# I Thermodynamics: Gas Expansion with Piston Friction

#### Chestermiller

Mentor
I'm thinking of the following comparison:

System 1: massless piston with friction force F, and a pile of small masses totaling M sitting on top of the piston initially

System 2: frictionless piston having mass m, such that mg for this system is equal to F for system 1, and the same pile of small masses totaling M sitting on top of the piston initially

The initial states of these two systems will be the same, but System 2 will experience an adiabatic reversible expansion and System 1 will experience an adiabatic irreversible expansion when the same pile of small masses is removed quasi-statically from the top of each piston.

#### Robert Davidson

I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.

#### Chestermiller

Mentor
I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.
I'll post this in a little while.

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.
So you would consider the same comparison as in my previous post, except with sudden removal of mass M at time zero?

#### Chestermiller

Mentor
The starting equations for the frictional analysis I alluded to in post #20 is:
$$dU=nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV\tag{1}$$and $$dU=nC_vdT=-\frac{f}{A}dV\tag{2}$$Dividing Eqn. 1 by $nC_vT$ yields:
$$\frac{dT}{T}=\frac{FdV}{AnC_vT}-\frac{R}{C_v}\frac{dV}{V}=\left[\frac{FV}{AnC_vT}-\frac{R}{C_v}\right]\frac{dV}{V}\tag{3}$$
In Eqn. 3, $$\frac{FV}{AnC_vT}=\frac{FV_0}{nC_vAT_0}\frac{(V/V_0)}{(T/T_0)}=\frac{FR}{AC_v}\frac{V_0}{nRT_0}\frac{(V/V_0)}{(T/T_0)}$$
But, from the ideal gas law, $$\frac{nRT_0}{V_0}=P_0$$
Therefore, $$\frac{FV}{AnC_vT}=\frac{F}{P_0A}\frac{C_v}{R}\frac{(V/V_0)}{(T/T_0)}\tag{4}$$
Substituting Eqn. 4 into Eqn. 3 then yields:
$$\frac{dT}{T}=\frac{C_v}{R}\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}=(\gamma -1)\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}\tag{5}$$
We now make the following dimensionless substitutions:
$$\bar{T}=T/T_0$$
$$\bar{V}=V/V_0$$
This then yields:
$$\frac{d\ln{\bar{T}}}{d\ln{\bar{V}}}=-(\gamma -1)\left[1-\frac{F}{P_0A}e^{(ln{\bar{V}}-\ln{\bar{T}})}\right]\tag{6}$$
This is the differential equation I integrated to get the results in post #20.
To get the work terms, I did the following: From Eqn. 2, if I divide by $nC_vT_0$ and integrate, I obtain:
$$\frac{\left[\int_{V_0}^V{\frac{f}{A}dV}\right]}{nC_vT_0}=1-\bar{T}$$
Similarly, for the integral of the gas pressure, I get:
$$\frac{\left[\int_{V_0}^V{PdV}\right]}{nC_vT_0}=(\gamma-1)\frac{F}{P_0A}(\bar{V}-1)+(1-\bar{T})\tag{7}$$

#### Robert Davidson

Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights plus atmospheric pressure.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob

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#### Chestermiller

Mentor
Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob
Bob,

Although I don't think that the two cases are really going to be comparable, I think each of them is interesting on its own. So please proceed and let's see what you come up with. (Please note that, in both these cases, the gas experiences a non-quasistatic deformation, and entropy is generated within the gas in each case).

Chet

#### Chestermiller

Mentor
I've been able to analytically solve the differential equation (Eqn. 6) of post #29 for the dimensionless temperature of the gas as a function of the dimensionless volume ratio. The result, which agrees with the numerical solution I presented in post #20 is:
$$\bar{T}=\frac{(1-\xi)}{\bar{V}^{(\gamma-1)}}+\xi \bar{V}$$where $$\xi=\frac{(\gamma-1)}{\gamma}\frac{F}{P_0A}$$

#### Robert Davidson

Hi Chet,

Haven’t forgot this. So far only tackled the easy case (system 2). Here’s what I have so far.

Going to assume dealing with ideal gas (air) with the following initial conditions and properties:

$P_{i}=10atm$, $T_{i}=300K$, $n=1$, $C_{v}=0.2053\frac{l.atm}{mol.K}$, $R_{u}=0.08205\frac{l.atm}{mol.K}$, $k=1.4$

Applying the ideal gas equation to get the initial volume:

$$V_{i}=\frac{nRT_{i}}{P_{i}}=\frac{(1)(0.08205)(300)}{10}=2.46 l$$

Weight is abruptly removed to reduce the external pressure by one half, or to 5 atm, and the gas allowed to expand non-quasi-statically at constant pressure until equilibrium is reached.

