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

[matt brown]

If you are comfortable with:

Qh = Ql +W

Are you comfortable with rearranging the terms:

W = Qh - Ql

That is the exact same equation. Just solved for W, Work.
Or would you like me to go through the detailed steps?

Sorry Fool, but I'm not comfortable with this entire post since it appears as a very slick method to prove conventional Carnot limit 'equation' which uses temperature extremes.

Let's say Q = energy vs 'heat'. If W = Qh - Ql then work equals high energy minus low energy, and Tom's idea that ALL additional energy above Ql can become work appears correct. Furthermore, the whole Carnot buzz is a maximum that only directly applies to common isothermal cycles and ignores common adiabatic cycles that dominate the world.

A much better approach is from work directly where Wnet = Wpos - Wneg and usually considered mere mechanical efficiency since system eff does not equal Wneg/Wpos.

[Tom Booth]

...
A much better approach is from work directly where Wnet = Wpos - Wneg and usually considered mere mechanical efficiency since system eff does not equal Wneg/Wpos.

How better?

How would the value of "negative work" for example be determined?

Logistically, negative work can only be a result of the conversion of heat into work. Work is work regardless if it is applied to overcoming atmospheric pressure, a conversion from heat into work is still taking place..

That some work output is "positive" and some work output is "negative" does not change the fact that it is work and no longer heat.

Further, "work" is equivalent to mechanical motion itself, i.e. the engine in operation, running.

If work is put into pushing the piston against an air spring pressure (or atmosphere) and atmospheric (or air spring) pressure pushes the piston back. It is still the engine that "loaded" the air spring, so that the "negative" work is derived from positive work and still results in motion. The return motion still revolves the flywheel.

So IMO, no such thing as this "negative work"

[VincentG]

W = Qh - Ql

Thinking in terms of work per cycle over time...

100-99 equals 1

2-1 equals 1

100-99 equals 2-1 ??

[Tom Booth]

W = Qh - Ql

Thinking in terms of work per cycle over time...

100-99 equals 1

2-1 equals 1

100-99 equals 2-1 ??

100 joules of heat in and 99 joules "waste heat" out

That means 1 joule was converted to work output.

2 joules in and one joule out as "waste" rejected

That leaves 1 joule "missing" that was converted to work.

Simple quantities of joules in and out.

In as heat and out as either work or heat.

Yes 100-99 = 2-1

Both result in 1 joule being converted to work.

Same as 89 joules of heat in with 88 out etc.

Simple accounting like a bank balance.

So far so good IMO.

It's when you try to toss in the Carnot limit that things get a bit less trivial.

[VincentG]

100 joules of heat in and 99 joules "waste heat" out

That means 1 joule was converted to work output.

2 joules in and one joule out as "waste" rejected

That leaves 1 joule "missing" that was converted to work.

Likely I'm too dumb to grasp these mathematical concepts and should do some formal studying. Doesn't 100-99 affect more change in a system than 2-1?

Sorry, not trying to derail the conversation, just wondering.

[Tom Booth]

100 joules of heat in and 99 joules "waste heat" out

That means 1 joule was converted to work output.

2 joules in and one joule out as "waste" rejected

That leaves 1 joule "missing" that was converted to work.

Likely I'm too dumb to grasp these mathematical concepts and should do some formal studying. Doesn't 100-99 affect more change in a system than 2-1?

Sorry, not trying to derail the conversation, just wondering.

That's OK.

Think of 2-1 like an LTD on heat of your hand or something at 50% efficiency

100-99 is a high temp engine with only 1% efficiency 99% waste heat.

[Tom Booth]

Anyway, so far the equations "fool" has posted in his detailed review are really only 1st Law conservation of energy, which as Matt pointed out,

Let's say Q = energy vs 'heat'. If W = Qh - Ql then work equals high energy minus low energy, and Tom's idea that ALL additional energy above Ql can become work appears correct.

The first law or conservation of energy does not bar 100% conversion efficiency, so, that is correct. Conservation of energy does not prohibit an "ambient heat engine". All it says is the books need to be balanced.

