it.
>> Too fast and you don't have enough time to transfer any significant
>> heat into/out of the air.

>
> I'm struggling (again) with this view. This seems a bit like the cart
> pushing the horse up a hill...


Think of a simple mass suspended on a spring. When you set it to
bouncing up/down, it has a predictable natural frequency. Same with the
air 'bouncing' back and forth in the tube between hot and cold.

You can force either system to oscillate at almost any frequency when
you apply an external force, but if the frequency of the external
'pushes' matches the natural frequency, you can get much larger movement.

So if we could get the external column of water to have the same natural
frequency as the internal acoustics, we might get some really
'interesting' results.

daestrom
 
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sno
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      10-14-2010, 06:46 PM
On 10/14/2010 11:40 AM, daestrom wrote:
> On 10/14/2010 9:20 AM, Morris Dovey wrote:
>> On 10/13/2010 6:19 PM, daestrom wrote:
>>> On 10/10/2010 3:37 PM, BobG wrote:
>>>> Lets assume that the hot side and the cool side heat and cool above
>>>> and below the avg temperature. Lets use ice water and boiling water.
>>>> (212 deg F + 32 deg F)/2= 122 deg F. So the experiment is to get a
>>>> cylinder full of 122 deg air, plunge it into the tank of boiling
>>>> water, and time how fast the heat rises. It seems to me that the
>>>> 'sweet spot' is when the temp gets 'half way' to the hot temp. The
>>>> time to cool to half way to the cold temp should be the same. Since
>>>> PV=nRT, when the T goes up, the P goes up. So we started with 15 psi
>>>> in the cylinder and the T rise in Rankine is (167+453)/(122+453)=1.08,
>>>> so the P rise is 15*1.08=16.2. That pressure on a piston of a certain
>>>> area will puch with a certain force. So far I've only used algebra! No
>>>> calculus needed!
>>>
>>> Quite right.
>>>
>>> Now, the next step is instead of plunging the whole thing in hot then
>>> cold water, just make the air inside move from a 'hot' area to a 'cold'
>>> area. In early Stirling engines, this was done with a 'displacer' a
>>> sort of slug positioned in the cylinder that would shuttle back and
>>> forth as the crankshaft turns. It would 'displace' the air in the hot
>>> end, forcing it around to the cold end. Then as the air contracts and
>>> sucked the piston inward, the turning crank would move the displacer and
>>> it would 'displace' the air in the cold end, forcing it back to the hot
>>> end.

>>
>> Ok - I'm with you this far...
>>
>> In my toy Stirling, the power piston and displacer pistons are pinned to
>> a flywheel in a crankshaft arrangement, with the displacer connection at
>> a 90° remove from the power connection.

>
> Right. The idea is to get the air moved to the opposite end and start it
> cooling/heating while the power piston is mid-stroke.
>
>>
>>> Now you get into heat-transfer across the tube and into/out of the air.
>>> If the engine is running really fast, you don't have a lot of time to
>>> cool/heat the air. That means less temperature rise/fall and lower
>>> Carnot efficiency (also less pressure difference between
>>> expansion/compression stroke).

>>
>> Now you've lost me. It seems to me that the speed of the engine is
>> determined by the rate at which heat is transferred into the hot head
>> and out of the cold head - not the other way around.
>>

>
> What I was trying to say is that heat transfer into/out of the air is a
> process that takes time. For a given engine speed, the air is only at
> the hot or cold end for a fixed amount of time. In that short time you
> have to get as much heat into /out-of the air as you can.
>
> But there's a limit as to just how fast you can transfer heat. So if the
> air is in the hot end for a short time, you can only get so much heat
> into it and that limits how hot the air can get.
>
> So it's a sort of feedback mechanism. If the engine develops a lot of
> torque and accelerates, it spins faster. But a faster engine cannot
> heat/cool the air as far and that reduces the amount of torque that can
> be developed. So the engine can't overspeed and if it slows down the
> opposite happens and you develop more torque.
>
> Yes, if you make the hot end hotter, then there is a larger temperature
> difference between the casing and the air, so heat transfers from the
> casing into the air faster. And that means the ultimate temperature that
> the air reaches is higher.
>
> It all works together.
>
>>> Remember, Carnot efficiency is based on
>>> the hottest and coldest temperature that the working fluid reaches, not
>>> the temperature of the flame or ultimate heat sink.

