# Thermodynamics - Piston-and-Steam-Table-Type Question

1. Apr 30, 2016

### LostStudent5

1. The problem statement, all variables and given/known data

An insulated cylinder containing water has a piston held by a pin. The water is initially saturated vapour at 65°C and a volume of 5L. The piston has a total mass of 10kg, its area is 0.003 m^2 and the atmospheric pressure is 100kPa. The pin is then released, allowing the piston to move. Determine the final pressure and temperature of the water assuming the process to be adiabatic.

2. Relevant equations

Sum of forces = 0
F = PA
W = mg

3. The attempt at a solution

At equilibrium:
P2 *A = mg + Patm *A
P2 = mg/A + Patm = 10*9.8/0.003 + 100*10^3 = 132666.6667 Pa = 1.326666667 bar

That's the first part done. Now I'm stuck at the second part - finding the final temperature of the water (T2). This is what I've got so far:

V1 = 5 L = 0.005 m^3 (given)
v1 = 6.201 m^3/kg (using the Steam Table)
m1 = V1/v1 = 0.005/6.201 = 0.00080632156 kg

P1 = 0.2501 (using the Steam Table)

However, I don't see how any of this is relevant to finding T2. Any help is much appreciated.

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Last edited: Apr 30, 2016
2. Apr 30, 2016

### rude man

For an adiabatic process, what is the pressure-volume relationship for an ideal gas? Assume a quasi-static process (slowly-moving piston).
Then, determine the new pressure and solve for V, then use the ideal gas law relating p, V and T.

3. Apr 30, 2016

### Staff: Mentor

This is an irreversible process, so the ideal gas law can't be used. Let V2 be the final volume. In terms of V2, how much work did the gas do? In terms of V2, what was the final specific volume? In terms of V2, how much work was done by the gas per unit mass of gas? In terms of V2, what was the final internal energy per unit mass of the gas? (The work done by the gas was the same as if the external pressure were constant at 1.327 bars).

4. Apr 30, 2016

### rude man

Thanks Chet!b
But if you then get the final U, don't you have to assume an ideal gas to get T?
rudy

5. Apr 30, 2016

### rude man

Oh wait, the quality will be between 0 and 100%, won't it? NM.

6. Apr 30, 2016

### Staff: Mentor

I think maybe it's going to be superheated.

7. Apr 30, 2016

### rude man

If it's already saturated vapor and then gets compressed, wouldn't it be either within the mixed-phase area (on a p-V curve, where p determines T) or even liquid?

Come to think of it, the orientation of the cylinder is not given but I assumed the piston was on top.

Should we pursue this privately?

8. Apr 30, 2016

### Staff: Mentor

No need. It will all play out when the problem is actually solved, such that the final specific volume and final specific internal energy are made good on.

9. Apr 30, 2016

### LostStudent5

Hey guys, thanks a lot for the replies!

Q-W = DeltaU
Q = 0 since the process is adiabatic
h = u+PV -> u = h-PV

Therefore:
W = -DeltaU = m(u1-u2) = m (h1 - P1V1 - h2 +P2V2) (1)

Is this right? h2 is unknown...

v2 = V2/m

From equation (1):
w = h1 - P1V1 - h2 +P2V2

Sorry I don't really understand this part. What I got is:

u2 = h2 - P2V2

but h2 is unknown.

I can confirm that the final state of water is in superheated vapour form, according to the answers.

10. Apr 30, 2016

### Staff: Mentor

Does your steam tables not have internal energies tabulated, only enthalpies? If so, see below.
No.

$W=(V_2-0.005)(132.7)$ kJ

$\Delta U=-W=-(V_2-0.005)(132.7)$

$\Delta H=-(V_2-0.005)(132.7)+132.7V_2-(25.01)(0.005)=(107.7)(0.005)$ kJ

Correct.
Consider this:

$$h_2=h_1+\frac{\Delta H}{m}=h_1+\frac{(107.7)(0.005)}{0.000806}$$

So, you now know the final pressure and the final enthalpy per unit mass.

11. May 1, 2016

### LostStudent5

It does, but it's in the form of uf and ug. I can work out u1 with the given data, but not u2 since its state is unknown.
This is the only part that I do not understand. Why is the pressure constant here? W=PΔV only applies to isobaric processes. P2 has a magnitude of 1.32 b while P1 has a magnitude of 0.2501 b, which I have already determined in the OP. Clearly there is a pressure difference so why are you using the formula for work for an isobaric process?

Anyway, I got the correct answer thanks to your help. Just don't understand the isobaric part!

12. May 1, 2016

### rude man

My question too, Chet! Why is pressure essentially constant and equal to the final equilibrium pressure? Otherwise I think I follow.

13. May 1, 2016

### Staff: Mentor

In an irreversible compression like this, the gas pressure within the cylinder is non-uniform spatially, and, in addition, viscous stresses within the gas contribute to the force per unit area exerted by the gas on the piston face. It is this force by the gas that does the work on its surroundings. So, the ideal gas law cannot be applied to get the gas force on the piston face.