This is all very straightforward so far. "Fool" has not yet, in his detailed, checked and carefully rechecked review, incorporated the second law into the mathematics.

That is where IMO things get convoluted.

Carnot's (or Kelvin/Clausius interpretation of Carnot really) theorem that the temperature difference limits efficiency does not REALLY conform with the simple accounting of conservation of energy.

I think, if "Fool" is actually really careful, thoughtful and thorough he may reach the same conclusions as I have before he's finished.

[matt brown]

...
A much better approach is from work directly where Wnet = Wpos - Wneg and usually considered mere mechanical efficiency since system eff does not equal Wneg/Wpos.

How better?

How would the value of "negative work" for example be determined?

Just a lame ass guess, but how about the same way "positive work" is determined...

Logistically, negative work can only be a result of the conversion of heat into work.

Work is work regardless if it is applied to overcoming atmospheric pressure, a conversion from heat into work is still taking place.

That some work output is "positive" and some work output is "negative" does not change the fact that it is work and no longer heat.

Huh...Wpos (aka expansion) is work output from heat vs Wneg (aka compression) is work input into heat.

Further, "work" is equivalent to mechanical motion itself, i.e. the engine in operation, running.

If work is put into pushing the piston against an air spring pressure (or atmosphere) and atmospheric (or air spring) pressure pushes the piston back. It is still the engine that "loaded" the air spring, so that the "negative" work is derived from positive work and still results in motion. The return motion still revolves the flywheel.

So an air spring has Wnet = 0 since Wpos = Wneg or...Wpos - Wneg = 0

So IMO, no such thing as this "negative work"

The bottom line is that no compression CYCLE can transform all 'heat' input into work since all compression cycles include Wneg. It's just where in the cycle this Wneg occurs.

[Tom Booth]

...
So an air spring has Wnet = 0 since Wpos = Wneg or...Wpos - Wneg = 0

....

The bottom line is that no compression CYCLE can transform all 'heat' input into work since all compression cycles include Wneg. It's just where in the cycle this Wneg occurs.

So, if you wind up a spring on a grandfather clock, the spring then does "Wneg". The clock still runs.

[matt brown]

...
So an air spring has Wnet = 0 since Wpos = Wneg or...Wpos - Wneg = 0

....

The bottom line is that no compression CYCLE can transform all 'heat' input into work since all compression cycles include Wneg. It's just where in the cycle this Wneg occurs.

So, if you wind up a spring on a grandfather clock, the spring then does "Wneg". The clock still runs.

Earth to Tom: winding the clock is Wneg aka work INPUT vs clock running is Wpos aka work OUTPUT.

[Tom Booth]

...
So an air spring has Wnet = 0 since Wpos = Wneg or...Wpos - Wneg = 0

....

The bottom line is that no compression CYCLE can transform all 'heat' input into work since all compression cycles include Wneg. It's just where in the cycle this Wneg occurs.

So, if you wind up a spring on a grandfather clock, the spring then does "Wneg". The clock still runs.

Earth to Tom: winding the clock is Wneg aka work INPUT vs clock running is Wpos aka work OUTPUT.

OK then the expansion stroke is Wneg as that "winds" the air spring and the compression stroke is Wpos.

Either way, it's my muscle putting in energy to wind the clock and either way it's HEAT that puts in the energy to run the engine.

...
How would the value of "negative work" for example be determined?

Just a lame ass guess, but how about the same way "positive work" is determined...

I should have phrased the question: how is "work negative" clearly differentiated?

The whole concept of this Wneg and Wpos is your own made up fiction . No such thing as far as I can see.

The source of energy is heat. Work, however you slice it is no longer heat. Heat from adiabatic compression is retained in the working fluid and reused the following cycle to help power the engine. Not "negative".

[Tom Booth]

Okay. Thanks.

If you are comfortable with:

Qh = Ql +W

Are you comfortable with rearranging the terms:

W = Qh - Ql

That is the exact same equation. Just solved for W, Work.
Or would you like me to go through the detailed steps?

