Super-Carnot Cycle graphic proof
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Super-Carnot Cycle graphic proof
Exactly 200 years after Carnot released "Reflections", it appears that I've (finally) sunk the widely held Carnot theorem with a simple GRAPHIC proof that will cause major controversy among academics (see forum link):
http://www.stirlingengineforum.com/view ... 072#p22072
Is this what Senft's mechanical effectiveness was aiming at ? I don't know, but it could be. After above post, I returned to Geogebra app and quickly found the secret sauce behind this 'magic'. Heads up guys, there are several ways this can play out and not limited to a single mech, so this should be amusing. Needless to say, I'm busy gaming the values...
I'll leave it to Fool to run the calcs while I drink a beer (or two)
http://www.stirlingengineforum.com/view ... 072#p22072
Is this what Senft's mechanical effectiveness was aiming at ? I don't know, but it could be. After above post, I returned to Geogebra app and quickly found the secret sauce behind this 'magic'. Heads up guys, there are several ways this can play out and not limited to a single mech, so this should be amusing. Needless to say, I'm busy gaming the values...
I'll leave it to Fool to run the calcs while I drink a beer (or two)
Re: Super-Carnot Cycle graphic proof
Grab a Bud light Matt, you may just be a "real American hero".
So why the Otto pitch if Stirling prevails?
So why the Otto pitch if Stirling prevails?
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Re: Super-Carnot Cycle graphic proof
Within the crazy 300-1050k cycle per graphics
(1) ideal Stirling has Carnot = .71
(2) ideal Otto without regen has Carnot = .25
(3) ideal Otto with regen has Carnot = .85
The common ICE cycles we're used to (Otto and Diesel) can't use regen due to PVT values where T after compression exceeds T after expansion (entire sink range is below entire source range).
The odd thing about this Otto graphic is that I could "connect the dots" (complete the cycle) with a 2:1 volume ratio. I'm now chasing a 300-450k scheme where geothermal becomes credible.
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Re: Super-Carnot Cycle graphic proof
Yikes, another botched post which should bematt brown wrote: ↑Sun Apr 21, 2024 6:36 pm
Within the crazy 300-1050k cycle per graphics
(1) ideal Stirling has Carnot = .71
(2) ideal Otto without regen has Carnot = .25
(3) ideal Otto with regen has Carnot = .85
(1) ideal Stirling has Carnot win = .71 eff
(2) ideal Otto without regen has Carnot win = .25 eff
(3) ideal Otto with regen has Carnot loss = .85 eff
IOW for this 300-1050k cycle, the Carnot Theorem imposes max eff = .71
Any eff above Carnot limit is often referred to simply as "Super-Carnot"
Tom should enjoy that ideal LTD could approach 1.0 eff via similar adiabatic cycle, but with meager PVT values. The only caveat is accounting for req'd regen which might be obscured via out-of-phase mech.
Re: Super-Carnot Cycle graphic proof
I'm a doof and missed the lower eff. of the Stirling. Ok so how do you regen closed cycle Otto?
I like the LTD eff. approaching 1.00, how does it get around Carnot? Does it have to do with near Cv operation?
I like the LTD eff. approaching 1.00, how does it get around Carnot? Does it have to do with near Cv operation?
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Re: Super-Carnot Cycle graphic proof
I'm working on it.
You nailed it again !!! It's the "one-one" (300k-Vmin) approach where Vmin would be Cv indeed.
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Re: Super-Carnot Cycle graphic proof
I tend towards common “Urieli speak” notation, but always add r to PVT whereby
Pr = pressure ratio
Vr = volume ratio
Tr = temperature ratio
I’m also going to use eff = efficiency to save pecking. OK, class, let’s begin…
Ideal Otto eff can be derived via a needless long method for any Doubting Thomas that likes getting lost in the weeds or via 1 – 1/(Vr^gamma-1) which reads eff = one minus one divided by the volume ratio to the power of gamma minus one, wherein “gamma” is the specific heat ratio Cp/Cv whereby Cp/Cv = 1.4 for air (and other diatomic gasses).
Ex: when Vr = 10 (think ICE) then ideal eff = 1- 1/(10^.4) = 1 – 1/(2.5119) = 1 - .3981 = .6019 which is commonly rounded off to simply .60 and well known by many gear heads.
When Vr = constant, then all Otto cycles have eff = constant, regardless of finite source and sink values. However, all this eff mumbo-jumbo only applies to common Otto cycles without regen. The following PV plot is mere Vr = 3 and much lower than century old Vr = 6, but even by Vr=3 regen potential is limited by Tsink range that overlaps Tsource range.