>>
>> Yes.
>>
>>> It's my understanding that the various 'acoustic' designs are actually
>>> using the inertia of the air itself as it is thrust up/in and expands
>>> out/down to keep moving in the same direction for a few moments. This
>>> 'piles up' some extra air in the hot/cold end for a moment as the piston
>>> goes through dead center. With that tiny 'pressure wave' of air moving
>>> back and forth, it acts something like a displacer. A tiny extra bit of
>>> air end up in the section being heated so that you get more push when
>>> expanding it than it took to compress it.

>>
>> The acoustic designs may take advantage of inertia as you describe, but
>> I'm having difficulty seeing that as a requirement for operation. It
>> seems to me that the pressure wave would be the overriding element.

>
> Well, perhaps I've mixed terms. Any pressure wave in a fluid travels at
> 'sonic velocity' so that's where the term 'acoustic' comes into play.
>
>>
>>> Trying to figure out the frequency of oscillation would be quite a
>>> calculation. The acoustic speed of the pressure wave (i.e. the speed of
>>> sound in air at those temperatures/pressures) is probably a key part. If
>>> you run too slow, the pressure wave would simply reflect off the
>>> cylinder end and carry the extra bit of air away from where you want it.
>>> Too fast and you don't have enough time to transfer any significant
>>> heat into/out of the air.

>>
>> I'm struggling (again) with this view. This seems a bit like the cart
>> pushing the horse up a hill...

>
> Think of a simple mass suspended on a spring. When you set it to
> bouncing up/down, it has a predictable natural frequency. Same with the
> air 'bouncing' back and forth in the tube between hot and cold.
>
> You can force either system to oscillate at almost any frequency when
> you apply an external force, but if the frequency of the external
> 'pushes' matches the natural frequency, you can get much larger movement.
>
> So if we could get the external column of water to have the same natural
> frequency as the internal acoustics, we might get some really
> 'interesting' results.
>
> daestrom


What about adding them together in series....each would add a little
more energy to the total system....

have fun....sno


--
Correct Scientific Terminology:
Hypothesis - a guess as to why or how something occurs
Theory - a hypothesis that has been checked by enough experiments
to be generally assumed to be true.
Law - a hypothesis that has been checked by enough experiments
in enough different ways that it is assumed to be truer then a theory.
Note: nothing is proven in science, things are assumed to be true.

 
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daestrom
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      10-15-2010, 11:17 PM
On 10/14/2010 1:32 PM, Jim Wilkins wrote:
> On Oct 14, 11:40 am, daestrom<daest...@twcny.rr.com> wrote:
>> On 10/14/2010 9:20 AM, Morris Dovey wrote:
>> ...
>>
>> Think of a simple mass suspended on a spring. When you set it to
>> bouncing up/down, it has a predictable natural frequency. Same with the
>> air 'bouncing' back and forth in the tube between hot and cold.
>>
>> You can force either system to oscillate at almost any frequency when
>> you apply an external force, but if the frequency of the external
>> 'pushes' matches the natural frequency, you can get much larger movement.
>>
>> So if we could get the external column of water to have the same natural
>> frequency as the internal acoustics, we might get some really
>> 'interesting' results.
>>
>> daestrom

>
> Here is a practical application:
> http://www.motorcycle.com/how-to/how...care-3423.html
>
> When I first encountered this, in print with formulas, I measured the
> exhaust on my dirt bike and found an exact match between theory and
> practice.
>
> In an engine that has internal friction and delivers power the "Q" of
> the resonance is very important. A resonator with a high Q vibrates
> easily at its tuned frequency but not at others. A lower Q allows it
> to operate over a wider frequency range, IOW it doesn't stall as soon
> as the load slows it down a bit.
>
> http://en.wikipedia.org/wiki/Q_factor
>


Exactly the point I was trying (not quite successfully) to state. Know
anywhere that has the formulae? I'd like to see what goes into the
calculation.

daestrom
 
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sno
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      10-16-2010, 01:50 AM

Morris...

Do not know if you have thought of this...but what about instead of
heating the air you heat the water to the point that steam is produced....

Am thinking along the lines of a pop-pop engine...

http://www.nmia.com/~vrbass/pop-pop/

You may have to put a simple valve in the circuit somewhere....

thank you for listening to my thoughts...have fun....sno


--
Correct Scientific Terminology:
Hypothesis - a guess as to why or how something occurs
Theory - a hypothesis that has been checked by enough experiments
to be generally assumed to be true.
Law - a hypothesis that has been checked by enough experiments
in enough different ways that it is assumed to be truer then a theory.
Note: nothing is proven in science, things are assumed to be true.

 
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