If comfortable, I'd like to combine that equation with efficiency. (Put W=Qh-Ql into the efficiency equation for W.):

W = Qh - Ql
into:
n = W/Qh

and get:
n = (Qh-Ql)/Qh

I can give more detail on that too.

Please. Any one may ask for more detail. It would please me to expand. I would rather not lose anyone at this point, or any point. Algebra is about skipping steps if you can. They can always be put back in if needed for clarity. I wasn't really comfortable with this stuff until having two years of college. I had to use small steps. I confess.

I know you said that you are being very thorough and giving this all a lot of thought, so I'm trying to be patient.

As I said. No problems or issues from me so far.

Even as Matt pointed out:

Let's say Q = energy vs 'heat'. If W = Qh - Ql then work equals high energy minus low energy, and Tom's idea that ALL additional energy above Ql can become work appears correct.

I assume your intention was not to prove my point for me, so,there is more to come?

The usual logic applied at this point is that the above equation is too liberal. "Everybody knows" 100% efficiency is impossible.

Therefore, looking around we find the Carnot Limit to save the day and restore order to the universe.

Therefore we can assume that:

η = 1 – (Qc / Qh). = 1 – (Tc / Th)

Which perhaps it has been noticed is the equation I chose for the title heading for this thread.

But, there is a minor problem with this. Well, a couple problems really, and maybe not so minor.

In the efficiency equation: η = 1 – (Qc / Qh)

Qh and Qc represent quantities of energy or joules of heat (or work) using Qc as the baseline or zero. In other words, starting at 300K (ambient temperature) we have added zero "heat" to Qc to get 300k, that is a given. we add, say, 100 joules so that Qc = 300K and Qh with tbe addition of 100 joules to the given 300 becomes 400K (by the addition of 100 joules to the 300 ambient).

Between 300K and 400K we have 100 Joules of available (or added) heat. If our engine converts that 100 joules to work we are back where we started at 300K (NOT zero K!)

Using Joules what is the result?

η = 1 – (Qc / Qh)
η = 1 – (Qc / 100 quantity of heat added in joules)

We know Qc = Qh - W

Suppose our engine converts 50 Joules into work and rejects 50 Joules to the sink. Then Qc = 50 Joules. (50 Joules now "added" or "rejected to" Qc )

Therefore:


η = 1 – (Qc / Qh)
η = 1 – (50 / 100)
η = 1 – 0.50
η = .5
Efficiency = 50%

All well and good, right?

Then we have, for example this blurb, which in this case is taken from the Wikipedia article on the Second Law of Thermodynamics, subheading: The thermodynamic temperature scale (absolute T):

The thermodynamic temperature scale (absolute T)

Efficiency of a Carnot engine is:

η = 1 – (Qc / Qh) = 1 – (Tc / Th)

The zero point of the thermodynamic scale is fixed as the T of the cold reservoir Tc at which η = 1 and hence Qc = 0 and W = Qh.

T must be the absolute scale because otherwise it may have -ve (negative) values, and hence the engine will perform work more than the heat given by the source. i.e. we would create E. from nothing, which is in contradiction to 1st law of thermo.

Thus, T = 0 is the lowest T in all scales i.e. the absolute zero.

In other words, there is more heat below 300K what if our engine could get at some of that and utilize, say 110 joules of heat, bringing the exhaust temperature down to 290K this would be a violation of the Second Law of Thermodynamics, NOT, I would venture to say, a violation of the first. It is not as stated: creating E from "nothing". There is plenty of kinetic energy in the air below 300K. That is not really "nothing" is it?

So using a bit of slight of hand we will shoe in the Carnot Limit by making a substitution in our equation:

Instead of:

η = 1 – (Qc / Qh)
η = 1 – (Qc = 0 + 50 joules "rejected"/100 joules supplied)
η = 1 – 50/100
η = 1 – 0.50
η = .5
Efficiency = 50%

We move Qc from 300K to Zero K

η = 1 – (50 joules / 100 joules)
η = 1 – (400-50)=350/(300+100)=400)
η = 1 – 350/400
η = 1 – 0.875
n = 0.125
Efficiency = 12.5%

The second law / Carnot limit advocate rubs his hands together and smacks his lips and says now that's much better.

