1st Law - Piston Cylinder Question

[KeJoSaBe]

Ok, I need help setting up the energy balance for the following problem:

 

I have a horizontal frictionless piston-cylinder device in which both sides of the piston are enclosed (i.e. no side is open to atmosphere).

Assume no viscous dissipation from the gas also)

Piston is initially held in place by a removable latch (imaginary) within the cylinder.

Initial conditions given on the left side of the piston is P = 8 bar, V = 0.1 m3

Initial conditions given on the right side of the piston is P= 1bar

Total Volume of the cylinder is 0.25m3

Fluid follows ideal gas EOS

Piston is 700 kg

Cross sectional area of the piston is 0.1 m2

No information is given on the temperature of the system or the surroundings other than to say we can assume this is an isothermal process

 

Question: When the latch is released, and the piston is free to move, what is the highest pressure reached on the right side of the piston and what would be the position of the piston at that pressure?

 

 

Initially, I took the piston itself as the system and performed a closed system energy balance on it. Doing that, I get the following:

dKE+dPE = dWs where KE is the kinetic energy of the piston as it moves, PE is potential energy of the piston and Ws is the work done at the boundary of the piston due to expansion/contraction of the gas on either side of the piston.

 

 

From here, I am getting lost in the details of this problem and I am starting to wonder if my basis is off. I just need a bit of help getting this started.

 

Thanks!

 

 

FYI - this question is actually from Elliott and Lira's Introductory Chemical Engineering Thermodynamics 1st edition P2.19 if you have the book

[studiot]

Why are you considering piston kinetic energy?

 

Start by describing what you think will happen once the piston is released.

This is important because there is a catch that I don't think you have picked up.

 

I think the problem (I don't know the book) must also specify adiabatic walls to the cylinder, but not an adiabatic piston?

Edited May 23, 2014 by studiot

[Endy0816]

Solutions for the 2nd edition, but you might be able to get something out of it.

 

http://www.egr.msu.edu/~lira/supp/index.htm

 

I would probably work the problem in terms of pressure equalization.

[studiot]

 

I would probably work the problem in terms of pressure equalization.

 

 

Will that work?

[Endy0816]

I was thinking Boyle's Law for a starting point. Only other one that I saw might apply was F=PA. Has been awhile now though, entirely possible I'm off base.

 

I normally just browsed the book in situations like this. Find the best fit equation(s). No gravity involved, no temperatures, friction, don't have to worry about atmospheric pressure, no piston thickness given nor anything about its connecting rod. Should be able to narrow it down at least.

[studiot]

But that piston has mass, and therefore momentum if it is moving.

 

The key is in the question I asked about what you think will happen.

.

[Fuzzwood]

If you consider pressure to be [Pa] = [N]/m²; [N] = m*a and both weight and cross section are given, then you might be able to convert the piston into an additional pressure.

[studiot]

 

If you consider pressure to be [Pa] = [N]/m²; [N] = m*a and both weight and cross section are given, then you might be able to convert the piston into an additional pressure.

 

 

Yes that's right but it still didn't answer my question about what actually happens.

 

I don't understand whether the OP is coming back or not. He was listed as still linked to the question for an hour after I posted my first hints.

 

I think his energy approach, properly implemented, would be easier to calculate, than Newton's laws.

[studiot]

Well it is disappointing for someone to come here for a few minutes post something and not come back, when others have put in significant effort.

 

This is a more interesting problem than most and I said there was a catch.So for the benefit of others who also looked at the problem I have set out my description of what I think happens.

 

It is tempting to assume that the piston is pushed along by the pressure difference until there is the same pressure in both chambers and that the system is then in equilibrium and the question is asking for this pressure.

 

But the question did not ask for this it asked for the maximum pressure in the right hand chamber.

 

The setup described leads to oscillation of the piston.

 

Initially, the difference in pressures between the faces of the piston gives rise to a force on the piston, held back by the latch.

 

Upon release of the latch the piston is accelerated by this force into the right chamber where the pressure is lower.

This compresses the gas in the right chamber, raising its pressure, and expands the gas in the left chamber, reducing its pressure.

 

As the piston travels the pressure difference diminishes and with it the force on the piston. So the acceleration continues until the chamber pressures are equal and the piston is now travelling at maximum velocity.

 

The piston continues to compress the right hand chamber and expand the left, suffering a decelerating force as a result of the now reversed pressure difference.

 

This deceleration now continues until the piston stops.

 

This is the point of maximum pressure in the right hand chamber, where the right hand chamber pressure is now greater than the left hand chamber pressure.

