The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

[Tom Booth]

There is a good reason why these equations are turning out with the values of 100 = 400.

Because according to the Carnot limit, the 100 joules of heat added is not just in some kind of RATIO. It says, mathematically that the heat supplied actually IS all the heat down to absolute zero because it tries to draw a false equivalence between heat and temperature.

This is exactly what I've been complaining about for years. The Carnot limit equation is a mathematical hodge podge of irrational nonsense, that for some reason people have assumed, just because it's been around for so long it must be right. Checked and rechecked and verified experimentally a thousand times. Rather than just some old obsolete hair brained theory that heat "falls down" the thermometric scale like water down a waterfall.

[Fool]

2+x=4

What does x equal?

X=5

2+5=4

7=4

The equation must be wrong.

Give me a break.

An equation isn't valid for any numbers you want to put in. An equation is only valid if use properly.

[Fool]

Cooling reduces the "repelling" forces that counteract the attractive force.

There is no repulsive force. Temperature changes the velocity. More velocity, more pressure. Purely MV collisions.

[Fool]

Later theoreticians seem to have dropped the "quantity" part, so the current "Carnot limit" equation has no "quantity" variable. Only Th and Tc. Nothing corresponding to Amps or "quantity of liquid" in a waterfall. Gallons per minute or anything else sensible

The equations I started with had quantity. "M" in the equation, Qbc = CvMTc

I demonstrated how M cancels out. Efficiency is therefore, not a function of quantity. Only temperature, or as Carnot originally put it potential. The proof, however, didn't assume that, and instead showed how it is all relevant and cancels in the final form.

Electric motors have the same efficiency formula based on electric potential V.

n ele = (Vh-Vr)/Vh

Vh = line voltage
Vr = IR voltage also known a the resistance voltage drop.

[Tom Booth]

2+x=4

What does x equal?

X=5

2+5=4

7=4

The equation must be wrong.

Give me a break.

An equation isn't valid for any numbers you want to put in. An equation is only valid if use properly.

It should at least be valid for the potential range of application, don't you think?

Or are you presupposing ONLY the presupposed Carnot values are "valid"?

Of course you are.

[Tom Booth]

Can we try some other more "realistic" (in your opinion) values?

How about

Thot = 600
Tcold = 300
Qhot = 150 joules

Anything wrong with those numbers?

Equations are supposed to balance right? Isn't that what the equal sign is for? Isn't that what mathematical precision is all about?

We assume some efficiency within a "reasonable" range. How about 25% ?

Take one of your equations chosen at random

Substituting your variables:

Oh wait.

Qhz is supposed to be Qcz + DQh

But 300 + 150 = 450 not 600

So why can't Thot be 600 ?

OK, so let's just use 450 shall we?

Thot = Qhz = 450 <--- arbitrary constraint
Tcold = 300 = Qcz
Qhot = 150 joules = DQh


Equation from derivation:

DQh = Qhz - Qhz(1-n)

Substitute values:

150 = 300 - 450(1 - 0.25)
150 = 300 - 450(0.75)
150 = 300 - 337.5
150 = -37.5

Or

150 = 300 - 450(0.75)
150 = -150(0.75)
150 = 112.5

Getting closer.

Maybe we could introduce some constant and call it a day.

Let's see, 150 - 112.5 = 37.5

Tom's constant is 37.5

150 = 112.5 + (Tom's constant)

That's better.

150 = 112 + 37.5

150 = 150

Perfect!

[Tom Booth]

Suppose we use my original values ?

Thot = Qhz = 600 (but also 450 ?)
Tcold = 300 = Qcz
Qhot = 150 joules = DQh

DQh = Qhz - Qhz(1-n)
150 = 600 - 600(1 - .25)
150 = 600 - 600(0.75)
150 = 600 - 450
150 = 150

Hey, worked out!

But elsewhere Qcz + DQh = Qhz

Qcz + DQh = Qhz
300 + 600 = 900

So Qhz can equal 450, 600 or 900

Try this again?

DQh = Qhz - Qhz(1-n)
150 = 900 - 900(1-.25)
150 = 900 - 900(.75)
150 = 900 - 675
150 = 225

Rats

I thought that was going to work out.

150 = 225 + Tom's constant ????

[Fool]

Thot = 600
Tcold = 300
Qhot = 150 joules

You forgot the units, 'Kelvin', for the temperatures.