In addition, in this problem, the gas force on the piston face clearly varies with time. So, how does one proceed? Well, we if we apply Newton's 2nd law to the piston, we obtain:
$$F(t)-mg-p_aA=m\frac{dv}{dt}$$where F(t) is the force exerted by the gas on the piston at time t, m is the mass of the piston, A is the piston cross sectional area, and $p_a$ is the air pressure. If we multiply this equation by the piston v=dx/dt, we obtain the mechanical energy balance equation: $$F\frac{dx}{dt}-mg\frac{dx}{dt}-p_aA\frac{dx}{dt}=mv\frac{dv}{dt}=m\frac{d(v^2/2)}{dt}$$
If we integrate this equation with respect to t, we obtain:
$$\int_{0}^{x(t)}F(x)dx=\left(\frac{mg}{A}+p_a\right)(V(t)-V(0))+m\frac{v^2(t)}{2}$$
The left hand side of this equation is the work done by the gas on the piston between time zero and time t, and the last term on the right hand side of this equation is the kinetic energy of the piston at time t. At relatively short times, the piston will overshoot the equilibrium position and oscillate up an down, so its kinetic energy will be varying with time. But, as time progresses, the oscillation of the piston will be damped out by viscous dissipation caused by the viscous stresses within the gas. (There is no need for actual friction between the piston and the cylinder to bring about this damping, and, in this analysis, we assume that the contact between the piston and the cylinder is frictionless). So, at very long times, we will have:
$$W=\int_{0}^{\infty}F(x)dx=\left(\frac{mg}{A}+p_a\right)(V(\infty)-V(0))$$
But this equation is exactly the same relationship we would have obtained if we had assumed that the gas force on the piston face were constant at its final equilibrium value throughout the entire expansion/compression.

14. May 1, 2016

### LostStudent5

Thank you so much Chest! Everything is clear as day now :)

15. May 1, 2016

### rude man

Or perhaps a simpler approach, math-wise at least:
we have
ma = Ap(t) - Apa - mg
where m = mass of piston
a = acceleration of piston, initially negative (call "up" the "+" direction)
A = area of piston
pa = atmospheric pressure
p(t) = instantaneous gas pressure after pin is removed at t=0.
Now we invoke d'Alembert's principle, which just states that the term ma can be considered another force ("inertia force") = -ma. Then Σforces = 0:
-ma + Ap(t) - Apa - mg = 0
where now -ma + Ap(t) is the total instantaneous force on the piston in the up direction, and always > 0; and
-Apa - mg is the force on the piston in the down direction, and always < 0.

i.e -ma + Ap(t) always equals Apa + mg, including when the piston finally comes to rest so that a = 0; then Ap(∞) = Ap2 = Apa + mg.

Therefore, the effective force, including the inertia force, in the up direction, due obviously entirely to the gas, is always Ap2.

Last edited: Sep 9, 2016
16. May 2, 2016

### Staff: Mentor

Hi rude man.

Maybe this works for you, but it doesn't work for me. In order for this to work for me, it still needs to show that the integral of the left hand side with respect to the displacement is equal to the integral of Ap(t) with respect to the displacement (the work W), which means that it needs to show that the integral of a(t) with respect to the displacement is zero. But that's basically what my analysis in post #13 showed.

Chet

17. May 2, 2016

### rude man

18. Sep 9, 2016

### rude man

5 months later, I just can't seem to let this one go:
OK, d'Alembert did not this time meet with universal approbation, although I still think my argument was valid.

m = piston mass
A = area of piston
pa = atmospheric pressure

Looking just at the piston, it starts with zero motion and zero net applied force (the pin offsetting gravity plus atmospheric pressure). Initial volume and pressure are V1 and p1.

Then, the pin is pulled and after some motion including possible oscillation the piston rests again at a new volume V2. Barring friction this means that the total work done on the piston by all forces (gas, atmosph. & gravity) must then be zero.

Therefore, the work done by the gas = W = -(work done by the outside), the "outside" being atmosphere plus gravity.
But the latter is merely (pa + mg/A)(V1 - V2) = -W
then ΔU = - W = (pa + mg/A)(V1 - V2) , and
ΔH = H2 - H1
= ΔU + (pa + mg/A)V2 - p1V1
= (pa + mg/A)(V1 - V2) + (pa + mg/A)V2 - p1V1
= V1(pa + mg/A - p1)
yielding H2 and T2 from the steam tables.

19. Sep 9, 2016

### Staff: Mentor

Yes. This all looks correct. But, one just can't say that, during the entire expansion, the force exerted by the gas on the piston is constant at $mg+p_aA$. My point was that, in the end, the work done by the gas is the same as if it were.

I think we are in agreement now?

20. Sep 9, 2016

### rude man

Chet, I always agreed with your derivation. I just tried to find a simpler derivation. And I still think the d'Alembert approach is also valid. But what's a few differing approaches between friends?