So what have we now?

We add 100 joules

12.5% of 100 equals 12.5 joules converted to work

100 - 12.5 = 87.5 joules "rejected" to the cold reservoir.

87.5 - 50 = 7.5 joules discrepancy.

7.5 joules of the supplied heat has gone missing.

What's going on here?

Maybe we can call this vanishing thermal energy "ENTROPY"!

What is the justification for this substitution

η = 1 – (Qc / Qh) = 1 – (Tc / Th) ?

Well, Carnot stated that the power of a heat engine depends only on the temperature difference or the "height of the fall".


So, we just make the two equations equivalent.

On what basis?

There is another problem though.

Qh and Qc are quantities of heat transfered in and out of the engine in joules. (And of course "work" in joules is subtracted from Qh to get Qc).

But what it T ?

T is temperature right?

Temperature is not a quantity of heat.

For example, my heat source could be a cup of hot water, a four quart pot of hot water or a bathtub full of hot water, all at the same 400 K temperature. Does each source of heat supply 400 joules?

So how do we just produce this equivalence between joules of heat and temperature without reference to actual quantity? Temperature is not a quantity of heat. Is it?

Does a cup of hot water supply the same quantity of heat as a bathtub full of hot water at the same temperature?

Well, OK, if you break it down into time frames or engine cycles and heat capacity and so forth the volume of the heat source becomes irrelevant, the engine can only take in a fixed quantity of heat per cycle for a given ∆T. That's OK in the abstract, but in some real world situation where we want to determine the actual efficiency of a real engine we need to be able to actually measure that somehow

We still have an unexplained drop in calculated efficiency from 50% down to 12.5% and 7.5 joules gone missing from the universe.

Energy cannot be created out of nothing but the inverse is true as well is it not?


I'm rather anxious to see how this is all sorted out.

Using η = 1 – (Qc / Qh).in the example I get 50% efficiency.

Using 1 – (Tc / Th) I get 12.5% efficiency

So how can this equation be true?

η = 1 – (Qc / Qh) = 1 – (Tc / Th) ?

It is just an unproven and unjustified supposition. It arbitrarily replaces the zero heat addition at 300k with absolute zero.

Why?

Well, it works out better in the minds of those who adhere to the Carnot theorem, and the Second Law. If you want to avoid the dastardly bugaboo of any possibility of 100% thermal efficiency or God forbid tapping into that vast reservoir of ambient heat, well it's a good mathematical slight of hand for accomplishing that, but in reality it violates conservation of energy.

How do we fix this?

ENTROPY!


But you know, to be honest, I don't think heat is a fluid, so heat does not really "fall down". From high to low temperature. Maybe efficiency depends on factors other than the temperature difference.

Has any of this ever been subject to experimental proof?

Experimentally, I don't seem to be finding the 87.5 joules of "rejected" heat per cycle that the Carnot limit says are supposed to be there.

I'm finding closer to the zero joules that the first equation suggests should be there if the engine is near 100% efficient.

The experimental results confirm η = 1 – (Qc / Qh) but do not confirm η = 1 – (Tc / Th)

Where Qh or Th = heat supplied in joules and Qc or Tc equals heat "rejected" in joules.

Where is the justification for saying that you cannot have 100% efficiency unless your exhaust temperature is at "absolute zero" when in actuality all the supplied heat has been used up if the exhaust is 300K, back down to ambient where we started before any heat was supplied?

I'm very anxious to see how you make sense of all this my friend.

Remember the definition of heat.

Heat is the transfer of energy.

At ambient, with everything at 300K there is no transfer of energy and so there is zero heat.

300K ambient then is the zero of our y intercept rather than 0K s it not?

That would reconcile all the differences.

The Wikipedia article stated:

The zero point of the thermodynamic scale is fixed as the T of the cold reservoir Tc at which η = 1 and hence Qc = 0 and W = Qh.