You put temperatures into an energy equation, and mixed them with heat. The reason it worked in the previous example is that DQh and DT were both 100. It had a one to one relationship between energy and temperature.

To correct your example you need to calculate quantity and the relationship between energy and temperature for your example numbers. And guessing at the efficiency won't work either. Efficiency has to be calculated from the temperatures given.

Qcz = CvMTc

DQh = CvMDT

DT = 600 K - 300 K = 300 K
DQh = Qhot = 150 J

Putting those valves into the equation :
DQh = Cv•M•DT = 150 = Cv•M•300

Dividing both sides by 300 :
150/300 = Cv•M•300/300

Rearranging, dividing, and cancelling :
Cv•M = 0.5 J/K

So :
Qcz = Cv•M•Tc = 0.5•300 = 150

And Qhz :
Qhz = Qcz + DQh = 150 J + 150 J = 300 J

n = (600 K - 300 K)/600 K = 0.5

DQh = Qhz - Qhz(1-n)
150 J = 300 J - 300 J •(1-.5)

150 J = 150 J

Try to remember to use the mass and type quantities being heated. Quantity was not discarded. It just cancels out of the efficiency equation leaving only temperatures. The other equations still require it. "M" Mass is the quantity. Cv, the Coefficient of heat, goes along with mass to get the quantity of heat and absolute internal energy, associated with temperature.

In other words energy quantity is measured by temperature and mass. Q=U=CvMT

Equations do equate, if used correctly. It's called the 'rigor' of mathematics and logic and science.

[Tom Booth]

Thot = 600
Tcold = 300
Qhot = 150 joules

You forgot the units, 'Kelvin', for the temperatures.

You put temperatures into an energy equation, and mixed them with heat. The reason it worked in the previous example is that DQh and DT were both 100. It had a one to one relationship between energy and temperature.

To correct your example you need to calculate quantity and the relationship between energy and temperature for your example numbers. And guessing at the efficiency won't work either. Efficiency has to be calculated from the temperatures given.

Qcz = CvMTc

DQh = CvMDT

DT = 600 K - 300 K = 300 K
DQh = Qhot = 150 J

Putting those valves into the equation :
DQh = Cv•M•DT = 150 = Cv•M•300

Dividing both sides by 300 :
150/300 = Cv•M•300/300

Rearranging, dividing, and cancelling :
Cv•M = 0.5 J/K

So :
Qcz = Cv•M•Tc = 0.5•300 = 150

And Qhz :
Qhz = Qcz + DQh = 150 J + 150 J = 300 J

n = (600 K - 300 K)/600 K = 0.5

DQh = Qhz - Qhz(1-n)
150 J = 300 J - 300 J •(1-.5)

150 J = 150 J

Try to remember to use the mass and type quantities being heated. Quantity was not discarded. It just cancels out of the efficiency equation leaving only temperatures. The other equations still require it. "M" Mass is the quantity. Cv, the Coefficient of heat, goes along with mass to get the quantity of heat and absolute internal energy, associated with temperature.

In other words energy quantity is measured by temperature and mass. Q=U=CvMT

Equations do equate, if used correctly. It's called the 'rigor' of mathematics and logic and science.

I did use Kelvin

You just now, though, assumed or "figured" Carnot efficiency 0.5 rather than my engine which was 0.25

a2+b2=C2 would be pretty useless if it only worked for equilateral triangles.

[Stroller]

a2+b2=C2 would be pretty useless if it only worked for equilateral triangles.

Pythagoras is spinning in his grave.

[Tom Booth]

Thot = 600
Tcold = 300
Qhot = 150 joules

You forgot the units, 'Kelvin', for the temperatures.

You put temperatures into an energy equation, and mixed them with heat. The reason it worked in the previous example is that DQh and DT were both 100. It had a one to one relationship between energy and temperature.

...

Well, to me it looks like you are putting "temperatures into an energy equation, and mixed them with heat" or that describes the Carnot efficiency equation itself which assumes temperature is a measure of the quantity of energy.

Anyway, previously I adhered to this "arbitrary constraint" and that didn't"t work out either. So in other words, if all your parameters adhere strictly to the Carnot efficiency formula and you are careful to color within all the lines and presuppose all the appropriate values predicted by the Carnot efficiency formula then the Carnot efficiency formula can be rearranged in various ways and after so many permutations you end up with the answer you already assumed from the begining.

[Tom Booth]

a2+b2=C2 would be pretty useless if it only worked for equilateral triangles.

