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Thermodynamic Piston problem.

by seang
Tags: piston, thermodynamic
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seang
#1
Oct10-06, 02:42 PM
P: 185
A cylinder contains 1kg of saturated H20 at 30C. This piston has a cross-sectional area of .065m^2, a mass of 40kg, and rests on stops. With the piston in this position, the volume of the H20 is .1m^3. The external atmospheric pressure is 94Kpa and g is 9.75m/s^2. Heat is transferred to the system until the cylinder contains saturated H2O vapor.

1. Sketch the problem on T-v and p-v diagrams.
2. What is the water temperature when the piston just moves off of the stops?

I've been stumped on part 2 for hours. Here's how far I am: The first thing I did was calculate the pressure at which the piston would move up off of the stops. Since at that exact moment the piston won't actually be moving, I can say that Psys*Apiston - Patm*Apiston - m(piston)*g = 0.

Solving, I get Psys = (Patm*Apiston + m(piston)*g)/Apiston. This calculation yields 100KPa.




Help?
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OlderDan
#2
Oct10-06, 02:59 PM
Sci Advisor
HW Helper
P: 3,031
I can't visualize the pistion you are describing. I assume it is completely full of water, but I don't get the geometry. What moves when the water evaporates? What doesn't move? Where are these "stops"
seang
#3
Oct10-06, 03:26 PM
P: 185
well here is about the best I can do away from home...

The H20 is in the bottom. The piston is frictionless thus the space between it and the cylinder. Hope that helps a bit.
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ccarlos300
#4
Oct16-10, 07:08 AM
P: 1
Thermodynamic Piston problem.

Hey if the H2O is at 100kpa and the piston is free to move with the expansion from saturated liquid to saturated vapor then the pressure is a constant and the temperature during the expansion will be the saturated temp. Due to the fact that the substance will be moving along the isobar in a p-t diagram and along the isotherm in the p-v diagram.

Hope that makes sense and helps


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