Entropy of Otto Cycle: Deriving s1=s2

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Discussion Overview

The discussion revolves around deriving the entropy values for the Otto cycle, specifically focusing on the isentropic compression phase where it is proposed that the entropy at states 1 and 2 (s1 and s2) are equal. Participants explore methods to calculate these entropies using temperature and pressure data, as well as the ideal gas properties.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant inquires about deriving the entropy values at the isentropic compression phase of the Otto cycle, given the compression ratio and other state variables.
  • Another participant questions the necessity of determining absolute entropies and their relevance in the context of the problem.
  • There is a suggestion to use the air properties table to find entropies for states 1 and 2 based on known temperatures and pressures.
  • One participant confirms the assumption that s1 equals s2 due to the isentropic nature of the process and seeks to derive s3 using a specific equation.
  • Concerns are raised regarding the air table typically providing relative entropy rather than absolute values, prompting a reminder to check the reference state used in the table.
  • Participants confirm the validity of the equation set up to solve for entropy at state 3.

Areas of Agreement / Disagreement

Participants generally agree on the assumption that s1 equals s2 during the isentropic process, but there is no consensus on the absolute values of entropy or the necessity of determining them. Multiple viewpoints regarding the use of the air properties table and its reference states are present.

Contextual Notes

Participants note the limitation that the air properties table may not provide absolute entropy values, which could affect the calculations and interpretations of the results.

stephen0507
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Homework Statement
Determine s1 and s2 given P1= 90.0, P2= 2260.7, T1=283.1, and T2= 711.2
Relevant Equations
P1= 90kpa
T1= 283.1K
v1/v2=10

P2=P1(v1/v2)^k= 90(10)^1.4= 2260.7kpa
T2=T1(v1/v2)^(k-1)= 283.1(10)^0.4= 711.2K
Hello, just wanted to ask regarding the otto cycle; if we were to find the entropy at the phase of isentropic compression and I was already able to derive the temperature 1 and 2 and the pressures at 1 and 2 and I also have the compression ratio of 10. How do I derive the entropy (s1 and s2)? given the fact that entropy at this phase is s1=s2.
 

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Are you sure you are being asked to determine the absolute entropies of the two states? Of what value is that?
 
Chestermiller said:
Are you sure you are being asked to determine the absolute entropies of the two states? Of what value is that?
They're asking for the datum of each phase it's more of an introductory class in thermodynamics so I guess they just want us to be familiar with the derivation of each point. So far, I was thinking of using the table of air properties to determine the entropies for each state since I have the temperatures and pressure of both phases 1 and 2. Would this be the right way to determine entropy (s) and internal energy (u)?
 
stephen0507 said:
They're asking for the datum of each phase it's more of an introductory class in thermodynamics so I guess they just want us to be familiar with the derivation of each point. So far, I was thinking of using the table of air properties to determine the entropies for each state since I have the temperatures and pressure of both phases 1 and 2. Would this be the right way to determine entropy (s) and internal energy (u)?
Is this course statistical thermodynamics?
 
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Are you familiar with the equation for an ideal gas: $$dS=C_p\frac{dT}{T}-R\frac{dP}{P}=C_v\frac{dT}{T}+R\frac{dV}{V}$$
 
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yes! I actually wanted to ask about this. So I was able to determine entropy from points 1 and 2 using the ideal gas table for air since the otto cycle goes through an isentropic process is it safe to assume that s1=s2? so the entropy of air I got in phase 1 is the same entropy for the second phase? If so I need to get s3 and I was wondering if I could do this:

s3-s2= Cpln(T3/T2)-Rln(P3/P2)

and solve for s3
 
stephen0507 said:
yes! I actually wanted to ask about this. So I was able to determine entropy from points 1 and 2 using the ideal gas table for air since the otto cycle goes through an isentropic process is it safe to assume that s1=s2? so the entropy of air I got in phase 1 is the same entropy for the second phase?
This is OK, but please understand that the air table typically does not give the absolute entropy. It typically gives the entropy relative to some reference state like 25 C and 1 atm. Check the table to see what the use as the reference state of zero entropy.
stephen0507 said:
If so I need to get s3 and I was wondering if I could do this:

s3-s2= Cpln(T3/T2)-Rln(P3/P2)

and solve for s3
Sure.
 
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Chestermiller said:
This is OK, but please understand that the air table typically does not give the absolute entropy. It typically gives the entropy relative to some reference state like 25 C and 1 atm. Check the table to see what the use as the reference state of zero entropy.
ok yes understood I have noticed this in different references too, Ill take note of this thank you so much,
 
Chestermiller said:
Sure.
Just to confirm I can use the equation that I set up to solve for entropy at 3 correct?
 
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stephen0507 said:
Just to confirm I can use the equation that I set up to solve for entropy at 3 correct?
Yes.
 
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Thank you so much, you helped me with this I will keep note of everything you mentioned. Thanks!
 

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