 

So the piston is forced to reverse its travel and pass again through the point of equal chamber pressures, this time compressing the left hand chamber above that of the right.

 

And so the oscillation cycle continues.

Edited May 25, 2014 by studiot

[KeJoSaBe]

My apologies for not keeping track of this post sooner. I appreciate all of the responses and effort put into this question!

 

I will try and explain my approach to this problem. Studiot, I reviewed my energy balance and agree that the kinetic energy term in my original equation is not needed here. My energy balance has become dPE = dWs where, PE and Ws are explained in my original post.

dWs = P_LdV_L - P_RdV_R = nRT_L/V_LdV_L - nRT_R/V_RdV_R (2)

where P_L and V_L are Pressure and Volume of the Left hand chamber while P_R and V_R are Pressure and Volume of the Right hand chamber.

Since this is an isothermal process, P_LV_L = constant at any point and is actually 800,000 N/m^2*0.1m^3 = 80,000 N*m (or Joules). Same can be said of P_RV_R = constant = 100,000 N/m^2*0.15m^3 = 15,000N*m (J).

 

From the Ideal Gas Law nRT_L = 80,000 J while nRT_R = 15,000 J (Since this is an isothemal process)

Then, dWs = 80,000/V_L*dV_L – 15,000/ V_R*dV_R

And, PE = Ws = 80,000*ln(V_L2/V_L1) - 15,000*ln(V_R2/V_R1)

But, since we already know V_L1 = 0.1 m^3 and V_R1 = 0.15 m^3 (Assuming the thickness of the piston has been taken into account when the question said the total volume of the cylinder is 0.25 m^3…I am assuming that is the total open space volume)

PE = Ws = 80,000*ln(V_L2/0.1) - 15,000*ln(V_R2/0.15)

And since we know that V_L+V_R = 0.25 m^3, then V_R2 = 0.25 -V_L2 and the equation becomes

PE = Ws = 80,000*ln(V_L2/0.1) - 15,000*ln((0.25-V_L2)/0.15)…

Now, I have 2 unknowns PE, and V_L2.

This is where I am stuck…..if I can get the Potential energy associated with this then I can calculate V_L2 which will allow me to calculate the position of the piston and Pressure (from the ideal gas law)……This is what I have done so far…please feel free to show me any mistakes in my thinking or, on the other hand what I might do to calculate the PE…seems I need to know the final position of the piston before I do that…..I will try to respond in a more timely manner this time around!

 

Thanks!

Edited May 28, 2014 by KeJoSaBe

[studiot]

 

if I can get the Potential energy associated with this

 

You seem to have written agreat deal, without comment on any of my thoughts about the problem.

 

Yes I agree the PE of the piston in its momentarily still position equals its potential energy.

 

This also equals the work done in compressing the gas in the right chamber minus the work done expanding the gas in the left, beyond the point where the chamber pressures are equal.

It should be elementary to calculate the point where the chamber pressures are equal.

[KeJoSaBe]

I agree with your assessment of the situation in your comments, however; I am not quite sure how to incorporate the oscillatory portion. I agree that to calculate the point where the chamber pressures are equal should be straight forward (however, I see that I calculate a value of 3.8 bars when chamber pressures are equal but, the book gives a final answer of 1.6 bars for the highest pressure in the right chamber).......so, I am still confused..

[studiot]

First I have a confession. I wrote this down the wrong way round by mistake in my post#2.

 

 

I think the problem (I don't know the book) must also specify adiabatic walls to the cylinder, but not an adiabatic piston?

 

 

It should of course be an adiabatic piston and non adiabatic (diathermal) walls.

 

 

OK I also get equal chamber pressures at 3.8 bar. It is, perhaps, a good idea to convert the bar to Pascal (N / m²⁾ so that work and energy calculations will come out in Joules.

 

Now since the expansions and contractions are isothermal it is easy to work out the isothermal work in each of the two sealed chambers to the equal pressure point.

 

since in this case all expansions and compressions are positive displacement ie there is no free expansion.

 

[math]W = PV\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) = PV\ln \left( {\frac{{{P_1}}}{{{P_2}}}} \right)[/math]

 

Where the Boyle's constant = PV for any pair of pressure and volume points, such as the given initial ones.

 

In an isothermal expansion of an ideal gas there is no change in internal energy. All the work done is balanced by heat drawn from or rejected to the surroundings.

 

This is why the cylinder walls must be non adiabatic, but the piston must be adiabatic, so there is no heat transfer between the chambers.

So the left hand chamber gas draws in heat from the surroundings equal to the work on the piston and the right hand chamber gas, whilst maintaining its temperature and internal energy constant.

