# Homework Help: ThermoDynamics: piston arrangement with stoppers

1. Feb 6, 2013

### sandpants

Right, I might be dumb, or whatever, but i've pondering for a while and it's annoying because I feel like there is not enough information (too many unknowns not enough equations)

1. The problem statement, all variables and given/known data
The piston/cylinder arrangement shown Fig Q3 contains air at a pressure of 2 bar and
a temperature of 450 ºC. The air is allowed to cool to an ambient temperature of
18ºC.
(a) Is the piston resting on the stoppers when the system reaches its final state?
[Show all working]
(b) What is the final pressure in the cylinder?
(c) If the cross-sectional area of the piston/cylinder is 2 m^2,calculate the work
transfer during the process.

Fig 3 is a simple piston/cylinder, 2m in length with stoppers at 1m.

2. Relevant equations
The properties of air may be related to one another by the equation
pV = mRT
where, p = pressure; V = volume; m = mass; R = specific gas constant; and T =
temperature.

W = -pdV, assuming isobaric. W= pvln(v1/v2) - assuming isothermal, which it isn't.

3. The attempt at a solution
Not told whether the process is isobaric or not. Clearly it is not isothermal and isometric/choric up UNTIL AND IF the piston rests on the stoppers.

So, is it REASONABLE to assume that this process is isobaric? The second question asks the final pressure, which make me hesitate whether isobaric is applicable in this case.

Could a process be non isobaric, non isothermal, non isochoric at the same time? (I don't know why not, but just in case).

If Isobaric - simply P1V1/T1 = P2V2/T2 where P1=P2
From that you can find that V ~ 1.6m^3
A is constant 2m^2, so L = 0.8 BUT the stoppers are at 1m. The piston stop at the stoppers, temperature keeps dropping, shouldn't this affect pressure?

Would it then be reasonable to split the process into 2 processes? One before it reaches the piston, and then the second after?

Or is there some other way?

2. Feb 6, 2013

### TSny

I take it that the axis of the cylinder is oriented vertically so that the pressure of the gas balances the weight of the piston (plus the force due to external atmospheric pressure). If so, can you see why the pressure would have to stay constant as the gas cools as long as the piston is above the stoppers?

I think you have the right idea about breaking it up into 2 processes. [EDIT: Drawing the entire process on a PV diagram should help in determining the work]

3. Feb 6, 2013

### sandpants

Drawing the PV diagram is part of the assessment, I omitted it, it was point d).

I see no evidence that this SHOULD be an isobaric, but without it, there just isn't enough information, unless Im forgetting more relevant equations.

Work is easy IF this is an isobaric. Considering that, as written in the OP, the piston does indeed stop at 1m and goes no further, the work would be equal to the change in volume times the constant pressure.

4. Feb 6, 2013

### TSny

That's correct.

I think you have to assume that the process is slow enough ("quasi-static") so that the system is very close to equilibrium at each point of the process. Otherwise, you really wouldn't be able to represent the process on a PV diagram (pressure and volume would not have well-defined values during the process.) So, if the process is quasi-static, then the piston has essentially zero acceleration. So, the net force on the piston must be zero as the volume decreases. The weight and external force of the atmosphere on the piston do not change, so what does that tell you about the force that the gas exerts on the piston?

5. Feb 6, 2013

### sandpants

I'm terrible at thermodynamics/fluids.

If I understand you correct, I would have to guess that because the gas exerts the force equal to the atmospheric force (and as you said, net force zero + no acceleration) the gas pressure tends to become the same as atmospheric pressure?

Thats just so subtle. And there is no mentioned atmospheric pressure.

6. Feb 6, 2013

### TSny

I was assuming that the cylinder is surrounded by the atmosphere. It doesn't have to be. The net force pushing down on the piston is the weight of the piston plus the external pressure (whatever it is) times the area of the piston. Before the piston reaches the stops, this total downward force is balanced by the force upward on the piston from the pressure of the gas in the cylinder. If the piston doesn't have any mass (or weight) then the gas pressure in the cylinder would match the external pressure. Or, if there is no external pressure (vacuum outside the cylinder), then there would have to be a weight of the piston to balance the upward force from the pressure of the gas. In the general case, the upward force from the gas pressure balances the total downward force equal to the sum of the weight of the piston and the external pressure times the area of the piston.

As long as the gas is essentially in equilibrium, these forces must balance. Since the net downward force on the piston doesn't change while the volume is decreasing, the upward force from the pressure of the gas also must remain constant. That's all you need to know. You don't need to know how much of the external force is due to the weight of the piston and how much is due to external pressure.

Once the piston hits the stops, the constant net downward force is balanced by the upward force of the pressure of the gas plus the upward force from the stops.

7. Feb 6, 2013

### sandpants

Nothing is mentioned on the weight of the piston. Our thermodynamics and fluid mechanics are separate modules. We typically do not consider weight/acceleration in thermodynamics.

I am fairly confident the weight of the piston is negligible. Nothing is mentioned on the surrounding atmosphere.

The only things that are mentioned is what I have wrote in the OP. Only 3 system parameters are known. Initial Pressure/temperature and final temperature. Initial volume is not mentioned but can be gotten from the figure and a hint from c) - 2m length (starting position of the piston) x 2m area.

I think there has to be a reasonable assumption that can be made. But without knowing at what rates do pressure and volume change - I really am lost.

8. Feb 6, 2013

### TSny

I'm pretty sure that you are to assume that the pressure of the gas remains constant as the volume decreases for the reasons I gave. Then think about the type of process that occurs once the piston hits the stoppers. That will help you find the final pressure.

9. Feb 6, 2013

### sandpants

Could you elaborate on what you said in your previous post?

Assuming the system is surrounded by vacuum, and the weight of the piston is negligible, but still high enough to cause it to go down, at a constant velocity, due to the equality of upward (system pressure) and downward (the "negligible" piston weight) forces.

So the sole reason it is safe to assume that pressure is constant is because the piston is not accelerating downwards?

10. Feb 7, 2013

Assuming air to be an ideal gas, the number of moles in this closed system is constant (you may say mass is constant) during the process. In the initial state the piston (mass less, friction less piston) is not moving, so is the case in the final state. The process simply consists of cooling the gas. It is a reversible (no friction) isobaric cooling process and P2 = P1= 2 bar (As the gas cools the volume decreases and piston is assumed to move with no friction, this is required only for mentally picturing the process and not needed for calculations). As the gas cools its internal energy, U, decreases and the decrease in U is equal to the heat lost to the atmosphere. No work interaction is involved in the process.

Using the ideal gas equation you can calculate the final volume and therefore, the position of the piston; you will find, that it does not rest on the stoppers located at 1 m from the bottom.

you got answers to all the Qs now.

11. Feb 7, 2013

### sandpants

Thats what I thought.

I just don't quite understand the reason to why it is safe to assume it's an isobaric process when the piston is above the stoppers.

12. Feb 7, 2013