Can One Reservoir Drive Cyclic Work in Classical Thermodynamics?

[mohd]

Classical thermodynamics Q's.
Hello !
I have some conceptual problems that are confusing me, hope someone could help.
Book Being used: Fundamentals of Engineering Thermodynamics. (Micheal J. Moran & Howard N. Shapiro).

1) Analytical Kelvin-Planck statement. (pg. 224)
W_cycle <= 0. ( Single Reservoir).

A) How come we have any cyclic work being done while only ONE single reservoir feeding the system? In fact, W_cycle doesn't make any sense in this scenario.

B) If we include the cold OR hot reservoir in the analysis, i.e. as a part of the system, wouldn't Kelvin-Planck be violated? ( or as in this case, since the cold reservoir is included, the cycle is interacting with the external hot reservoir and the included cold reservoir.)

C) Let’s assume that we have a reversible cycle. Furthermore, the cycle is being fed some heat Q and delivering some work W.
Now, Kelvin-Planck statement is satisfied by having hot/cold reservoirs. Now, to investigate the cycle, the first law will be used. dE= DQ-DW , dE= 0 ( cycle). DQ=DW , integrating, Q=W. What I don’t understand is that why is the W_rev = 0 in this case ?

D) Kelvin-Planck statement is sort of a common sense, for to complete a cycle, a cycle is needed to be restored to its initial state. However, how is this same for Carnot efficiency? To be more exact, according to Carnot findings, the only things affecting a reversible cycle are the T_{cold res.} and T_{hot res.}, however, I can NOT seem to see this intuitively. Is there an intuitive approach to understand this?

2) In finding the coefficient of performance of any refrigerator, we consider Q leaving the cold reservoir and W_in. However, to find any efficiency we just require E_out / E_in. Why in refrigerators case we consider different scenarios? Is it just that we avoid the confusion resulting in having efficiencies higher than 1 ?

3) COP_H.P. = COP_ref +1
Is there any physical interpretation for (+1) ? or it’s just the work of derivations ?

4) Clausius statement (pp. 241-242).
Cyclic integral of (DQ/T) = -S_gen
Shouldn’t this integral cover ,only, entropy generation due to heat and NOT due to irreversibilities.(pg. 242)

5) Is there any nice article about the violation of Kelvin-Planck and how that violation would lead to Clausius statement violation ? I couldn't really grasp the whole picture.

Thanks !

 

[Aaron Barnard]

Answer:
1) A) The cyclic work in a single reservoir system is done by the system itself. It is zero or negative, depending on the type of process that is occurring. B) No, the Kelvin-Planck statement will still be satisfied as long as the total cyclic work done is less than or equal to zero. C) The work done in a reversible cycle is equal to the heat added, so Wrev = 0. D) The Carnot efficiency is an expression of the maximum theoretical efficiency of a reversible cycle. Intuitively, it can be thought of as the ratio of the maximum amount of useful work that can be extracted from a given amount of heat energy.

2) The coefficient of performance of a refrigerator is a measure of how efficiently the refrigerator can transfer heat from one medium (such as air) to another (such as a liquid refrigerant). In order to calculate this, the amount of heat leaving the cold reservoir must be taken into account, as well as the work input. This is different from the efficiency, which is simply the ratio of the output energy to the input energy.

3) The +1 in COPH.P. = COPref +1 represents the additional work that must be done in order to maintain the heat pump cycle. This work is necessary to move heat from the cold reservoir to the hot reservoir.

4) The Clausius statement applies to any type

 

[astrokreng]

Hello!

I completely understand your confusion and I will try my best to help clarify some of these concepts for you. Classical thermodynamics can be quite complex and it's important to have a strong understanding of the fundamentals in order to fully grasp these concepts.

1) The analytical Kelvin-Planck statement states that in a system with only one reservoir, the maximum amount of work that can be done is 0. This means that in a single reservoir system, no work can be done because there is no temperature difference to drive the process. This statement is not meant to be applied to a real-life scenario, but rather as a theoretical concept to understand the limitations of thermodynamic systems.

B) Including both the hot and cold reservoir in the analysis would indeed violate the Kelvin-Planck statement. This is because the cold reservoir would be acting as a source of work, which is not allowed in this statement.

C) In a reversible cycle, the work done is equal to the heat absorbed. This is because in a reversible process, there is no loss of energy due to heat transfer. This is why Wrev=0 in this case.

D) The Carnot efficiency is based on the fact that a reversible cycle is the most efficient possible. In a Carnot cycle, the only things affecting the efficiency are the hot and cold reservoir temperatures. This is because in a reversible process, all energy is converted to work and there are no losses.

2) The coefficient of performance (COP) is a measure of the efficiency of a refrigeration system. In order to calculate the COP, we consider the heat absorbed and the work input. This is because in a refrigerator, the goal is to remove heat from the cold reservoir, which requires work input. The efficiency, on the other hand, is a measure of how well the system converts energy into work. In a refrigerator, the work input is not the only form of energy, as there is also heat removal from the cold reservoir. This is why we consider different scenarios in calculating the COP and efficiency.

3) The (+1) in the equation COPH.P. = COPref +1 represents the work input required for a heat pump to operate. This is because in a heat pump, work must be input in order to transfer heat from a colder reservoir to a hotter one.

4) The Clausius statement states that the integral of heat divided by temperature is equal to the entropy generated. This does include entropy generation due to irre