Recent content by Dale

Your perceived effort is a very poor measurement of force. It is exceptionally difficult for humans to do constant velocity motions or to accurately gauge the force applied during different movements. The perceived effort will also involve the motion of your own body, not just the external...

Why would it? There isn't a limit of energy, there is a fixed energy. Energy is conserved. It cannot appear, but it also cannot disappear. Think about it. When you look at a star you are looking at an object that is as bright as the sun. But it doesn't hurt your eyes. It is radiating about the...

No. There is no inertial frame where the energy of a pulse of light is zero.

So it sounds like you want adiabatic, not isothermal. Consider a gas. As a gas expands adiabatically it reduces its temperature. And as it compresses adiabatically its temperature increases. So an adiabatic process is not isothermal. The same happens in a liquid, but just to a smaller...

This is self contradictory. If you want T2=T3 (isothermal) then it cannot be thermally insulated (adiabatic). You have to choose.

There would be a huge difference. By Noether’s theorem a conservation law implies a symmetry in the laws of physics. So if you violate the conservation law then you must not have the corresponding symmetry. In this case, the conservation of energy is associated with time translation symmetry...

Then Occham would complain, but I guess since he is dead you can ignore him if you want.

This is much better than before. So let’s look at what you have and what you are still missing (you keep jumping to the end without completing the middle) Process 1-2 appears to be isobaric heating given the graph, but you never specifically state it. Please confirm? Similarly process 3-4...

So now we are back to using water? You can find this information at: https://webbook.nist.gov/cgi/fluid.cgi?ID=C7732185&Action=Page Here is the isobaric data at 0.1 MPa...

Trying to make a cycle with an incompressible fluid is complicating things Again, Why would a non-compressible fluid flow at all?

So now you do not want to use an incompressible fluid? You want to use water?

This is not nearly as difficult as you pretend. On the theoretical side, take any tetrad and integrate the timelike vector field. Those integral curves are the positions. On the experimental side, devices for measuring position can easily be purchased. With a clear theoretical meaning and a...

For an incompressible fluid there is no temperature drop. The efficiency of an ideal Brayton cycle is given by $$1-\frac{T_1}{T_2}=1-\left(\frac{P_1}{P_2}\right)^{(\gamma-1)/\gamma}$$ Since for an incompressible fluid ##\gamma=1## the efficiency is zero, the temperatures and pressures are equal...

That is fine. It will just slow things down so that the power output will be reduced. Why would a non-compressible fluid flow at all? There is a really good reason why existing engines don’t usually use incompressible fluids.

This is false. These cycles are well understood. They don’t work this way.