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Thermodynamic Piston problem. |
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| Oct10-06, 02:42 PM | #1 |
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Thermodynamic Piston problem.
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? |
| Oct10-06, 02:59 PM | #2 |
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Recognitions:
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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"
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| Oct10-06, 03:26 PM | #3 |
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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. |
| Oct16-10, 07:08 AM | #4 |
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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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