- #36
Signature
- 19
- 0
Chestermiller said:For the initially hot slab in the return part of the cycle, this is correct.For the initially cold slab in the return part of the cycle, this is correct.Yes, for the initially hot slab and the initially cold slab in the return part of the cycle, this is correct. For the hot slab and the cold slab over the entire cycle, if we add the entropy change for the irreversible equilibration part of the cycle (positive) with the entropy change in the reversible return part of the cycle (negative), they add up to zero (as they must for a cycle).
These equations for the hot slab and the cold slab in the reversible part of the cycle are incorrect. They should read
$$\int {\frac{dq}{T_I}}=ΔS_H$$
and
$$\int {\frac{dq}{T_I}}=ΔS_C$$
The integral of dq/T along a reversible path determines the change in entropy.
In the reversible return part of the cycle, we are free to do whatever is necessary to return the two slabs to their original state. It certainly can't happen spontaneously. So we separate them, and return them each to their original states by contacting each of them with a continuous sequence of constant temperature reservoirs at slightly different temperatures. So the system is not isolated during the return part of the cycle.
For the irreversible part of the cycle, we already showed that the integral of dq/TI is zero. For the reversible (return) part of the cycle, we have now shown that the integral of dq/TI is negative. So, for the entire cycle, the entropy change of our system is zero, and the integral of dq/TI for our system is negative.
Chet
Nice sir. Now I understood the concept.. It's very good in having such interactions.
Now I need to get some idea on another example.
Suppose we are having an engine whose piston and cylinder housings are made of an imaginary material whose thermal diffusivity is zero so that the heat produced during combustion of fuel will not be absorbed by the piston and cylinder housings but only some amount of heat can be taken as work from the piston.
Will this engine's efficiency can be equal to the efficiency of engine operating in carnot cycle with the same working temperatures?