Tag Info

New answers tagged

0

Reversibility implies that the entropy change of the universe is 0. An isentropic process need not necessarily be reversible, provided that the entropy of the surroundings is increased. For an irreversible process, heat can be removed from the system in order to make it isentropic (since $dQ < TdS$. As heat is removed, you will have an isentropic ...


0

Entropy and irreversibility are related to each other. Irreversibility is the one because of which process takes place and due to this energy degradation takes place that is exergy (availability) decreases. And basically irreversibility(I) is defined as I = To * ( del s ) That is when you multiply change in entropy of universe with the reference ...


0

The difference is that one expansion is quasi-static (the reversible one) while the other is spontaneous because of a dramatic change of the external constraints (the irreversible one). In the quasi-static case, you start off indeed in the state where gas pressure equates external pressure. An external operator then slightly decreases the outside pressure ...


1

Landau damping seems mysterious because it is derived from completely reversible equations, yet it gives a "damping" behaviour. But the funny thing is that when you go through the equations you find even an "antidamping" exponentially growing solution$^1$! The reason for this is that the Landau damping isn't really a damping caused by particle collisions. ...


1

Of course thermodynamics says things about differences in entropy between reversible and irreversible processes. In fact, we can analyse how irreversible a process was and what to do in order to, for instance, extract more work from it. That's a very important part of thermodynamics. But, no, there are no true reversible processes — at least for daily ...


0

In fact, $\delta Q/T$ is the entropy of $\delta Q$, and if $\delta Q$ is considered as the heat energy in transfer, it follows that $\delta Q/T$ is the entropy in transfer. For the inequality \begin{align}\oint dS \ge \oint \frac{\delta Q}{T}.\end{align} Whether reversible or not does not need to be considered for $\delta Q/T$ in that the entropy production ...


0

One should never write $d S_S= \delta Q_{E\to S}/T_S,$ unless it is a reversible process in the system, as well, because the temperature $T$ is whatever temperature the heat is supplied and that is the temperature of the environment (heat reservoir), presumably behaving independently of the amount of heat it supplies. The reservoir is always assumed to ...


0

Heat is defined as a transfer from the environment to the system. No subscripts are needed for that purpose. I think you would have to be very clear about the specific situation for the other processes that you mention before there can be any discussion about them. Friction where? Turbulence implies a non-equilibrium condition...


1

The Stirling cycle as you describe it is not reversible. The transfer of heat from thermal reservoirs along paths 4->1 and 2->3 is not a reversible process, because heat is being transferred between two objects at different temperatures. To reverse the process, you would need to spontaneously transfer heat from a colder to hotter reservoir, which violates ...



Top 50 recent answers are included