A piston-and-cylinder device initially contains 0.8 m3 of saturated water vapor at 250 kPa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of 300 kPa is required to move it. Heat is now slowly transferred to the steam until the volume doubles. Show the process on a PV diagram with respect to saturation lines and determine...
a.) The final temperature
b.) The work done during this process
c.) The total heat transfer
9 comments:
Looking at the problem a friend of mine assumed that the final pressure was 300kPa, doing so works out in the end. I'm wondering how you know to assume that? It makes sense to me that without the distance the piston rose given the pressure is unlikely to be greater than 300kpa, but I can see it being between 250 and 300. How can you tell?
Think of it might all the other problems. The weight/friction of the piston will allow the volume to expand while keeping the pressure constant at 300kPa. It's a little wierd becuase it stops in there but I'm pretty sure thats the right way to think about it.
Now my question: On part c, I'm not quite sure how to preceed, I think I'd apply the Q + W = deltaE, but I'm not quite sure how. A little help would be appreciated.
What I did for C was take the 2 stages and calculate the heat transfer for both and add them up in the end. Step 1-2 has O work and no change in KE or PE so you just have to look up values and find the deltaU. For step 2-3 you need to include work and a new delta U. Its how I did it.
The problem to me is that the pressure is not constant. It increased from 250 to 300kPa, I don't know how to tell where the increasing stopped.
Katie 6:12
This is a 2-step process. The first step is isochoric because the pressure inside the cylinder is not great enough to lift the piston.
The 2nd step is isobaric because once the piston is lifted by the gas in the cylinder and begins to "float" free, it will always exert the same force/area or pressure on the gas inside the cylinder.
The explanation offerred by dkress is correct.
dkress 7:53 PM
1st Law: Q - W = deltaE
That is Q MINUS W, not PLUS with our sign convention.
Any change in kinetic or potential energies a are negligible, so deltaE is really just deltaU. Look up the U^ values in the steam tables.
Katie returned the favor by correctly explaining this at 9:07 PM
Katie 9:07 PM
The pressure started at 250 kPa and increased throughout step 1-2 until P reached 300kPa. At this point, the upward force on the piston due to pressure inside the cylinder is exactly equal tothe downward force on the piston due to air pressure outside the cylinder AND the weight of the piston.
In step 2-3, as more heat is transferred into the water in the cylinder, it expands. The volume increases, but the pressure remains 300 kPa.
To whom it may concern ...
Now I understand your question.
If the volume doubles in step 2-3 then the piston MUST have floated up off of the stops, right ! The stops just keep the piston from falling down. I think a sketch would have been helpful on this problem. I always make one. But a BIG part of this course is translating problem statements into good sketches.
If the piston is floating, then the forces must balance as I described in the blog. The pressure inside the cylinder MUST be 300 kPa in order to make the piston float.
so whats the answer to the whole question?
The answers are 662 degC, 240 kJ and 1213 kJ, but your answers may vary depending on your source for thermodynamic data.
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