Thermodynamics: Insulated Box Partition Question

In summary, a rigid, thermally insulated vessel divided into two compartments undergoes a change in pressure and temperature when the partition is removed. Using the 1st Law of Thermodynamics, it can be determined that the internal energy remains constant and the specific volume doubles. Since the initial state is superheated steam, the final state must also be a superheated vapor. By interpolating in the saturated steam tables, it can be determined that the final temperature would be 400 degrees Celcius and the final pressure would be 50 bars. However, there is no one right way to determine the specific temperature and pressure, and interpolation may need to be done in multiple steps.
  • #1
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.
 
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  • #2
If you check the saturated steam tables, there are no states of vapor in which the internal energy is as high as 2835. So certainly, all mixtures of saturated liquid and saturated vapor are going to have internal energies less than 2835. Therefore, the final state must be a superheated vapor. If it were an ideal gas, then the final temperature would be 400 C and the final pressure would be 50 bars. So this is the region of the saturated steam tables where you should be looking. You need to interpolate to find the combination of temperature and pressure in this vicinity at which the internal energy is 2835 and the specific volume is 0.0528.
 
  • #3
Chestermiller said:
If you check the saturated steam tables, there are no states of vapor in which the internal energy is as high as 2835. So certainly, all mixtures of saturated liquid and saturated vapor are going to have internal energies less than 2835. Therefore, the final state must be a superheated vapor. If it were an ideal gas, then the final temperature would be 400 C and the final pressure would be 50 bars. So this is the region of the saturated steam tables where you should be looking. You need to interpolate to find the combination of temperature and pressure in this vicinity at which the internal energy is 2835 and the specific volume is 0.0528.

Thank you for your response. When looking in the saturated tables, I see that I can interpolate in 2 ways (either by looking at adjacent pressure or adjacent temperature) Is there a 'proper' way of deciding between the two or would they come out the same?

At the moment, I am thinking about doing:
1. Interpolate to find the corresponding temperature for the specific volume at 50 bar
2. Using our temperature to find the u value at that temperature (at 50 bar)
- this turns out to be lower than required

Would the following step be suitable?
3. We need to interpolate between pressure values. Therefore, will need to find the u value at the same temperature for the next pressure step up. Then I will need to interpolate between the 2 u values at the same temperature, but different pressures

Many thanks
 
  • #4
There's no one right way of doing this. I would try something like this.
1. Interpolate on U to get the corresponding temperatures at the bounding pressures (40 bars and 60 bars).
2. Interpolate to get the specific volumes at the interpolated temperatures
3. Interpolate on specific volume to get the temperature and pressure at the desired specific volume.
 

FAQ: Thermodynamics: Insulated Box Partition Question

What is the purpose of an insulated box partition in thermodynamics?

An insulated box partition is used to create a barrier between two substances or systems in order to prevent heat transfer between them. This allows for controlled experiments and analysis of thermodynamic properties.

How does an insulated box partition affect the transfer of heat?

An insulated box partition acts as an insulator by reducing or preventing the transfer of heat from one side of the partition to the other. This is due to the low thermal conductivity of the insulating material used in the partition.

Can an insulated box partition be used to maintain a constant temperature?

Yes, an insulated box partition can be used to maintain a constant temperature by creating a thermal equilibrium between the two sides of the partition. This is because the partition prevents heat transfer, allowing the temperature to remain constant.

What are some common materials used to make insulated box partitions?

Some common materials used to make insulated box partitions include foam, fiberglass, polystyrene, and vacuum-sealed panels. These materials have low thermal conductivity and are effective in preventing heat transfer.

How does the thickness of an insulated box partition affect its effectiveness?

The thicker the insulated box partition, the more effective it will be in preventing heat transfer. This is because a thicker partition will have a greater distance for heat to travel through, making it more difficult for heat to transfer from one side to the other.

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