But our zero of heat (energy transfer) is when Qc equals Qh and both are at 300K.

Where Qh=Qc there is zero "heat" when Q is defined as heat transfer, or the quantity of heat transfered over a certain time interval or cycle.

[Tom Booth]

To arrive at the "desired" result then, we cannot view heat as a transfer of energy. Or merely the transfer of 100 joules.

We would have to say when transferring 100 joules at 300K to make 400K we are also transferring 300K of implicit heat existing below 300K.

In other words, to fill a container with a fluid, to raise the fluid level from 300 milliliters to 400 milliliters we have to start with the container already filled to the 300 milliliter mark.

Then to use "ALL" the fluid in the container we have to drain the container all the way, not just the 100 milliliters added.

So according to the "Carnot efficiency limit" for 100% efficiency, if we add 1 milliliter to our 300 milliliters we will need to drain all 301 milliliters for 100% efficiency, not just the 1 milliliter added.

It is implicit in the Carnot efficiency limit that heat is a fluid, so naturally 100% efficiency means draining or using up ALL of the contents of our "hot reservoir" down to the last milliliter.

If heat is energy, and we add 1 Joule of heat then we get 100% efficiency when that 1 joule is fully converted into work.

The Carnot limit formula would have that if we add one joule of heat per cycle, then for 100% efficiency we would need to utilize 301 joules per cycle so as to bring the temperature down to zero degrees Kelvin.

I'm not sure how anyone ever considered this reasonable.

[Fool]

VincentG, you are on the correct track. It does reflect a greater change, for the same outcome. Thanks for the help.

Matt, I think it is a very slick explanation of the Carnot theorem . Thanks for the compliment. I'm trying to use nothing more than linear algebra. If one were to stay solely in the work theory, it wouldn't be thermodynamics. Think mechanical equivalence of heat. You should know by now that you can't beat the second law...

The equation W = Qh - Ql, applies to all processes, and the equations: n=1-Qh/Ql=1-Tl/Th only applies to cyclic process. So yes efficiency can be higher for single non-cyclic processes. The first law allows that. The second doesn't. People have trouble swallowing the second law because the first allows 100% and the cyclic nature of the second doesn't. Call it back-work, irreversibility, heat pump COP greater than one, increase in entropy, etc...

Bumpkin, the current thermodynamic theory is splitting the definition of heat away from internal energy. Molecular motion is becoming the definition for internal energy and the related temperature.

U = Qin + Win + Chemical energy release + etc... Also minus energy out...

It took me years of challenge/skeptical research to finally understand that. Heat flows into a system and is converted to internal energy (Temperature), similar to the flow of force times distance (Work) into a flywheel and becoming kinetic energy (velocity), or into rasing a weight and becoming potential energy (height). The potential for heat to flow is temperature.

We don't talk of a flywheel containing work. We, now, are beginning to, not allow talking of a mass containing heat. Mass contains internal energy. That can be kinetic/.5mv^2, potential/mgh, or TPmCvR thermal energy with Temperature being the measure.

Tom, temperature doesn't equal heat or internal energy. Temperature is 'linerly' dependent on internal energy, and internal energy is 'linerly' dependent on temperature, sort of. Since there are other ways of adding energy to a system than heat, such as work and chemistry. One needs to look at the changes to a system from only the added heat. We can do that because all energy is equivalent. We can account for other forms by a simple energy balance equation. Please let me start from just heat transfer.

First, the base line and starting point can be different. Physically, temperature can't go below zero Kelvin, that is the absolute baseline. Pressure can't go negative, zero Pascals. Both of those can have a starting point for a problem that is higher. STP, Standard Temperature and Pressure is typical.

Linearly dependent means that a dependent variable say, Y, changes with another independent variable say, X, graphed onto a straight line. The equation for it is of the form:

Y = mX + b
Or:
Q=CvT + k

b and the substitute k are the Y intercepts. Meaning when X is Zero, Y will equal b, or k. b in the first equation, k in the second.