Pythagoras is spinning in his grave.

There is of course, no such thing as an equilateral right triangle. Which is the point.

There is no such thing as a "perfect" Carnot engine either

An algebraic formula should be able to accept any "real" parameters from the domain to which the equation is supposed to apply that actually exist in the real world or what's the point?

[Tom Booth]

Begining of derivation:

Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both
The term 'n' applies to the engine regardless of 'Qcz' base heat amount or reference point. It will have the same efficiency burning the same energy at the same temperatures but with different reference points. That allows us to look at the total energy in the system. So Qhz should have the same relation ship to Qcz that DQh has to DQc, in other words, 'n' is the same. Basically this says that Qcz can be calculated from 'n' and Qhz,
or:
Qcz = Qhz•(1-n)
...

I'm pretty much fine with everything up to this point, but the statement: " 'n' applies to the engine " seems specious.

What engine?

Where does this formula Qcz = Qhz•(1-n) come from?

As far as I can see it emerges out of this paragraph of rather dubious rationalizations and unproven, unverifiable suppositions.

The pre-existing internal energy (Qcz) must (or "should") equal this same energy plus the supplied energy (Qcz+DQh = Qhz) multiplied by the "efficiency" of the engine.

Right off the bat "efficiency" has already been determined based on the temperature difference in conformity with the Carnot efficiency limit assumption.

That is not a "derivation" it's an arbitrary insertion of a rationalization that sets the "efficiency" in stone in accordance with the assumptions represented by the Carnot efficiency formula.

[Tom Booth]

Begining of derivation:

Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both
The term 'n' applies to the engine regardless of 'Qcz' base heat amount or reference point. It will have the same efficiency burning the same energy at the same temperatures but with different reference points. That allows us to look at the total energy in the system. So Qhz should have the same relation ship to Qcz that DQh has to DQc, in other words, 'n' is the same. Basically this says that Qcz can be calculated from 'n' and Qhz,
or:
Qcz = Qhz•(1-n)
...

Why go any further? You've already produced the Carnot formula by fiat. Just slightly disguised or rewritten in a slightly different form with renamed variables.

Say we have our heat source Th = 600 and Tc = 300

Plug those values into the first equation presented in your "derivation".

Qcz = Qhz•(1-n)

Divide both sides by Qhz (solve for n)

n = Qcz/Qhz = 300/600 = 50%

[Tom Booth]

Begining of derivation:

Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both
The term 'n' applies to the engine regardless of 'Qcz' base heat amount or reference point. It will have the same efficiency burning the same energy at the same temperatures but with different reference points. That allows us to look at the total energy in the system. So Qhz should have the same relation ship to Qcz that DQh has to DQc, in other words, 'n' is the same. Basically this says that Qcz can be calculated from 'n' and Qhz,
or:
Qcz = Qhz•(1-n)
...

Why go any further? You've already produced the Carnot formula by fiat. Just slightly disguised or rewritten in a slightly different form with renamed variables.

Say we have our heat source Th = 600 and Tc = 300

Plug those values into the first equation presented in your "derivation".

Qcz = Qhz•(1-n)

Divide both sides by Qhz (solve for n)

n = Qcz/Qhz = 300/600 = 50%

n is "the same" ?

In the real world, yes. "efficiency" is a property of the engine. "regardless of 'Qcz' base heat amount".

But is that the case with the Carnot "efficiency" formula?

Is it the case with your trojan horse, stealth, masked, rewritten Carnot equation ?

Well, suppose we change the "height of the fall"

What about an LTD running on hot coffee at maybe 360°K; 300°K ambient ? Will the "efficiency be "the same"?

n = Qcz/Qhz = 300/360 = 8.333...%

Carnot formula?

η = Tc/Th = 300/360 = ?

Our "engine" is still "burning" the same energy, blah blah whatever.... Who are you trying to kid?

As I've said over and over, the so called "Carnot efficiency", (never actually produced by Carnot, but who knows who, maybe Kelvin), so-called "efficiency" has nothing to do with ANY actual engine efficiency. It's nothing but the temperature difference (and a little number juggling to turn it into a percentage).

​Qcz = Qhz•(1-n)

Exactly the same thing right out of the gate.

25 lines or so of more number juggling mock calculations and Wallah!!!!

n=(Th-Tc)/Th <<<The final solution.

Right back where you started where you slipped in your "trojan horse" or as Matt would say switcheroo.