Now, consider 3 Otto cycles with regen potential, cycles A-B-C where all have Vr=2 whereby eff=.25 when no regen…
Otto A with 300-600k cycle with 400-600k source and 300-455k sink
Otto B with 300-800k cycle with 400-800k source and 300-600k sink
Otto C with 300-1050k cycle with 400-1050k source and 300-800k sink
WnegA=WnegB=WnegC and effA=effB=effC
But putting other values in descending order
WposC > WposB > WposA
So how can effC = effB = effA ???
Despite Trange sourceC > sourceB > sourceA…so is Trange sinkC > sinkB > sink A
In effect, without regen, as the Trange of source increases and produces more Wpos, the Trange of sink increases and “outstrips” source increases somewhat, whereby eff = constant.
Interesting, as the Trange of source increases, the Trange of sink increasingly overlaps the Trange of source whereby the regen potential increases (wink-wink Fool). Adding regen reduces sink heat as follows: A = 1/4 , B = 2/3 , C = 4/5
In this manner C goes from eff = .25 with no regen to eff = .85 with regen and proves Carnot Theorem bogus. I probably lost most guys with this hack, so I’ll cancel the test on Friday (and to think this is the nickel tour).
Pr = pressure ratio
Vr = volume ratio
Tr = temperature ratio
I’m also going to use eff = efficiency to save pecking. OK, class, let’s begin…
Ideal Otto eff can be derived via a needless long method for any Doubting Thomas that likes getting lost in the weeds or via 1 – 1/(Vr^gamma-1) which reads eff = one minus one divided by the volume ratio to the power of gamma minus one, wherein “gamma” is the specific heat ratio Cp/Cv whereby Cp/Cv = 1.4 for air (and other diatomic gasses).
Ex: when Vr = 10 (think ICE) then ideal eff = 1- 1/(10^.4) = 1 – 1/(2.5119) = 1 - .3981 = .6019 which is commonly rounded off to simply .60 and well known by many gear heads.
When Vr = constant, then all Otto cycles have eff = constant, regardless of finite source and sink values. However, all this eff mumbo-jumbo only applies to common Otto cycles without regen. The following PV plot is mere Vr = 3 and much lower than century old Vr = 6, but even by Vr=3 regen potential is limited by Tsink range that overlaps Tsource range.
Now, consider 3 Otto cycles with regen potential, cycles A-B-C where all have Vr=2 whereby eff=.25 when no regen…
Otto A with 300-600k cycle with 400-600k source and 300-455k sink
Otto B with 300-800k cycle with 400-800k source and 300-600k sink
Otto C with 300-1050k cycle with 400-1050k source and 300-800k sink
WnegA=WnegB=WnegC and effA=effB=effC
But putting other values in descending order
WposC > WposB > WposA
So how can effC = effB = effA ???
Despite Trange sourceC > sourceB > sourceA…so is Trange sinkC > sinkB > sink A
In effect, without regen, as the Trange of source increases and produces more Wpos, the Trange of sink increases and “outstrips” source increases somewhat, whereby eff = constant.
Interesting, as the Trange of source increases, the Trange of sink increasingly overlaps the Trange of source whereby the regen potential increases (wink-wink Fool). Adding regen reduces sink heat as follows: A = 1/4 , B = 2/3 , C = 4/5
In this manner C goes from eff = .25 with no regen to eff = .85 with regen and proves Carnot Theorem bogus. I probably lost most guys with this hack, so I’ll cancel the test on Friday (and to think this is the nickel tour).
Re: Super-Carnot Cycle graphic proof
Bringing the following over from isothermal thread :
1 to 2, isentropic expansion:
W = -Apha•N•R•Th•((V2/V1)^(1-Gama)-1)
W12 = -2.5•3•8.314•1050•((2/1)^(1-7/5)-1)
W12 = 15,853.684 J
Back work 3 to 4 isentropic compression:
W34 = -5976.875
Work total cycle :
Wtc = W12+W34 = 9,876.81 J
This agrees with Matt.
Heat put into regenerator. The regenerator operates from T4 to T2, or 400 K to 800 K. Delta-T = DTr = 400 K. Isochoric heat addition and rejection :
DQ = Cv•N•R•DT
DQr23a = 5/2•8.314•3•(400-800) = -24,942 J
Also heat back in from regenerator :
DQr41a = 5/2•8.314•3•(800-400) = 24,942 J
Heat discarded 2 to 3 'b' to cold plate :
DQd23b = 5/2•8.314•3•(300-400) = -6,235.5 J
Heat added from 4 to 1 'b', from hot plate, AKA DQh :
DQh41b = 5/2•8.314•3•(1050-800) = 15,588.75 J
Efficiency = Wtc / DQh41b
n = 9,876.81 / 15,588.75 = 0.6336
Or 63.36%.