If the starting point is zero Kelvin, a chunk of mass will have zero internal energy (molecular motion). That is true of all mass, metal, plastic, water, helium, nitrogen, oil, wood, etc...

The Y intercept will be zero, b or k will be zero.

If you look at a chunk of air at 300 K, it will have a starting point internal energy of Ql = 300•Cv + k. If you raise it up to 400 K, it will have an internal energy of Qh = 400•Cv + k.

Qh - Ql = delta Q = (400•Cv + k) - (300•Cv + k)
Delta Q = 400•Cv + k - 300•Cv - k

The k's cancel, Cv is pulled out:

Delta Q = Cv•(400-300)

The starting point doesn't matter it becomes:

Delta Q = Cv•DeltaT

It becomes dependent only on the slope, Cv.
The Y intercept, k equals zero, disappeared too.

One more thing. The total value of internal energy equals the heat added from absolute zero Kelvin so the equation:

Q = Cv•T + k

Reduces to:

Q = Cv•T

Again it is a function of only the Slope. Cv or 1/Cv doesn't matter, as the slope, Cv, cancels out when put into the efficiency equation. That means the maximum thermodynamic efficiency doesn't depend on working fluid. Real fluids reduce the efficiency, as do many many other aspects of "real" machines. No real process improves efficiency, except above other worse real processes.

Side note. What is negative work? The ancients didn't believe negative numbers could possibly be useful as they weren't real. Show me a negative apple. It is a tally sheet method. Farmers put apples into cold storage, Tlow. Why is it always apples? LOL. If farmer A puts in 100 apples the tally reads 100. If he then removes 10, the tally shows 90. If farmer B puts in ten apples, then removes 20, his tally will show negative ten apples. Cold storage can't show a negative ten apples, it is just a tally sheet entry.

The same with work or heat. Positive work increases a flywheel's velocity. Negative work slows a flywheel's velocity.

If the flywheel is pushing a car up a hill the internal kinetic energy is being converted to potential energy. That is negative kinetic energy and positive potential. If that process goes on long enough the same exact force, down hill, will stop the flywheel and the negative energy will reverse and become positive energy and the flywheel will store that energy by spinning in the opposite direction. The tally sheet for that energy change will be positive and negative at times, depending on how the direction is defined.

Like having a starting point for pressure, and defining lower 'guage' pressure to be negative. The absolute pressure will still be positive, but the tally sheet will have negative pressure on it. Like the cold storage having positive apples in it but the tally sheet shows negative apples for some farmers.

And yes, it is a made up fiction, but very useful for mathematically keeping track of what direction energy flows, without having to add extra variables like Qhin, Qhout. Qhout=-Qhin, so to speak. In engineering a variable is defined as positive in one specific direction, after the analysis is finished, if it has a negative Value, that just means it, a force or motion, is in the opposite direction. Don't get me started on the square root of minus one, another mind numbing useful obscure mathematical tally sheet method.

Please take pause to note that I'm not mentioning Carnot or his theorem. I'm starting from accepted first law equation, and agreed efficiency equations. I'm now trying to add observed thermodynamic relationships of temperature, heat flow, and internal energy concepts. Delta heat = Delta Temperature. Trying to add the linear equation for the energy contained, verses, absolute temperature. Heat potential verses absolute temperature. Confusing apples with farmers...

Ps, for the record, your math didn't seem to add up correctly. I'm not going to analyze it. It seemed like you were trying to prove that a single process could be more efficient than for complete cycle. So I'll leave alone. At this point all I'm striving for is to show that the Y intercept for the equation Q verses T is Zero, regardless of where an analysis is started.

[Fool]

Oh! And Matt, if the world is dominated by adiabatic processes, how does heat get into an engine? It doesn't.

Imo, the world is dominated by real processes, neither adiabatic, nor isothermal. But it is limited by temperature and pressure extremes. Those extremes will dictate maximum efficiencies. Those maximums will be further eroded by reality. The processes will be between adiabatic and isothermal and material and design constraints. Smiles