I haven't found any reasonable explanation to why we say that most work is delivered in quasi-static operations. Is it just that because they are frictionless or are there any other plausible arguments?
Adiabatic. It is reversible. No entropy production. All work in converted into the inner energy
In quasi -static process, the system is in equilibrium at every step and there are no dissipative forces. So no other way to spend energy other than doing work or heat transfer. So temperature difference is fixed, the whole of the remaining energy can be converted to work.
Mr. Mitra's explanation is on track. The notion of a quasi-static path in classical thermodynamics is one that moves through states in equilibrium (e.g. each point in the trajectory between states is within the space of thermodynamic equilibria, that is each point corresponds to a set of thermodynamic state variables). The terminology apparently refers to allowing sufficient time to maintain equilibrium at each point (e.g. almost static). While reversible processes are quasi-static, the converse apparently does not follow- for example, arbitrarily slow heat flow across a spatial gradient (2D or 3D) can have each point/neighborhood in the thermal conductor in a local equilibrium, but be irreversible (deep connection to irreversible thermodynamics).
The question is probably related to thermodynamic efficiency - an you have got the exact answer from our colleagues. From the practical viewpoint, however, it should be taken into account that quasi-equilibrium processes tend to be infinitely slow and it is also assumed that there are no other losses, only those ascribed to the non-equlibrium nature of the process. In practice, however, you cannot slow down infinitely the process to improve the thermodynamic efficiency and you always will have other heat etc. losses in the process. Therefore, from the engineering aspect you have to optimize your process in terms of e.g. energy consumption or profit. Even in this case, however, on has to know what is the theoretical upper limit not to chase impossible goals.
Mr. Banhegyi, please be aware that the term losses is no precise thermodynamic variable. As pointed out above in connection with the example, the sum of internal and external entropy *can* increase in in a quasi-static process (e.g. a quasi-static process can be pure "loss" as in the "heat leak" cited). As for speed, Carnot Efficiency, etc., they are process specific, as may be seen by contemplating extraction free energy from black body radiation i in a multispectral PV conversion scenario.
You are absolutely correct and more precise than my comment. I only wanted to point out that while thermodyanmic efiiciency is important from the theoretical point of view, if one is interested in practical optimization, other aspects of the process may become more important than the quasi-equilibrium pathway and conditions of the given process.
Dear Saransh,
Yours is a very important question!
Formally, if you consider a closed system, the 1st Principle can be expressed as an equality, the 2nd is an inequality. It becomes an equality when the process is reversible. Among other things, this is essential for Nernst law and the definition of electrode potentials Because it requires a few special characters, I am attaching the equations, etc. I found that the Wikipedia discussion of this "principle of maximum work" is excellent.
All the best,
.
J-L M. Abboud
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