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Master1022
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Homework Statement
A thermally insulated, rigid vessel is divided into two equal compartments. One contains steam at 100 bar and 400 degrees Celcius, and the other is evacuated. The partition is removed. Calculate the resulting pressure and temperature.
(Please let me know if this is the wrong section)
Homework Equations
1st Law of Thermodynamics: [itex] Q - W = \Delta U [/itex]
The Attempt at a Solution
from the 1st law, Q = 0 (insulated) and W = 0, thus internal energy will remain constant.
At the start
Initially, at 100 bar, the saturation temperature is 310.96 deg Celcius and thus we need to look in the superheated tables.
we find that the specific internal energy for steam @ 100 bar and 400 deg C is: u = 2835.8 kJ/kg and the specific volume is: v = 26.408 [itex] \times 10^{-3} m^3/kg [/itex]
After the partition is removed
we know that the internal energy remains the same and also mass is conserved, thus [itex] u_{f} = 2835.8 kJ/kg [/itex].
Given that the volume doubles, I thought that the specific volume would double, thus [itex] v_{f} = 2 \times v [/itex]
For this type of problem, the textbook then calculates the dryness fraction/ quality. However, given that the steam is superheated, I am unsure of what to do. Once the book has a dryness fraction, it goes on to guess pressure and temperature values that will yield the same internal energy.
I have just thought of using the new specific volume to try and pinpoint the conditions, but how would I know where to start interpolating? The v value lies in a number of areas...
Thanks in advance.