 

Similarly the right hand chamber gas accepts work in compression and rejects an equal amount of heat to the surroundings to maintain its constant temperature and internal energy.

 

Therefore the difference in the work done by the left hand chamber gas and the work accepted by the right hand chamber gas equals the work done on the piston in accelerating it from rest and moving it to the equal chamber pressure position.

 

This must therefore be equal to the energy of the piston, which at maximum velocity is all KE. This is the energy that further compresses the gas in the right hand chamber until the piston is stationary, at which point the right hand chamber is at maximum compression.

 

This presents a problem since you say the book claims the maximum pressure to be around half that of the equal chamber pressure.

Are you sure it did not ask for the minimum?

 

Perhaps you could post the question verbatim?

Edited May 29, 2014 by studiot

[KeJoSaBe]

Thanks for spending the time on this studiot, I really do appreciate it! The following is the question verbatim:

 

"A 700 kg Piston is initially held in place by a removable latch inside a horizontal cylinder. The totally frictionless cylinder (assume no viscous dissipation from the gas also) has an area of 0.1 m^2; the volume of the gas on the left of the piston is initially 0.1 m^3 at a pressure of 8 bars. The pressure on the right of the piston is initially 1 bar, and the total volume is 0.25 m^3. The working fluid may be assumed to follow the ideal gas equation of state. What would be the highest pressure reached on the right side of the piston and what would be the position of the piston at that pressure? (a) assume isothermal; (b) What is the kinetic energy of the piston when the pressures are equal? (partial answer 1.6 bars)"

 

I am starting to think that the answer given at the end of the question is incorrect. I cannot reproduce this not matter how I try to...a maximum pressure of 1.6 bar in the right hand chamber simply makes no sense, based on the given information.

 

Also, After further review of my work in post #10 I realize there is a mistake in my original energy balance. Where I have PE (which is meant to represent change in potential energy) = Ws = 80,000*ln(V_L2/0.1) - 15,000*ln((0.25-V_L2)/0.15). This equation is true but, I failed to realize at the time that the change in PE from the initial state to the final state (i.e. the point at which the piston is furthest the to the right it can go...where KE=0 and PE2=PE1) is 0. I realized this after reading over your explanations...thanks again! Therefore, my equation above Should be PE = 0 = Ws = 80,000*ln(V_L2/0.1) - 15,000*ln((0.25-V_L2)/0.15). With this in mind, I can easily calculate the volume of the left hand chamber by trial and error (ends up being 0.2488 m^3). THis implies that the volume of the right chamber at this point is 0.0012 m^3 and from the ideal gas law, the pressure at this point is 125 bar......seems like a lot but, it makes a lot more sense to me than the answer given in the book.

 

For part (b) I simply PE1 = KE = Ws = 80,000*ln(V_L2/0.1) - 15,000*ln((0.25-V_L2)/0.15) where the initial KE is equal to 0. V_{L2 (Volume of the left hand side)} can be found based on the information supplied and the fact that PL = PR.

 

Hopefully this makes sense.

[studiot]

I will see what figures I come up with over the weekend.

 

You do not have to arrive at the chamber volumes by trial and error. Boyles law works even better.

 

Because the cylinder must have constant cross section the volume is proportional to the piston displacement. You can easily calculate the piston start position geometrically.

Boyles law again will give the equal pressure volumes (or any other) and thus the piston displacement.

So calculating the piston position at any moment is easy.

If you like you can form an equation for the work done on the piston in terms of the pressure difference force and the displacement, but it is a quadratic fraction that must be integrated.

 

As regards the book answer, perhaps there was an error which is why Endy couldn't find it in his later edition of the book in post#3?

 

One final thought.

 

Say the piston was 0.1 metres long then its volume is (0.1 x 0.1) m^{3 (note the superscript and subscript icons in the toolbar here)} = 0.01m³

 

So I wonder what the authors think this piston is made of? Gold is heavy stuff of density 19000kg/m³, so my gold piston has a mass of 190kg.

 

Any longer piston would make nonsense of the rest of the calcs since the cylinder is only 2.5 m (plus piston) long.

 

Edited May 31, 2014 by studiot

[studiot]

I ran a little spreadsheet to test values.

 

If there was a misprint so that the lefthand chamber pressure was 1.8 bar instead of 8 bar then the equal chamber pressure would be 1.32 and the max right chamber pressure I make 1.45 bar, pretty close to the 1.6 bar stated.

A left hand pressure of 0.8 or 2.8 diverges further from that offered.