Thermodynamics Problem: Cylinder-Piston System with Friction

[MathError]

Homework Statement

Find the expression of work during the compression of a gas in presence of friction froce between piston and cylinder

Relevant Equations

P_e=external pression;
P_i=pression of the gas inside the cylinder (uniform pression);
F_f=friction force;
L=work on the system
L_a=friction work
Q=exchanged heat
E_t=total energy;
mass of piston=0;
A=area of the piston

Hi guys,
First of all I'm sorry for my bad english I'll try to be as clear as possible.
I have tried to solve this problem to understand the First Law of Thermodinamics: Q+L=ΔE_t
In fact I know L (in the current convention) is the work which the envirorment does on the system but I don't understand which contributes I have to include in L.
let we start from simple case: in this case F_f=0, so there is no friction force between piston and cylinder. In the hypothesis of quasistatic process the resulting force on the piston is zero so:
P_e=P_i.
The work made by envirornment on the system can be calculated as:
L=P_e*A*Δx=P_e*ΔV=P_i*ΔV
If there is friction force the problem becomes more complicated.
First all since the resulting force on the piston is still zero, in the hypothesis of compression we can find that:
P_e=P_i+F_f/A;
Also, there are two friction forces: the first one which is applied on the piston and it's considered in the previous equation and a second one which is applied on the cylinder but which doesn't produce work because it's applied to fixed points.
Now we have to choose our system: we can consider only the gas (A) or system which includes gas and piston (B).

Case A
In this case, in my opinion, the force on the system is:
F=P_i*A
In fact external pressure and friction force aren't applied to the surface of control of the system but to the piston.
As you can see in the photo, the the resulting work is:
L=P_e*ΔV+L_a

Case B
Including the piston in the system, the external forces applied to the surface of control are:
F=P_e*A-F_f
In this case the resulting work is different from the previous one and it values:
L=P_e*ΔV-L_a

I think both the results are wrong, anyway it's impossibile the two works are different because I only changed the point of view of the system. I hope I have been enough clear and

you can give me an help.
Thanks you

 

[Chestermiller]

For a detailed discussion and analysis of this kind of problem, please see the following Physics Forums thread (especially my posts in this thread): https://www.physicsforums.com/threads/thermodynamics-gas-expansion-with-piston-friction.963282/

 

[MathError]

Thanks Chestrmiller for your answer. I have read the discussion and I think I have understood better the problem.
First of all I'm happy to see I have made a few good observations, friction forces on the cylinder didn't do work because cylinder is fixed. Then also work expression seems correct in both cases.
However I made a few mistakes in the chooice of the system. In particular I think the two systems are the same system indeed. When I consider gas only or both gas and piston, it's the same case since piston has to be considered part of surface of control.
Anyway the biggest error is not consider friction heat absorbed by system during compression.
So I have repeated the exercise and I have got these results.
However I'have still have a doubt. If I look to the work which envirorment do on system I have the following cases:
L=P_i*ΔV (Gas Only) (1)
L=P_e*ΔV (Gas+Piston+Cylinder) (2)
At a frist sight I could think that work done in first case is lower than in second one indeed in first the envirorment surrenders heat to system so that lost energy is equal in the two cases and the First Law becomes:
P_e*ΔV=ΔE_tot (3)
It's clear the first member of this eqaution is a kind of work and it's tha same in both cases ("it's the work we are looking for") so I would like to know what the difference is between the work which figures in the First Law (3) and the work written in the equations (1) and (2) which are different.

 

[Chestermiller]

Thanks Chestrmiller for your answer. I have read the discussion and I think I have understood better the problem.
First of all I'm happy to see I have made a few good observations, friction forces on the cylinder didn't do work because cylinder is fixed. Then also work expression seems correct in both cases.
However I made a few mistakes in the chooice of the system. In particular I think the two systems are the same system indeed. When I consider gas only or both gas and piston, it's the same case since piston has to be considered part of surface of control.
Anyway the biggest error is not consider friction heat absorbed by system during compression.
So I have repeated the exercise and I have got these results.
However I'have still have a doubt. If I look to the work which envirorment do on system I have the following cases:
L=P_i*ΔV (Gas Only) (1)
L=P_e*ΔV (Gas+Piston+Cylinder) (2)
At a frist sight I could think that work done in first case is lower than in second one indeed in first the envirorment surrenders heat to system so that lost energy is equal in the two cases and the First Law becomes:
P_e*ΔV=ΔE_tot (3)
It's clear the first member of this eqaution is a kind of work and it's tha same in both cases ("it's the work we are looking for") so I would like to know what the difference is between the work which figures in the First Law (3) and the work written in the equations (1) and (2) which are different.

I don't understand what you are asking. All three of your equations are correct. Eqn. 1 applies to the system consisting of the gas only, while Eqns. 2 and 3 apply to the system consisting of the combination of gas and piston.

 

[MathError]

I have understood almost each point of the problem. My dubst is that equation (1) and (2) could pull to think the two works are different and they are indeed. However in the first case envirorment work is lower but thare is also heat trasmission.
Thank you for the help