Thermodynamic Reversible/Irreversible Closed Systems

In summary, the problem involves heating a 2 m3 rigid storage tank containing 10 kg of steam from 300 to 400 C using a hot reservoir at 500 C. The amount of heat transferred and the change in entropy of the steam, surroundings, and universe must be determined. The steam is assumed to be a real fluid following the steam tables in Appendix A.III. To solve the problem, the specific volume of the steam is kept constant and the change in internal energy is determined using the superheat tables for 300 and 400 C. The difference in internal energies is the heat transfer. The process is considered irreversible because the heat cannot be returned to the reservoir without work. The entropy change of the steam can be determined
  • #1
Nah346
5
0
I am completely lost with this problem. How would I go about breaking it up into manageable components that so I won't get lost in the problem?

Homework Statement



A 2 m3 rigid storage tank containing 10 kg of steam is heated from 300 to 400 C by transferring heat from a hot reservoir at 500 C. Determine the total amount of heat transferred to accomplish this change. Determine the entropy change of the steam, the surroundings, and the universe. Assume steam is a real fluid following the steam tables in Appendix A.III.

Homework Equations





The Attempt at a Solution

 
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  • #2
This problem is a bit vague. Is the steam saturated or superheated at 300 C? If you assume it is saturated, you can determine its internal energy and its specific volume. Because the container is rigid, the specific volume remains constant when you heat because both mass and volume are constant. Once again go to the superheat tables for 400 C and determine the pressure and internal energy at the known specific volume. The difference in internal energies is the heat transfer. This should get you started.
 
  • #3
Thanks for the response. There seems to be no indication in the problem statement whether or not the steam is saturated or superheated. The topic I am currently covering in class is Reversible/Irreversible Closed Systems and Unsteady Flow Systems. Do you know what part of the problem statement indicates reversiblity or irreversibility and how that concept applies here?
 
  • #4
The situation is irreversible because there is no way you can get the heat from the rigid container back into the reservoir (returning container and reservoir to initial states)without supplying work. Think of two blocks at different temperatures coming into contact and are isolated. They come to a common temperature. But they cannot be returned to their initial states without the addition of work. Heat will not move from cold to hot without work (heat pump).

You can determine the entropy change of the steam when heated from 300 C to 400 C from steam tables. You can also determine the heat transfer needed from internal energy changes based on steam tables. Now assuming the reservoir temperature does not change (big reservoir), you can determine the entropy change (decrease) of the reservoir. Comparing that to the increase in entropy of the container contents, you can determine the change in the entropy of the universe for this process. As you know it must increase because the process is irreversible.
 
  • #5
This is not my specialty but look at the following:

By looking at a T-v diagram (T = 300C and 400C and v = 0.2m^3/kg) for H2O you can see that you're always in the superheated steam region. From those two data points you can look up the change in internal specific energy Δu which, since W = 0 gives you Δq (1st law). Multiply by your actual specific volume gives you your total ΔQ.

Changes in entropy and enthalpy are read right off your steam tables.

My steam tables list T and v for a certain p, spaced 0.2 MPa apart, so I would have to interpolate a bit but I got p(300C) ~ 1.2 MPa and p(400C) ~ 1.6 MPa from which two tables I got Δu ~ 160 kJ/kg. Use the 1st law to get Δq and then scale to your actual specific volumer.

The reservoir at 500C loses entropy ΔS = ΔQ/Tr J/K where Tr is reservoir temperature. The universe ΔS is ΔS of the steam - loss of reservoir entropy.
 

FAQ: Thermodynamic Reversible/Irreversible Closed Systems

What is a thermodynamic reversible/irreversible closed system?

A thermodynamic reversible/irreversible closed system is a physical system that does not exchange matter with its surroundings, but can exchange energy in the form of heat and work. A reversible system is one in which all processes can be reversed without any change in the system or its surroundings, while an irreversible system involves some form of irreversibility, such as friction or heat transfer through a finite temperature difference.

What is the difference between a reversible and an irreversible closed system?

The main difference between a reversible and an irreversible closed system is the presence of irreversibilities, which lead to a loss of usable energy in the system. In a reversible system, all processes can be reversed without any loss of usable energy, while in an irreversible system, some form of irreversibility leads to a decrease in usable energy.

How does a reversible/irreversible closed system affect the laws of thermodynamics?

The laws of thermodynamics still apply to both reversible and irreversible closed systems. However, the presence of irreversibilities in an irreversible system can lead to a violation of the second law of thermodynamics, which states that the total entropy of an isolated system will always increase over time.

What are some examples of reversible/irreversible closed systems?

A reversible closed system could be a perfectly insulated gas chamber, where the gas molecules can expand and contract without any loss of usable energy. An irreversible closed system could be a car engine, where the combustion of fuel and the production of heat lead to a decrease in usable energy.

What are the practical applications of reversible/irreversible closed systems?

Reversible/irreversible closed systems have many practical applications, such as in engines, refrigerators, power plants, and chemical reactions. Understanding the reversible and irreversible processes in these systems is crucial in optimizing their efficiency and minimizing energy losses.

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