Tom, I don't know how I can be any more consistent than the definition:
Qcz = M•Cv•Tc
You are confusing other things being equivalent to it, such as:
Qcz = M•Cv•Tc being the base, or starting point, or even the energy contained in the gas, or the internal energy, or energy at P1, V1, and T1, or the energy of M•Cv•(Tc-0 K)
The definition starts it. The equivalents follow. They are all the same thing and extremely consistent because they are all equal to :
Qcz = M•Cv•(Tc Kelvin)
You also seem to be confusing Qcz with the path to get to Qcz.
If you define a point part way up a hill, the point is the same point regardless of what it looks like coming from different directions. From higher up it will look like a point down in the valley. From lower it will look like a point way up on a hill. Regardless it will always be at the same height, as in the case of Qcz, at the same energy level. Climbing there from your current point will require or provide work accordingly to the fixed Qcz level and your current level. And will be a function of Qcz. If you are headed to it, there is no escaping it's influence.
On other words, once starting with Qcz, it will dictate returning there and climbing/(work input) or descending/(heat rejection) to get there, after, the maximum amount of work has been extracted from the expansion stroke.
Tom Booth wrote:In a real engine, judging by actual PV diagrams and some basic Newtonian laws of motion. At P2 heat, or if you prefer "internal energy" from the expanding gas has been converted into motion/velocity which continues to carry the piston so that pressure drops below atmospheric pressure.
Piston speed, 1/2MV^2, is the result of work out from the gas, as P3 reaches P1 the speed is at a maximum. If it passes that point the inside pressure drops below the outside pressure and the piston begins to slow. This means the working gas is reabsorbing energy from the piston's momentum.
In other words energy returned to the gas is less energy out. The maximum out is at P3 = P1. As I stated.
Tom Booth wrote:These statements or assumptions are contrary even to the Carnot cycle
True. I was describing a cycle that had constant volume heat addition, followed by adiabatic expansion. A lot less efficient than Carnot. Less than 25% at those temperatures.
Although the above is a valid theoretical description, let me redescribe it with a more Carnot cycle like description, LOL.
Start at Qcz, V1, P1, T1=Tc. Compress the gas, work input, (At a loss of piston momentum.), until the gas is at Th. V2, P2, T2=Th. Smallest volume.
Expand the gas while adding heat to maintain Th, (Yes from a hotter than Th source), positive work output, until P3 = P1 for maximum work out, V3, P3=P1, T3=T2=Th. T3 still equals Th.
Expand the gas further at a loss of work out. I.E., work goes back into the gass as reduced pressure. Inward piston force. Having to fight inward force by pulling out or slowing of the piston's, or flywheel's, velocity. This will be continued until T4=T1. Volume will be much larger and pressure will be much lower. V4, P4, T4=T1=Tc
Compress the gas maintaining the temperature Tc (Yes, this will require a much lower Temperature than Tc.). This is the trip back to Qcz, the starting point. The gas will heat up if DQc is not rejected. That is often called the heat of compression. This process is a work into the gas heat out/rejected stroke. It is powered by the outside air pressure being higher than the inside pressure. This energy gain is used up by the energy loss in the previous stroke, and the following stroke for a net amount of zero. This means the buffer pressure effect can be ignored. It's just as if the engine were in a vacuum, the same net zero help from outside pressure, in regard to work out only. This plus the next stroke (The same as the first stroke.) Are the "back work" that Matt calls it. This stroke returns the engine's internal energy back to Qcz, and why there is a battle against the internal pressure, associated with Qcz. Back to V1, P1, T1=Tc.
Another way to look at Qcz=M•Cv•Tc is, it is the pressure energy that is opposing the compression back to V1, P1, and, T1, from a larger volume. Qcz isn't used but, it is fought all the way back home, to the starting point. The starting point isn't at the bottom of a valley. It is 300 K up the hill. It dictates a pressure level. It is Qcz=M•Cv•Tc as well as other things are equal to it. Do not confuse the direction of definition, with equivalents.
This is a lot easier to follow on a PV diagram for an ideal Carnot, or Stirling, Cycle.