Carnot efficiency:
(1050-300)/1050 = 0.7143
Or 71.43%
Any questions?
Analyzing:matt brown wrote: ↑Sat Apr 20, 2024 8:10 pmThere are so many factors to consider when scheming ECE, but nixing heated working cylinder is near top of list. Recent zero-zero buzz reminded me of similar one-one buzz regarding PVT values that approach one, similar LTD. Here's an Otto scheme that's under the radarFool wrote: ↑Fri Apr 19, 2024 11:05 am I think it goes along with your experiments. Somewhere someone made the claim that Stirling Engine expansion is "more adiabatic". I think that is misleading.
https://en.m.wikipedia.org/w/index.php? ... prov=rarw1
"This article is about the adiabatic Stirling cycle. "
Very misleading.
Without heat going into the gas somewhere during the expansion stroke, power to weight will greatly suffer.
Improving the heat input, especially during expansion seems like a very good thing to strive for.
Otto PV Geo.png
Sorry for the extreme values, but Geogebra app doesn't allow scale changes. Basic scheme is low volume ratio Otto where
(1) 2:1 volume ratio is still common Stirling ratio
(2) "zero point" isobar to allow 'buffer' compression
(3) both charge AND buffer pressure are above ambient
Despite this crazy 300-1050k cycle (due to app limits) check out the regen potential where the 2-3 isochor is 800k >>> 300k vs 4-1 isochor 400k >>> 1050k. Indeed, when approaching one-one values, things change, and an Otto cycle may include substantial regen. Here, 80% of normal sink heat can be regen to input (400-800k "sink" to 400-800k "source") which drastically alters 'normal' Otto eff.
Meanwhile, consider similar Stirling cycle
Stirling PV Geo.png
Note similarities and differences between this Stirling and previous Otto, especially same volume and thermal ratios. Ha, this Otto is still 3/4 the output of vaulted Stirling.
1 to 2, isentropic expansion:
W = -Apha•N•R•Th•((V2/V1)^(1-Gama)-1)
W12 = -2.5•3•8.314•1050•((2/1)^(1-7/5)-1)
W12 = 15,853.684 J
Back work 3 to 4 isentropic compression:
W34 = -5976.875
Work total cycle :
Wtc = W12+W34 = 9,876.81 J
This agrees with Matt.
Heat put into regenerator. The regenerator operates from T4 to T2, or 400 K to 800 K. Delta-T = DTr = 400 K. Isochoric heat addition and rejection :
DQ = Cv•N•R•DT
DQr23a = 5/2•8.314•3•(400-800) = -24,942 J
Also heat back in from regenerator :
DQr41a = 5/2•8.314•3•(800-400) = 24,942 J
Heat discarded 2 to 3 'b' to cold plate :
DQd23b = 5/2•8.314•3•(300-400) = -6,235.5 J
Heat added from 4 to 1 'b', from hot plate, AKA DQh :
DQh41b = 5/2•8.314•3•(1050-800) = 15,588.75 J
Efficiency = Wtc / DQh41b
n = 9,876.81 / 15,588.75 = 0.6336
Or 63.36%.
Carnot efficiency:
(1050-300)/1050 = 0.7143
Or 71.43%
Any questions?
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Re: Super-Carnot Cycle graphic proof
Awesome job Fool, you're hired !!! Did you catch my blunder ?
back story : I don't waste much time trying to beat Carnot. This Otto cycle evolved while gaming low volume Otto cycles with Rocky Horror sound track playing in background (via my youtube feed due to upcoming 50 yr anniversary). This inspired me to go waaaay out and I found this cycle, whereupon the Time Warp took over with "just a jump to the left" and I saw the Holy Grail - lol.
Using short equation for Otto eff, all finite work values are irrelevant. So, I stared at this PV plot and pondered massive backwork ratio "a thing apart" vs massive regen potential raising thermal eff. My blunder was "summing" eff via regen from sink value vs regen from source value (thanks Rocky). Thus, via sink values .25 + .60 = .85 eff vs via source values .25 (650/250) = .65 eff which becomes .63 eff with more precision (this Otto base eff = .2421). This is exactly the kind of blunder Tom continually makes with his folksy analogies using addition and subtraction vs multiplication and division.
