Conceptual problem - Thermodynamics

[H. S.]

Not homework per se, but this is giving me so much problems in their resolution: long story short, dealing with the 1st Principle, which would be the proper way to note down energetic exchanges in a closed system? I'm coming here from handling hydraulic systems with springs attached, and one or several spaces where ideal gases get compressed and exchange heat and work with the exterior.

The teacher has advised us to use two equations in the fashion of:

(1) dQ+dW = dE = dU+dEm

(Q=Heat, W=Work, dE=Overall energy exchange, dU=Internal energy exchange, dEm=Mechanical energy exchange)

(2) dEm = dEc+dEp = dW-dEi+pdV+xdX

(dEc=Kinetic energy, dEp=Potential energy, dW=Work, dEi=Friction losses, pdV=Expansion/Compression Work (of the system, if deformable), xdX=any other deforming work)

Problem with this is calling twice in both equations work, ESPECIALLY when in several places both terms equate to different works, exerted by either the system on the environment or the other way around. When replacing dEm in (1) with the equivalence in (2), sometimes work clears out, sometimes it doesn't!

And I believe the problem comes from the own teacher taking into account other work equations that do not figure in this scheme. EG: W = S(F·dx), given any force acting in or upon the system.

It confuses me terribly and it's giving me a headache even in relatively simple questions. I think the teacher should put it all in a single place in either of the equations but I'm so clueless I don't know even how to fix this so it works for me.

Anybody would care to clear this up for me?

 

[Chestermiller]

Let's stake a step back. Suppose you have a body that you exert a force on, and there is no heat added and no change in its internal energy U. Do you have any trouble accepting the idea that the work done by the force translates into a change in kinetic and/or potential energy? That is, for this limited situation,:

dW=dE_c+dE_p

Try to think of dW as the work being done on the boundaries of the closed system.

Chet

 

[H. S.]

Let's stake a step back. Suppose you have a body that you exert a force on, and there is no heat added and no change in its internal energy U. Do you have any trouble accepting the idea that the work done by the force translates into a change in kinetic and/or potential energy? That is, for this limited situation,:

dW=dE_c+dE_p

Try to think of dW as the work being done on the boundaries of the closed system.

Chet

That is easy to picture, and it would leave all thermodynamic variables unaltered as we would be simply moving the system along a force field if we understood work that way. I don't know if this is preceding further explanation, but answering your question I can understand that, yes.

 

[Chestermiller]

In the example I described, the body is rigid, and there are no deformations taking place within the body. In this case, all the work goes into kinetic and potential energy change. But, suppose the work applied at the boundary of the body also causes the body to deform, such as by decreasing (or increasing) in volume (P-V work done on the boundary), or by shearing (to generate heat viscously. Deformational work (at the boundary of the closed system), over and above work to accelerate the body as a whole (kinetic energy) or raise it against gravity (potential energy), causes the internal energy of the body to increase (or decrease). So part of the work applied at the boundary can change the kinetic energy and the potential energy of the closed system, and part of it can change the internal energy. To do work on the boundary of the body, you apply forces over the boundary, and the way in which the work resolves into deformational work and into work to change kinetic- and potential energy depends on the specific problem being solved.

Chet

 

[paranormal]

I understand your frustration with trying to understand and apply the principles of thermodynamics. It can be a complex and confusing subject, but with practice and a deeper understanding, it can become clearer.

Firstly, it's important to note that the first principle of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that in a closed system, the total energy remains constant.

Now, let's break down the equations provided by your teacher. In equation (1), we see that the total energy exchange (dE) is equal to the sum of the internal energy exchange (dU) and the mechanical energy exchange (dEm). This makes sense, as the internal energy of a system can change due to heat transfer or work done on the system. The mechanical energy, on the other hand, includes both kinetic and potential energy, which can also change due to work done on the system.

In equation (2), we see that the mechanical energy exchange (dEm) can also be expressed as the sum of kinetic energy (dEc) and potential energy (dEp). This equation also takes into account other forms of work, such as expansion/compression work and friction losses. This is where things may become confusing, as there are multiple forms of work involved in a closed system.

To address your concern about calling work twice in both equations, it's important to understand that work done by the system (dW) and work done on the system (dEi) are different concepts. Work done by the system refers to the work done by the system on its surroundings, while work done on the system refers to the work done by the surroundings on the system. In the context of thermodynamics, these two forms of work are necessary to consider in order to accurately calculate the total energy exchange in a closed system.

Finally, it's important to note that the equations provided by your teacher may not cover all forms of work that could be involved in a closed system. The equation W = S(F·dx) that you mentioned is an example of a work equation that is not included in these equations. This equation is typically used to calculate the work done by a variable force on a system, which may not be applicable in all situations.

In summary, the key to understanding and using these equations effectively is to have a clear understanding of what each term represents