Estimation of Work in Gas Compression - Thermodynamic

In summary, the thermodynamic process is an isothermal compression where the volume of a gas changes (reducing it), the walls of the compressor are refrigerated (trying to keep the initial temperature in the system the all time), and the gas that is being compress raise its temperature a little bite because the compression process, despite the walls of the compressor are refrigerated.
  • #1
Kasper_NYC
26
0
What kind of thermodynamic process is the compression of a gas when the volume varies reducing it, and the temperature increase, and there is an exchange of temperature with the exterior (the walls of the compressor are refrigerated, but I assume that the gas will increase its temperature a little bite because the compression).

I am trying to find the formula in order to calculate the needed Work to compress 1Kg of CO2 (molecular weight 44.01g/mol), at 278.15K, from 0.067 liters at 30 atm of pressure, to 0.057 liters at 35 atm.

I though the thermodynamic process could be something close to an isothermal process, because the compressor is refrigerated, trying to keep the inicial temperature, and the estimation that I got for an isothermal process is:

From the formula W = n * R * T * Ln (Vb / Va)

W = Watios
n (number of moles) = 22.72
R (Constant) = 0.082
T (Kelvin degrees) = 278.15 K
Vb = 0.067 liters
Va = 0.057 liters

W = 22.72 * 0.082 * 278.15 * Ln (0.067/0.057)

W = 22.72 * 0.082 * 278.15 * 0.16 = 82.91W

Please, is it this estimation right? What it could be the right Thermodynamic process?

Best and thanks.

PS - I apologize for my English, I am learning it.
 
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  • #2
The problem isn't solvable without knowing the specific path taken between the two states. Assuming that the path is isothermal, the problem is solvable and you have set up the calculations correctly. However, remember to check the units on all the calculations to make sure they cancel correctly. Work will be in joules (J), not watts (W) because watts are a measure of power (work/time).
 
  • #3
Thanks Ygggdrasil, it is true, I mess up that Work is measured in Joules. Such a mistake! It is usual in me.

I have been thinking about your answer, I have reviewed again all the information that I have about Thermodynamics Processes, but still I can’t figure out what kind of thermodynamic process would be a compression where the volume of a gas changes (reducing it), the walls of the compressors are refrigerated (trying to keep the initial temperature in the system the all time), and the gas that is being compress raise its temperature a little bite because the compression process, despite the walls of the compressor are refrigerated.

The supposed I have posed it is not Isobaric (the pressure changes), not Isochoric (the volume changes), not Adiabatic (the system is refrigerated), and I don’t think it is adiabatic at all…. So, what kind of thermodynamic process could it be? I can’t say if this is a Polytropic Process, because still I don’t understand well what a Polytropic Process means.

Best and thanks.
 
  • #4
In theory, compressing a gas slowly in a refrigerated container will give you an isothermal compression. All the heat generated by the process will be drawn away by the refrigerated walls.
 
  • #5
Ygggdrasil,

By chance reviewing my book of physics I have found an exercise about Isothermal Expansion; and the R constant is “8.31 J/mol K”, and not “0.082 l atm/K mol” as I set the first time, because “n” in the equation refers to the number of moles.

Also I did another mistake, Vb refers to the final volume and Va should be the initial one… so the valor of the Ln will be negative, what means that we have to put “energy” in the system to work.

Finally:

W = 22.72 * 8.31 * 278.15 * -0.16 = -8402.5 J , what is a very different valor from the one I got initially.

Best and thanks.
 

Related to Estimation of Work in Gas Compression - Thermodynamic

1. What is the purpose of estimating work in gas compression?

The purpose of estimating work in gas compression is to determine the amount of energy required to compress a gas from a lower pressure to a higher pressure. This is important in the design and operation of gas compression systems, as it helps to optimize efficiency and cost-effectiveness.

2. How is work typically estimated in gas compression?

Work in gas compression is typically estimated using the thermodynamic principles of the ideal gas law and the first law of thermodynamics. This involves calculating the change in enthalpy and internal energy of the gas as it is compressed, and using this information to determine the work required.

3. What factors affect the amount of work required in gas compression?

The amount of work required in gas compression is affected by several factors, including the initial and final pressures, the type of gas being compressed, and the efficiency of the compression process. Other factors such as temperature, volume, and gas composition may also play a role.

4. How can work estimation in gas compression be improved?

Work estimation in gas compression can be improved by using more accurate and sophisticated models, taking into account real-world factors such as non-ideal gas behavior and heat transfer. Additionally, regular maintenance and optimization of compressors can help to reduce the work required for compression.

5. What are the limitations of work estimation in gas compression?

There are several limitations to work estimation in gas compression, including the assumptions made in thermodynamic models, the complexity of real-world gas compression processes, and the accuracy of data and measurements. Additionally, external factors such as changes in ambient temperature or gas supply can affect the accuracy of work estimations.

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