Fool Re- wrote: ↑Wed Mar 20, 2024 7:36 am
Question: How does Qh an Qc become Q's all the way to Zero, and then, Th and Tc?
Answer:
Setting of parameters, mathematical definitions, and descriptions:
Let Th be the temperature of the hot plate/reservoir.400K
Let Tc be the temperature of the cold plate. 300 K
Let Qh be Delta Qh = DQh, the heat added per cycle. 100 J
Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible. Determined by Derivation and Calculation once 'n' is known.
Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.
Let Qcz be the heat added from zero K all the way to Tc. That is true but confusing, so, Qcz is the internal energy of the engine's gas mass at Tc. It is the base energy. It neither used, nor rejected. It is the absolute energy level at Tc contained the the engine gas mass.
DQh is added to Qcz, to get to Qhz at Th. Qcz is the temperature determined energy content of the engine's gas mass at Tc.
Qhz will become the heat at Qcz plus the extra heat added, or Qcz + DQh to get to Th.
Mathematical definitions follow, also true:
Qcz = M•Cv•Tc
Qhz = M•Cv•Th = Qcz + DQh
Noted equation:
Qhz-Qcz = DQh = DT• Cv•M (Noted equation)
Where:
DT = Delta T = Th-Tc
M = Mass
Cv = specific heat.
Where M is the mass of the working gas contained in the engine, and Cv the heat capacity. Cv example units: Joules per Kelvin per Kilogram of gas contained in the engine, or J/(K•Kg). M is chosen, in collaboration with Cv, to have a one to one relationship between heat energy and Temperature. M•Cv=1. The derivation here will show that it cancels out later. So the actual numbers won't matter, "works for all".
Starting with the first law, "Energy is conserved", and the definition of efficiency 'n':
W = Work output from the working gas.
W = DQh - DQc
n = efficiency = W/DQh = (DQh - DQc) / DQh
Begining of derivation:
Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both sides:
Qhz = Qcz + DQh (Noted equation)
Qhz-Qcz = Qcz-Qcz + DQh
Canceling and rearranging the terms gives:
DQh = Qhz -Qcz (DQh equation)
Multiplying both sides by n:
n•DQh = n•(Qhz-Qcz)
Dividing both sides by DQh:
n = n•(Qhz - Qcz) / DQh (#1)
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)
Substituting that equivalence for Qcz into the 'DQh' equation:
DQh = Qhz -
Qcz (DQh equation)
Qcz =
Qhz•(1-n)
Substituting and distributing:
DQh = Qhz -
Qhz(1-n)
DQh = Qhz - Qhz + n•Qhz
Subtracting:
DQh = n•Qhz (#2)
Combining #1 and #2
n = n•(Qhz - Qcz) /
DQh (#1)
DQh =
n•Qhz (#2)
#2 into #1:
n = n•(Qhz - Qcz) /
n•Qhz
The n's on the right side, of the equals sign, cancel becoming one, and rewriting:
n = (Qhz - Qcz) / Qhz (#3)
Equation #3 shows that DQh and DQc have now become Qhz and Qcz. In retrospect, it seems logical that the efficiency profile should be the same regardless of absolute scale. Substituting in Qhz and Qcz straight across for DQh and DQc makes sense, and would have been faster. That was the mathematical derivation/proof. Note also, that it tends to maximize efficiency by equating DQc with Zero. DQc is not zero, it is just the amount of inevitable heat rejection and with maximum work out.
Now using the equations above for Qhz and Qcz:
Qhz = M•Cv•Th (The equations above)
Qcz = M•Cv•Tc (The equations above)
Substituting the above two lines into equation #3:
n = (Qhz - Qcz) / Qhz (#3)
Here:
n = (M•Cv•Th - M•Cv•Tc) / (M•Cv•Th)
Rearranging and removing the distributed 'M•Cv':
n=M•Cv•(Th-Tc) / (M•Cv•Th)
Canceling 'M•Cv' top and bottom because they become one:
n=(Th-Tc)/Th <<<The final solution.
That shows the logical step by step progression for the mathematical derivation or proof.
Basically we took relative scale Delta heat:
n=(DQh-DQc)/DQh
Converted it to absolute temperature scale heat:
n=(Qhz-Qcz)/Qhz
Then converted it to temperature calculated heat:
n=(Th-Tc)/Th
They are all the same value of n, because they are all the same heat-in heat-out ratios. They are just three different ways of obtaining the same thing.
It's logical to think that higher temperature differences produce higher pressure differences, that make greater power to weight and size ratios, ease of construction, operation, and efficiency increases.
1/10 degree temperature, would be like trying to move a piston with 1/10 of a psi. How can that possibly be as efficient as trying to move a smaller piston with 100 psi.
It would take a piston 1000 times larger in area to develop the same force. Power and efficiency sucking bad bulk would be the down fall. I hope this makes sense in regard to the Carnot Theorem.