Proof that in Brayton reverse cycle heat is negative

In summary: P1V1 - P2V2) + (P2V2 - P3V3) = P1V1 - P3V3 = nRT1 - nRT3 = nR(T1 - T3) = nR(T3 - T2) + nR(T1 - T4) Therefore, we can see that the total heat transfer in the reverse Brayton cycle is equal to the difference between the heat absorbed from the cold reservoir and the heat rejected to the hot reservoir. Since the heat rejected is greater than the heat absorbed, we can conclude that Qtot is negative, and the heat pump cools the system. In summary, we can prove that
  • #1
Avishai
1
0

Homework Statement


I was given the Brayton reverse cycle and asked to prove that the total heat is negative (hence the heat pump cools the system).
I was to assume that all the steps are reversible.

Homework Equations


An ideal diatomic gas.
1st step: adiabatic compression.
2nd step: isobaric heat.
3rd step: adiabatic expansion.
4th step: isobaric heat absorption.
I managed to get to the equation:
Qtot = Cp*((T3-T2) + (T1 - T4)

The Attempt at a Solution


Since I have two adiabtes there are only two places in the cycle where Q can change - the two isobars I summed the two heata calculated from the Energy differential and managed to get to the equation above for total heat but didn't manage to prove that Qtot<0
 
Physics news on Phys.org
  • #2


Hello,
Thank you for sharing your question about the Brayton reverse cycle. I would approach this problem by first understanding the basic principles of the Brayton cycle and the assumptions made in this problem.

The Brayton cycle is a thermodynamic cycle that describes the operation of a gas turbine engine. It consists of four processes: adiabatic compression, isobaric heat addition, adiabatic expansion, and isobaric heat rejection. In a reverse Brayton cycle, the direction of these processes is reversed, with heat being absorbed from a cold reservoir and rejected to a hot reservoir.

In order to prove that the total heat is negative in this cycle, we must consider the first and second laws of thermodynamics. The first law states that energy cannot be created or destroyed, only transferred or converted from one form to another. The second law states that in any thermodynamic process, the total entropy of the system and its surroundings will either remain constant or increase.

In this problem, we are assuming that all the steps are reversible, which means that there is no entropy generation in the system. This also means that the heat transfer in each process is equal to the work done, as described by the equation Q = W.

Using this information, we can analyze each step in the reverse Brayton cycle to understand the direction of heat transfer and work done. In the adiabatic compression process, no heat is transferred, and work is done on the system, resulting in a decrease in temperature and an increase in pressure. In the isobaric heat addition process, heat is absorbed from the cold reservoir, resulting in an increase in temperature and pressure, and work is done by the system. In the adiabatic expansion process, no heat is transferred, and work is done by the system, resulting in a decrease in temperature and pressure. Finally, in the isobaric heat rejection process, heat is rejected to the hot reservoir, resulting in a decrease in temperature and pressure, and work is done on the system.

By summing up the heat transfer and work done in each process, we can calculate the total heat transfer in the reverse Brayton cycle as:

Qtot = Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4

= (P1V1 - P2V2) + (P3V3 - P4V4)

 

1. What is Brayton reverse cycle heat?

Brayton reverse cycle heat is a thermodynamic process in which a working fluid is compressed, cooled, and expanded in a cyclical manner to transfer heat from a cold reservoir to a hot reservoir.

2. How is heat measured in Brayton reverse cycle heat?

The heat transfer in Brayton reverse cycle is typically measured in terms of the change in enthalpy (heat content) of the working fluid as it goes through the cycle.

3. How is negative heat possible in Brayton reverse cycle?

In Brayton reverse cycle, the working fluid is compressed and cooled, resulting in a decrease in its enthalpy. This decrease in heat content is then used to transfer heat from a cold reservoir to a hot reservoir, resulting in a negative heat transfer.

4. What are the applications of Brayton reverse cycle heat?

Brayton reverse cycle heat is commonly used in air conditioning and refrigeration systems, as well as in power generation processes such as gas turbines.

5. What are the advantages of using Brayton reverse cycle heat?

Some advantages of Brayton reverse cycle heat include its efficiency, as it allows for heat transfer between two reservoirs at different temperatures, and its versatility, as it can be applied in various industries and processes.

Similar threads

  • Other Physics Topics
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
Replies
12
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Back
Top