Thermodynamics; Finding Work of a Piston cylinder Device

[Sara1]

Homework Statement

An insulated piston-cylinder device contains 5 kg of water at 100oC with a quarter of the mass is vapor. The water is compressed reversibly to 900 kPa. Find the work done during this process. Let the ambient temperature to be 27oC.

Homework Equations

I can use the energy balance equation

The Attempt at a Solution

I am not sure how to continue and find the work of an irreversible process. What equation shall i take to find the work in a piston device?

 

[rude man]

If you have a mixture of liquid water and water vapor at 100 deg C, what must p be?
Since you're told p is increased to 900 kPa, what does that imply about the liquid-vapor mixture?

Add the fact that liquid water is incompressible so that ∫pdV = 0 once the "quality" of the mixture reaches 0%, think latent heat of vaporization.

Or, you can do pΔV.

(Obviously, the ambien temp. is immaterial since the piston is insulated.)

 

[Sara1]

How do i use the latent heat to find the work? Do i calculate to find the h or perhaps find the specific volume then multiply by the mass to find the volume. then i can use the equation of work which is:
w=p(v2-v1)

 

[rude man]

You'll have to pardon me here. I need to figure this out correctly myself first. I apologize and hope someone else will stepin here in the meantime.nI will get back as soon as I can.

 

[rude man]

How do i use the latent heat to find the work? Do i calculate to find the h or perhaps find the specific volume then multiply by the mass to find the volume. then i can use the equation of work which is:
w=p(v2-v1)

Never mind using enthalpy. That was a blunder on my part. Enthalpy doesn't change since no heat is added or subtracted from the mixture.

So, compute the volume of the steam, given its mass and temperature and knowing it's saturated, and use pΔV. Of course, V = 0 when you're done appplying the work - why?

 

[Sara1]

why will the volume be equal to zero?

 

[rude man]

why will the volume be equal to zero?

By volume V I meant the volume of steam. Well, think about what's happening. You're apllying work to squish the steam until it's gone. The clue as to why it's gone rests in the fact that the final pressure is upped to 900 kPa. What is the pressure when you just removed the last bit of steam?

 

[Sara1]

Won't the final pressure be 900?

 

[rude man]

Yes, but remember I pointed out that water in liquid form is incompressible? If it's incompressible, what is the work done in increasing pressure on it?

 

[Sara1]

Oh, so shall i use the equation Q=m*cv*(T2-T1) instead? The t1 is given and i can find the t2 from the pressure=900kpa

 

[rude man]

Sara, I have to apologize again. I'm not sure what the exact siuation is. I thought p amd T stayed cosntant while the steam was compressed until it was gone. Now I have grave doubts about that. And my statement that H stays constant would then be invalidated, sinve H can change isentropically if p changes, which I'm beginning to think it does.

I apologize again, and suggest you keep the post & hopefully somebody smarter will help us both! Meanwhile I am not giving up & will let you know asap.

.

 

[I like Serena]

I'm still working through the problem myself and I was kind of hoping to see the solution here.

Since no heat is exchanged, I'm drawing the conclusion that the total entropy must remain the same.

From the steam tables, the entropy and the internal energy can be calculated in the initial state.

In the final state, we're looking at either another liquid-vapor combination, or a condensed liquid.
Either way, the entropy must be the same, from which the final temperature and masses can be deduced.

With the internal energy in the final state, you can calculate the work done, which is the difference in internal energies.

 

[rude man]

I'm still working through the problem myself and I was kind of hoping to see the solution here.

Since no heat is exchanged, I'm drawing the conclusion that the total entropy must remain the same.

From the steam tables, the entropy and the internal energy can be calculated in the initial state.

In the final state, we're looking at either another liquid-vapor combination, or a condensed liquid.
Either way, the entropy must be the same, from which the final temperature and masses can be deduced.

With the internal energy in the final state, you can calculate the work done, which is the difference in internal energies.

That sounds much better! Thanks! I will muse on that.

 

[rude man]

The problem as I see it is that we know the final p, but not T. If the water is supercooled then there is no T to go along with p as there would be in the unsaturated region.

At this point, if I had to give an answer I would argue that p and T don't change until the vapor has all been compressed into liquid, and after that there is no more work done since the liquid is incompressible. That would reduce the problem to finding the volume of the vapor at onset of compression (easy), then calling that V and going with W = Vp, p being cosntant at the initial p.

I'm just worried about enthalpy "not changing". Seems like the enthalpy of the water-vapor mixture should be higher than the enthalpy of just the water, even with T still constant. That would dictate a change in p, otherwise H couldn't change, since dH = dQ - Vdp and dQ = 0.

.

 

[I like Serena]

Assuming that entropy is constant (dQ=TdS=0), my calculations show that the system ends up in the liquid-vapor region with a much smaller volume and T=175 ^oC.

So pressure and temperature increase, while volume decreases.
Furthermore entropy remains constant, while internal energy and enthalpy both increase.

 

[rude man]

Assuming that entropy is constant (dQ=TdS=0), my calculations show that the system ends up in the liquid-vapor region with a much smaller volume and T=175 ^oC.

I'm still wrestling with a problem though.
I would expect that the work done on the piston (equal to the change in internal energy) would be roughly equal to the heat absorbed by the part of the vapor that condensates.
But I've got a discrepancy there that is just big enough that I don't want to simply ignore it.

My problem with that is that condensation represents a loss rather than a gain in U (or Q, but there is no Q in this situation). Heat is not absorbed by the vapor, it's liberated, isn't it?

Consider the part of the refrigeration cycle where the compressed vapor at the high-p point, and saturated, is reduced in V at constant T and p (isothermally). During this part of the cycle, heat is expelled to the higher T reservoir as the vapor is compresed to a pure (saturated) liquid. So should you worry about that?

I'd say that if in your analysis came to an end-point isentropically where p = 900 kPa and there was still some vapor left, then you've solved the problem, using ΔU = W. I would not agree with that were there no vapor left. Thank you for your willingness to jump in!

 

[I like Serena]

Yes, I realized that heat was released by the condensation instead of absorbed, so I was looking at it the wrong way around.
That's why I edited my previous post.

 

[rude man]

Yes, I realized that heat was released by the condensation instead of absorbed, so I was looking at it the wrong way around.
That's why I edited my previous post.

Well, I think you went the right way with using isentropy to get to (p=900kPa, T=175C) with quality > 0. I think that was very clever.

 

[I like Serena]

Thanks. :)

 

[Sara1]

I figured out how to solve it as well. Since the process is reversible, then the piston is adiabatic, then the process is isentropic. Thus, s1=s2. We can Find the work using the equation -W=m(U2-U1)
The u1 is determined from state one from the given information. To find U2, we must realize that s1=s2 so at the given pressure and s2 we can find the state, and then find the corresponding u2.
Then we calculate and solve for work.

 

[rude man]

I figured out how to solve it as well. Since the process is reversible, then the piston is adiabatic, then the process is isentropic. Thus, s1=s2. We can Find the work using the equation -W=m(U2-U1)
The u1 is determined from state one from the given information. To find U2, we must realize that s1=s2 so at the given pressure and s2 we can find the state, and then find the corresponding u2.
Then we calculate and solve for work.

Yes, that's what the fellow (?) who likes Serena did. Congrats.