Thermodynamics - Regenerator Efficiency

In summary, the conversation discusses a question about the efficiency of a regenerator in a Regenerative Brayton cycle. The efficiency is given as 0.75 and the regenerator takes in air at h2=636.109 kJ/kg and expels it at h3=590.1625 kJ/kg. The air then travels through a combustion chamber and to a turbine. The question also mentions a need to find the enthalpy at h6 in order to solve the problem.
  • #1
ConnorM
79
1

Homework Statement


Hey I just have a question about regenerator efficiency. I have a problem where the efficiency of the regenerator in a Regenerative Brayton cycle is 0.75, the regenerator takes air in from a compressor which I called h2=636.109 kJ/kg, air then leaves the regenerator and is now at h3=590.1625 kJ/kg. The air then travels through a combustion chamber and then to a turbine. After this the air heads back to the regenerator at an enthalpy of h5=574.847 kJ/kg. Once the air passes through the regenerator is is expelled to the ambient at some T6 and h6. I am thinking if I can find the h6 I can simply interpolate to find T6.
If some of my numbers do not make sense here is the question,

http://imgur.com/dnjS1MF

Homework Equations


e = regenerator efficiency = 0.75

e = (h6 - h5) / (h6s - h5)

From this website, http://www.mhtl.uwaterloo.ca/courses/me354/lectures/pdffiles/c8.pdf, on page 8 I have used this equation, ( The numbers used in the following equation correspond to the picture here )

http://imgur.com/dnjS1MF

e = (h3 - h2) / (h5 - h2)

The Attempt at a Solution



I have already solved the question up to this point but don't know what to do for this part. From the equation above h6s is assuming the system is isentropic and there is no change in enthalpy between the exit and entrance to the regenerator, could I simply use

h2 = h6s

and since I know h2 I can solve for h6 ?
 
Last edited:
  • #3
ConnorM said:

Homework Statement


Hey I just have a question about regenerator efficiency. I have a problem where the efficiency of the regenerator in a Regenerative Brayton cycle is 0.75, the regenerator takes air in from a compressor which I called h2=636.109 kJ/kg, air then leaves the regenerator and is now at h3=590.1625 kJ/kg.

This is the part I don't understand. According to your calculations, the regenerator is acting like a refrigerator, since the air exiting the compressor has a higher enthalpy than when it exits the regenerator and enters the combustor. It seems to me that in order to increase cycle efficiency, the air leaving the regenerator should have a higher enthalpy, which energy it has recovered from the exhaust gases, and which energy would normally be lost when the exhaust leaves the engine.

In other words, why are you using compressed intake air to heat the exhaust?
 

What is thermodynamics?

Thermodynamics is the branch of physics that deals with the study of heat and its relation to other forms of energy, such as work. It also deals with the laws governing the behavior of thermodynamic systems.

What is a regenerator in thermodynamics?

A regenerator is a device that is used in thermodynamic systems to store and release heat energy. It is typically made up of a series of channels or passages that allow a working fluid to flow through and absorb or release heat.

What is regenerator efficiency?

Regenerator efficiency is a measure of how effectively a regenerator can store and release heat energy. It is calculated by dividing the heat transfer in and out of the regenerator by the maximum possible heat transfer between the same temperature ranges.

How does a regenerator improve efficiency?

A regenerator improves efficiency in thermodynamic systems by reducing the amount of heat that is wasted. It does this by storing heat from the hot side of the system and transferring it to the cold side, which reduces the amount of external heat that needs to be added.

What factors affect regenerator efficiency?

The efficiency of a regenerator depends on several factors, including the design and size of the regenerator, the type of working fluid used, the temperature difference between the hot and cold sides, and the flow rate of the working fluid. Additionally, the regenerator material and its thermal conductivity can also impact efficiency.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
518
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
  • Materials and Chemical Engineering
Replies
23
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
3K
Back
Top