Fool wrote: ↑Wed Jan 03, 2024 1:51 am
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.