I only found my blunder while messing with a small matrix for 2 cycles where x=regen range, y=source range, and z=sink range. Only after staring at these values in little squares, did the factor concept kick in and I found "my passage back to the place I was before".
back story : I don't waste much time trying to beat Carnot. This Otto cycle evolved while gaming low volume Otto cycles with Rocky Horror sound track playing in background (via my youtube feed due to upcoming 50 yr anniversary). This inspired me to go waaaay out and I found this cycle, whereupon the Time Warp took over with "just a jump to the left" and I saw the Holy Grail - lol.
Using short equation for Otto eff, all finite work values are irrelevant. So, I stared at this PV plot and pondered massive backwork ratio "a thing apart" vs massive regen potential raising thermal eff. My blunder was "summing" eff via regen from sink value vs regen from source value (thanks Rocky). Thus, via sink values .25 + .60 = .85 eff vs via source values .25 (650/250) = .65 eff which becomes .63 eff with more precision (this Otto base eff = .2421). This is exactly the kind of blunder Tom continually makes with his folksy analogies using addition and subtraction vs multiplication and division.
I only found my blunder while messing with a small matrix for 2 cycles where x=regen range, y=source range, and z=sink range. Only after staring at these values in little squares, did the factor concept kick in and I found "my passage back to the place I was before".
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Re: Super-Carnot Cycle graphic proof
Here's another 300-1050k Otto cycle, but where Vr = 1.5 and 'wider' isochors allow more regen.
Due to paltry Vr, this Otto cycle has
base eff = .15
base sink = 300-900k
base source = 350-1050k
however, isochors overlap 350-900k thereby
potential regen = 550k
potential sink = 50k
potential source = 150k
whereby potential eff = .70 via .15(700/150) or "initial eff (initial source/final source) = final eff"
In this manner, regen is .78 source or 1-(150/700) and achieves Carnot as a rounding error (Carnot = .71) without isothermal processes. Kinda interesting that this Vr = 1.5 cycle still has 2/3 the output of similar Vr = 2 cycle.
Due to paltry Vr, this Otto cycle has
base eff = .15
base sink = 300-900k
base source = 350-1050k
however, isochors overlap 350-900k thereby
potential regen = 550k
potential sink = 50k
potential source = 150k
whereby potential eff = .70 via .15(700/150) or "initial eff (initial source/final source) = final eff"
In this manner, regen is .78 source or 1-(150/700) and achieves Carnot as a rounding error (Carnot = .71) without isothermal processes. Kinda interesting that this Vr = 1.5 cycle still has 2/3 the output of similar Vr = 2 cycle.
Re: Super-Carnot Cycle graphic proof
Thanks for exercise. I think I caused permanent serious brain damage. LOL
I'm so used to the "perfect" Stirling regeneration that it blinded me to the trick for quite some time. I don't think many people in the world would pick it up.
Do I get my honorary PHD in MESE now! Mechanical Engineering Stirling Engines. LOL
Thanks again. May your favorite grog flow freely, and cause zero unwanted side effects. Grins!
I'm so used to the "perfect" Stirling regeneration that it blinded me to the trick for quite some time. I don't think many people in the world would pick it up.
Do I get my honorary PHD in MESE now! Mechanical Engineering Stirling Engines. LOL
Thanks again. May your favorite grog flow freely, and cause zero unwanted side effects. Grins!
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Re: Super-Carnot Cycle graphic proof
Degree issued, but here's some more stuff to add to your post grad work...
(1) did you notice your work calcs had Wpos nearly equal final input while Wneg was nearly equal final sink ?
(2) I twice showed initial eff vs final eff in eye opening relationship that I missed
where I had…initial eff (initial source/final source) = final eff, but I missed the relationship and I still missed this after my little matrix comping 2 cycles revealed factor relationship (eyes roll)
(3) AFTER my 300-1050k Otto D cycle post with Vr=1.5 some ‘similar’ values vs 300-1050k Otto C cycle caught my eye big time, but I still failed relationship until I rewrote the expression and…added Otto D values
whereupon, I finally found the ‘relationship’ A(A’) = B(B’) when
A = initial source
A’ = initial eff
B = final source
B’ = final eff
(4) which I quickly recognized akin Brayton regen potential
Note max regen has min source input equal exp Tdrop and min sink equal comp Trise
I searched various forms of AA’ = BB’ for regen, but found nothing, so maybe I just stumbled across the simple method vs typical long method.
And to think that I was the guy asking if you had ever studied low volume ratio Braytons. Nothing like the dog chasing his tail…
Re: Super-Carnot Cycle graphic proof
No. I've only worked those you've seen here. I come here for casual reading. You and VincentG have been treated special. Tom too.
Sitting down having to think hurts my brain. LOL
Tom has worn me out. For the time being.
Sitting down having to think hurts my brain. LOL
Tom has worn me out. For the time being.