Since the final pressure is known, we can (1) determine the relationship between the final temperature and volume by applying the general gas equation to the initial and final states and (2) substitute the relationship into the first law equation for a constant pressure adiabatic process.

Applying the general gas equation:

$$\frac{P_{i}V_{i}}{T_{i}}= \frac{P_{f}V_{f}}{T_{f}}$$

$$T_{f}=\frac{P_{f}V_{f}T_{i}}{P_{i}V_{i}}$$

$$T_{f}=\frac{(5)(V_{f})(300)}{(10)(2.46)}=61V_{f}$$

Applying the first law for an adiabatic process:

$$\Delta U=-W$$

$$nC_{v}(T_{f}-T_{i})=-P_{f}(V_{f}-V_{i})$$

Substituting $T_{f}=61V_{f}$ and the other known values

$$(1)(0.2053)[61V_{f}-300]=-5(V_{f}-2.46)$$

Which gives us $V_{f}=4.22l$ and $T_{f}=257K$, and the work done is

$$W=5(4.22-2.46)l.atm=892j$$

Comparing to the final volume and work done for a reversible adiabatic expansion with the same initial conditions and same final pressures:

$$P_{i}V_{i}^{k}= P_{f}V_{f}^{k}$$

$$V_{f}=6l$$

$$W_{rev}=\frac{(P_{f}V_{f}-P_{i}V_{i})}{1-k}$$

$$W_{rev}= 13.5 l.atm=1368j$$

Which, as expected, is greater than the irreversible work.

The next step is to tackle system 1, which adds piston friction. I would like to bring system 2 to the same final pressure and then compare the work on the surroundings to system 2. I recall that in your analysis of the quasi-static expansion with friction it was necessary to apply an external tensile force to overcome static friction in order to reach the same volume as the non-friction system. It appears that will again be the case here, except it would be in order to reach the same final pressure.

I was thinking about a possible alternative approach, such as assuming the surface of the cylinder at some point changes to a smooth service prior to reaching the maximum static friction force so that the expansion could be completed, albeit without friction work done at the end. But that would probably create more complications than it's worth.

#### Chestermiller

Mentor
I approached this problem differently but, of course, arrived at the same results. My focus was on the final temperature, rather than the final volume.
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)=P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$
Solving for $T_f$ then yields:
$$T_f=\left[\frac{1}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{P_f}{P_i}\right]T_i$$For $\gamma=1.4$ and $P_f/P_i=0.5$, this yields $T_f=0.8571T_i$. And, for $T_i=300\ K$, $T_f=257.1K$.

The work is given by: $$W=-\Delta U=-nC_v(T_f-T_i)=-(1)(2.5)(8.314)(257.1-300)=892\ J$$
It'll be interesting to compare the approach and results you get for system 2 with those that I obtained.

#### Chestermiller

Mentor
Here's an interesting paradox for you. As $P_f/P_i\rightarrow 0$, $$\frac{T_f}{T_i}\rightarrow\frac{1}{\gamma}$$and then $$W\rightarrow nC_vT_i\left(1-\frac{1}{\gamma}\right)>0$$But how can this be if $P_f/P_i\rightarrow 0$ implies free expansion.

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#### Robert Davidson

Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados!! (Ha Ha).

#### Chestermiller

Mentor
Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados!! (Ha Ha).
Don't laugh. You and I both know that you actually will be thinking about it, even on the beaches of Barbados.

#### Chestermiller

Mentor
To complete this thread, here is the beginning of my solution for system 2 in which there is a friction force F.

Equation for the initial force balance on the piston: $$P_{gi}A-F-P_{atm,i}A=0$$where A is the piston cross sectional area, F is the friction force, $P_{gi}$ is the initial equilibrium gas pressure, and $P_{atm,i}$ is the initial atmospheric pressure.

Equation for the force balance on the piston during the expansion: $$F_g-F-\frac{P_{atm,i}}{2}A=0$$
where $F_g$ is the force that the gas exerts on the inside piston face during this irreversible expansion.

#### Robert Davidson

Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?

#### Chestermiller

Mentor
Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?
I find this kind of confusing. We are saying that, initially, there is no friction force, but, as soon as we drop the external pressure, friction kicks in at full force during the entire time that the piston is sliding. And it's still present at the very end. I guess this would be OK, under our assumption that the coefficients of static and kinetic friction are equal. But, during the expansion, the friction force would have to be less than 5 atm times the piston area, or the gas would not be able to expand. OK. Let's see where that takes us.

Chet

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"Thermodynamics: Gas Expansion with Piston Friction"

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