alt.energy.renewable

>"Bruce Richmond" wrote in message
>news:80a6a0ff-1a9c-4a1d-9cb2-d273494c2658@c65g2000hsa.googlegroups.com...
>On Apr 9, 4:46 pm, "daestrom"
>wrote:
>> "Morris Dovey" wrote in message
>>
>> news:47FA1F1C.B502DEF@iedu.com...
>>
>>
>>
>>
>>
>> > Bruce Richmond wrote:
>>
>> >> On Apr 6, 7:14 pm, Morris Dovey wrote:
>> >> > daestrom wrote:
>>
>> >> I saw your web page about building the fluidyne engine a while back
>> >> and after reading the mention that it was a Stirling engine didn't
>> >> give it any more thought to the details of how it works. I Know how
>> >> Stirling engines work so that was a I figured I needed to know. But
>> >> after looking again I realize the principals may be the same but the
>> >> details are different.
>>
>> >> The diagrams shown are close to the standard ones but are still a bit
>> >> confusing to most that see them the first time. The reason I say
>> >> close is that line 2-3 on the p-v chart should be shorter than line
>> >> 4-1. With them the same length you have the same heat going out as
>> >> coming in and no energy going out as work.
>>
>> > Umm, yes. Sometimes it's amazing how obvious some things are
>> > after someone else points 'em out. :-) Thanks.
>>
>> > Methinks I need to begin thinking about some P/V/T
>> > instrumentation for these things - about a year or two ago. Ouch!
>> > - nothing I have comes close to fast enough for sampling in real
>> > time.
>>
>> >>http://www.cse.iitk.ac.in/~amit/courses/371/abhishe/main2.html
>>
>> > That looks more reasonable than what I drew. I'll redraw.
>>
>> >> Your explaination of the rise in air pressure pushing the water down
>> >> on the hot side doesn't agree with the diagram either. While it takes
>> >> pressure to push the water down, the chart shows the pressure
>> >> decreasing from 1-2. Actually the chart shows the pressure rise from
>> >> 4-1 with the volume held constant. The volume is then allowed to
>> >> expand from 1-2 while more heat is added to hold the temp constant,
>> >> despite the drop in pressure.
>>
>> > Good. This was an area of uncertainty for me as I wrote. When
>> > I've finished writing this, I need to spend a bit of time
>> > re-reading and and digesting this.
>>
>> The water going 'down' on the hot side *does* occur while the pressure is
>> rising. This is line 4-1. But the water in the *outlet* tube doesn't rise
>> until you get to line 1-2. You have to understand that the water in the
>> three columns are not oscillating 'in phase'. The 'power stroke' is from
>> 1-2 and that is when the 'outlet tube' water level is rising. But the
>> water
>> in the 'hot side' is BDC at that point (i.e. it already dropped during
>> 4-1).
>
>I find it hard to believe that the pressure of the air trapped in the
>loop is rising as the water goes down on the hot side. The trapped
>air is acting as a spring. If the water on the cold side is hardly
>moving and the water on the hot side is going up, the air in the loop
>is compressed and its pressure rises. When that pressure overcomes
>the inertia of the water moving up it then pushes the water back
>down. The water rising in the cold tube may keep the pressure from
>dropping as quickly as it would without that contribution, but the
>volume moving in the cold tube is too small to keep the pressure
>rising. The only time I can see when the pressure would be rising
>while the water on the hot side is going down is at the very beginning
>of the drop when the cold side is at its maximum velocity and the hot
>side is just starting to move after changing direction. But that will
>be a very short portion of the stroke after which the pressure will be
>falling.
>

Thinking some more about it, I may have to admit 'I misspoke' :-)

The output column and the *cold* leg are the two columns that are shifted
out of phase by 90 degrees in time. I believe you're right that when the
outlet water is falling, the hot water is rising (180 out of phase if you
want). When the outlet falls, if the tube's 'Tee' were in the middle and
symmetrical, the water would rise in both the 'hot' and 'cold' at the same
rate. The gas would compress in both sides equally. But because of the
asymmetry of the tubing, the branch leading to the cold side does not flow
as quickly, so the level in the cold side does not rise as fast. So more of
the gas is pushed out of the hot side than the cold side and the result is
that more of the gas is compressed under 'cold' conditions.

As the output column reaches BDC and the gas pressure reaches near maximum,
the water in the 'cold side' starts to rise because of the difference in
static head between the 'hot' and 'cold' water columns. Just as this water
gets moving from hot side to cold side, the overall pressure starts to force
water back up the output column.

So now the situation reverses. If the two sides were symmetrical, water
would leave both sides at the same rate towards the output column. But
because they are not, more of the water from the 'hot' side leaves and more
of the gas finds its way to the 'hot' side while expanding.

But there are still some interesting points to be made. What's the best way
to get this time-shift / phase-difference? As Morris found from
experimenting, a restriction in the tube will slow the rise/fall of the cold
side and boost output some. But carried to the extreme (just shutting off
the cold side completely) would make the engine stop. Instead, leave the
pathway fully open but put a long 'loop' in the line between the 'Tee' and
the cold side. The inertia in the water would give you a time-delay without
the losses of a restriction in the line. Sort of like increasing the
capacitance in an RC circuit instead of the resistance, you still get a
larger time-constant, but less I^2R losses. But for best results, the
'loop' should be just two straight sections with a turn-around, not a spiral
coil of tubing. That way the liquid's momentum isn't dissipated against the
spiral wall.

You were right that a higher gas pressure in *any* Stirling engine will lead
to more output. To do that with this type of engine, you obviously need a
higher column of water in the outlet pipe. But the higher mass of water
would slow down the 'stroke'. And since one still needs the cold leg to be
delayed, you would have to increase the time constant of that leg at the
same time. Of course there are practical limits to how tall you can make
it, but still there's a lot of room for experimentation.

daestrom